source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34297 | # Conditional Probability with Circuits
I am very sorry for the myriad of probability questions but I have a dam hard time trying to understand these questions:
This is the question:
Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is at least one closed path when the relays are activated. The relays may malfunction and not close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of .9.
The question is: What is the probability that current will flow when the relays are activated?
Answer:
1 - (0.1)^3
I don't understand why you are multiplying 0.1 * 0.1 * 0.1 and not adding 0.1 + 0.1 + 0.1. Since if any one of the three are open then the current will flow so isn't this technically a union which means you add?
• It's very simple: the multiplication principle. The relays are independent of each other. – Parcly Taxel Oct 12 '16 at 4:26
• @ParclyTaxel Oh ok. Thanks for pointing that out. So if they are dependent of each other then you would add them? – CapturedTree Oct 12 '16 at 5:23
## 2 Answers
Yes , it is a union . The current will flow when at least one of the relays will pass it through .
However , you do not simply add because the events are not disjoint . This then requires using the Principle of Inclusion and Exclusion (PIE) to avoid over-counting common outcomes .
The following is multiple choice question (with options) to answer.
A relay has a series of 5 circuits in a line. The even-numbered circuits are control circuits; the odd are buffer circuits. If both a control circuit and the buffer circuit immediately following it both fail in that order, then the relay fails. The probability of circuit one failing is 3/8; circuit two, 1/9; circuit three, 3/10; circuit four, 3/4; and circuit five, 2/5 .What is the probability that the relay fails? | [
"9/80",
"3/10",
"303/800",
"97/300"
] | D | The first circuit doesn't matter.
Prob(relay fails) = 1 - Prob(relay succeeds)
Prob(2+3 work) = 1 - 1/30 = 29/30
Prob(4+5 work) = 1 - 3/10 = 7/10
Prob(relay fails) = 1 - Prob(2+3 work AND 4+5 work) = 1 - (29/30)(7/10) = 1 - 203/300 = 97/800
D |
AQUA-RAT | AQUA-RAT-34298 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 240 m long passed a pole in 24 sec. How long will it take to pass a platform 650 m long? | [
"65 sec",
"89 sec",
"100 sec",
"150 sec"
] | B | Speed = 240/24 = 10 m/sec.
Required time = (240 + 650)/10 = 89 sec
ANSWER:B |
AQUA-RAT | AQUA-RAT-34299 | • But in this problem $a^2$ is not necessarily the largest square that divides $n$. – Lone Learner May 2 '12 at 19:25
• @LoneLearner But my proof in the linked thread doesn't use that. Rather, it uses only that $\rm\:c\:$ is squarefree (necessarily true when its cofactor $\rm\:b^2\:$ is a maximal square divisor of $\rm\:n = b^2 c,\:$ else $\rm\:d^2\:|\:c,\ d > 1\:$ $\Rightarrow$ $\rm\:(bd)^2\:|\:n,\ bd > b,\:$ contra maximality of $\rm\:b).$ – Bill Dubuque May 2 '12 at 19:49
For existence, let $a$ be the largest integer, in the usual ordering, such that $a^2$ divides $n$. If $n=a^2q$, then $q$ must be square-free.
For uniqueness, call a positive integer bad if it has two different decompositions $a^2 c$ and $b^2 d$, where $c$ and $d$ are square-free, and $a$ and $b$ are positive. If there are bad positive integers, let $M$ be the smallest bad one.
If $a$ and $b$ are not relatively prime, we can produce a bad positive integer smaller than $M$. So $a$ and $b$ are relatively prime.
We show that $a^2$ and $b^2$ are relatively prime. There are various approaches. One I like is that there exist integers $x$ and $y$ such that $ax+by=1$. Cube both sides. We get $$a^2(ax^3+3x^2by)+b^2(3axy^2+by^3)=1,$$ which says that $a^2$ and $b^2$ are relatively prime.
The following is multiple choice question (with options) to answer.
Which of the following does NOT have a divisor greater than 1 that is the square of an integer? | [
"28",
"32",
"49",
"62"
] | D | Prime Factorization
28=2*2*7=2^2*7. Includes a perfect square of 4.
32=2*2*2*2*2=4^2*2. Includes the perfect square of 16.
49=7*7. Is a perfect square.
62=2*31. No perfect square here, so a possibility.
25= 5*5. Is a perfect square.
All but 62 have perfect squares as a factor so D is the correct answer.
28=2*2*7=2^2*7. Perfect square of 4.
32=2*2*2*2*2=4^2*2. Perfect square of 16.
49=7*7. Perfect square.
62=2*31. No perfect square.
25= 5*5. Perfect square. |
AQUA-RAT | AQUA-RAT-34300 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
A 340-liter solution of Kola is made from 88% water, 5% concentrated Kola and the rest is made from sugar. If 3.2 liters of sugar, 10 liter of water and 6.8 liters of concentrated Kola were added to the solution, what percent of the solution is made from sugar? | [
"6%.",
"7.5%.",
"9.2%.",
"10.5%."
] | B | Denominator:
340+10+3.2+6.8=360
Numerator:
340(1-.88-.05)+3.2
340(0.07)+3.2
23.8+3.2
27
Ratio:
27/360=3/40
Answer: B |
AQUA-RAT | AQUA-RAT-34301 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
7 men, 5 women and 8 children were given as assignment of distributing 2000 books to students in a school over a period of 3 days. All of them distributed books on the 1st day. One of the 2nd day 2 women and 3 children remained absent and on the 3rd day 3 men and 5 children remained absent. If the ratio of the no. of books distributed in a day by a man, a woman & a child was 5 : 4 : 2 respectively, a total of approximately how many books were distributed on the second day? | [
"340",
"450",
"600",
"650"
] | D | Sol. Let the books distributed by man, a woman and a child be 5x, 4x and 2x respectively.
∴ No. of books distributed in 1st day
= 7×5x+5×4x+8×2x=71x
No. of books distributed in 1Ind day
=7×5x+3×4x+5×2x=57x
And no. of books distributed in IIIrd day
=4×5x+5×4x+3×2x=46x
71x + 57x + 46x = 2000, x = 2000/174
57x=2000/174×57=650
D |
AQUA-RAT | AQUA-RAT-34302 | Now $25^{-1}=48$, since $25*48=1200=1(mod\ 109)$. So we have -
$x=48*3=35(mod\ 109)$
-
Hi thanks for the reply, can you please explain how you got Now 25−1=48, since 25∗48=1200=1(mod 109)? – KaliKelly Aug 22 '10 at 6:34
To calculate modular inverse you need to use your extended Euclid's algorithm. (The procedure is there in the Wikipedia link.) – KalEl Aug 22 '10 at 6:38
Let me know if you have trouble understanding why $48=25^{-1}$. (The reason it is so is 25*48=109*11+1.) – KalEl Aug 22 '10 at 7:29
Okay I've used EEA, I got 25(48) - 109(11) = 1. I googled how to do it following this: mast.queensu.ca/~math418/m418oh/m418oh04.pdf Which was what you got... in one line, while I took a dozen lines. BTW, how do I use the fancy math formatting on my posts? – KaliKelly Aug 22 '10 at 7:40
The fancy formatting is done by a component called MathJax, which is basically a TeX formatter using Javascript. So for example 25^{−1} in-between two dollar signs looks like looks like $25^{-1}$ automatically when you post. You can google TeX formatting to learn how it works, and for anything on this site which interests you, you can right-click the mathematical equation to "view source". – KalEl Aug 22 '10 at 9:30
I meant this as a comment to the discussion after Student's answer but it seems that I don't have the option (reputation too low?) so I'll post it as an answer. Sorry.
The following is multiple choice question (with options) to answer.
-24*29+1240=? | [
"-544",
"584",
"544",
"345"
] | C | => -24*(30-1)+1240;
=> -(24*30)+24+1240;
=> -720+1264=544.
Correct Option: C |
AQUA-RAT | AQUA-RAT-34303 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? | [
"14",
"25",
"63",
"74"
] | C | Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = = 63.
Ans: C |
AQUA-RAT | AQUA-RAT-34304 | $=1+(-\frac12+\frac12)+(-\frac13+\frac13)+...+(-\frac{1}{n-1}+\frac{1}{n-1})-\frac1n$
$=1-\frac{1}{n}$
$=\frac{n-1}{n}$
Hint: Most of the terms in $\sum_{i=2}^n 1/(i-1)$ and $\sum_{i=2}^n 1/i$ are the same. Write out a few examples for small $n$ and you'll see how it works.
The following is multiple choice question (with options) to answer.
What is the sum of the numbers between 1 and 12, inclusive? | [
"99",
"65",
"79",
"78"
] | D | Sol. add the numbers between 1 and 12.
answer is D, 78. |
AQUA-RAT | AQUA-RAT-34305 | a height. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . Preview. Using Properties of Parallelograms 1.40 B.4.58 C.3.69 D. Cannot be solved Parallelogram, triangles etc; The surface area and the volume of pyramids, prisms, cylinders and cones; About Mathplanet; SAT. By: Learn Zillion. Add to My Library . SURVEY . Substitute the area and the height h by their values in the above equation and solve for BC length of BC = 1000 / [ 15 sqrt (2) / 2] = 94.28 feet (rounded to two decimal places). 2X. If two lines are parallel, then its slopes will be equal. Our mission is to provide a free, world-class education to anyone, anywhere. The rules of a 30-60-90 are as follows: 60° x. x√3. You could do this length right over here as the base. If we have a quadrilateral where one pair and only one pair of sides are parallel then we have what is called a trapezoid. The formula is: Because the parallelogram has an angle of 60 degrees you can create a 30-60-90 triangle to find the height. Students are given a variety of parallelograms where the side lengths are algebraic expressions. Back to Geometry Calc Next to Interactive Parallelogram. So, x = 4 and y = 3. To find the area of a parallelogram, use the formula area = bh, where b is the length of the parallelogram and h is the height. So the area here is also the area here, is also base times height. It doesn't matter which one, as long as one of the angle's arms is the base. 7. It can be shown that opposite sides of parallelogram must be congruent , that opposite angles are also congruent , and that consecutive angles are supplementary . To find the missing coordinate of a parallelogram, we use one of the following methods. All courses. Designed with Geometer's Sketchpad in mind . A parallelogram with base b and height h can be divided into a trapezoid and a right triangle, and rearranged into a rectangle, as shown in the figure to the left. Find the altitude. Area of a parallelogram.
The following is multiple choice question (with options) to answer.
The area of a parallelogram is 128sq m and its altitude is twice the corresponding base. Then the length of the base is? | [
"8",
"9",
"5",
"3"
] | A | 2x * x = 128 => x
= 8
Answer:A |
AQUA-RAT | AQUA-RAT-34306 | arithmetic, boolean-algebra, integers
Prove that for $A, B \in \mathbb{Z}$, $A + B$ $= (A\operatorname{\&}B) + (A \mid B)$ $= (A \oplus B) + 2(A\operatorname{\&}B)$ where $\operatorname{\&}$ is bitwise AND, $\mid$ is bitwise OR and $\oplus$ is bitwise XOR. (It is reasonable to assume that the bit representation of integers is two's complement.)
