source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35597 | # 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.
The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 10, which of the following represents the maximum possible value of the largest integer? | [
"4",
"6",
"20",
"10"
] | C | Let the distinct number be A,B,C, and D
Its given A > B > C > D
also A + B + C + D =40 and A + B = 10 means C + D = 30
Since the question ask for the largest possible number we should choose the least value for A and B,C. So D should be 20. If D=20 then C = 10 > A,B <10
Answer : C |
AQUA-RAT | AQUA-RAT-35598 | 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 with a speed of 23 kmph in still water. If the stream flows at 9 kmph, then the speed in downstream is? | [
"32",
"26",
"20",
"87"
] | A | M = 23
S = 9
DS = 23 + 9
=32
Answer: A |
AQUA-RAT | AQUA-RAT-35599 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup? | [
"7/11",
"6/7",
"1/5",
"2/7"
] | C | Explanation :
Let, the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water then quantity of water in new mixture
=>3−(3x/8)+x litres.
Quantity of syrup in new mixture =5−5x/8 litres.
Therefore,
=>3−(3x/8)+x=5−5x/8.
=>5x+24=40−5x.
=>x=8/5.
So part of the mixture replaced,
=>(8/5)×(1/8)=1/5.
Answer : C |
AQUA-RAT | AQUA-RAT-35600 | # Sitting $n$ men and $n$ woman around a table having the wedding couple seated together
The problems says
At the wedding of John and Mary there are $n$ men and $n$ women. In how many ways they can sit at a round table, so that no two men is next to each other and John and Mary sit together?
I have some doubts because of the answer to his problem, which doesn't match the one I got.
My attempt:
Let suppose that John and Mary are already seated and that people begin sitting from the side of Mary. We have two cases. One where the person next to Mary is a man and the other when it is a woman.
1st Case:
We can set the men en $n!/n = (n-1)!$ different ways.
The are $n-1$ gaps to put the women (since next to a man are Mary and John together. So there are $(n-1)!$ different ways to sit the women.
By the rule of product we have that there are $[(n-1)!]^2$ number of ways of sitting n men and n women beginning with a man next to Mary.
2nd Case (A woman sits next to Mary):
By the same reasoning as above we find that there are $[(n-1)!]^2$ number of ways of sitting them.
Finally by the rule of sum we have that there is a total of $2[(n-1)!]^2$ number of ways of sitting n men and n women having John and Mary always seated together.
What I got matches the answer of the textbook but I feel the reasoning I followed wasn't that right. The fact of having assumed that ''The are $n-1$ gaps to put the women..'' doesn't convince me. Certainly next to the first man is Mary (that is, the gap of that side is occupied) but nothing is telling that after putting the last men there will be no gap between him and John, so there may be anyway $n$ gaps to put the women
• Surely you mean "wedding" couple rather than "weeding"? :) – hypergeometric Nov 1 '16 at 17:56
• @hypergeometric Ups. Yep, I mean that (: – Jazz Nov 1 '16 at 18:03
The following is multiple choice question (with options) to answer.
Three photographers, Lisa, Mike and Norm, take photos of a wedding. The total of Lisa and Mikes photos is 80 less than the sum of Mike's and Norms. If Norms photos number 10 more than twice the number of Lisa's photos, then how many photos did Norm Take? | [
"40",
"50",
"150",
"80"
] | C | L+M=M+N-80 / N=2L+10
80=M+N-L-M
80=N-L
80=2L+10-L
70=L
2(70)+10=150
C |
AQUA-RAT | AQUA-RAT-35601 | # 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.
The average of seven numbers is 27. If one number is excluded, the average becomes 25. The excluded number is | [
"25",
"27",
"30",
"39"
] | D | Sol.
Therefore excluded number
= (27 × 7) - ( 25 × 6)
= 189 – 150
= 39.
Answer D |
AQUA-RAT | AQUA-RAT-35602 | ## Sunday, May 23, 2010
### GR9768.63: Intersections
63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-
Note: the answer provided in the Answer Sheet is D, 3 points of intersection.
result from wolframalpha
#### 13 comments:
1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.
2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!
3. Proper result from wolfram:
The following is multiple choice question (with options) to answer.
In the xy-coordinate plane, the graph of y = -x^2 + 9 intersects line L at (p,-5) and (t,-7). What is the least possible value of the slope of line L? | [
"6",
"-1",
"-2",
"-6"
] | B | We need to find out the value of p and L to get to the slope.
Line L and Graph y intersect at point (p,-5). hence, x= p and Y=-5 should sactisfy the graph. soliving
5 = -p2 +9
p2 = 4
p = + or - 2
simillarly point (t,-7) should satisfy the equation. hence x=t and Y=-7.
-7 = -t2+9
t = +or - 4
considering p = -2 and t =4, the least slope is (-7+5)/(4-2) = -1
IMO option B is correct answer. |
AQUA-RAT | AQUA-RAT-35603 | bruce
the jet will fly 9.68 hours to cover either distance
bruce
Riley is planning to plant a lawn in his yard. He will need 9 pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs$3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound? Vonna Reply 33.336 Robinson Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have? mikayla Reply 2q=n ((2q).05) + ((q).25) = 8.75 .1q + .25q = 8.75 .35q = 8.75 q = 25 quarters 2(q) 2 (25) = 50 nickles Answer check 25 x .25 = 6.25 50 x .05 = 2.50 6.25 + 2.50 = 8.75 Melissa John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?
7-$10 21-$5
Robert
Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.)
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hour longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
8mph
michele
16mph
Robert
3.8 mph
Ped
16 goes into 80 5times while 20 goes into 80 4times and is 4mph faster
Robert
The following is multiple choice question (with options) to answer.
A rectangular-shaped carpet remnant that measures x feet by y feet is priced at $30. What is the cost of the carpet, in dollars per square yard? (9 square feet = 1 square yard) | [
"30xy",
"270/(xy)",
"xy/9",
"xy/30"
] | B | xy sq ft = $30
1 sq ft = $30/xy
multiplying by 9 on both side
9 sq ft = $270/xy
or 1 sq yard = $270/xy
Hence B. |
AQUA-RAT | AQUA-RAT-35604 | root we have to have an x for which x^2 = -3. As -9 < 0, no real value of x can satisfy this equation. An equation in one unknown quantity in the form ax 2 + bx + c = 0 is called quadratic equation. This means that x = s and x = t are both solutions, and hence they are the roots. For example: Then the root is x = -3, since -3 + 3 = 0. Solving quadratic equations by completing square. The solution of quadratic equation formulas is also called roots. If any quadratic equation has no real solution then it may have two complex solutions. One example is solving quadratic inequalities. In most practical situations, the use of complex numbers does make sense, so we say there is no solution. Sometimes they all have real numbers or complex numbers, or just imaginary number. An example of a quadratic function with only one root is the function x^2. This is the case for both x = 1 and x = -1. This means to find the points on a coordinate grid where the graphed equation crosses the x-axis, or the horizontal axis. Coefficients A, B, and C determine the graph properties and roots of the equation. Learn all about the quadratic formula with this step-by-step guide: Quadratic Formula, The MathPapa Guide; Video Lesson. Let α and β be the roots of the general form of the quadratic equation :ax 2 + bx + c = 0. In the above formula, (√ b 2-4ac) is called discriminant (d). So let us focus on it. Lastly, we had the completing the squares method where we try to write the function as (x-p)^2 + q. The most common way people learn how to determine the the roots of a quadratic function is by factorizing. The root is the value of x that can solve the equations. $$\frac{-1}{3}$$ because it is the value of x for which f(x) = 0. f(x) = x 2 +2x − 3 (-3, 0) and (1, 0) are the solutions to this equation since -3 and 1 are the values for which f(x) = 0. Solutions or Roots of Quadratic Equations . Example 5: The quadratic equations x 2
The following is multiple choice question (with options) to answer.
The roots of the equation 3x2 - 12x + 10 = 0 are? | [
"rational and unequal",
"complex",
"real and equal",
"irrational and unequal"
] | D | Explanation:
The discriminant of the quadratic equation is (-12)2 - 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
ANSWER IS D |
AQUA-RAT | AQUA-RAT-35605 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
A father is four times as old as his son. In twenty years, he'll be twice as old. How old are they now? | [
"5,40",
"10,40",
"20,40",
"10,30"
] | B | If F is age of father and s is age of son
f=4s
f+20 = 2*(s+20)
4s+20 = 2s+40
2s=20
s=10
F=40
ANSWER:B |
AQUA-RAT | AQUA-RAT-35606 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. | [
"48.55",
"42.25",
"50",
"51.25"
] | A | Explanation:
Average weight of 16 boys = 50.25
Total weight of 16 boys = 50.25 × 16
Average weight of remaining 8 boys = 45.15
Total weight of remaining 8 boys = 45.15 × 8
Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)
Total boys = 16 + 8 = 24
Average weight of all the boys = (50.25×16)+(45.15×8)/24
=(50.25×2)+(45.15×1)/3
=(16.75×2)+15.05
=33.5+15.05
=48.55
Answer: Option A |
AQUA-RAT | AQUA-RAT-35607 | solution: solution to get the median: the mean compared to sets a mean questions with solutions b Question! To new CBSE exam Pattern, MCQ questions for Class 10to help you are presented 6 ICSE.. A- a ; each being equal to the common difference we will help you sum must 12! Step 3: the given mean of a set of numbers deep understanding of maths.... Mean μ ' and the new standard deviation σ MCQ questions for Class 10to help.... To get the mean compared to sets a, a, b and C. a given data set you. Is 8 of all 5 numbers about Harmonic progression formula for nth term, sum of 5! Multiply all data values by a constant k and Calculate the mean been designed to test for understanding! Μ ' and the new standard deviation σ understanding of maths concepts b – =. Averages problem is a vey easy Question so the sequence is 2, 3 8... Mean is also called average of 56, 41, 59, 52, 42 and.... ' and the new standard deviation σ exercises on calculating the mean of given sets and word problems on mean. Set, you may need to determine the possible values of the 3 numbers in. Aggarwal Solutions Class 10 Chapter 9 - Benefits the workers earn a salary of Rs 6 to get the for. Set has a mean μ ' and the new standard deviation of the page 7 +11 ) /5 5.6! Using the average of the given classes will give the sum is 18 Divide by! Their sum must be 12, so you have x+y+z=12 ⇒ a = A- a ; each being equal the. To sets a, b are in A.P x in the middle solution: Question 21 properties... One of the 5 numbers using the average of 56, 41, 59, 52, and..., a, b and C. a given data set a = 3 if r1, r2 r3... B are in A.P a ) Calculate the mean example 2: Compute sum of set! So you have x+y+z=12 a complete solution 2 # the mean questions with solutions mean (... Lower part of the 20 people and Harmonic mean the minimum number 1! Between the first and second must be equal to the difference second and third, giving the x-y=y-z! =
The following is multiple choice question (with options) to answer.
