source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35997 | 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 moving at 80 kmph and 70 kmph in opposite directions. Their lengths are 150 m and 100 m respectively. The time they will take to pass each other completely is? | [
"4 sec",
"5 sec",
"6 sec",
"7 sec"
] | C | 70 + 80 = 150 * 5/18 = 125/3 mps
D = 150 + 100 = 250 m
T = 250 * 3/125 = 6 sec
Answer:C |
AQUA-RAT | AQUA-RAT-35998 | the area and volume of a sphere, using simple geometric arguments. The surface area is the surface part of a three-dimensional object. Find out what's the height of the cylinder, for us it's 9 cm. Volume of a cylinder is (to find half, divide by 2): Vcylinder = Abase * height In a cylinder, the base is a circle and the area of a circle can be found: Acircle = pi * radius2 After you find the. The area of some circle is known and equals 15. 809625 Cubic Inches The volume of a cylinder 5 inches in diameter and 6 inches high is 117. What is the formula for finding the volume of a cylinder? V = π r 2 h 3. The volume of the cylinder is the amount of space inside the cylinder. Visually Understanding Area of a Circle and Volume of a Cylinder. Then click Calculate. Don't forget the two end bits: Total Surface Area. 0021m^3 Please Help or try and Solve this for me, Thankyou Very Much. For the rectangular solid, the area of the base, B, is the area of the rectangular base, length × width. Volume of a Right Circular Cylinder. Volume of a cylinder. Pi is the ratio of a circle's circumference to its diameter: 3. Im doing Maths about 'Volume' one of the questions are finding the volume of a semi-cylinder. The volume of three cones is equal to the volume of one cylinder with the same base and height. The cylinder has two bases, the base has radius r. The length of each cylinder (h) is 15,000 m. Surface Area of a Cylinder= 2πr² + 2πrh. Surface area and volume of a cylinder. Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. Area of a Trapezoid. To find the volume of a cylinder we multiply the base area (which is a circle) and the height h. In this example, r and h are identical, so the volumes are πr 3 and 1 ⁄ 3 π r 3. Browse by Radius in Meters. For engine's, this would be: Swept
The following is multiple choice question (with options) to answer.
The radius of a cylindrical vessel is 5cm and height is 3cm. Find the whole surface of the cylinder? | [
"251.4 sq cm",
"220 sq cm",
"440 sq cm",
"132 sq cm"
] | A | r = 5 h = 3
2πr(h + r) = 2 * 22/7 * 5(8) = 251.4
ANSWER:A |
AQUA-RAT | AQUA-RAT-35999 | Here is the situation: My friend and I are at an impasse. I believe I'm correct, but he's so damn stubborn he won't believe me. Also, I'm not the most articulate at explaining things. Hopefully some of you guys can help me explain this to him in a way he'll understand. Here is the problem:
A DVD is either at his parent's house or his own. The probability that it's at his house is 30%. If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room. What is the % chance the DVD is on the porch?
My friend says you take 90% of 30% which is 27 and that is the % chance it's on the porch. Is this correct? I don't believe so.
I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall. Location inside his house won't change those odd because the porch and the living room are both part of the house. If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.
Now, if you rephrase the question and ask, "The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%. If there are three places it could be, then there is 33.333% chance it's on the porch. Even if it's a 90% chance it's at his house, if there are only three places it can be, it remains the same.
I think the correct way of answering the question is: There is a 30% chance the DVD is at my house. If it is at my house, there is a 90% chance it's on my porch. They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house.
Is this correct or am I wrong? And regardless, please give me your explanation.
The following is multiple choice question (with options) to answer.
In a apartment, 30% of the people speak English, 20% speak Hindi and 13% speak both. If a people is selected at random, what is the probability that he has speak English or Hindi? | [
"1/30",
"2/30",
"10/27",
"2/90"
] | C | P (E) = 30 / 100 , P (H) = 20 / 100 and P (E ∩ H) = 13 / 100 =
P (E or H) = P (E U H)
= P (E) + P (H) - P (E ∩ H)
= (30 / 100) + (20/100 - (13 / 100) = 37/100 = 10/27
C |
AQUA-RAT | AQUA-RAT-36000 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
If 36 men can do a piece of work in 25 hours, in how mwny hours will15 men do it? | [
"22",
"38",
"60",
"88"
] | C | Explanation:
Let the required no of hours be x. Then
Less men , More hours (Indirct Proportion)
\inline \fn_jvn \therefore 15:36 ::25:x \inline \fn_jvn \Leftrightarrow (15 x X)=(36 x 25) \inline \fn_jvn \Leftrightarrow \inline \fn_jvn x=\frac{36\times 25}{15}=60
Hence, 15 men can do it in 60 hours.