You have interpreted item 23 so that it is almost correct.
If we allow $A$ or $B$ to be negative, however, it is hard/impossible to assign/determine the sign of $A\operatorname{\&}B$, $A\mid B$ or $A\oplus B$ if we use the natural/standard representation of a number in base $2$. A different method that can represent signed integers without using the sign $\pm$ has to come into play. You selected two's complement. Then means you should also specify the length of a binary representation as well as restrict the magnitude of $A$ and $B$ to avoid overflow. That kind of specification and restriction is unlikely to be the intention of Rich Schroeppel, the author of this standalone item.
On the other hand, the equalities hold for all integers if integers are represented by two's complement with infinite length, as mentioned by hobbs and the Wikipedia article Two's complement and 2-adic numbers.
Let $A$ and $B$ be nonnegative integers
Here is a simpler way to interpret item 23. The only change is $A$ and $B$ are natural numbers (including $0$) instead.
For all $A, B \in \Bbb N$, $\ A + B$ $= (A\operatorname{\&}B) + (A\mid B)$ $= (A \oplus B) + 2(A\operatorname{\&}B)$ where $\operatorname{\&}$ is bitwise AND, $\mid $ is bitwise OR and $\oplus$ is bitwise XOR.
The following is multiple choice question (with options) to answer.
If a(a + 2) = 99 and b(b + 2) = 99, where a ≠ b, then a + b = | [
"-2",
"-5",
"-8",
"-10"
] | A | i.e. if a = 9 then b = -11
or if a = -11then b = 9
But in each case a+b = -11+9 = -2
Answer: A |
AQUA-RAT | AQUA-RAT-34307 | As before, the numerator can be simplified to P(at least 2 red), since if there are 2 red then you know that there is at least 1 red or green.
P(at least 2 red) = P(2 red) + P(3 red)+P(4 red) + P(5 red). Note that it also equals 1 - (P(0 red)+P(1 red)). We'll use the latter form as it's less work:
P(0 red) = Permut(24,5)/Permut(32,5), where "Permut(A,B)" means the permutation of A things B at a time:
$P(0\ red)= \frac {24 \times 23 \times 22 \times 21 \times 20}{32 \times 31 \times 30 \times 29 \times 28} = 0.211$
P(1 red) = Permut(24,5)xPermut(8,1)/Permut(32,5) x C(5,1):
$P(1\ red)= \frac {24 \times 23 \times 22 \times 21 \times 8} {32 \times 31 \times 30 \times 29 \times 28} \times 5 = 0.422$
Therefore
$P(at\ least\ 2\ red) = 1 - 0.211-0.422 = 0.367$
Now for the denominator: P(at least 1 red or green) = 1 - P( no red or green)
$1 - P(no\ red\ or\ green) = 1-\frac {16 \times 15 \times 14 \times 13 \times 12}{32 \times 31 \times 30 \times 29 \times 28} = 0.978$
So - the probability that there are at least 2 red apples given that there is at least 1 red or green apple is:
$\frac {0.367}{0.978} = 0.375.$
13. ## Re: Probability with a "known"
The following is multiple choice question (with options) to answer.
A box contains 10 apples, 9 of which are red. An apple is drawn from the box and its color is noted before it is eaten. This is done a total of n times, and the probability that a red apple is drawn each time is less than 0.5. What is the smallest possible value of n? | [
"3",
"4",
"5",
"6"
] | D | When you choose (and then eat) the first apple, the probability of that apple being red is 9/10. So if we do the activity 1 times, the probability of it being red is 9/10.
For 2 times, it is (9/10)*(8/9)
For 3 times, it is (9/10)*(8/9)*(7/8)
You can notice that the numerator of the first term cancels with the denominator of the second. So we can see that the probability becomes 0.5 when the last term is 5/6 & it becomes less than 0.5 when the last term is 4/5.
9 accounts for n=1, so 4 will account for n=6,
Answer.D |
AQUA-RAT | AQUA-RAT-34308 | Torsional stiffness k = dT / dφ Nm/deg: www. Example - Beam with Uniform Load, English Units. That’s because for every 1 kill you have 1 death, or exactly average. Calculate the value for load torque, load inertia, speed, etc. The accuracy of the selected models is evaluated with test data generated from a testing program detailed herein, which load tested full-scale reinforced concrete T-beams. It is also required to find slope and deflection of beams. To specify units, go to the Units sheet and describe your own units of Force, Distance, Inertia, Modulus and Deflection. Modulus of Elasticity: 210000 (N/mm 2) Moment of Inertia: 271188 (mm 4) Perpendicular distance from the neutral axis: 38 (mm) Material weight. rectangular or square hollow sections) and rolled angles and Tees. separated so that inertia can be calculated for each part that moves as the Servo Motor rotates. The one-way slab is deflected in one way direction and primary reinforcement is placed in one direction whereas. The resulting moment of inertia or centre of gravity when placing one or multiple loads can now be easily calculated. Calculating vehicle inertia Knowing the inertia of a vehicle, especially around the yaw axis is very useful for dynamic simulations such as ChassisSim. 2 Calculate the density the wheel and hub (you may completed the formula in the pre-lab already). Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Moment of Inertia : angular motion, define and derive the moment of inertia of a body, definte radius of gyration, define and use inertia torque, … Download [257. Polar Moments of Inertia. I = π (d o 4 - d i 4) / 64 ≈ 0. The current version (5. The designer sent me an excel spreadsheet for calculating moments with tubes. J (cm4): Torsion constant. Downloads Online Books & Manuals Engineering News Engineering Videos Engineering Calculators. This free cross section property tool calculates moment of inertia, polar moment of inertia and second moment of inertia for various shapes. 27 Moment of Inertia -
The following is multiple choice question (with options) to answer.
The toll T, in dollars, for a truck using a certain bridge is given by the formula T = 1.50 + 1.50( x − 2), where x is the number of axles on the truck. What is the toll for an 18-wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles? | [
" $ 2.50",
" $ 3.00",
" $ 3.50",
" $ 6.00"
] | D | Number of wheels in truck = 18
Number of wheels on its front axle =2
number of wheels remaining = 16
number of axles remaining axles = 16/4 = 4
Total number of axles = 5
T = 1.50 + 1.50( x − 2)
= 1.50 + 1.5*3
= 1.5 +4.5 = 6 $
Answer D |
AQUA-RAT | AQUA-RAT-34309 | ### Show Tags
13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
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The following is multiple choice question (with options) to answer.
If two of the four expressions x + y , x + 7y , x - y , and 7x - y are chosen at random, what is the probability that their product will be of the form of x^2-(by)^2, where b is an integer? | [
"1/3",
"1/8",
"1/2",
"1/6"
] | D | Total number of expressions if 2 are multiplied = 4C2 = 4!/2!*2! = 6
now we have to find the expressions that are in form of x^2 - (by)^2
one is (x+y)(x-y) = x^2 - (1y)^2
If you see any other combination we are always going to have one term of xy since 7x and 7y are there, so there can be only one type of combination like this.
therefore, probablity is 1/6
Answer : D |
AQUA-RAT | AQUA-RAT-34310 | Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios. I used that to write a second equation.
$\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}$
Simplifying terms and clearing denominators leads to $4x=36-3y$ and $3y=12-x$, respectively.
A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true. Specifically, the $\Delta ABP : \Delta DAP = 3:4$ ratio is true everywhere along the line 4x=36-3y. This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint. The same is true for the second line and constraint.
I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.
Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines. There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.
The question asks for the area of $\Delta ABP = \frac{1}{2}*12*x = 6*8 = 48$.
PROBLEM VARIATIONS
Just two extensions this time. Other suggestions are welcome.
The following is multiple choice question (with options) to answer.
Triangle ATriangle B are similar triangles with areas 2268 units square and 2527 units square respectively. The ratio of there corresponding height would be | [
"9:10",
"18:19",
"23:27",
"13:17"
] | B | Let x be the height of triangle A and y be the height of triangle of B.
since triangles are similar, ratio of area of A and B is in the ratio of x^2/y^2
Therefore, (x^2/y^2)=2268/2527
(x^2/y^2)=(18*18*7)/(19*19*7)
(x^2/y^2)=18^2/19^2
x/y=18/19
Ans=B |
AQUA-RAT | AQUA-RAT-34311 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Rajesh calculated his average over the last 24 tests and found it to be 76. He finds out that the marks for three tests have been inverted by mistake. The correct marks for these tests are 87, 79 and 98. What is the approximate percentage difference between his actual average and his incorrect average? | [
"0",
"9",
"8",
"7"
] | A | No Change
Incorrect value is: 78, 97, 89
correct values are: 87, 79, 98
difference between correct and incorrect value is= 9 + 9 -18=0
Answer:A |
AQUA-RAT | AQUA-RAT-34312 | kinematics
and consequently
$$v[~\tau~] = u - h~\frac{g}{2~u} = u~\text{Exp}[~-\tau~\frac{g}{2~u}~].$$
Therefore also: $v[~\tau = 0~] = u$, as required.
The following is multiple choice question (with options) to answer.
Given that -1 ≤ v ≤1, -2 ≤ u ≤ -0.5 and -2 ≤ z ≤ -0.5 and w = vz/u, then which of the following is necessarily true ? | [
"-0.5 ≤ w ≤ 2",
"-4 ≤ w ≤ 4",
"-4 ≤ w ≤ 2",
"-2 ≤ w ≤ 0.5"
] | B | Explanation :
We try to get an upper bound on the value of w.
After taking a look at the options we figure it out that if we take v=-1,z=-2 and u=-0.5.
we get the lower bound as -4.
If we take v=1, z=-2 and u=-0.5 we have the upper bound as 4.
Hence, the right option is 2.ie -4 ≤ w ≤ 4
Answer : B |
AQUA-RAT | AQUA-RAT-34313 | homework-and-exercises, energy, earth, space-travel, estimation
We need to get there in a reasonable amount of time (e.g., 1 million years)
We don't need to account for ecosystem survival on Earth
We don't need to worry about over-acceleration breaking Earth up (including water)
We don't need to worry about stopping upon arrival at Proxima Centauri If you want to do it in 1 million years then your basic problem is kinetic energy.
The gravitational potential energy of the Earth around the Sun is $-GM_{\odot}M_{E}/(1au) = -5.3\times 10^{33}$ J.
To get to Proxima Centauri within 1 million years requires a relative velocity of at least 1.27 km/s.
So I'm not sure how exact an answer you need. The main issue is how fast is it moving relative to the Sun.
The relative motion of Proxima Centauri and the Sun, is 22 km/s radially towards the Sun and 24 km/s tangentially. Thus while the radial velocity towards the Sun eliminates the need to do anything but escape the Sun's gravitational well, the tangential component must be matched.
Hence my rough answer would be - enough energy to escape the potential well of the Sun plus enough kinetic energy to track the tangential velocity of Proxima Cen.
This amounts to $5.3 \times 10^{33} + 1.7\times10^{33} = 7\times10^{33}\ J$.
That should see a collision take place in about 60,000 years.