The sum of the mean, the median, and the range of the set {5,7,9} equals which one of the following values? | [
"11",
"17",
"14",
"12"
] | B | Here Mean => 5+7+9/3 => 6
median => 7
and range => 9-5 => 4
hence sum => 6+7+4=> 17
Answer: B |
AQUA-RAT | AQUA-RAT-35608 | 1 Answer
Your decomposition of the problem is valid, and only works because those two divisors are co-prime (there is no number bigger than $$1$$ dividing both divisors). This means that if a number is divisible by $$3$$ and $$4$$ it is automatically divisible by $$12$$, and you can check each condition independently – which you did.
• Like if the factors were 6 and 3 then it is not necessary that 18 would divide the numbers, but 12 would? – BJKShah Mar 13 at 13:10
• @BJKShah Indeed. – Parcly Taxel Mar 13 at 13:11
The following is multiple choice question (with options) to answer.
If x is divisible by 4 and 6, which of the following must divide evenly into x?
I. 2
II. 35
III. 70 | [
"I only",
"I and II",
"II and III",
"II only"
] | A | If x is divisible by 4,6 means it will be = or > 12, 24,48 etc...
That are not divisible by 35, 70.
So, the answer is A |
AQUA-RAT | AQUA-RAT-35609 | # Thread: vertices of a square, coordinate geometry and differentiation
1. ## vertices of a square, coordinate geometry and differentiation
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
For the first question I observed that I have 4 unknowns, the two coordinates of each point. So I figured I should determine 4 equations that would help me find these; I used the following remarks to define these equations:
The gradient of AC will be perpendicular to the gradient of OB,
The gradient of OA will be perpendicular to the gradient of AB
The gradient of OC will be perpendicular to the gradient of CB,
The distance between A and C will be equal to the distance between O and B
However I’m not sure that this is the right (or at least, most effective) way to proceed as I’ve ended up with a mess of equations, often ending up quadratic, that appear to be leading me nowhere. Am I on the right track, have I used the wrong equations or am I just using a totally incorrect method?
2. Originally Posted by red and white kop!
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
Assuming that the square has a side parallel to the two x-axis and $p>0~\&~q>0$ then
$A0,q)~\&~Cp,0).$
Is the second question part of the first?
3. The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
The following is multiple choice question (with options) to answer.
In the graph below, no axes or origin is shown. If point B's coordinates are (9,3), which of the following coordinates would most likely be A's? | [
"(17, -2)",
"(10, 6)",
"(6, 8)",
"(-10, 3)"
] | B | wherever be the axis as per question y coordinate of point A will be greater than 3 and X coordinate will be lower than 9.
B rules.. |
AQUA-RAT | AQUA-RAT-35610 | } \left( \dfrac { 2y+b }{ a-2y } \right)$$$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right)$$Maths
The following is multiple choice question (with options) to answer.
2y - x = 2xy . If x and y are integers, which of the following could equal y? | [
"2",
"1",
"0",
"3"
] | C | Plug in the answer choices in the equation from the question stem.
A) y = 2 >>> 4-x = 4x >>> No value of x will satisfy this, not even 0. POE
B) y = 1 >>> 2 - x = 2x >>> Same, POE
C) y = 0 >>> -x = 0 >>> x This is the answer
Answer C |
AQUA-RAT | AQUA-RAT-35611 | ### 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.
A began business with 12500 and is joined afterwards by B with 37500. When did B join, if the profits at the end of the year are divided equally? | [
"8 months",
"9 months",
"10 months",
"7 months"
] | A | Let B join after x months of the start of the business so that B’s money is invested for (12 – x) months.
∴ Profit ratio is 12 × 12500 : (12 – x) × 37500
or 12 : 3(12 – x)
Since profit is equally divided so
12 = 3(12 – x) or x = 8. Thus B joined after 8 months.
Answer A |
AQUA-RAT | AQUA-RAT-35612 | 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.
Two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively. When they meet, it is found that one train has traveled 60 km more than the other one. The distance between the two stations is? | [
"km",
"km",
"km",
"km"
] | B | 1h ----- 5
? ------ 60
12 h
RS = 16 + 21 = 37
T = 12
D = 37 * 12 = 444
Answer :B |
AQUA-RAT | AQUA-RAT-35613 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
By selling 100 pencil, a trader gains the cost of 40 Pencil. Find his gain percentage? | [
"26 1/3%",
"51 1/3%",
"40%",
"53 1/3%"
] | C | C
40%
Let the CP of each pencil be Rs. 1.
CP of 100 pens = Rs. 100
Profit = Cost of 40 pencil = Rs. 40
Profit% = 40/100 * 100 = 40% |
AQUA-RAT | AQUA-RAT-35614 | Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.
• All times are GMT -7. The time now is 09:21 AM.
The following is multiple choice question (with options) to answer.
at three different signals,light change after 48 sec,72 sec,108 sec respectively.if at morning 8:20:00 change together,then at what time again they will change together? | [
"8:27:12",
"8:27:24",
"8:27:36",
"8:27:48"
] | A | l.c.m of 48,72,108=7min.12sec
again they will change at=8:27:12
answer A |
AQUA-RAT | AQUA-RAT-35615 | 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 360 meter long is running at a speed of 45 km/hour. In what Ɵme will it pass a bridge of 140
meter length. | [
"20 seconds",
"30 seconds",
"40 seconds",
"50 seconds"
] | C | Explanation:
Speed = 45 Km/hr = 45*(5/18) m/sec
= 25/2 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
=500
∗
225=40seconds
Answer: C |
AQUA-RAT | AQUA-RAT-35616 | 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.
Two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 18 kmph and 21 kmph respectively. When they meet, it is found that one train has traveled 60 km more than the other one. The distance between the two stations is? | [
"468",
"444",
"676",
"767"
] | A | 1h ----- 3
? ------ 60
12 h
RS = 18 + 21 = 39
T = 12
D = 39 * 12 =468
Answer: A |
AQUA-RAT | AQUA-RAT-35617 | (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.
A boy pays Rs. 369 for an article marked at Rs. 600, by enjoying two successive discounts. If the first discount is of 25%, how much should be the second discount? | [
"18",
"99",
"27",
"26"
] | A | Explanation:
First discount = 25% of 600 = Rs. 150.
Thus, the reduced price = 600 – 150 = Rs. 450.
Since the person actually paid Rs. 369, the value of the second discount must be equal to Rs. 81 (450 – 369).
Let the second discount be x
Thus, we get, 81 = x of 450
( 81/450)*100 = 18%
ANSWER: A |
AQUA-RAT | AQUA-RAT-35618 | # 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 dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent? | [
"20%",
"25%",
"30%",
"15%"
] | B | 800 --- 200
100 --- ? => 25%
ANSWER:B |
AQUA-RAT | AQUA-RAT-35619 | Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink] 21 Apr 2017, 03:36
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The following is multiple choice question (with options) to answer.
If x, y, and d are positive numbers such that 10*x/(x+y)+ 20*y/(x+y)=d and if x is less than y, which of the following could be the value of d? | [
"10",
"12",
"15",
"18"
] | D | Answer choice C: d= 15
(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.
We need y to be larger, so let's try answer choice D: d=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.
The correct answer is D. |
AQUA-RAT | AQUA-RAT-35620 | # Runner's High (Speed)
I find the following mind-boggling.
Suppose that runner $$R_1$$ runs distance $$[0,d_1]$$ with average speed $$v_1$$. Runner $$R_2$$ runs $$[0,d_2]$$ with $$d_2>d_1$$ and with average speed $$v_2 > v_1$$. I would have thought that by some application of the intermediate value theorem we can find a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$. This is not necessarily so!
Question. What is the smallest value of $$C\in\mathbb{R}$$ with $$C>1$$ and the following property?
Whenever $$d_2>d_1$$, and $$R_2$$ runs $$[0,d_2]$$ with average speed $$Cv_1$$, then there is a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$.
The following is multiple choice question (with options) to answer.
An athlete runs 200 meters race in 36sec. His speed is? | [
"20km/hr",
"15km/hr",
"30km/hr",
"25km/hr"
] | A | speed = 200/36 = 50/9 m/sec
= 50/9 * 18/5 km/hr = 20km/hr
Answer is A |
AQUA-RAT | AQUA-RAT-35621 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed at which a man can row a boat in still water is 15 km/hr. If he rows downstream, where the speed of current is 3 km/hr, how many seconds will he take to cover 90 meters? | [
"12",
"16",
"17",
"18"
] | D | The speed of the boat downstream = 15 + 3 = 18 km/hr
18 km/hr * 5/18 = 5 m/s
The time taken to cover 90 meters = 90/5 = 18 seconds.
The answer is D. |
AQUA-RAT | AQUA-RAT-35622 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is: | [
"4:5",
"3:5",
"6:8",
"3:5"
] | A | third number be x.
first no = 120% of x = 6x/5
second no = 150% of x = 3x/2
ratio in first two nos = (6x/5:3x/2)=> 4:5
ANSWER A |
AQUA-RAT | AQUA-RAT-35623 | 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 thrice as efficient as B and is, therefore, able to finish a piece of work 10 days earlier than B. In how many days A and B will finish it together? | [
"14 days",
"5 days",
"8 days",
"4 days"
] | D | WC = 3:1
WT = 1:3
x 3x
1/x – 1/3x = 1/8
x = 16/3
3/16 + 1/16 = 1/4 => 4 days
ANSWER:D |
AQUA-RAT | AQUA-RAT-35624 | 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 speeds past a pole in 15 sec and a platform 150 m long in 25 sec, its length is? | [
"100",
"150",
"160",
"225"
] | D | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 150)/25 = x/15 => x = 225 m.