To refine this, I think you need to take account of the Earth's 30 km/s motion around the Sun and to what extent you can use this kinetic energy to aid you. This depends on the direction in which you need to propel the Earth. Currently Proxima Cen has an ecliptic latitude of 45 degrees, but this will change rapidly, depending on the proper motion direction. However, at best, this could knock about $2.7\times10^{33}\ J$ off the answer in the most propitious circumstances.
The following is multiple choice question (with options) to answer.
Mars travels around the sun at a speed of approximately 15 miles per second. The approximate speed is how many miles per hour? | [
"48,000",
"51,000",
"54,000",
"57,000"
] | C | 15 * 3600 = 54,000
The answer is C. |
AQUA-RAT | AQUA-RAT-34314 | -
We can use a simple sieve to find these numbers in $O(x \log \log x)$ time. I went ahead and compiled my solution to make it as fast as possible.
PrimesUpTo = Compile[{{n, _Integer}},
Block[{S = Range[2, n]},
Do[
If[S[[i]] != 0,
S[[2i+1 ;; -1 ;; i+1]] *= 0;
],
{i, Sqrt[n]}
];
Select[S, Positive]
],
CompilationTarget -> "C",
Parallelization -> True,
RuntimeOptions -> "Speed",
CompilationOptions -> {"InlineCompiledFunctions" -> True, "InlineExternalDefinitions" -> True}
];
F = Compile[{{x, _Integer}, {n, _Integer}},
Block[{S = Range[x], primes = PrimesUpTo[Prime[1223]]},
Do[
If[S[[p]] != 0,
S[[p ;; -1 ;; p]] *= 0;
],
{p, primes}
];
Select[S, Positive]
],
CompilationTarget -> "C",
Parallelization -> True,
RuntimeOptions -> "Speed",
CompilationOptions -> {"InlineCompiledFunctions" -> True, "InlineExternalDefinitions" -> True}
];
Here's timings of all functions so far:
F[10000, 1223] // AbsoluteTiming
(* {0.001991, {1, 9929, 9931, 9941, 9949, 9967, 9973}} *)
fn[10000, 1223] // AbsoluteTiming
(* {0.007860, {1, 9929, 9931, 9941, 9949, 9967, 9973}} *)
fn2[10000, 1223] // AbsoluteTiming
(* {0.004625, {1, 9929, 9931, 9941, 9949, 9967, 9973}} *)
The following is multiple choice question (with options) to answer.
Find a a prime number from below options? | [
"84",
"97",
"98",
"100"
] | B | Clearly, 97 is a prime number.
B |
AQUA-RAT | AQUA-RAT-34315 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 160 kilometers? | [
"4",
"5.5",
"6.5",
"7.5"
] | A | Each 40 kilometers, 1 gallon is needed. We need to know how many 40 kilometers are there in 160 kilometers?
160 ÷ 40 = 4 × 1 gallon = 4 gallons
correct answer A |
AQUA-RAT | AQUA-RAT-34316 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
On a summer camp, 25% of the campers paid $120 each, 30% paid $80 each and the rest paid $65 each. What percentage of the total amount paid came from the campers who paid $80? | [
"18%",
"21%",
"35%",
"33.3%"
] | C | let the total no of campers be 100
total amount = (25*120)+(30*80)+(40*65) = 3000+2400+2600= $8000
required percentage = (28/80)*100 =35%
C |
AQUA-RAT | AQUA-RAT-34317 | physical-chemistry
Title: Solution with salts Does a solution of water and NaCl increase its volume of the same amount of the volume of salt added? Thank you
Does the ions Na+ and Cl- taken separately ,occupy more space than the binded molecule? Suppose we take $100$g of an $x$% by weight solution of sodium chloride, so we have $x$g of salt and $100-x$g of water. The volume of $x$g of salt is:
$$ V_S = x/\rho_S $$
where $\rho_S$ is the density of solid salt. Likewise the volume of the water is:
$$ V_W = (100-x)/\rho_W $$
Suppose when we dissolve salt in water the volumes just add i.e.
$$ V_\text{total} = V_S + V_W = x/\rho_S + (100-x)/\rho_W $$
then the density of our $x$% salt solution would be:
$$ \rho_\text{sol}(x) = \frac{100}{x/\rho_S + (100-x)/\rho_W} \tag{1} $$
The density of salt is $2.165$g/cm$^3$, and we'll take the density of water to be $1$g/cm$^3$, so we can use equation (1) to calculate what the density would be if the volumes just added and we can compared this with the experimentally measured density. I did this in Excel and got:
$$\begin{matrix}
x & Equation (1) & Experimental & Constant Volume\\
0 & 1.000 & 1.000 & 1\\
0.5 & 1.003 & 1.002 & 1.005\\
1 & 1.005 & 1.005 & 1.01\\
2 & 1.011 & 1.013 & 1.02\\
3 & 1.016 & 1.020 & 1.03\\
4 & 1.022 & 1.027 & 1.04\\
5 & 1.028 & 1.034 & 1.05\\
The following is multiple choice question (with options) to answer.
A bottle contains a certain solution. In the bottled solution, the ratio of water to soap is 3:4, and the ratio of soap to salt is four times this ratio. The solution is poured into an open container, and after some time, the ratio of water to soap in the open container is halved by water evaporation. At that time, what is the ratio of water to salt in the solution? | [
"3:8",
"3:9",
"3:4",
"3:12"
] | D | Water:soap = 3:4
Soap:Salt=12:16
=> For 12 soap, salt = 16
=> For 4 Soap, salt = (16/12)*4 = 16/12=64/12=18/3=6/1
So, water:soap:salt = 3:4:6 = 3:4:6
After open container, water:soap:salt = 1.5:4:6
So, water:salt = 1.5:6 = 3:12
ANSWER:D |
AQUA-RAT | AQUA-RAT-34318 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Sari and Ken climb up a mountain. At night, they camp together. On the day they are supposed to reach the summit, Sari wakes up at 07:00 and starts climbing at a constant pace. Ken starts climbing only at 09:00, when Sari is already 700 meters ahead of him. Nevertheless, Ken climbs at a constant pace of 500 meters per hour, and reaches the summit before Sari. If Sari is 50 meters behind Ken when he reaches the summit, at what time did Ken reach the summit? | [
"13:00",
"13:30",
"14:00",
"15:00"
] | C | Both Sari and Ken climb in the same direction.
Speed of Sari = 700/2 = 350 meters/hr (since she covers 700 meters in 2 hrs)
Speed of Ken = 500 meters/hr
At 8:00, distance between Ken and Sari is 700 meters. Ken needs to cover this and another 50 meters.
Time he will take = Total distance to be covered/Relative Speed = (700 + 50)/(500 - 350) = 5 hrs
Starting from 9:00, in 5 hrs, the time will be 14:00
Answer (C) |
AQUA-RAT | AQUA-RAT-34319 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
Look at this series: 2, 1, (1/2), (1/4), ... What number should come next? | [
"(1/3)",
"(1/8)",
"(2/8)",
"(1/16)"
] | B | Answer: Option B
This is a simple division series; each number is one-half of the previous number.
In other terms to say, the number is divided by 2 successively to get the next result.
4/2 = 2
2/2 = 1
1/2 = 1/2
(1/2)/2 = 1/4
(1/4)/2 = 1/8 |
AQUA-RAT | AQUA-RAT-34320 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A cyclist rides a bicycle over a route which is 1/3 uphill, 1/3 level, and 1/3 downhill. If the cyclist covers the uphill part of the route at the rate of 12 km per hour and the level part at the rate of 18 km per hour, what rate in km per hour would the cyclist have to travel the downhill part of the route in order to average 18 km per hour for the entire route? | [
"24",
"30",
"36",
"42"
] | C | Let V be the speed on the downhill section.
Let D be the distance of each of the three equal sections.
Total time T = T1+T2+T3 and 3D/T = 18 km/hour
3D / (D/12 + D/18 + D/V) = 18
1/6 = 1/12 + 1/18 + 1/V
1/V = 1/36 and so V = 36 km/h
The answer is C. |
AQUA-RAT | AQUA-RAT-34321 | acid-base, everyday-chemistry, ph
Title: How do you calculate percentage acidity? I am looking to calculate the percentage acidity of different drinks for the purposes of devising cocktail recipes. According to Liquid Intelligence by David Arnold, the optimum acidity level is 0.7-0.9, based on analysis of classic cocktails. How might I calculate the acidity percentage level of say lemonade (pH 2.9)? After looking at the link you posted, it seems "% acidity" refers to the mass percentage of citric acid in lemon juice, which is approximately 6%.
The percent acidity of your lemonade will depend on the percentage of lemon juice that constitutes it.
Let:
$X_{J/L}$ = the mass fraction of lemon juice in lemonade.
$X_{C/J}$ = the mass fraction of citric acid in lemon juice.
$m_{C/L}$ = mass of citric acid in lemonade
$m_L$ = total mass of lemonade
If you know how much water and lemon juice constitutes your lemonade, you can calculate "percent acidity" with:
$$m_L\;X_{J/L}\;X_{C/J}=m_{C/L}$$
Solving for $X_{C/J}$:
$$X_{C/J}=\frac{m_{C/L}}{m_L\;X_{J/L}}$$
Which is equivalent to:
$$X_{C/J}=\frac{X_{C/L}}{X_{J/L}}$$
Finally, solving for $X_{C/L}$:
$$X_{C/L}=X_{C/J}\;X_{J/L}$$
For example, if you mix 150g of lemon juice with 850g of water, the total mass of the mixture will be 1000g, and ${X_{J/L}}$=0.15, ${X_{C/J}}$=0.06
So "%acidity" for your lemonade would be:
The following is multiple choice question (with options) to answer.
Solution P is 20 percent lemonade and 80 percent carbonated water by volume; solution Q is 45 percent lemonade and 55 percent carbonated water by volume. If a mixture of PQ contains 67.5% percent carbonated water, what percent of the volume of the mixture is P? | [
"25%",
"40%",
"50%",
"60%"
] | C | 67.5% is 12.5%-points below 80% and 12.5%-points above 55%.
So the ratio of solution P to solution Q is 1:1.
Mixture P is 1/2 = 50% of the volume of mixture PQ.
The answer is C. |
AQUA-RAT | AQUA-RAT-34322 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percent of profit did he make on the original price? | [
"72%",
"82%",
"12%",
"22%"
] | C | Original price = 100
CP = 80
S = 80*(140/100) = 112
100 - 112 = 12%
Answer: C |
AQUA-RAT | AQUA-RAT-34323 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
The ratio between the present ages of P and Q is 6:7.If Q is 4 years old than P,what will be the ratio of the ages of P and Q after 4 years ? | [
"3:4",
"3:5",
"4:3",
"None of these"
] | D | Solution
Let P's age and Q's age be 6x years and 7x years respectively.
Then,7x-6x = 4 ⇔ x = 4.
∴ Required ratio = (6x + 4) : (7x + 4)=28 : 32 =7: 8. Answer D |
AQUA-RAT | AQUA-RAT-34324 | Your best bet in a situation like this, where there are relatively few numbers between $50$ and $100$ which are divisible by $7$, is to just list them, and count the number of entries on the list. (Of course, you'd like a more efficient means if the interval over which multiples of a given number span is very very large!) Note: When computing the number of integers between $50$ and $100$ that are divisible by both $7$ and $11$, there is only one such number: $7 \times 11 = 77$.