Answer: Option D |
AQUA-RAT | AQUA-RAT-35625 | # 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 shopkeeper purchased 80 kg of potatoes for Rs. 550 and sold the whole lot at the rate of Rs. 9 per kg. What will be his gain percent? | [
"22.6 %",
"30.9 %",
"35.6 %",
"39.5%"
] | B | C.P. of 1 kg = 550/80 = Rs. 6.875
S.P. of 1 kg = Rs. 9
Gain % = 2.125/6.875 * 100 = 212.5/6.875 = 30.9 %
Answer: B |
AQUA-RAT | AQUA-RAT-35626 | You may often has to use both these equation to get to the answer. Solution: To find the pattern, look closely at 24, 28 and 32. 3 and a6 = 0. Work Together • Using the sequence given at the right. Each term (except the first term) is found by multiplying the previous term by 2. Teaching & Learning Plan: Geometric Sequences theterm number - position of the term in. To make work much easier, sequence formula can be used to find out the last number (Of finite sequence with the last digit) of the series or any term of a series. C program to print geometric progression series and it's sum till N terms. Substitute 24 for a 2 and 3 for a 5 in the formula a n = a 1 ⋅ r n − 1. Let {b sub k} be a geometric sequence. A geometric progression is a sequence of numbers (also called terms or members) where the ratio of two subsequent elements of the sequence is a constant value. Given the explicit formula for a geometric sequence find the common ratio, the term named in the problem, and the recursive formula. The first term, the last term and the number of terms. The 3rd, 4th and 7th terms of the arithmetic sequence are the first three terms of a geometric sequence. (Total 2 marks) 2. Let’s work to get the next sequence. The explicit formula is an = 22(4) n± 1. A geometric sequence, or geometric progression, is a sequence of numbers where each successive number is the product of the previous number and some constant r. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. Meaning, the common difference of the sequence is five. The general formula of a Geometric Sequence found from the general sequence; the reciprocal with a positive power as the fractional root of the base. Writing Terms of Geometric Sequences. n is our term number and we plug the term number into the function to find the value of the term. This answer discusses finite differences and other handy techniques for solving this sort of problem. The Sum of the First n terms of an Geometric Sequence For a Geometric Sequence whose first term is a1 and whose common ratio is r where r≠0,1,−1 the sum Sn of the first n terms. (a) Show that every term of a geometric sequence with non-negative terms, except the first term
The following is multiple choice question (with options) to answer.
We are given the positive numbers a,b,c, which form an arithmetic progresion in the given order. We know that a+b+c=9. The numbers a+1,b+1,c+3 form a geometric progression in the given order. Find c. | [
"5",
"6",
"7",
"8"
] | A | Solution:
Since a,b,c form an arithmetic progression, c=b+d=a+2d and a+b+c=a+a+d+a+2d=3a+3d=9, so a+d=b=3 and a+c=6=>c=6−a. Since a+1,b+1,c+3 form a geometric progression, we have (b+1)2=(a+1)(c+3), or 16=(a+1)(6−a+3)=(a+1)(9−a)
16=9a+9−a2−a
a2−8a+7=0, therefore a=7 or a=1. But if a=7, c=6−a=6−7=−1<0 and by definition c is positive, so a=7 is not a solution. Therefore a=1 and c=6−a=6−1=5.
Answer A |
AQUA-RAT | AQUA-RAT-35627 | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
The following is multiple choice question (with options) to answer.
36 identical chairs must be arranged in rows with the same number of chairs in each row. Each row must contain at least three chairs and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible? | [
"2",
"4",
"5",
"6"
] | C | Three conditions have to be satisfied.
1. The number of students per row has to be at least 3.
2. Number of row has to be at least 3.
3. Equal number of students has to be seated in a row.
The following arrangements satisfy all 3 conditions.
Arrangement 1: 3 students to a row; 12 rows.
Arrangement 2: 4 students to a row; 9 rows.
Arrangement 3: 6 students to a row; 6 rows.
Arrangement 4: 9 students to a row; 4 rows.
Arrangement 5: 12 students to a row; 3 rows.
You will observe that the number of students in a row is a factor of 36.
So, an alternative and faster approach is to list down factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
And then start from 3 and quickly find out if the number of rows is at least 2.
Both the conditions are satisfied for the following factors : 3, 4, 6, 9, and 12. i.e., 5 arrangements.
Option C is answer |
AQUA-RAT | AQUA-RAT-35628 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
A and B start a business with Rs.16000 and Rs.32000 respectively. Hoe should they share their profits at the end of one year? | [
"1:2",
"3:4",
"2:5",
"3:7"
] | A | They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
16000 : 32000 => 1:2
ANSWER:A |
AQUA-RAT | AQUA-RAT-35629 | # How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
My try:
Let $A_2, A_3, A_5, A_7$ be the set of numbers between 1 and 1,000 that are divisible by 2, 3, 5, and 7 respectively. I used the inclusion-exclusion formula for $|A_2\cup A_3\cup A_5\cup A_7|= |A_2|+|A_3|+|A_5|+|A_7|-|A_2\cap A_3|-|A_2\cap A_5|-|A_2\cap A_7|-|A_3\cap A_5|-|A_3\cap A_7|-|A_5\cap A_7|+|A_2\cap A_3\cap A_5|+|A_2\cap A_3\cap A_7|+|A_2\cap A_5\cap A_7|+|A_3\cap A_5\cap A_7|-|A_2\cap A_3\cap A_5\cap A_7| = 500+333+200+142-166-100-71-66-47-28+33+23+14+9-4 = 772$
And the result I received was - 772.
I would appreciate if you could confirm my method and result, and I'd be happy to see a different, more elegant approach.
• It should be “2 , 3, 5 or 7”. – user371838 Jan 1 '18 at 13:01
• I am afraid there may be an elegant solution if you insist with and, and not so, if it is or. – user371838 Jan 1 '18 at 13:02
• It should be or. My apologies – Moshe King Jan 1 '18 at 13:03
• I haven't checked the numbers, but your inclusion-exclusion argument should work. – Paul Aljabar Jan 1 '18 at 13:10
The following is multiple choice question (with options) to answer.
How many numbers up to 300 and 700 are divisible by 2, 3 and 7 both together? | [
"9",
"10",
"11",
"12"
] | A | (700 – 300)/42 = 9 22/42
=> 9 Numbers
ANSWER:A |
AQUA-RAT | AQUA-RAT-35630 | # In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty?
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.
My attempt:-
First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.
Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.
Total number of ways $= 60\cdot9 = 540$.
Where am I going wrong ?
• When you place the last two balls, you are over counting. Your answer better be less than $243$. Oct 1 '17 at 10:51
There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are $$\binom{3}{1}2^5$$ ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is $$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$ by the Inclusion-Exclusion Principle.
Where am I going wrong?
The following is multiple choice question (with options) to answer.
In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes? | [
"152",
"3^5",
"125",
"217"
] | B | The first letter can be posted in any of the 3 post boxes. Therefore, we have 3 possibilities.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Each of the 5 letters has 3 possibilities because we can post any number of letters in all of the boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is
3 * 3 * 3 * 3 * 3 = 3^5
Ans: B |
AQUA-RAT | AQUA-RAT-35631 | 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 are running in opposite directions with the same speed. If the length of each train is 120 m and they cross each other in 12 sec, then the speed of each train is? | [
"30",
"31",
"36",
"25"
] | C | Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 => x = 10
Speed of each train = 10 m/sec.
= 10 * 18/5 =- 36 km/hr.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35632 | 12:14 AM
@MatheinBoulomenos nice
12:25 AM
quick question... IF we have eight people showing up for free concert tickets and we want to figure out how many ways can exactly 3 of them get tickets, isn't that just 8 choose 3 or $\binom{8}{3}$?
yes
ok so if we wanted to find out how many subsets of size k are there from a set of size n isn't it just $\binom{n}{k}$ or am I missing something here since we need different subsets of size k ... maybe it's $\binom{n}{k_{1}}\binom{n-k_{1}}{k_{2}}$ and so forth?
12:48 AM
it is just $\binom n k$
ah got it. I've been typing a lot and staying up late these past couple of days X_X
so my thinking is like weeeeee
oh oh ... it's Demonark
3 hours later…
whooaaa
2 hours later…
5:57 AM
do you know tensor product?
In above argument, we can conclude that $K_1K_2$ is spanned by $\alpha_i\beta_j$ over F, because, closed set under addition and multiplication of $\sum\alpha_i\beta_j$ is a field, right?
@LeakyNun no!
there's no need to shout that
ok :)
@Silent yes
6:02 AM
they span K_1 K_2 because they include the generators $\alpha_i$, $\beta_j$
so the ring they generate contains $F[\alpha_i, \beta_j]$
which is $K_1 K_2$
@Silent Yeah
Well, $\sum a_{ij}\alpha_i\beta_j$
sorry
It's a field, it contains $\alpha$ and $\beta$, and anything that contains $\alpha$ and $\beta$ contains $\sum a_{ij}\alpha_i\beta_j$
Therefore it's the field generated by $\alpha$ and $\beta$
The following is multiple choice question (with options) to answer.
There are k-2 members in a certain band, including Jim and Ellen. Two members are to be selected to attend the Grammy awards ceremony. If there are 15 possible combinations in which Jim and Ellen are not selected, what is the value of k? | [
"8",
"9",
"10",
"11"
] | C | There are k-2 members in the band, and k-4 members without Jim and Ellen.
(k-4)C2 = 15
(k-4)(k-5)/2 = 15
(k-4)(k-5) = 30 = 6*5
k = 10
The answer is C. |
AQUA-RAT | AQUA-RAT-35633 | ## 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.
Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter? | [
"177",
"132",
"176",
"278"
] | B | 2 * 22/7 * 14 = 88
88 * 1 1/2
=Rs.132
Answer:B |
AQUA-RAT | AQUA-RAT-35634 | ### Show Tags
23 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?