If, however, the question asked how many such numbers are divisible by $7$ OR $11$, then you'd need to list/count
(1) those divisible by $7$,
(2) those divisible by $11$;
(3) those divisible by both $7$ and $11$: any multiple of $7 \times 11$ in the given range;
Then add the number of entries on list (1) to those on (2), and then subtract the number of entries on (3) from that sum. (Otherwise, without subtracting, you'd be counting, e.g. 77, twice, since it would appear on both (1) and (2).)
In general, if the lower bound of the interval under consideration is $L$, and the upper bound of the interval under consideration is $U$, then the smallest multiple of a positive integer $n$ that is larger than or equal to $L$ is $L/n$, and if the result is not an integer, round up to the next integer (say $k_{min}$). On the other hand, the largest multiple of $n$ that is less than or equal to $U$ is $U/n$, but if the result is not an integer, round down (drop off the remainder), say $k_{max}$. Then, count all multiples of $n$ such that $L \leq nk_{min} < nk_i < nk_{max} \leq U$, where $nk_i$ are all the multiples of $n$ between the least and greatest integers in the given interval [$L, U$].
-
The following is multiple choice question (with options) to answer.
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: | [
"1008",
"1015",
"1022",
"1032"
] | B | Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Answer: Option B |
AQUA-RAT | AQUA-RAT-34325 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train is moving at a speed of 132 km/hr. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metres long? | [
"7½ sec",
"10 sec",
"12 ½ sec",
"15 sec"
] | A | Speed of train = 132 *(5/18) m/sec = 110/3 m/sec.
Distance covered in passing the platform = (110 + 165) m = 275 m.
Time taken =275 *(3/110) sec =15/2 sec = 7 ½ sec
Answer : A. |
AQUA-RAT | AQUA-RAT-34326 | • With conditional probability $\frac{1}{2+1}=\frac{1}{3}$ person B finishes the group task before A finishes the individual task and person A takes an additional expected $\frac{1}{2}$ hours to finish, with an overall probability $\frac{3}{7}\times \frac{1}{3}=\frac{1}{7}$ and an overall expected time of $\frac{2}{7}+\frac{1}{3}+\frac{1}{2}=\frac{47}{42}$ hours
• With conditional probability $\frac{2}{2+1}=\frac{2}{3}$ A finishes the individual task before person B finishes the group task and together they take an additional expected $\frac{1}{2}$ hours to finish the group task, with an overall probability $\frac{3}{7}\times \frac{2}{3}=\frac{2}{7}$ and an overall expected time of $\frac{2}{7}+\frac{1}{3}+\frac{1}{2}=\frac{47}{42}$ hours
As a check, the overall probabilities add up to $1=\frac{8}{35}+\frac{12}{35}+\frac{1}{7}+\frac{2}{7}$.
So the overall expected time is $\frac{8}{35}\times\frac{142}{105} +\frac{12}{35}\times \frac{83}{70}+\frac{1}{7}\times \frac{47}{42}+\frac{2}{7}\times \frac{47}{42} = \frac{251}{210}$ hours. This is about $71$ minutes and $43$ seconds, slightly more than your result.
The following is multiple choice question (with options) to answer.
WINK, Inc. follows a certain procedure that requires two tasks to be finished independently in order for a job to be done. On any given day, there is a 1/8 probability that task 1 will be completed on time, and a 3/5 probability that task 2 will be completed on time. On a certain day, what is the probability that task 1 will be completed on time, but task 2 will not? | [
"1/20",
"3/40",
"13/40",
"7/20"
] | A | P(1 and not 2)=1/8*(1-3/5)=1/20.
Answer: A. |
AQUA-RAT | AQUA-RAT-34327 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When n is divided by 5 the remainder is 3. What is the remainder when (n + 2)^2 is divided by 5? | [
"0",
"1",
"2",
"3"
] | A | n = 5x+3, for some integer x
(n+2)^2=(5x+5)^2=5y, for some integer y
When we divide this by 5, the remainder is 0.
The answer is A. |
AQUA-RAT | AQUA-RAT-34328 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man sells a horse for Rs.800 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse? | [
"287",
"279",
"267",
"860"
] | D | CP = SP + 1CP = SP - g
800 + x = 980 - 2x
3x = 180 => x
= 60
CP = 800 + 60
= 860
Answer:D |
AQUA-RAT | AQUA-RAT-34329 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
The ratio between the speeds of two Bullet trains is 7:6. If the second Bullet train runs 440 kms in 4 hours, then the speed of the first train is: | [
"89.25 km/hr.",
"37.25 km/hr.",
"128.3 km/hr.",
"94.25 km/hr."
] | C | Let the speed of two bullet trains be 7x and 6x km/hr.
Then, 6x = (440/4) = 110
x = (110/6) = 18.33
Hence, speed of first bullet train = (7 x 18.33) km/hr = 128.3 km/hr.
C |
AQUA-RAT | AQUA-RAT-34330 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
In a Private Company 20 Nos. staffs are employed. In a Festival season their Boss give festival allowance to all. He says his accountant to calculate for 30 days @ 100. Also he says balance payment will give to driver and cleaning staff and to be adjust from petty cash. He had given Rs.65000/- to the accountant. How much amount will taken from Petty cash? | [
"1000",
"3000",
"3500",
"2500"
] | A | For 20 staffs: 300*100*20 = 60000
for driver and cleaning staff: 300*100*2 =6000
From petty Cash : 1000
Answer is 1000 |
AQUA-RAT | AQUA-RAT-34331 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m? | [
"Rs.4082.87",
"Rs.4082.42",
"Rs.4082.97",
"Rs.4082.40"
] | D | Length of the first carpet
= (1.44)(6) = 8.64 cm
Area of the second carpet
= 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m
= (12.96)(7) sq m
Cost of the second carpet
= (45)(12.96 * 7) = 315 (13 - 0.04)
= 4095 - 12.6
= Rs.4082.40
Answer:D |
AQUA-RAT | AQUA-RAT-34332 | # Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have $$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$ There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B The answer for this question should be $0.35$, I am not sure where I did wrong.
I just figured maybe I should do this instead
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?
• Possible duplicate of Expected values of a dice game with a 30-sided die and a 20-sided die. – samjoe Mar 29 '17 at 6:00
• this is a different question – szd116 Mar 29 '17 at 6:05
• No, look properly. The first part of the question is same as yours, just it asks the probability of A winning. – samjoe Mar 29 '17 at 6:11
• I read it, but still couldn't figure out this problem – szd116 Mar 29 '17 at 6:13
• Look the answer to first part is 7/20 right? Now look at answer to that question. – samjoe Mar 29 '17 at 6:23
The following is multiple choice question (with options) to answer.
A gambler has won 40% of his 30 poker games for the week so far. If, all of a sudden, his luck changes and he begins winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week? | [
"20",
"25",
"30",
"35"
] | C | Let x be the number of additional games the gambler needs to play.
0.4(30) + 0.8x = 0.6(x+30)
0.2x = 6
x = 30
The answer is C. |
AQUA-RAT | AQUA-RAT-34333 | # Find the number of ways so that each boy is adjacent to at most one girl.
There are $7$ boys and $3$ girls who need to be lined up in a row. Find the number of ways so that each boy is adjacent to at most one girl.
In simple terms the situation demands that any distribution of the type $$...GBG...$$ must not come into play.
First of all the total number of arrangements are $10!$ and we can actually find a complement of those situations which we don't want.
In order to calculate the number of ways in which the wrong position can be true, I considered $GBG$ to be kind of a single package. The number of ways to make this package are:$${7 \choose 1} \cdot {3 \choose 2} \cdot 2!$$, now considering this package and $6$ boys plus the $1$ girl left, we can permute them all in 8! ways [$6$ boys, $1$ girl and our "package"], thus making the total to be $${7 \choose 1} \cdot {3 \choose 2} \cdot 2! \cdot 8! \tag{1}$$
Things seem to be tractable hithero, but as I was writing this questions I saw one problem in my argument: The cases containing the configurations $GBGBG$ have been possible counted several times thus $(1)$ is not giving the correct number of ways to be subtracted.
Can we anyhow make some changes in this approach and find the solution?
• Well, inclusion-exclusion lets you first subtract the cases in which at least one boy is alongside two girls so long as you add back the cases in which at least two boys are alongside two girls. – lulu Apr 11 '18 at 16:59
• I suppose the boys are distinguishable? That would be a fair supposition I guess... – Fimpellizieri Apr 11 '18 at 17:10
• @Fimpellizieri Certainly! – Harsh Sharma Apr 11 '18 at 17:12
• Do they self-identify in accordance with the genders assigned to them? – Strawberry Apr 12 '18 at 9:09
## 7 Answers
The following is multiple choice question (with options) to answer.
Find the no.of ways of arranging the 10 boys around a round tabele? | [
"9!",
"10!",
"8!",
"7!"
] | A | Ans.(9!)
Sol. Total number of persons = (10-1)! =9! |
AQUA-RAT | AQUA-RAT-34334 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Angelo and Isabella are both salespersons. In any given week, Angelo makes $560 in base salary plus 8 percent of the portion of his sales above $1,000 for that week. Isabella makes 10 percent of her total sales for any given week. For what amount of weekly sales would Angelo and Isabella earn the same amount of money? | [
"23,500",
"24,000",
"25,500",
"26,500"
] | B | Let the weekly sales of both = x
560+(x−1000)8/100=10/100x
x = 24000
ANSWER:B |
AQUA-RAT | AQUA-RAT-34335 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
1600 men have provisions for 28 days in the temple. If after 4 days, 300 men leave the temple, how long will the food last now? | [
"25 1/2 days",
"29 1/2 days",
"28 1/2 days",
"27 1/2 days"
] | B | 1600 ---- 28 days
1600 ---- 24
1200 ---- ?
1600*24 = 1300*x
x =29 1/2 days
ANSWER:B |
AQUA-RAT | AQUA-RAT-34336 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
When I was born, my mother was 23 years of age. After 6 years, when my sister was born, my father was 34 years of age. what is the difference between the ages of my parents? | [
"17 years",
"11 years",
"6 years",
"5 years"
] | D | When age of Son=0 ,Mother=23
After 6 years, Son=6 ,Mother=23+6=29 ,Father=34
So difference between ages of Mother & Father=34-29=5
ANSWER:D |
AQUA-RAT | AQUA-RAT-34337 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is | [
"35",
"45",
"55",
"65"
] | A | Explanation:
Number is (5*27) - (4*25) = 135-100 = 35
Answer: Option A |
AQUA-RAT | AQUA-RAT-34338 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
At a certain college, 70 percent of the students taking physical education class who were injured reported their injury, while 20 percent of the students who were injured in physical education class didn't report their injury. What percent of the students taking physical education class were injured? | [
"87.5%",
"60.5%",
"40%",
"75.5%"
] | A | Statement: 70% of students taking Pysical Education class (PE) Admitted to being Injured (AI). Of the students who took PE who were Injured (I), 20% declined to admit injury (notAI). [Note: A Venn Diagram is useful to visualize the problem.]
Solution: Sample space is 100%.
PE = 100%
AI/PE = 70%
notAI/I = 20% => 100% - 20% => AI/I = 80%
I/PE = (70)/(80)% = 80%
Answer: A |
AQUA-RAT | AQUA-RAT-34339 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 200 m long is running with a speed of 54 kmph. In what will it pass a car whose speed at 36 kmph in the direction opposite to that in which the train is going | [
"8 sec",
"10 sec",
"12 sec",
"14 sec"
] | A | Explanation:
Speed of the train relative to man = (54 + 36) kmph
= 90 ×5/18 m/sec = 25 m/sec.
Time taken by the train to cross the car = Time taken by it to cover 200 m
at 25 m / sec = (200 ×1/25) sec = 8 sec
Answer: Option A |
AQUA-RAT | AQUA-RAT-34340 | Kudos [?]: 53125 [5] , given: 8043
Re: problem solving question on ratios [#permalink] 16 Dec 2010, 13:47
5
KUDOS
Expert's post
2
This post was
BOOKMARKED
spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134
Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$
$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.
$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).
The following is multiple choice question (with options) to answer.
If the ratio of ages of two friends is 4:3 and sum of their ages is 70 years then what is the age of older friend? | [
"30 years",
"35 years",
"40 years",
"25 years"
] | C | Sum of thier ages is 70 in ratio 4:3
So 40+30 gives us 70. Wich is 4:3 so, older guy is having 40.
ANSWER:C |
AQUA-RAT | AQUA-RAT-34341 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
In a group of 100 cars, 47 cars do not have air conditioning. If at least 53 cars have racing stripes, what is the greatest number of cars that could have air conditioning but not racing stripes? | [
"45",
"47",
"49",
"51"
] | B | Lets assume AC=53(includesonly AC carsandcars with AC and racing stripes)
lets assume RS(racing stripes)>=53(includescars with AC and racing stripesandonly racing stripes).
Now since we want to maximize(only AC) we have to see to it thatcars with AC and racing stripesis minimal(assume 0) But since RS>=53.. we have to assign atleast 6 tocars with AC and racing stripes.
Hence AC=53-6=47.
The answer is B |
AQUA-RAT | AQUA-RAT-34342 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 = | [
"6,030,053",
"6,030,054",
"6,030,055",
"6,020,030"
] | D | Interesting problem.
I think key is to notice that all the given answer choices differs in last two digits. Therefore, our entire focus should be to figure out how the given terms contribute to last two digits of total.
1000^2 -> 00
1001^1 -> 01
.
.
.
1004^2 -> 16
Total -> *30
Answer D. |
AQUA-RAT | AQUA-RAT-34343 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
The simple interest at x% for x years will be Rs. x on sum of: | [
"x",
"100x",
"[100/x]",
"[x/100]"
] | C | Sol.
Sum = [100 * S.I. / R * T] = Rs. [100 * x / x * x] = Rs. [100/x].
Answer C |
AQUA-RAT | AQUA-RAT-34344 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
The standard deviation of a normal distribution of data is 2, and 3 standard deviations below the mean is greater than 45. What is a possible value for the mean of the distribution? | [
"52",
"47",
"48",
"49"
] | A | The standard deviation ({SD}) = 2;
3 standard deviations below the mean is greater than 45:
{Mean} - 3*{SD} > 45;
{Mean} - 6 > 45;
{Mean} > 51.
Answer: A. |
AQUA-RAT | AQUA-RAT-34345 | 1. All 6 dice show even numbers.
2. 4 dice show even and 2 dice show odd.
3. 2 dice show even and 4 dice show odd.
4. All 6 dice show odd.
The probability of cases 1 and 4 is $(\frac{1}{2})^6$ and the probability of cases 2 and 3 is $\binom{6}{2}(\frac{1}{2})^6=15(\frac{1}{2})^6$ (because we need to order the dice which show odd numbers.) Add them up and we get the probability is $$32(\frac{1}{2})^6=\frac{1}{2}.$$
The problem is about fair dice.
Whether the number of dice is 6 (even) or 7 (odd) $Pr = \dfrac12$
The logic is based on parity (odd/even). Since the probability that any particular throw is even or odd is equal at $\dfrac12$
(a) For an even number of dice, if you interchange odd and even throws, the parity remains the same, thus there will always be an equal number of odd and even sums.
EEEEEE OOOOOO even
EEEEEO OOOOOE odd
EEEEOO OOOOEE even
EEEOOO OOOEEE odd
(b) For an odd number of dice, every such interchange changes the parity, but by symmetry, again there will be an equal number of odd and even sums.
EEEEEEE even OOOOOOO odd
EEEEEEO odd OOOOOOE even
EEEEEOO even OOOOOEE odd
EEEEOOO odd OOOOEEE even
All outcomes are equally likely. Half the outcomes will have an odd number of odd dice faces and half of the outcomes will have an even number of odd dice faces. The outcomes with an odd number will have an odd total and the outcomes with an even number of odd face will be even. As there are equal numbers of outcomes for each case the probability for each case is 50%.
The following is multiple choice question (with options) to answer.
Four 6 faced dice are thrown together. The probability that all the three show the same number on them is? | [
"1/32",
"1/216",
"1/33",
"1/38"
] | B | The three dice can fall in 6 * 6 *6 * 6 = 1296 ways.
Hence the probability is 6/1296
= 1/216
Answer: B |
AQUA-RAT | AQUA-RAT-34346 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is? | [
"3/7",
"3/2",
"3/5",
"3/4"
] | B | Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train = 27 x meters, and length of the second train = 17 y meters. (27 x + 17 y) / (x + y) = 23 ==> 27 x + 17 y = 23 x + 23 y ==> 4 x = 6 y ==> x/y = 3/2.Answer: B |
AQUA-RAT | AQUA-RAT-34347 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Michele invests 300 of her money in a mutual fund that quickly goes sour, losing 25% of its value each day for a week. Which of the following COULD be the value of Michele’s investment at the close-of-business any day that week? | [
"$40.10",
"$133.00",
"$250.05",
"$1575.00"
] | A | Suppose she invests n$ in the starting then in the subesequent days the money left after the day will be
n/2, n/4, n/8, n/16, n/32
So, answer will be A |
AQUA-RAT | AQUA-RAT-34348 | # How many numbers made out of 5 distinct digits contain 4 and are even
This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.
I distinguished some cases (not sure if I distinguished in a 'smart' way)
• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.
The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.
This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?
I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.
The following is multiple choice question (with options) to answer.
How many 4 digit numbers can be formed with the digits 0, 1, 2, 3, 4, 5, 6 and 4? | [
"a. 220",
"b. 249",
"c. 432",
"d. 216"
] | D | Taking Question with it's same language as given
To make number divisible by 5, the unit digit should be either 0 or 5 only
Case 1: If unit digit is fixed as zero
With two6s The choice to fill the remaining three digits = 3C2 x 5 = 15
With all remaining 3 digits different, The choice to fill the remaining three digits = 6 x 5 x 4= 120
Total Such cases = 120+15 = 135
Case 2: If unit digit is fixed as Five
With two6s and one0The choice to fill the remaining three digits = 2 [6605 or 6065]
With two6s and without0The choice to fill the remaining three digits = 3C2 x 4 = 12
With all remaining 3 digits different, and one0The choice to fill the remaining three digits = 2 (ways to place zero) x 5 x 4= 40
With all remaining 3 digits different, and without0The choice to fill the remaining three digits = 3 x 5 x 4= 60
Total Such cases = 2+12+40+60 = 114
Total numbers = 135+114 = 216
ANSWER OPTION D |
AQUA-RAT | AQUA-RAT-34349 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
In a games hour 4 different types of players came to the ground? cricket 11, hokey 15, football 21, softball 15. How many players are present in the ground? | [
"70",
"52",
"62",
"49"
] | C | total number of players= 11+15+21+15= 62
Answer is C |
AQUA-RAT | AQUA-RAT-34350 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
Director
Joined: 13 Mar 2017
Posts: 703
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WE: Engineering (Energy and Utilities)
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Sushil got thrice as many marks in English as in Science. His total marks in English, Science and Maths are 180. If the ratio of his marks in English and Maths is 1:2, find his marks in Science? | [
"19",
"77",
"66",
"55"
] | A | S:E = 1:3
E:M = 1:2
------------
S:E:M = 1:3:6
1/10 * 190 = 19
Answer: A |
AQUA-RAT | AQUA-RAT-34351 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
An automobile parts supplier charges $25 per package of gaskets. When a customer orders more than 10 packages of gaskets, the supplier charges 4/5 the price for each package in excess of 10. During a certain week, the supplier sold 60 packages of gaskets. If 30 percent of the gaskets went to Company X, 15 percent to Company Y, and the rest to Company Z, what was the total amount, in dollars, that the parts supplier received in payment for the gaskets? | [
"1305",
"1375",
"1345",
"1415"
] | C | $25 per packet of gasket in case a customer orders less than 10
in case a customer orders > 10 price per gasket = 25*4/5=20
a certain week the supplier sold 60 gasket
1.he sold 30 % of the gaskets to X = 18 gaskets = 25*10 + 20 *8 = 250 + 160 = 410
2.he sold 15 % of the gaskets to Y = 9 gaskets = 25*9= 225
3.he sold remaining 55% to z =33 gaskets = 25*10 =250 + 20*23=710
thus ,total money earned
410+225+710 = 1345
Answer is C |
AQUA-RAT | AQUA-RAT-34352 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
A car going at 20 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.? | [
" 40",
" 50",
" 53",
" 55"
] | A | Let Car A = car that starts at 9 AM
Car B = car that starts at 9:10 AM
Time for which car A travels at speed of 20 m per hour = 1.5 hours
Distance travelled by Car A = 20 *1.5 = 30 miles
Since Car B catches up Car A at 10:30 , time = 80 mins = 4/3 hour
Speed of Car B = 30/(4/3) = 40 miles per hour
Answer A |
AQUA-RAT | AQUA-RAT-34353 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row 6 kmph in still water. When the river is running at 2 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man? | [
"5.32",
"5.7",
"5.76",
"5.74"
] | A | M = 6
S = 2
DS = 8
US = 4
x/8 + x/4 = 1
x = 2.66
D = 2.66 * 2 = 5.32 Answer: A |
AQUA-RAT | AQUA-RAT-34354 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A is three times as fast as B and working together, they can complete a work in 6 days. In how many days can B alone complete the work? | [
"11 days",
"16 days",
"13 days",
"24 days"
] | D | A= 3B
A + B= 3B+ B= 4B
These 4B people can do the work in 6 days, which means B can alone do the work in 4*6=24 days.
Answer : D |
AQUA-RAT | AQUA-RAT-34355 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in scheme B? | [
"6400",
"2778",
"2699",
"2789"
] | A | Let the sum invested in scheme A be Rs. x and that in scheme B be Rs. (13900 - x). Then,
(x * 14 * 2)/100 + [(13900 - x) * 11 * 2]/100 = 3508
28x - 22x = 350800 - (13900 * 22)
6x = 45000 => x = 7500
So, sum invested in scheme B = (13900 - 7500) = Rs. 6400.