Aman
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
24 Jun 2019, 03:25
pkloeti wrote:
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.
Does that approach hold up in general? Bunuel VeritasPrepKarishma
Thanks a lot for the feedback!
I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
_________________
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
04 Aug 2019, 10:54
i tried like this:
Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85
The following is multiple choice question (with options) to answer.
Mala and Usha shared their water from a bottle. If mala and usha both drank for the same amount of time, but usha only drank 2/10 of the total bottle, what was the ratio of mala and usha speed? | [
"1:4",
"2:5",
"4:1",
"3:5"
] | C | The answer is suppose to be C. 4:1. It's from the GMATPrep
Answer:option C |
AQUA-RAT | AQUA-RAT-35635 | newtonian-mechanics, kinematics, free-fall
Title: Free fall equations I've been looking for this on the internet for a long time, but nowhere explains it clearly, and I need to explicitly calculate this and not other things:
If I have the height at which an object is, and it falls, how do I calculate at what speed and at what height it will be at x time? If the object starts falling from rest at height $h$, then its velocity at time $t$ later is
$$\vec v=-gt\hat z$$
and its vertical position is
$$z=h-\frac12gt^2$$
where $g$ is the magnitude of the constant and uniform downward gravitational acceleration. This neglects air drag, and assumes that the gravitational field is the same everywhere.
These equations come from integrating the kinematical equation for free-fall,
$$a_z=\frac{d^2z}{dt^2}=-g,$$
and applying the initial conditions that $z=h$ and $\vec v=0$ at $t=0$.
The following is multiple choice question (with options) to answer.
The velocity of a falling object in a vacuum is directly proportional to the amount of time the object has been falling. If after 5 seconds an object is falling at a speed of 84 miles per hour, how fast will it be falling after 12 seconds? | [
"204 miles per hour",
"212 miles per hour",
"216 miles per hour",
"1080 miles per hour"
] | A | Since Velocity is proportional to Time
Velocity =k*Timewhere k is the constant of proportionality
Time= 5 seconds
Velocity = 5k = 84 miles per Hour
i.e. k = 17
i.e. The relation between Velocity and Time becomes
Velocity =17*Time
Time= 12 seconds
Velocity = 17*12 = 204 miles per hour
Answer: Option A |
AQUA-RAT | AQUA-RAT-35636 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
One farmer takes 12 hours to plow his field, but another farmer takes 8 hours to plow the same field with an equal tractor. How long would it take them together to plow the field? | [
"12.5 hours",
"4 hours",
"4.8 hours",
"5 hours"
] | C | First we must find the least common denominator.
1/12 + 1/8 =
2/24+3/24=5/24
They can plow the field 5/24 per hour. Now to find the actual time, Lets use T for time as a variable.
1/T = 5/24
T/1 = 24/5 = 4.8 hours
Answer is C |
AQUA-RAT | AQUA-RAT-35637 | # String probability (with conditional prob and combinations)
I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.
SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.
Q1. What is the probability that it contains exactly one letter that is a vowel?
A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)
Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?
A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))
SET2: How many distinct permutations of the letters in "letters"...
Q1. Begin with two vowels?
A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...
Q2. Begin with two e's or end with two t's?
A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.
Q3. Have the vowels together?
A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.
Thank you!
The following is multiple choice question (with options) to answer.
Set #1 = {J, K, L, M, E}
Set #2 = {K, L, M, N, O, P}
There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel? | [
"1/6",
"1/8",
"1/3",
"1/4"
] | C | At least questions are best solved by taking the opposite scenario and subtracting it from 1. Probability of choosing no vowel from set 1 is 4/5 and set 2 is 5/6. Multiply these to get 2/3. Therefore, probability of picking at least one vowel = 1-2/3=1/3.
ANSWER : C |
AQUA-RAT | AQUA-RAT-35638 | # 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.
Bobby bought 2 shares, and which he sold for $106 each. If he had a profit of 120% on the sale of one of the shares but a loss of 47% on the sale of the other share, then on the sale of both shares Bobby had... | [
"a profit of 10",
"a profit of 8",
"a loss of 38",
"a loss of 10"
] | C | Cost of the first stock: 106/2.2=48.18, so profit from it 106-48.18=57.18
Cost of the second stock: 106/0.53=200, so loss from it 106-200=-94
Overall loss 57.18-94=-36.82
Answer: C. |
AQUA-RAT | AQUA-RAT-35639 | python, python-3.x, datetime
if sun == 1:
first_sunday = np.busday_offset(start, 0, roll='forward', weekmask='Sun')
last_sunday = np.busday_offset(end, 0, roll='preceding', weekmask='Sun')
sun_count = np.busday_count(first_sunday, last_sunday, weekmask='Sun') + 1
for i in trange(sun_count, desc="Sunday"):
touch.touch(file_name + " " + datetime.strptime(str(first_sunday), '%Y-%m-%d').strftime('%m-%d-%Y') + "." + file_type)
first_sunday += np.timedelta64(7, 'D')
if mon == 1:
first_monday = np.busday_offset(start, 0, roll='forward', weekmask='Mon')
last_monday = np.busday_offset(end, 0, roll='preceding', weekmask='Mon')
mon_count = np.busday_count(first_monday, last_monday, weekmask='Mon') + 1
for i in trange(mon_count, desc="Monday"):
touch.touch(file_name + " " + datetime.strptime(str(first_monday), '%Y-%m-%d').strftime('%m-%d-%Y') + "." + file_type)
first_monday += np.timedelta64(7, 'D')
if tue == 1:
first_tuesday = np.busday_offset(start, 0, roll='forward', weekmask='Tue')
last_tuesday = np.busday_offset(end, 0, roll='preceding', weekmask='Tue')
tue_count = np.busday_count(first_tuesday, last_tuesday, weekmask='Tue') + 1
for i in trange(tue_count, desc="Tuesday"):
The following is multiple choice question (with options) to answer.
If 1st October is Sunday, then 1st November will be | [
"Saturday",
"Thursday",
"Wednesday",
"Tuesday"
] | C | Explanation :
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1st November = (Sunday + 3 odd days) = Wednesday
Answer : Option C |
AQUA-RAT | AQUA-RAT-35640 | 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? | [
"1",
"61",
"63",
"64"
] | C | We need to select 5 men from 7 men and 2 women from 3 women.
Number of ways to do this
= 7C5 × 3C2
= 7C2 × 3C1 [∵ nCr = nC(n-r)]
= 21 * 3 =63
C |
AQUA-RAT | AQUA-RAT-35641 | at ½. Also, there are ""_5C_3= (5!)/(3!2!)=10 ways to get exactly 3 tails. The probability is 2/19. The probability of getting zero tails three times in a row is 1/2 x 1/2 x 1/2. What is the odds of getting beat on a 50/50 situation 6 times in a row? For the sake of the math lets assume that it is exactly 50 Coin flip situations - Gambling and Probability - Probability Theory Forum. So each toss of a coin has a ½ chance of being Heads, but lots of Heads in a row is unlikely. The odds of a fair coin landing on heads 5 times in a row are roughly 3 in 100. But is that really unusual?. Example: A coin and a dice are thrown at random. For example, the odds of becoming a movie star are only 1. Example 1: Coin and Dice. There is some information in knowing the outcome of the coin toss, but not as much as for a fair coin, because we already know that it will probably be heads. Perfect Skip Lists, continued • Nodes are of variable size: -contain between 1 and O(log n) pointers • Pointers point to the start of each node (picture draws pointers horizontally for visual clarity) • Called skip lists because higher level lists let you skip over many items 2 10 15 16 31 71 96 2 31 31 22 15 31 96 15 31 96. I flip a coin and it comes up heads. In other words, if you only want to flip ONE heads in a row, you'd have a 1 out of 2 raised to the power of ONE. Thus, the probability for each individual toss, regardless of what came before, is 50/50. The Probability That The Coin Will Come Up Heads On The Next Flip Is Greater Than 50%, Since It Appears That We Are In A Streak Of "heads. Why don't you take a penny, and try to get 4 heads in a row by flipping? Then see what the fifth flip gives you. So each toss of a coin has a ½ chance of being Heads, but lots of Heads in a row is unlikely. The top right entry (1,7) is the probability of getting 6+ heads/tails in a row in 200 flips or fewer, assuming a fair coin. Many events can't be predicted with total certainty. (c)
The following is multiple choice question (with options) to answer.
What is the probability of getting exactly three heads on five flips of a fair coin? | [
"1/32",
"3/32",
"1/4",
"5/16"
] | D | 5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10
Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32
Total probability = 1/32 x 10 = 5/16
Answer:D |
AQUA-RAT | AQUA-RAT-35642 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
The average of first seven multiples of 8 is: | [
"9",
"16",
"15",
"32"
] | D | Explanation :
(8(1+2+3+4+5+6+7)/7
= 8x28/7
= 32
Answer : D |
AQUA-RAT | AQUA-RAT-35643 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be? | [
"11",
"99",
"89",
"10"
] | D | lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
Answer:D |
AQUA-RAT | AQUA-RAT-35644 | 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 and B can complete a job in 30 and 20 days respectively. They start working together and B leaves 5 days before the work is finished. In how many days is the total work finished? | [
"15",
"6",
"5",
"4"
] | A | A and B can do the job in 30 and 20 days individually, so together they will take 30*20/(30 + 20) = 600/50 = 12 days.
The work that A can do in 5 days is 1/6th. So, A left when 5/6th of the work was done or A left after (5/6)*12 = 10 days
Total time taken = 10 + 5 = 15 days.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35645 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
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 1.2 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man? | [
"5.98",
"8.98",
"5.76",
"5.87"
] | C | M = 6
S = 1.2
DS = 7.2
US = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
D = 2.88 * 2 = 5.76
Answer:C |
AQUA-RAT | AQUA-RAT-35646 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
The length of rectangle is thrice its breadth and its perimeter is 80 m, find the area of the rectangle? | [
"432",
"300",
"252",
"992"
] | B | 2(3x + x) = 80
l = 30 b = 10
lb = 30 * 10 = 300
Answer:B |
AQUA-RAT | AQUA-RAT-35647 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
_________________
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Posts: 12145
Followers: 538
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of six numbers is 9. If 3 is subtracted from each of four of the numbers, what is the new average? | [
"1.5",
"2",
"7",
"4"
] | C | Sum of 6 numbers = 6*9 =54
If 3 is subtracted from each of four of the numbers , we subtract 3*4=12 from the total sum
Sum of 6 number after subtracting 3 from each of four of the numbers = 54 - 12 = 42
New average = 42/6 = 7
Answer C |
AQUA-RAT | AQUA-RAT-35648 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in upstream is 50 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"10 kmph.",
"17 kmph.",
"15 kmph.",
"11 kmph."
] | C | Speed of the boat in still water = (50+80)/2 = 65 kmph. Speed of the stream = (80-50)/2
= 15 kmph.
Answer:C |
AQUA-RAT | AQUA-RAT-35649 | 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 and B run a 1 km race. If A gives B a start of 50m, A wins by 14 seconds and if A gives B a start of 22 seconds, B wins by 20 meters. Find the time taken by A to run 1 km. | [
"222",
"100",
"287",
"2976"
] | B | To solve these type of questions, always keep in your mind that, the ratio of the speeds of two contestents never change.
A gives B a start of 50 m means, A runs 1000 m and B runs only 950. By the time A reaches the target, B has to take 22 seconds to reach the target.
ab=1000950−14b=9801000−22bab=1000950−14b=9801000−22b
50,000 -1100b = 46550 -686b
Solving we get b = 25/3
Now Assume A's speed = x
1000950−14(25/3)=x25/31000950−14(25/3)=x25/3
x = 10
So x takes 1000/10 = 100 seconds.
Answer:B |
AQUA-RAT | AQUA-RAT-35650 | #### romsek
MHF Helper
Ah. I see. The book doesn't say that it equals 0.0333, as I mentioned in my OP, it just asks to find the decimal version to the fourth decimal. So instead of it repeating endlessly, it just stops at 0.0333. So the book's answers are "1/30; 0.0333."
What I didn't get, was how to make sense of the fact that 333/10000 and 1/30 both equal 0.0333, even though they aren't equivalent fractions, and if 333/10000 would be an incorrect answer for the fractional part of the question.
$\dfrac 1{30} \neq 0.0333$
$\dfrac 1{30} = 0.033333333333333333333333333333333333 \dots$ with 3's going on forever
1 person
The following is multiple choice question (with options) to answer.
The value of 1 + 1/(4 x 3) + 1/(4 x 32) +1/(4 x 33) up to four places of decimals is ? | [
"1.1202",
"1.1203",
"1.1204",
"None of these"
] | B | Answer
Given expression = (108 + 9 + 3 +1) / 108
= 121 / 108
= 1.1203
Correct Option: B |
AQUA-RAT | AQUA-RAT-35651 | # Find the Remainder when $792379237923$…upto 400 digits is divided by $101$?
Find the Remainder when $792379237923\ldots$upto 400 digits is divided by $101$?
MyApproaach
when ($792379237923\ldots$400 digts)/$101$=
I learned this approach that I have to calculate(let say U)=Is the sum all of all the alternate groups starting with the rightmost
and (let say)Th=Is the the sum all of all the alternate groups starting with the second rightmost
Rem(U-Th)/$101$=?
But I am not following how to calculate U and Th
Can anyone guide me how to approach this problem?
Let $N$ be the number. Then we really want to find $N \pmod {101}$.
Note that $N=7923\cdot10^{396}+7923\cdot10^{392}+\cdots+7923$.
Next note that $7923\equiv 45 \pmod {101}$
Also, for example, $10^{100}=100^{50}\equiv(-1)^{50}\equiv 1 \pmod {101}$
We get the same result for each term, and there are $100$ of these terms so $N\equiv 45\cdot 100\equiv 56$.
Since $100\equiv-1$ mod $101$, your number, mod $101$ is:$$-79+23-79+23-\cdots$$ which is $100$ copies of $-56$. And mod $101$, that makes $56$.
The following is multiple choice question (with options) to answer.
A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ? | [
"9",
"7",
"6",
"5"
] | A | On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.
Answer is A. |
AQUA-RAT | AQUA-RAT-35652 | +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}
The following is multiple choice question (with options) to answer.
(540 - ?) × 1/7 + 12 = 25 | [
"A)227",
"B)449",
"C)32",
"D)600"
] | B | Explanation:
=> (540 - ?) × 1/7 + 12 = 25
=> (540 - ?) × 1/7 = 25 - 12 = 13
=> 540 - ? = 13 × 7 = 91
=> ? = 540 - 91 = 449
Answer: Option B |
AQUA-RAT | AQUA-RAT-35653 | (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.
A housewife wishes to purchase three articles A, B and C from a sum of Rs. 200. The unit prices of the articles A, B and C are Rs. 20, Rs. 35 and Rs. 25 respectively. If she spends the entire amount by purchasing 5 numbers of articles of type C, what is the ratio of the number of articles purchased of type A to that of, type B? | [
"1 : 2",
"2 : 1",
"1 : 1",
"None of these"
] | B | Explanation:
After spending Rs. 125 (25 * 5) for articles of type C, the housewife is left with Rs. 75 (200 – 125). Since this amount has to be spent in totality, she must have purchased 2 articles of type A (equivalent to Rs. 40) and 1 article of type B (equivalent to Rs. 35). Thus, the required ratio is 2 : 1.
ANSWER B |
AQUA-RAT | AQUA-RAT-35654 | +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.
What is the value of (12 + 22 + 32 + 42 + ----- + 102) | [
"2237",
"269",
"268",
"385"
] | D | Explanation:
(12 + 22 + ….. + n2) = (1/6) n(n + 1) (2n + 1)
Here, n = 10
Therefore,
(12 + 22 + ….. + 102) = (1/6) 10 (10 + 1) ( 2 x 10 + 1)
= (1/6) x 10 x 11 x 21
= 385
ANSWER: D |
AQUA-RAT | AQUA-RAT-35655 | They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
A money lender lent a total of $1900 to two of his customers. He charged at the rate of 5%p.a. to one of them and 6% p.a. to the other. If he earned an average of 5.67% on $1900, how much did he lend to each of the two customers? | [
"700; 1100",
"1273; 627",
"1000; 800",
"1200; 800"
] | B | The method given above is the one given in the book.
However the method I used was
(1*x*5)/100 + (1900-x)*6/100 = 1900*5.67/100
Simplifying we get x=627
B |
AQUA-RAT | AQUA-RAT-35656 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
There are three numbers. 5/7th of the first number is equal to 48% of the second number. The second number is 1/9th of the third number. If the third number is 1125, then find 25% of the first number? | [
"168",
"84",
"42",
"21"
] | D | Let the first number and the second number be F and S respectively.
5/2 F = 48/100 S ----> (1)
S = 1/9 * 1125 = 125
(1) => 5/7 F = 48/100 * 125
=> F = 84
25% of F = 1/4 * 84 = 21
ANSWER:D |
AQUA-RAT | AQUA-RAT-35657 | materials, energy
With these 3 we can find $f$ from (4): 0.25
Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$
For totals in kg/h, simply take the units given to find the factor by which to multiply:
$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$
(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$
(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.
The following is multiple choice question (with options) to answer.
A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.1000, what is the reduced price for kg? | [
"s.50",
"s.46",
"s.49",
"s.41"
] | A | 1000*(25/100) = 250 ---- 5
? ---- 1
=> Rs.50
Answer: A |
AQUA-RAT | AQUA-RAT-35658 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
A customer using a certain telephone calling plan pays a fee of $25 per month, and then receives a discount of 40% on the regular charge for all calls made to country W. If calls to country W are regularly charged at $1.60 per minute for the first 3 minutes, and $0.80 per minute for each minute thereafter, what is the maximum the customer could have saved over regular prices if he was charged for 1 hour of calls made to country W in a certain month? | [
"$8.75",
"$12",
"$13.40",
"$17.40"
] | C | Discount = 40% of spending
We want to maximise discount so we will need to maximise spending. We do that by assuming that 60 calls were made of 1 min each because the first minute is the most expensive.
Max discount = (40/100)* 60 * 1.6 = (40/100)*96
This would be slightly less than 40. About 38 to 39.
Saving = (Slightly less than 40) - 25 = Slightly less than 15 charged in W country
Answer (C) |
AQUA-RAT | AQUA-RAT-35659 | 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 161? | [
"1090",
"2060",
"3080",
"4030"
] | D | 100 + 102 +...+ 160 = 31*100 + (2+4...+60) = 31*100 + 2(1+2+...+30) =
31*100 + 2(30)(31)/2 = 130*31 = 4030
The answer is D. |
AQUA-RAT | AQUA-RAT-35660 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language? | [
"321",
"65",
"120",
"100"
] | D | 1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million
i.e. n3 = 106 (1 million = 106)
=> n3 = (102)3 => n = 102 = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
Ans: D |
AQUA-RAT | AQUA-RAT-35661 | ###### 5
$x^{44}x^{17}$
Solution
\begin{aligned}[t] x^{44}x^{17}\amp=x^{44+17}\\ \amp=x^{61} \end{aligned}
###### 6
$(a^4b^2)(a^7b)$
Solution
\begin{aligned}[t] (a^4b^2)(a^7b)\amp=a^{4+7}b^{2+1}\\ \amp=a^{11}b^3 \end{aligned}
###### 7
$(3x^6)(-2x^5)$
Solution
\begin{aligned}[t] (3x^6)(-2x^5)\amp=(3 \cdot -2)x^{6+5}\\ \amp=-6x^{11} \end{aligned}
###### 8
$(4y^{12})\left(\frac{y^{12}}{4}\right)$
Solution
\begin{aligned}[t] (4y^{12})\left(\frac{y^{12}}{4}\right)\amp=\left(4 \cdot \frac{1}{4}\right)y^{12+12}\\ \amp=1 \cdot y^{24}\\ \amp=y^{24} \end{aligned}
###### 9
$(x^2)^3$
Solution
\begin{aligned}[t] (x^2)^3\amp=x^{2 \times 3}\\ \amp=x^6 \end{aligned}
###### 10
$(t^4)^2$
Solution
\begin{aligned}[t] (t^4)^2\amp=t^{4 \times 2}\\ \amp=t^8 \end{aligned}
###### 11
$(w^3)^2$
Solution
The following is multiple choice question (with options) to answer.