Answer:A |
AQUA-RAT | AQUA-RAT-34356 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shop keeper marked 30% above the cost price and offered 20% discount then find it's net profit? | [
"2%",
"2.5%",
"3.7%",
"4.6%"
] | B | Net profit = 30-20 + (30*(-25)/100) = 2.5%
Answer is B |
AQUA-RAT | AQUA-RAT-34357 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
Each of the positive integers a and c is a four-digit integer. If each of the digits 0 through 8 appears in one of these three integers, what is the minimum possible value of the sum of a and c? | [
"3768",
"3789",
"3850",
"3825"
] | D | According to the stem we should use the digits 0 through 8 to construct 2 four-digit integers, so that their sum is as small as possible.
To minimize the sum, minimize the thousands digits of a and c, so make them 1 and 2.
Next, minimize hundreds digits. Make them 3 and 4 .
Next minimize 10's digit place by 5 and 6
Use the remaining digits (7,and 8) for units digits.
So, a would be 1357,and c would be 2468.
1357 + 2468 =3825
Answer: D. |
AQUA-RAT | AQUA-RAT-34358 | 2. How many ways can one choose three distinct characters to be placed into the forms from above.
3. Watch out for any duplication or overcounting.
There are two options (using Joe's suggestion in the comments):
• $\{a,b,c,c,c\}$. We can choose $3$ of the $5$ positions for the $c$'s, then there are two options for the $a$ and $b$ ($a$ first and then $b$ or the other way around). This gives $$2\binom{5}{3}=20$$ ways to write $a$, $b$, and $c$.
• $\{a,b,b,c,c\}$. We can choose $1$ of the $5$ positions for the $a$, and then $2$ of the remaining $4$ positions for the $b$'s. This gives $$5\binom{4}{2}=30$$ ways to write $a$, $b$, and $c$.
Now, there are $10\cdot 9\cdot 8=720$ ways to choose three digits for the positions $a$, $b$, and $c$. So, one might expect $$720\cdot (20+30)=36000$$ arrangements. The problem with this is that there is some overcounting since $abbcc$ and $accbb$ result in the same arrangement when the numbers in $b$ and $c$ are also flipped. This occurs for $a\leftrightarrow b$ in the first option and $b\leftrightarrow c$ in the second form. Therefore, we've overcounted by a factor of $2$ and there are $18000$ total arrangements. (Note that $720\cdot 30/2=10800$ matching @Green's approach).
The following is multiple choice question (with options) to answer.
In how many ways can the letters of the CHEATER be arranged | [
"20160",
"2520",
"360",
"80"
] | B | Explanation:
As we can see the letter "E" is twice in given word, so Required Number
=7!/2!=7∗6∗5∗4∗3∗2!/2!=2520
Option B |
AQUA-RAT | AQUA-RAT-34359 | the letter B occurs 3 times. Like a "super wildcard". Not all of those are distinguishable, because some words occur more than once on a single line. (d) type and case of letters, spacing between letters and punctuation marks; (e) joining words together or separating the words does not make a name distinguishable from a name that uses the similar, separated or joined words; (f) use of a different tense or number of the same word does not distinguish one name from another;. So a reasonable guess would be that total comparable settlements would be 150 times$24 million in cash, or $1. Swap these numbers as reverse the subset. 2% more than the actual collections in January 2019, and$35 million or 1. We couldn't distinguish among the 4 I's in any one arrangement, for example. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. We can use the following formula, where the number of permutations of n objects taken k at a time is written as n P k. Permutation. 4 billion increase in dividends paid, and a $324 million decrease in proceeds from the issuance of common stock, offset in part by a$3. P (10,3) = 720. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. From a 4 billion Indian Rupee gaming industry in 2007, to a 62 billion Indian Rupees industry in 2019, gaming in India has certainly caught the eye of consumers and proves to be a valuable market today. ership as an arrangement that leverages the uniqueness of Dutch law to avoid taxes and prevent a hostile takeover attempt. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. One of these code words, the 'start signal' begins all the sequences that code for amino acid chains. Microsoft Excel. com, the international travel industry has grown from 528 million tourist arrivals in 2005 to 1. The spots can also be divided by the total of legs, ears, eyes and tail to leave a remainder of 6. In the Text section, click the WordArt option. Letter Arrangements in a Word Video. 4 percent of loan participations
The following is multiple choice question (with options) to answer.
All of the stocks on the over the counter market are designated by either a 4 letter or a 5 letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes? | [
"2 (26)^5",
"26(26)^4",
"27(26)^4",
"26(26)^5"
] | C | In 4-digit code {XXXX} each digit can take 26 values (as there are 26 letters), so total # of 4-digits code possible is 26^4;
The same for 5-digit code {XXXXX} again each digit can take 26 values (26 letters), so total # of 5-digits code possible is 26^5;
Total: 26^4+26^5=26^4(1+26)=27∗26^4
Answer: C. |
AQUA-RAT | AQUA-RAT-34360 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 201? | [
"7650",
"9250",
"12,650",
"14,250"
] | A | 100 + 102 + ... + 200 =
100*51 + (2+4+...+100) =
100*51 + 2*(1+2+...+50) =
100*51 + 2(50)(51)/2 =
100*51 + 50*51 = 150(51) = 7650
The answer is A. |
AQUA-RAT | AQUA-RAT-34361 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
Find the surface area of a 10 cm x 4 cm x 3 cm brick. | [
"84 sq.c.m.",
"124 sq.cm",
"164 sq.cm.",
"180 sq.cm."
] | C | Sol.
Surface area = [2(10 x 4 + 4 x 3 + 10 x 3)] cm2
= (2 x 82) cm2
= 164 cm2.
Answer C |
AQUA-RAT | AQUA-RAT-34362 | ### Show Tags 23 Aug 2014, 11:31 kanusha wrote: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at$0.60 per glass on the first day, what was the price per glass on the second day?
The following is multiple choice question (with options) to answer.
If X boxes of orange juice cost 300 cents, how much will Y boxes cost in dollars? | [
"Y*3/x",
"2Y/x",
"5Y/x",
"Y*4/x"
] | A | X boxes of orange juice cost 300 cents or 3 dollars.
cost of orange juice = 3/x
Cost of Y orange juices = Y*3/x
Hence Answer : A |
AQUA-RAT | AQUA-RAT-34363 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains of length 350 m and 250 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 36 kmph and 36 kmph. After how much time will the trains meet? | [
"27/7 sec",
"20/7 sec",
"5 sec",
"21/7 sec"
] | C | They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (36 + 36)*5/18 = 4*5 = 20 mps.
The time required = d/s = 100/20 = 5 sec.
Answer : C |
AQUA-RAT | AQUA-RAT-34364 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Due to 10% decrease in the price of sugar and John can buy 5kg more sugar in Rs100 , then find the CP of sugar? | [
"Rs. 2(1/9)",
"Rs. 2(2/9)",
"Rs. 2(1/7)",
"Rs. 3(2/9)"
] | B | Here r = 10 % ,x = 100 and A = 5 kg
Actual price of sugar = 10*100/((100-10 )*5) = Rs. 2(2/9)
B |
AQUA-RAT | AQUA-RAT-34365 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find out the wrong number in the series.
7, 8, 18, 57, 228, 1165, 6996 | [
"8",
"18",
"57",
"228"
] | D | The series is
7*1 + 1 = 8
8*2 + 2 = 18
18*3 + 3 = 57
57*4 + 4 = 232
232*5 + 5 = 1165
1165*6 + 6 = 6996
so 228 is wrong
ANSWER:D |
AQUA-RAT | AQUA-RAT-34366 | ### Show Tags
05 Sep 2017, 11:01
2
In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hoursat 60 mph. What was his average speed for the whole trip?
(A) 50
(B) 53.5
(C) 55
(D) 56
(E) 57.5
The total distance traveled from City A to City B is 50*1 + 60*3 = 230
Since John drove for 4 hours to travel this distance,
the average speed of the whole trip is $$\frac{Distance}{Time} = \frac{230}{4}$$ = 57.5(Option E)
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Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink]
### Show Tags
17 Jan 2019, 03:38
How does this work using the harmonic mean?
$$\frac{1/50+3/60}{4}$$=$$\frac{7/100}{4}$$=57.1
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink] 17 Jan 2019, 03:38
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The following is multiple choice question (with options) to answer.
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 10 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? | [
"13",
"13.5",
"14",
"14.5"
] | D | Ans is D
Given d_ab = 2*d_bc
let d_ab = d and d_bc = x so d=2x
for average miles per gallon = (d+x)/((d/10)+(x/18)) = 14.5 (formula avg speed = total distance/ total time) |
AQUA-RAT | AQUA-RAT-34367 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Aishwarya’s mom was 28 years of age at the time of her birth, while her mom was 20 years old when her 2 years younger sister was born. The variance between the parents ages is : | [
"10 years",
"11 years",
"15 years",
"9 years"
] | A | A
6 years
Mom’s age when Aishwarya’s sister was born = 20 years.
Dad’s age when Aishwarya’s sister was born = (28 + 2) years = 30 years.
Needed Variance = (30 – 20) years = 10 years.
Answer is A |
AQUA-RAT | AQUA-RAT-34368 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
Two members of a club are to be selected to represent the club at a national meeting. if there are 91 different possible selections of the 2 members, how many members does the club have? | [
"20",
"14",
"40",
"57"
] | B | I have done this problem by substituting numbers
We have n!/2!(n-2)! = 91
or n(n-1)/2 = 182 or (n-14)(n+13) =0
n=14 satisfies the equation
B |
AQUA-RAT | AQUA-RAT-34369 | Circular motion question
1. Nov 10, 2005
donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
=.942/.000216 = 4361.11 rad/s
Does this approach look right?
2. Nov 10, 2005
BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
r = radius
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
The following is multiple choice question (with options) to answer.
A circular rim A having a diameter of 36 inches is rotating at the rate of x inches/min. Another circular rim B with a diameter of 84 inches is rotating at the rate of y inches/min. What is the value of y in terms of x, if both the rims reach their starting positions at the same time after every rotation. | [
"3x/7",
"4x/5",
"7x/5",
"5x/7"
] | A | t = S1/V1 = S2/V2
or, 36/x = 84/y
or, y = 36x/84 = 3x/7 (Answer A) |
AQUA-RAT | AQUA-RAT-34370 | If you want to show ##H_n## gets greater than some integer M, you just have to pick n big enough to generate 2M 1/2's. I think that's probably n = 2##^{2M}##.
The following is multiple choice question (with options) to answer.
When positive integer H is divided by positive integer B, the result is 4.35. Which of the following could be the reminder when H is divided by B? | [
" 13",
" 14",
" 15",
" 16"
] | B | the remainder will be obtained from the decimal part when H is divided by B i.e. 0.35
0.35 = 35/100 = 7/20 so possible remainders are 7,14,21,28. Only option B-14 satisfies this
PS: for B-14 H=174 and B=40 |
AQUA-RAT | AQUA-RAT-34371 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
FEDERAL bank offers an interest of 5% per annum compounded annually on all its deposits. If $10,000 is deposited, what will be the ratio of the interest earned in the 4th year to the interest earned in the 5th year? | [
"1:5",
"625:3125",
"100:105",
"100^4:105^4"
] | C | Hi Bunuel,
Here is my approach: is this correct?
FEDERAL bank offers an interest of 5% per annum compounded annually on all its deposits
Interest earned in 4 year= 10000(1+0.05)^4
Interest earned in 5 year= 10000(1+0.05)^5
Ratio= {10000(1.05)^4}/{10000(1.05^5)} =>1.05^4/1.05^5 =>1/1.05 Multiplied by 100 in both numerator and denominator gives 100:105
Hence Ans:C |
AQUA-RAT | AQUA-RAT-34372 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
If the length of the sides of two cubes are in the ratio 4:1, what is the ratio of their total surface area? | [
"4:1",
"6:1",
"8:1",
"16:1"
] | D | Let x be the length of the small cube's side.
The total surface area of the small cube is 6x^2.
The total surface area of the large cube is 6(4x)^2=96x^2.
The ratio of surface areas is 16:1.
The answer is D. |
AQUA-RAT | AQUA-RAT-34373 | -
– Lucian Apr 11 '14 at 17:14
Java program for proving this
-
Thanks for the code, that's a useful method of verification, but I'm trying to find a solid method to verify by hand that it is the smallest and all other $x$ of the same form are the only solutions; I was worried that feeling my way through by intuition I'd somehow miss some possible values of $x$. – user142340 Apr 11 '14 at 17:37
No problem, I didn't know how to do this by hand, but I knew I could contribute with this code and a solution to this particular problem. – Kristoffer Ryhl Apr 11 '14 at 17:56
$2\!-\!5 = \color{#c00}{-3} = 8\!-\!11\,$ so $\,x\equiv \color{#c00}{-3}\,$ mod $\,5,11\!\iff\! x\equiv -3\equiv 52\pmod{55},\,$ and $\,52\equiv1\pmod 3$
-
The following is multiple choice question (with options) to answer.
What is the leastvalue of x. So that 45x09 is divisible by 3? | [
"0",
"5",
"2",
"6"
] | A | The sum of the digits of the number is divisible by 3, then the number is divisible by3.
4 + 5 + x + 0+ 9 = 18 + x
Least value of x may be 0
Therefore 18 + 0= 18 is divisible by 3.
A |
AQUA-RAT | AQUA-RAT-34374 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Find the value of (X) in the given equation ?
35% of 1500 + X = 45% of 4200 – 320 | [
"910",
"980",
"1012",
"1045"
] | D | Explanation:
35% of 1500 + X = 45% of 4200 – 320
(35/100 * 1500) + X = (45/100 * 4200) – 320
525 + X = 1890 – 320
X = 1890 – 320 – 525
X = 1890 – 845
X = 1045
ANSWER: D |
AQUA-RAT | AQUA-RAT-34375 | Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have
\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}
And so we have our case where
\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}
The following is multiple choice question (with options) to answer.
A basketball coach will select the members of a five-player team from among 8 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter? | [
"1/9",
"1/6",
"2/9",
"5/14"
] | D | you can but that will require more cases and will be time consuming.
straight forward way is to group J and P as always present on the team and as order does not matter so we just need to find the total number of ways to select rest three players = 6c3
total number of ways of selecting 5 players out of 8 = 8c5
probability = 6c3/8c5 = 5/14
D |
AQUA-RAT | AQUA-RAT-34376 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A truck driver drove for 2 days. On the second day, he drove 3 hours longer and at an average speed of 15 miles per hour faster than he drove on the first day. If he drove a total of 900 miles and spent 21 hours driving during the 2 days, what was his average speed on the first day, in miles per hour? | [
"25.0",
"30.3",
"34.2",
"40.5"
] | C | Day 1
2t+3=21
t=9
Day 2
t+3
9+3=12
9r+12(r+15)=900
r=34.2
Answer:C |
AQUA-RAT | AQUA-RAT-34377 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The ratio of the number of ladies to gents at a party was 1:2 but when 2 ladies and 2 gents left, the ratio became 1:3. How many people were at the party originally? | [
"22",
"18",
"12",
"66"
] | C | x, 2x
(x-2):(2x-2) = 1:3
3x-6 = 2x-2
x = 4
x+2x = 3x
=> 3*4 = 12
Answer: C |
AQUA-RAT | AQUA-RAT-34378 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
On a certain road, 10% of the motorists exceed the posted speed limit and receive speeding tickets, but 30% of the motorists who exceed the posted speed limit do not receive speeding tickets. What percent of the motorists on that road exceed the posted speed limit? | [
"10.5%",
"12.5%",
"14.3%",
"22%"
] | C | suppose there are X motorists.
10% of them exceeded the speed limit and received the ticket, i.e. X/10.
Again, suppose total no. of motorists who exceeded the speed limit are Y.
30% of Y exceeded the speed limit but didn't received the ticket, i.e. 3Y/10.
it means 7Y/10 received the ticket.
hence, 7Y/10 = X/10
or Y/X=1/7
or Y/X * 100= 1/7 * 100= 14.3%
C |
AQUA-RAT | AQUA-RAT-34379 | python, python-3.x, calculator
Title: Taxi fare calculator, trip recorder, and fuel calculator I solved this practice problem in preparation for school exams.
The owner of a taxi company wants a system that calculates how much money his taxis take in one day.
Write and test a program for the owner.
Your program must include appropriate prompts for the entry of data.
Error messages and other output need to be set out clearly and understandably.
All variables, constants and other identifiers must have meaningful names.
You will need to complete these three tasks. Each task must be fully tested.
TASK 1 – calculate the money owed for one trip
The cost of a trip in a taxi is calculated based on the numbers of KMs traveled and the type of taxi that you are traveling in. The taxi company has three different types of taxi available:
a saloon car, that seats up to 4,
a people carrier, that seats up 8,
a mini-van, that seats up to 12.
For a trip in a saloon car the base amount is RM2.50 and then a charge of RM1.00 per KM. For a trip in a people carrier the base amount is RM4.00 and a charge of RM1.25 per KM. For a trip in a mini- van the base amount is RM5.00 and a charge of RM1.50 per KM. The minimum trip length is 3KM and the maximum trip length is 50KM. Once the trip is complete a 6% service tax is added.
TASK 2 – record what happens in a day
The owner of the taxi company wants to keep a record of the journeys done by each one of his taxis in a day. For each one of his three taxis record the length of each trip and the number of people carried. Your program should store a maximum of 24 trips or 350km worth of trips, whichever comes first. Your program should be able to output a list of the jobs done by each one of the three taxis.
TASK 3 – calculate the money taken for all the taxis at the end of the day.
At the end of the day use the data stored for each taxi to calculate the total amount of money taken and the total number of people carried by each one of the three taxis. Using the average price of RM2.79 per litre use the information in the table below to calculate the fuel cost for each taxi:
The following is multiple choice question (with options) to answer.
A cab driver 5 days income was $300, $150, $750, $200 , $600. Then his average income is? | [
"$350",
"$375",
"$400",
"$425"
] | C | avg = sum of observations/number of observations
avg income=(300+150+750+200+600)/5 = 400
Answer is C |
AQUA-RAT | AQUA-RAT-34380 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is? | [
"2898",
"277",
"500",
"297"
] | C | Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Answer: C |
AQUA-RAT | AQUA-RAT-34381 | 3. Sep 26, 2011
### Hioj
The following assumes $g=10\dfrac{m}{s}$ for simplicity.
We have
$$\dfrac{1}{2}mv^{2}=mgh$$
and so
$$h=\dfrac{v^{2}}{2g}=\dfrac{1}{20}$$ as $v^{2}=1\dfrac{m}{s}$ for the first box.
For the second box, it must be true that the kinetic energy with the end speed $v$ equals the kinetic energy it starts off with $(\dfrac{1}{2}mv_{0}^{2})$ plus the gained potential energy. Then
$$\dfrac{1}{2}mv^{2}=\dfrac{1}{2}mv_{0}^{2}+mgh$$
$$\Longrightarrow v^{2}=v_{0}^{2}+2gh=1+\dfrac{2g}{20}=1+1=2$$
$$\Longrightarrow v=\sqrt{2}.$$
Correct?
Could this be done any easier?
The following is multiple choice question (with options) to answer.
Assume the equation U = mgh. If m is quadrupled and h is halved, by what factor does U change? | [
"3",
"2",
"4",
"1/2"
] | B | The equation is U = mgh.
If m is quadrupled and h is halved, the equation becomes:
U = (4m)*g*(h/2)
U = 2mgh
Initially U was mgh and now it is 2mgh, therefore U increases by a factor of 2.
Answer = B = 2 |
AQUA-RAT | AQUA-RAT-34382 | c a b c u v c
7 5 8 7 1 2 7 = 7
13 7 15 13 1 3 13 = 13
19 16 21 19 2 3 19 = 19
31 11 35 31 1 5 31 = 31
37 33 40 37 3 4 37 = 37
43 13 48 43 1 6 43 = 43
49 39 55 49 3 5 49 = 7^2
61 56 65 61 4 5 61 = 61
67 32 77 67 2 7 67 = 67
73 17 80 73 1 8 73 = 73
79 51 91 79 3 7 79 = 79
91 19 99 91 1 9 91 = 7 * 13
91 85 96 91 5 6 91 = 7 * 13
97 57 112 97 3 8 97 = 97
103 40 117 103 2 9 103 = 103
109 95 119 109 5 7 109 = 109
127 120 133 127 6 7 127 = 127
133 23 143 133 1 11 133 = 7 * 19
133 88 153 133 4 9 133 = 7 * 19
139 69 160 139 3 10 139 = 139
151 115 171 151 5 9 151 = 151
157 25 168 157 1 12 157 = 157
163 75 187 163 3 11 163 = 163
169 161 176 169 7 8 169 = 13^2
181 104 209 181 4 11 181 = 181
193 175 207 193 7 9 193 = 193
199 56 221 199 2 13 199 = 199
211 29 224 211 1 14 211 = 211
217 208 225 217 8 9 217 = 7 * 31
217 87 247 217 3 13 217 = 7 * 31
223 168 253 223 6 11 223 = 223
229 145 264 229 5 12 229 = 229
241 31 255 241 1 15 241 = 241
247 203 275 247 7 11 247 = 13 * 19
247 93 280 247 3 14 247 = 13 * 19
259 155 299 259 5 13 259 = 7 * 37
259 64 285 259 2 15 259 = 7 * 37
271 261 280 271 9 10 271 = 271
277 217 312 277 7 12 277 = 277
283 192 325 283 6 13 283 = 283
301 136 345 301 4 15 301 = 7 * 43
301 279 319 301 9 11 301 = 7 * 43
307 35 323 307 1 17 307 = 307
313 105 352 313 3 16 313 = 313
331 320 341 331 10 11 331 = 331
337 272 377 337 8 13 337 = 337
343 37 360 343 1 18 343 = 7^3
349 111 391 349 3 17 349 = 349
361 185 416 361 5 16 361 = 19^2
The following is multiple choice question (with options) to answer.