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 5|+ 4 is equal to: | [
"–7",
"7",
"20",
"12"
] | C | x^2 - 8x + 17 = |x-5|
RHS can be -ve or +ve
x^2 - 9x + 22 = 0
x^2 - 7x + 12 = 0
x= 11,4,3,2
We test all 3 values in original equation, all ok.
Thus, Sum = 11 + 4 +3 + 2= 20
Ans (C) |
AQUA-RAT | AQUA-RAT-35662 | You might want to memorize the following logarithms, which can then allow you to approximate most others quite well:
(In base $10$)
$\log2=0.30$
$\log3=0.48$
$\log5=0.70$
$\log7=0.85$
Now, just remember the following logarithm properties: $\log{ab}=\log{a}+\log{b},$ $\log{\frac{a}{b}}=\log{a}-\log{b}$, and $\log_ab=\frac{\log_{10}{a}}{\log_{10}{b}}$. Using these identities, you can approximate, relatively, precisely, most logarithms.
Example 1: $\log_2{36}=\log_2{3}+\log_2{3}+\log_2{2}+\log_2{2}=\frac{\log3}{\log2}+\frac{\log3}{\log2}+1+1=\frac{0.48}{0.30}+\frac{0.48}{0.30}+1+1=1.60+1.60+1+1=5.20$
The actual answer is around $5.17$, but this is just an approximation.
Example 2: $\log_5{100}=\log_5{5}+\log_5{5}+\log_2{5}+\log_2{5}=1+1+\frac{\log2}{\log5}+\frac{\log2}{\log5}=1+1+\frac{0.30}{0.70}+\frac{0.30}{0.70}=1+1+0.43+0.43=2.86$
The actual answer is, when rounded to $2$ decimal places, $2.86.$
The following is multiple choice question (with options) to answer.
If log10 2 = 0.3010, the value of log10 80 is: | [
"2.65",
"1.903",
"3.05",
"6.487"
] | B | Explanation:
log10 80 = log10 (8 x 10)
= log10 8 + log10 10
= log10 (23 ) + 1
= 3 log10 2 + 1
= (3 x 0.3010) + 1
= 1.9030.
answer B |
AQUA-RAT | AQUA-RAT-35663 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
In a partnership between A, B and C. A's capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C's share is Rs.130, what is B's capital? | [
"3377",
"2788",
"27999",
"11750"
] | D | 200 + 130 = 330
800 - 330 = 470
200 ---- 5000
470 ---- ? => 11750
Answer: D |
AQUA-RAT | AQUA-RAT-35664 | This lets them bring a 15th bar and catch the train with less than 5 minutes to spare.
It is not possible to move all 16 bars.
Consider the total distance that each person moves, in each direction, carrying each possible number of bars. For example $$R_2^+$$ is the total distance that Rod moves forward while carrying 2 bars. We can set up a system of equations. Both Rod and Lia must move a net distance of $$1$$ mile forwards: $$R_0^+ - R_0^- + R_1^+ - R_1^- + R_2^+ - R_2^- = 1 \\ L_0^+ - L_0^- + L_1^+ - L_1^- = 1$$ Both of them must spend less than $$370$$ minutes moving (note we have taken the reciprocals of their speeds in minutes per mile): $$20(R_0^+ + R_0^-) + 30(R_1^+ + R_1^-) + 60(R_2^+ + R_2^-) \le 370 \\ 30(L_0^+ + L_0^-) + 40(L_1^+ + L_1^-) \le 370$$ Finally, the total net distance moved by all the bars must be $$16$$ miles forwards: $$R_1^+ - R_1^- + 2(R_2^+ - R_2^-) + L_1^+ - L_1^- = 16$$ And of course each distance must be nonnegative.
Next (thanks to RobPratt) we multiply these equations by $$\frac{1}{2}$$, $$\frac{3}{7}$$, $$\frac{1}{40}$$, $$\frac{1}{70}$$, and $$-1$$ respectively, and sum them, yielding:
The following is multiple choice question (with options) to answer.
John is traveling to a meeting that is 24 miles away. He needs to be there in 30 minutes. How fast does he need to go to make it to the meeting on time? | [
"25 mph",
"37 mph",
"48 mph",
"49 mph"
] | C | Well 48mph. C. Time*rate=distance --> 0.5*rate=24 --> rate=48 |
AQUA-RAT | AQUA-RAT-35665 | 28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
_________________
Senior Manager
Joined: 23 Oct 2010
Posts: 318
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
### Show Tags
28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
The following is multiple choice question (with options) to answer.
What is the units digit of (493) (915) (381) (756) (22) | [
"0",
"1",
"4",
"5"
] | B | Just multiply the digits in the units place for each term and you will get the answer. It should be 0. you got a 5 as a unit digit and an even number term. so the multiplication of this will definitely yield a 0. Answer has to be 0.
I also tried it using the calculator and the answer is 1.
IMO B. |
AQUA-RAT | AQUA-RAT-35666 | 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 covers a distance of 12 km in 10 min. If it takes 6 sec to pass a telegraph post, then the length of the train is? | [
"186 m",
"176 m",
"120 m",
"178 m"
] | C | Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec
= 20 m/sec.
Length of the train = 20 * 6
= 120 m.
Answer:C |
AQUA-RAT | AQUA-RAT-35667 | ### 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.
Tomy started a software business by investing Rs.50,000. After 6months, Nanda joined her with a capital of $80,000. After 3years, they earned a profit of $24,500. What was Simran's share in the profit? | [
"$8,500",
"$9,500",
"$10,500",
"$11,500"
] | C | Tomy : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
Tomy 's share = $24500 x 3/7 = $10,500.
C |
AQUA-RAT | AQUA-RAT-35668 | 2-4-8-6
2-4-8-6
2-4
The next digit in the pattern will be 8, which will belong to $$2^{11}$$.
In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base.
So $$652^{11}$$ will end in $$8,1896782^{11}$$ will end in 8, and so on…
A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits.
Units digit 3:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…
Units digit 4:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
The pattern here is 4, 6, 4, 6, 4, 6, and so on…
Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:
$$1545^{23} = ……..5$$
$$1650^{19} = ……..0$$
$$161^{28} = ………1$$
Hope you get the point.
Units digit 7:
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = ….1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)
7^5 = ….7 (Now multiply 1 from above by 7 to get 7 as the units digit)
7^6 = ….9
The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…
Units digit 8:
8^1 = 8
8^2 = 64
8^3 = …2
8^4 = …6
8^5 = …8
8^6 = …4
The following is multiple choice question (with options) to answer.
If n = 7^11 – 7, what is the units digit of n? | [
"0",
"1",
"4",
"6"
] | D | Always divide the power (incase 11) by 4 and use the remainder as the new power. The question now becomes 7^3 - 7. Now 7^3 has last digit 3. Since 7^11 (or for that matter 7^3) is greater than 7, we subtract 7 from 13 ( the 10+3 --> 10 has come from carry over from the tenth place). thus 13 - 7 = 6 is the answer. Option D |
AQUA-RAT | AQUA-RAT-35669 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue? | [
"33",
"37",
"28",
"36"
] | C | Explanation:
Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.
We may consider the two cases as under:
Case I:
Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40
Case II:
Number of persons between A and C
= (8 - 6) = 2
Clearly number of persons in the queue = (3+1+2+1+21) = 28
Now, 28 < 40. So, 28 is the minimum number of persons in the queue.
Answer: C) 28 |
AQUA-RAT | AQUA-RAT-35670 | # 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.
By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 6 p.c? | [
"38028",
"38000",
"32977",
"37600"
] | D | CP + CP/8 = 45000
CP = 40000
SP = 40000*(94/100)
= 37600
Answer:D |
AQUA-RAT | AQUA-RAT-35671 | "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.
A number when divided by a divisor leaves a remainder of 21. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor? | [
"12",
"13",
"34",
"37"
] | D | Let the number is N, the divisor = D,
I will make the two equations-
N = xD+21
2N = yD+11
where x and y are integers
Solving them: D(y-2x) = 34
as D is also integer and 34 is a prime number, the D should be 34 to satisfy the above equation.
Hence answer is 'D' |
AQUA-RAT | AQUA-RAT-35672 | $62 = 95% of$P or
$62 = 95/100$P so
$P =$6200/95 = $65.263 This is what your calculator is doing. Penny Hi Steve. The reason for this is that 65.263 - 5% = 62. It is a matter of perspective. If the regular price on a dress shirt is$50 and I put it on sale, selling it for 20% off, then you would rightly expect the selling price to be $40. But if I subsequently raise the price by 20%, then the new price is$48, not the original \$50.
This is because 50 - 50(0.20) = 40 + 40(0.25).
It becomes even more noticeable with higher markups like 80% (try it).
So when you mark up things manually, you are taking a 5% of the original cost and adding it to the original cost to make a new price. So your markup is 5% of the original cost. When you use the "markup" button on the calculator, you are using the manufacturer's definition of markup, which is to do it differently. The "markup" button says that 5% of the final price should be markup.
Stephen La Rocque >
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
The following is multiple choice question (with options) to answer.
A dress on sale in a shop is marked at $D. During the discount sale its price is reduced by 65%. Staff are allowed a further 60% reduction on the discounted price. If a staff member buys the dress what will she have to pay in terms of D ? | [
"0.15D",
"0.16D",
"0.65D",
"0.14D"
] | D | Effective discount = a + b + ab/100 = - 65 - 60 + (-65)(-60)/100 = -86
Sale price = D * ( 1 - 86/100)
Sale price = .14 * D
Answer (D) |
AQUA-RAT | AQUA-RAT-35673 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A bus travel 5/7 of its speed covers 42 km in 1 hr 40 min48 sec. What is theactual speed of the bus ? | [
"25 km/hr",
"35 km/hr",
"40 km/hr",
"42 km/hr"
] | B | Time = 1 hr 40 min 48 sec = 1hr +4060hr+483600hr=1+23+175=12675hrdistance = 42 kmspeed=distancetime=42(12675)=42×75126⇒57 of the actual speed = 42×75126⇒actual speed = 42×75126×75=42×1518=7×153=7×5=35 km/hr
B |
AQUA-RAT | AQUA-RAT-35674 | # 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.