3, 7, 15, 31, 63, ? | [
"89",
"127",
"142",
"158"
] | B | Each number in the series is the preceding number multiplied by 2 and then increased by 1.
Answer : B. |
AQUA-RAT | AQUA-RAT-34383 | of their probabilities, and that if the probability of two events A and B occurring together is the product of their probabilities, the two events are independent. Find P(A I ). If A and B are two independent events and P(A)=(3)/(6) and P(AcapB)=(4)/(9) then the value of P(B) will be. That is, if P(A|B) = P(A), and vice versa. A A∩B B B A. A and B are two events. I had a student bring this up today, I know 0 cannot be divided by zero. Given two events A and B, from the sigma-field of a probability space, with the unconditional probability of B (that is, of the event B occurring) being greater than zero, P(B) > 0, the conditional probability of A given B is defined as the quotient of the probability of the joint of events A and B, and the probability of B:. Events may or may not be independent; according to the definition, two events, A and B, are independent iff P(A∩B) = P(A) P(B). 30 and P(B) =. Drawing out a red ball with dots therefore represents a complementary event relative to the combination of Events A and B. 4 and P(B) = 0. When events A, B are independent, the probability of both happening can be computed by saying the event A happen first with P(A) and the event B happens afterwards with P(B). Intuitively I think it is because the. If A and B are independent events, then the events A and B’ are also independent. May 04,2020 - A and B are two independent events. In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is P(B). The intersection of those would then be equal to P(A), wouldn't it? This means that the formula would be like this: P(A)=P(A)*P(B). Given two spinners (this sort of thing) that each have the numbers 1, 2, and 3 (in place of
The following is multiple choice question (with options) to answer.
The probability that event B occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event A will occur? | [
"0.05",
"0.45",
"0.15",
"0.5"
] | B | P(A or B) = P (A) + P(B) - p(a n b)
0.6= 0.4 + P(A) - 0.25
P(A) = 0.45
Ans : B |
AQUA-RAT | AQUA-RAT-34384 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
There are 7 magazines lying on a table; 4 are fashion magazines and the other 3 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected? | [
"1/2",
"7/8",
"32/35",
"11/12"
] | B | answer is B
Total probability=8C3=48
4C3 +4C2*4C1+4C1*4C2=4+24+24=42
therefore the probability that at least one of the fashion magazines will be selected=42/48=7/8
B |
AQUA-RAT | AQUA-RAT-34385 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
Jamshid can paint a fence in 50 percent less time than Taimour can when each works alone. When they work together, they can paint the fence in 4 hours. How long would it take Taimour to paint the fence alone? | [
"6 hours",
"8 hours",
"12 hours",
"24 hours"
] | C | I believe the answer is C. Please see below for explanation.
if Jamshid can paint a dence in 50 percent less time then Taimour we can infer the following rate J = 2T
if working together they can do the job in 8 hours we can infer 1 = 2T+T * 4 => 1/12
Working alone Taimour can do the job in 1 = 1/12 * hours => 12
Answer C |
AQUA-RAT | AQUA-RAT-34386 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A case of 12 rolls of paper towels sells for $9. The cost of one roll sold individually is $1.What is the percent R of savings per roll for the 12-roll package over the cost of 12 rolls purchased individually? | [
"9%",
"11%",
"15%",
"25%"
] | D | Cost of 12 paper towels individually = 1*12=12
Cost of a set of 12 paper towels = 9
Cost of one roll = 9/12 = 3/4 = 0.75
Savings per roll = 1-.75 = 0.25
% of savings is R= .25/ 1 *100 = 25% D is the answer. |
AQUA-RAT | AQUA-RAT-34387 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Annika hikes at a constant rate of 12 minutes per kilometer. She has hiked 2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 40 minutes. If Annika continues east, then turns around and retraces her path to reach the start of the trail in exactly 40 minutes, for how many kilometers total did she hike east? | [
"3.625",
"3.5",
"3",
"3.04"
] | D | Set up two R x T =D cases.
1. 1/12 km/ min x T = 2.75
from which T= 33 mins.
We know total journey time now is 40 +33 =73.
The rate is the same ie 1/12km/min.
set up second R x T =D case.
1/12 km/min x 73 = 6.08 km
Now the total journey would be halved as distance would be same in each direction. 6.08/2 =3.04
D. |
AQUA-RAT | AQUA-RAT-34388 | # Difference between revisions of "2021 AMC 10B Problems/Problem 3"
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
## Solution 2 (Fast but Not Rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Solution 3
The following is multiple choice question (with options) to answer.
A bulldog received 1,600 votes in a bulldog competition, giving him 20 percent of the votes. What percent of the remaining votes would he have needed to receive in order to win 32 percent of the total votes? | [
"10%",
"12.5%",
"15%",
"17.5%"
] | C | There were 80% of the votes remaining and he needed 12%/80% = 3/20 = 15%
The answer is C. |
AQUA-RAT | AQUA-RAT-34389 | 713, 731, 799, 841, 851, 899
Next come a pair of twins, a 99, then another pair of twins, then another 99! 713 and 731 are twins because they have the same digits, with the 1 and 3 reversed. 731 we have already seen: it is from the same 17-sequence as 527 and 629. 713 is $729 - 16 = 27^2 - 4^2 = (27-4)\cdot (27+4)$. 799 is again from the 17-sequence beginning with 289.
841 and 851 are twins because they have the same digits except the 4 and the 5, which are consecutive. 841 is $29^2$, and 851 is $900 - 49 = 30^2 - 7^2 = (30 - 7) \cdot (30 + 7)$. Finally we have 899 which is $900 - 1 = 30^2 - 1^2 = (30 - 1) \cdot (30 + 1)$.
901, 943, 961, 989
I haven’t thought of a nice story to tell about these—I think of the last three as sort of “sporadic”, but there’s only three of them so it’s not that hard. Someone else could probably come up with a nice mnemonic.
901 in any case is not too hard to remember because it’s a twin with 899, and it’s also the end of the 17-sequence that started with 289.
943 is $23 \cdot 41$. It’s $1024 - 81 = 32^2 - 9^2$, but unlike some of the other differences of squares I’ve highlighted, I doubt this will actually help me remember it.
961 is $31^2$. I think it’s cute that $169$, $196$, and $961$ are all perfect squares.
Last but not least, $989 = 23 \cdot 43$. If you happen to know that $33^2 = 1089$ (I sure don’t!), then this is easy to remember as $33^2 - 10^2 = (33-10) \cdot (33+10)$.
Once again
The following is multiple choice question (with options) to answer.
If DREAM is coded as 78026 and CHILD is coded as 53417,how can LEADER be coded ? | [
"102078",
"102708",
"102087",
"102780"
] | B | from given two words L=1 E=0 A=2 D=7 E=0 and R=8
ANSWER:B |
AQUA-RAT | AQUA-RAT-34390 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
A number when divided by 4, gives 40 as quotient and 0 as remainder. What will be the remainder when dividing the same number by 3 | [
"A)1",
"B)3",
"C)4",
"D)6"
] | A | Explanation:
P ÷ 4 = 40
=> P = 40 * 4 =160
P / 3 = 160/ 3 = 53, remainder = 1
Answer: Option A |
AQUA-RAT | AQUA-RAT-34391 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A person can swim in still water at 4 km/h. If the speed of water 2 km/h, how many hours will the man take to swim back against the current for 6km? | [
"3",
"6",
"8",
"9"
] | A | M = 4
S = 2
US = 4 - 2 = 2
D = 6
T = 6/2 = 3
Answer:A |
AQUA-RAT | AQUA-RAT-34392 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
In honor of the year 2014, a donor has purchased 2016 books to be distributed evenly among the elementary schools in a certain school district. If each school must receive the same number of books, and there are to be no books remaining, which of the following is NOT a number of books that each school could receive? | [
"18",
"36",
"42",
"55"
] | D | All options other than 'D' are factors of 2014. Hence, 'D' is the answer |
AQUA-RAT | AQUA-RAT-34393 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
The ratio of Gomati’s and Rashmi’s ages is 3 : 5 respectively. After ten years this ratio will become 2 : 3. What is Rashmi’s age in years? | [
"50",
"40",
"60",
"Cannot be determined"
] | A | Quicker Method : After 10 years, the ratio of Gomati and Rashmi’s ages is 2 : 3. We can also write 2 : 3 as 4 : 6. Now, difference in the ratio is I in both the cases.
Therefore, 1 = 10
∴ 5 → 5 × 10 = 50 years.
Answer A |
AQUA-RAT | AQUA-RAT-34394 | # Calculate probability from normal distribution WITHOUT calculator
Susan commutes daily from her home to her office. The average time for a one-way trip is $24$ minutes with a standard deviation of $3.8$ minutes. Assume that the trip time follows a normal distribution.
(f) A trip to a client's office from her home takes $30$ more minutes than twice the time to her own office. Let $W$ be the time for a trip to the client's office.
Find the probability that a trip to the client's office takes more than $1$ hour but less than $1.5$ hours.
There are seven parts to this question, but I'm confused on just this part.
What I have tried so far:
So I know that $E(Y) = 24$, $V(Y) = 3.8^2$, and $SD(Y) = 3.8$
I also know that $W = 2Y + 30$
I've also calculated:
$$E(W) = 2E(Y) + 30 = 78$$
$$V(W) = 2^2V(Y) \approx 57.76$$
But at this point, I'm not sure how to proceed. I suppose I'd be looking for $P(60 < W < 90)$, but I'm not sure how to go about calculating this.
The following is multiple choice question (with options) to answer.
a man leaves office daily at 7 pm, a driver with car comes from his home to pick up him from office and bring back home.one day he gets free at 5:30 pm....,instead of waiting for car ,he starts walking towards his home,on the way he meets the car and returns home by car,he reaches home 20 minutes earlier than usual. In how much time does he reach home usually ? | [
"1 hr and 20 mins",
"1 hr and 30 mins",
"1 hr and 40 mins",
"1 hr and 10 mins"
] | A | Since the car has met the person 20 minutes before hand, it has actually saved 10 mins of journey (to and fro)
since the man has started 1.30 hrs before and car has met him 10 mins before actual time he takes to reach daily is 1 hr and 20 mins
ANSWER:A |
AQUA-RAT | AQUA-RAT-34395 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
In Richie's company 60% of the employees earn less than $50,000 a year, 60% of the employees earn more than $40,000 a year, 11% of the employees earn $43,000 a year and 5% of the employees earn $49,000 a year. What is the median salary for the company? | [
"43.0",
"45.5",
"46.0",
"49.0"
] | A | In Richie's company 60% of the employees earn less than $50,000 a year, 60% of the employees earn more than $40,000 a year, 50 and 51 employee will be each 43K. hence median = (43K+43k)/2=A |
AQUA-RAT | AQUA-RAT-34396 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Souju's age is 1/4th of her father's age. Souju's father's age will be twice Tharak's age after
10 years. If Tharak's eight birthdays was celebrated two years before, then what is Souju's present age. | [
"6.5",
"7.5",
"8.5",
"8"
] | B | tharak's present age =8+2=10
souju's father's age after 10 years =2(10+10)=40
souju's father's present age =40−10=30
tharak's present age =1/4 *30=7.5
ANSWER:B |
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