A sum of money becomes triple itself in 5 years at simple interest. How many years will it become six times at the same rate? | [
"12 1/2 years",
"12 1/8 years",
"17 1/2 years",
"12 3/2 years"
] | A | 100 ---- 200 ---- 5
200 ---- 5
100 ---- 2 1/2
------------------
600 ---- 12 1/2 years
Answer:A |
AQUA-RAT | AQUA-RAT-35675 | -
@ amwhy i only used 4,4 once can you please tell me what you mean by 4,4 twice? – MethodManX Jul 5 '13 at 17:49
When you multiply by $2$: to account for counting the permutations of $(4, 1), (4, 2), (4, 3), {\bf (4, 4)}, (4, 5), (4, 6)$. When multiplying by $2$, you are counting the following as well: $(1, 4), (2, 4), (3, 4), {\bf (4, 4)}, (5, 4), (6, 4)$ – amWhy Jul 5 '13 at 17:53
Doubling the first list, to account for these permutations, gives you $12$, but we need to throw out one of the counted $(4, 4)$, giving you $11$ pairs in which at least one $4$ appears. – amWhy Jul 5 '13 at 17:59
@ amwhy okay thanks! – MethodManX Jul 5 '13 at 18:04
You're welcome, MethodMan! – amWhy Jul 5 '13 at 18:06
Making a list and carefully counting is a good idea. Even though it is not really necessary in this case, we will look at things in a more abstract way. Let $A$ be the event "sum is $5$" and let $B$ be the event "at least one $4$." Depending on whether you are in a counting mood or in a probability mood, we have $$|A\cup B| =|A|+|B|-|A\cap B|,\tag{1}$$ or $$\Pr(A\cup B) =\Pr(A)+\Pr(B)-\Pr(A\cap B).\tag{2}$$
In (1), we are using $|X|$ to denote the number of elemements in the set $X$. Your course may use a different notation.
The nice thing about the above formulas is that they can help organize the calculations.
The following is multiple choice question (with options) to answer.
4+4 | [
"8",
"2",
"1",
"5"
] | C | C |
AQUA-RAT | AQUA-RAT-35676 | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
The following is multiple choice question (with options) to answer.
If sum of three numbers in A.P is 33 and sum of their squares id 491, then what are the three numbers. | [
"5,11,17",
"7,11,15",
"9,11,13",
"3,11,19"
] | D | a + (a + d ) + ( a + 2d ) = 3( a + d ) = 33,
a + d = 11, or second term = 11 , first term = 11– d ,
Then ( 11- d ) 2 + 11 2 + ( 11 + d ) 2 = 491
2 d 2 = 491 - ( 3 * 121 ) = 491 - 363 = 128
d 2 = 64, d = 8, a = 3,
ANSWER:D |
AQUA-RAT | AQUA-RAT-35677 | 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.
How many number are immediately preceded and immediately followed by different numbers?
7 7 7 5 7 5 7 5 7 7 7 7 5 7 5 7 5 7 7 7 7 7 5 7 5 | [
"2",
"3",
"4",
"5"
] | D | five
ANSWER:D |
AQUA-RAT | AQUA-RAT-35678 | (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 exhibition was conducted for 4 weeks. The number of tickets sold in 2nd work week was increased by 20% and increased by 16% in the 3rd work week but decreased by 20% in the 4th work week. Find the number of tickets sold in the beginning, if 1392 tickets were sold in the last week | [
"6",
"35",
"120",
"1250"
] | D | Answer: 1250
Explanation:
let initially A ticket has been sold.
So now in 2nd week 20% increases so
A × 120100120100
In 3rd week 16% increases so
A × 120100120100 × 116100116100
In 4th week 20% decrease so
A × 120100120100 × 116100116100 × 120100120100 = 1392
A = 1250
Answer:D |
AQUA-RAT | AQUA-RAT-35679 | algorithms, algorithm-analysis, logic, discrete-mathematics, induction
Title: how can i prove the following algorithm? Exp(n)
if n == 1
return 2
If n == 0
Return 1
End If
If n%2==0
temp = Exp(n/2)
Return temp × temp
Else //n is odd
temp = Exp((n−1)/2)
Return temp × temp × 2
End if
how can i prove by strong induction in n that for all n ≥ 1, the number of multiplications made by Exp (n) is ≤ 2 log2(n).
ps: Exp(n) = 2^n or Power(2, n) To compute $\exp(1)$ it does $0$ multiplications, since it enters the if and returns 2.
Assume that, for all $k<n$, to compute $\exp(k)$ it does no more than $2\log_2(k)$ multiplications.
To compute $\exp(n)$ we have two cases. Either $n$ is even and the number of multiplication is $1$ more than the number done for $\exp(n/2)$, or $n$ is odd and the number of multiplications is $2$ more than the number of multiplications done for $\exp((n-1)/2)$.
In the first case we have that the number of multiplications is no more than $2\log_2(n/2)+1=2\log_2(n)-2+1\leq 2\log_2(n)$. In the second case the number of multiplications is no more than $2\log_2((n-1)/2)+2=2\log_2(n-1)\leq 2\log_2(n)$.
Therefore, the number of multiplications to compute $\exp(n)$ is no more than $2\log_2(n)$.
By induction, this holds for all positive integers $n$.
The following is multiple choice question (with options) to answer.
Let exp(m,n) = m to the power n. If exp(10, m)= n exp(2,2) where m and n are integers then n= ___? | [
"24",
"25",
"26",
"27"
] | B | 10^m=n*(2^2)
(2*5)^m=n*(2^2)
(2^m)*(5^m)=n*(2^2) now equate both side we get m=2 and (5^m)=n
(5^2)=n=>n=25
ANSWER:B |
AQUA-RAT | AQUA-RAT-35680 | to solve for the other. Logic To Calculate Percentage Difference Between 2 Numbers. Sale Discount Calculator - Percent Off Mortgage Loan Calculator - Finance Fraction Calculator - Simplify Reduce Engine Motor Horsepower Calculator Earned Value Project Management Present Worth Calculator - Finance Constant Acceleration Motion Physics Statistics Equations Formulas Weight Loss Diet Calculator Body Mass Index BMI Calculator Light. The commonly used way is to go from right to left, which gives us a positive number. Posted: (1 week ago) Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Just subtract the past value from the current value. Click the calculate button 3. Calculate percentage difference between two columns I have a input text file in this format: ITEM1 10. The percentage difference between the two values is calculated by dividing the value of the difference between the two numbers by the average of the two numbers. First, you need Excel to subtract the first number from the second number to find the difference between them. Listing Results about Calculate Variance Between Two Numbers Real Estate. To write an increase or decrease as a percentage, use the formula actual increase or decrease original cost × 100%. The growth rate can be listed for real or nominal GDP. This difference needs to be divided between the first number (the one that doesn't change). We then append the percent sign. Follow 52 views (last 30 days) Show older comments. Enter an old number in cell A1 and a new number in cell B1. The percent difference formula or the percent difference equation of two numbers a and b is: ((a - b) / (a+b)/2) × 100, where a > b Calculate the percentage difference between the numbers 35 and 65. The first step to the equation is simple enough. Step 1: Calculate the difference (subtract one value from the other) ignore any negative sign. With a Difference From, Percent Difference From, or Percent From calculation, there are always two values to consider: the current value, and the value from which the difference should be calculated. A percentage variance, aka percent change, describes a proportional change between two numbers, an original value and a new value. It has more of an impact when you say, "There was a 50 percent increase in attendance at the concert compared to last year," versus when you say, "There were 100 more people at the concert this year than
The following is multiple choice question (with options) to answer.
Difference of 2 numbers is 100. If 5% of one number is 10% of the other number, find the addition of2 numbers? | [
"150",
"300",
"210",
"290"
] | B | Let the numbers be x and y
5% of x = 10% of y
x = 2y
x-y = 100
2y - y =100
y = 100
x = 2y = 200
x+y = 300
Answer is B |
AQUA-RAT | AQUA-RAT-35681 | • Thank you for taking the time to help me!! How can you say that $x$ and $y$ are co-prime? – user342661 Aug 6 '16 at 2:46
• Suppose $x$ and $y$ are not co-prime. Then let $x = gw$ and $y=gz$, where $g \neq 1$. Then note that $c=xd=gwd$,and $m = yd = gzd$. Note that $gd$ divides both $x$ and $y$ and is greater than $d$ because $g>1$. This contradicts the fact that $d$ is the gcd of $m$ and $c$. Hence $x$ and $y$ are co-prime. – астон вілла олоф мэллбэрг Aug 6 '16 at 2:51
• You are welcome. – астон вілла олоф мэллбэрг Aug 6 '16 at 4:16
Assuming integers for all variables, and $c' = c/d$, $m' = m/d$ :
\begin{align} ac \equiv bc \pmod m &\iff \exists k ~~ ac - bc = km \\ % &\iff \exists k ~~ ac'd - bc'd \equiv km \\ % &\iff \exists k ~~ a - b \equiv (k/c')(m/d) \end{align}
So we have to establish that $k / c'$ is an integer. From $\gcd(c, m) = d$ we can infer $\gcd(c', m') = 1$, so
$$ac - bc = km$$ $$ac'd - bc'd \equiv k(m'd)$$ $$c'(a - b) = km'$$
So $c'$ divides $k$, so $k/c'$ is an integer.
The following is multiple choice question (with options) to answer.
If y = 30p, and p is prime, what is the greatest common factor of y and 24p, in terms of p? | [
"p",
"2p",
"6p",
"8p"
] | C | y = 30p = 2*3*5*p
24p = 2^3*3*p
The greatest common factor of 30p and 20p is the product of all the common prime factors, using the lower power of repeated factors. The greatest common factor is 2*3*p = 6p
The answer is C. |
AQUA-RAT | AQUA-RAT-35682 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be | [
"22",
"24",
"26",
"20"
] | C | Explanation :
Let number of hens = h and number of cows = c
number of heads = 48
=> h + c = 48 ---(Equation 1)
number of feet = 140
=> 2h + 4c = 140
=> h + 2c = 70 ---(Equation 2)
(Equation 2) - (Equation 1) gives
2c - c = 70 - 48
=> c = 22
Substituting the value of c in Equation 1, we get
h + 22 = 48
=> h = 48 - 22 = 26
i.e., number of hens = 26
Answer : C |
AQUA-RAT | AQUA-RAT-35683 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
It takes 8 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 8 students start painting at 9:00 am and one student, working at the same rate, is added per hour starting at 2:00 pm? | [
"5:10 pm",
"5:30 pm",
"5:50 pm",
"6:10 pm"
] | C | Each student paints at a rate of 1/80 of the house per hour.
In 5 hours, the 8 students can paint 40/80 of the house.
From 2 pm to 3 pm, 9 students paint another 9/80 for a total of 49/80.
From 3 pm to 4 pm, 10 students paint another 10/80 for a total of 59/80.
From 4 pm to 5 pm, 11 students paint another 11/80 for a total of 70/80.
12 students can paint the remaining 10/80 of the house in 10/12 of an hour = 50 minutes.
The house is completed at 5:50 pm.
The answer is C. |
AQUA-RAT | AQUA-RAT-35684 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
If books bought at prices ranging from Rs. 200 to Rs. 350 are sold at prices ranging from Rs. 300 to Rs. 425, what is the greatest possible profit that might be made in selling eight books ? | [
"23267",
"2677",
"1800",
"2778"
] | C | Explanation:
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
Answer: C) 1800 |
AQUA-RAT | AQUA-RAT-35685 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 8 stores in town that had a total of 23 visitors on a particular day. However, only 12 people went shopping that day; some people visited more than one store. If 8 people visited exactly two stores each, and everyone visited at least one store, what is the largest number of stores anyone could have visited? | [
"2",
"3",
"4",
"5"
] | C | 8 people visited 2 stores each for 16 visits.
To maximize the number of stores that one person visited, let's assume that 3 people visited 1 store each. The number of remaining visits is 23 - 16 - 3 = 4, which is the maximum that one person could have visited.
The answer is C. |
AQUA-RAT | AQUA-RAT-35686 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
A car traveling at a certain constant speed takes 4 seconds longer to travel 1 km than it would take to travel 1 km at 80 km/hour. At what speed, in km/hr, is the car traveling? | [
"70",
"72",
"74",
"75"
] | C | Time to cover 1 kilometer at 80 kilometers per hour is 1/80 hours = 3,600/80 seconds = 45 seconds;
Time to cover 1 kilometer at regular speed is 45 + 4 = 49 seconds = 49/3,600 hours = 1/74 hours;
So, we get that to cover 1 kilometer 1/74 hours is needed --> regular speed 74 kilometers per hour (rate is a reciprocal of time or rate=distance/time).
Answer: C |
AQUA-RAT | AQUA-RAT-35687 | (A) 10
(B) 45
(C) 50
(D) 55
(E) 65
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
15 Nov 2017, 16:22
9
3
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
We can create the following equation:
Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things
150 = 0.6(150) + 0.5(150) + 0.3(150) - D - 2(5) + 5
150 = 90 + 75 + 45 - D - 10 + 5
150 = 205 - D
D = 55
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
29 May 2017, 05:24
8
5
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
This can be solved using Venn Diagram (refer the attachment)
AC = 60% of 150 = 90
Sunporch = 50% of 150 = 75
SP = 30% of 150 = 45
The following is multiple choice question (with options) to answer.
One night a certain hotel rented 3/5 of its rooms, including 2/3 of their air conditioned rooms. If 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned? | [
"50%",
"55%",
"60%",
"65%"
] | A | The rooms which were not rented is 2/5
The AC rooms which were not rented is (1/3)*(3/5) = 1/5
The percentage of unrented rooms which were AC rooms is (1/5) / (2/5) = 1/2 = 50%
The answer is A. |
AQUA-RAT | AQUA-RAT-35688 | 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.
How many seconds will a train 150 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph? | [
"28 sec",
"23 sec",
"29 sec",
"30 sec"
] | D | D
D = 150 + 150 = 300
S = 36 * 5/18 = 10 mps
T = 300/10 = 30 sec |
AQUA-RAT | AQUA-RAT-35689 | 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 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train? | [
"65 kmph",
"28 kmph",
"54 kmph",
"26 kmph"
] | C | D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5
= 54 kmph
Answer:C |
AQUA-RAT | AQUA-RAT-35690 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
In 2008, a certain factory produced 40% more widgets than it did in 2007, and its production in 2009 was 140% of its production in 2008. By approximately what percent would its production need to decrease the following year for the factory to produce the same number of widgets it did in 2007? | [
"33%",
"44%",
"49%",
"60%"
] | C | Let P be the original level of production in 2007.
Let x be the rate of production in 2010 compared to 2009.
x*1.4*1.4*P = P
x = 1/1.96 = 0.51 which is a decrease of 49%.
The answer is C. |
AQUA-RAT | AQUA-RAT-35691 | Since 1000 = 999 + 1, 100 = 99 + 1, and 10 = 9 + 1, this equation can be rewritten
$$999 d_{1}+d_{1}+99 d_{2}+d_{2}+9 d_{3}+d_{3}+d_{4}=3 k$$
Rearranging gives
$$d_{1}+d_{2}+d_{3}+d_{4}=3 k-999 d_{1}-99 d_{2}-9 d_{3}$$
$$=3 k-3\left(333 d_{1}\right)-3\left(33 d_{2}\right)-3\left(3 d_{3}\right)$$
We can now factor a 3 from the right side to get
$$d_{1}+d_{2}+d_{3}+d_{4}=3\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$
since $$\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$ is an integer, we have shown that $$d_{1}+d_{2}+d_{4}+d_{4}$$ is divisible by 3.
(2) Assume $$d_{1}+d_{2}+d_{3}+d_{4}$$ is divisible by $$3 .$$ Consider the number $$d_{1} d_{2} d_{3} d_{4} .$$ As remarked above,
$$d_{1} d_{2} d_{3} d_{4}=d_{1} \times 1000+d_{2} \times 100+d_{3} \times 10+d_{4}$$
so
The following is multiple choice question (with options) to answer.
The number of times 99 is subtracted from 1111 so that the remainder is less than 99 is | [
"10",
"11",
"12",
"13"
] | B | Sol.
Let it be n times.
Then, (1111 - 99n) < 99.
By hit and trial, we find that n = 11.
Answer B |
AQUA-RAT | AQUA-RAT-35692 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
From a group of 3 women and 3 girls, 4 are to be randomly selected. What is the probability that equal numbers of women and girls will be selected? | [
"3/5",
"1/5",
"2/5",
"4/5"
] | A | Using the first example, here is the probability of THAT EXACT sequence occurring:
wwGG = (3/6)(2/5)(3/4)(2/3) = 36/360 = 1/10
Each of the other 5 options will yield the exact SAME probability....
wGwG = (3/6)(3/5)(2/4)(2/3) = 36/360 = 1/10
So we have 6 different options that each produce a 1/10 chance of occurring.
6(1/10) = 6/10 = 3/5
Final Answer:
A |
AQUA-RAT | AQUA-RAT-35693 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The sale price sarees listed for Rs.150 after successive discount is 20% and 10% is? | [
"187",
"120",
"108",
"178"
] | C | 150*(80/100)*(90/100)
= 108
Answer:C |
AQUA-RAT | AQUA-RAT-35694 | 3,5,-5... The first term in the sequence of #'s shown above is 3. Each even # term is 2 more than the previous term and each odd # term, after the first, is -1 times the previous term. For example, the second term is 3+2, and the
8. ### Math *URGENT
Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646? 3. Determine
9. ### maths
The fifth term of an arithmetic sequence is 23 and the 12th term is 72. What is the value of the 10th term. Which term has a value of 268.
10. ### Sj
Mathematics......Under number patterns-Geometric series If a question goes.determine the expression for the nth term of the following sequence if the a) 4th term is 24 and the 7th term is 192 in a geometric sequence.what formula
More Similar Questions
The following is multiple choice question (with options) to answer.
In a certain sequence, each term except for the first term is one less than twice the previous term. If the first term is 1.5, then the 3rd term is which of the following? | [
"−1.5",
"−1",
"0",
"3"
] | D | First = 1.5
Second = 2*1.5-1 = 2
Second = 2*2-1 = 3
Answer: option D |
AQUA-RAT | AQUA-RAT-35695 | # 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.
The C.P of 15 books is equal to the S.P of 18 books. Find his gain% or loss%? | [
"16 2/3% loss",
"16 2/7% loss",
"16 8/3% loss",
"96 2/3% loss"
] | A | 15 CP = 18 SP
18 --- 3 CP loss
100 --- ? => 16 2/3% loss
Answer:A |
AQUA-RAT | AQUA-RAT-35696 | # What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour?
This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!
-
As a Frenchman, 80% chance in 1 hour seems really high to me, would be rather 20% chance in 1 day. – Benoit Jun 4 '12 at 13:18
80% is about right in Oslo. – sshow Jun 4 '12 at 19:56
Where I'm from, between the hours of 3PM - 6PM, M-F, the probability of getting a ticket and towed is 100% within 10 minutes. – user32990 Jun 5 '12 at 8:41
It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$.
The following is multiple choice question (with options) to answer.
The cost to park a car in a certain parking garage is $12.00 for up to 2 hours of parking and $1.75 for each hour in excess of 2 hours. What is the average (arithmetic mean) cost per hour to park a car in the parking garage for 9 hours? | [
" $1.09",
" $2.69",
" $2.25",
" $2.37"
] | B | Total cost of parking for 9 hours = 12$ for the first 2 hours and then 1.75 for (9-2) hours = 12+7*1.75 = 24.25
Thus the AVERAGE parking price = 24.25/9 = 2.69 $
B is the correct answer. |
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