Answer: C) 60 |
AQUA-RAT | AQUA-RAT-36001 | ### Show Tags
13 Nov 2017, 20:21
2
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
### Show Tags
13 Nov 2017, 22:31
2
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
### Show Tags
11 Dec 2017, 14:28
3
1
Hi All,
We're given the equations for 3 lines (and those equations are based on 4 unknowns: 2 variables and the 2 'constants' A and B):
Y = (A)(X) - 5
Y = X + 6
Y = 3X + B
The following is multiple choice question (with options) to answer.
In the xy-plane, a line has slope 3 and x-intercept 4. What is the y-intercept of the line? | [
"-12",
"-3",
"0",
"3"
] | A | Let the line be represented by a general equation y=mx+b, where m = slope (3) and b=y intercept. We are also given the value of x-intercept 4.
Theory : y intercept represents the point on the line where the x=0, and x intercept represents the point on the line where the y=0.
Putting these values in the equation : 0 = 3*4 + b => b = -12. Hence A. |
AQUA-RAT | AQUA-RAT-36002 | 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.
5 men and 5 women participated in a interview. If two people get the job, how many chances there will be at least a man is
selected? | [
"30",
"32",
"35",
"31"
] | C | 5 women and 5 men. Two people is slelected
It is always better to subtract the total from none in case of at least one problems.
Total ways = 10C2 = 45
Total ways without any man = 5C2 = 10
Hence ways in which at least one man will be present = 45 - 10 = 35
Correct option: C |
AQUA-RAT | AQUA-RAT-36003 | A triangle with two sides 6 and 8 units long is inscribed in a circle. If the third side is a diameter, find the length of the diameter.
Solution
A triangle with one side the diameter is a right triangle.
The lengths of the sides this triangle are 6 : 8 : h. This is a multiple of a
3 : 4 : 5 right triangle, so the length of the hypotenuse that is a diameter is 10.
Find the values of x, y and z.
Solution
The circle is the sum of the arcs, so start by finding 3z°.
360° = 110° + 130° + 54° + 3z° so 3z° = 66° and z = 22.
There are two methods to find the other two angles.
Method 1
The measures of inscribed angles are equal to half the sum of the opposite arcs.
Using the opposite arcs:
2x° = (\dfrac{1}{2})(66° + 110°)
so 2x° = (\dfrac{1}{2})176° and x = 44.
6y° = (\dfrac{1}{2})(66° + 54°)
so 6y° = 60° and y = 10.
Method 2
Since the inscribed polygon is a quadrilateral, opposite angles of the quadrilateral add to 180°.
The measure of inscribed
C = (\dfrac{1}{2})(130° + 54°) = 92°.
Add the opposite angles.
2x° + 92° = 180° so 2x° = 88° and x = 44.
The measure of inscribed
B = (\dfrac{1}{2})(130° + 110°) = 120°.
6y° + 120° = 180°
so 6y° = 60° and y = 10.
A square is inscribed in a circle of diameter 20. Determine the ratio of the area of the circle to the area of the square.
Solution
Sketch the figure.
You need to find the area of the circle and the square.
Be careful! Since diameter = 20,
r = 10, and the area of the circle is A = πr2 = 100π.
The following is multiple choice question (with options) to answer.
A right triangle PQR has sides such that the longest side PR is twice the length of side PQ and thrice the length of side QR. If PR = 6, what is the area of the triangle? | [
"6",
"36",
"3",
"2"
] | C | Since PR is twice of PQ, so PQ=3 and QR=2 since PR is thrice of QR. Area of right triangle equals PQ*QR/2. So the correct answer is C. |
AQUA-RAT | AQUA-RAT-36004 | The base length of an isosceles triangle is not enough to determine the triangle:
$\hspace{2cm}$
-
Given the base langth $a$ any $b>\frac a2$ constitutes a valid leg length.
-
You will always need three data to determine a triangle. It can be any combination of angles, lengths, heights...
In your example, since you know it's an isosceles triangle, you'll have three data once you define an angle or a length apart from the base's length, because you know both angles at the base's ends are equal.
So, if you have the segment AB being the base of an isosceles triangle, once you know either angle C, height from base to C, length BC or length AC, you'll have everything you need.
-
"It can be any combination of angles, lengths, heights..." wrong. How long are the sides on any given equilateral triangle? You have three pieces of data - 3 angles, each of 60°, but you don't have anywhere near enough information to determine the length of the sides. -1 (if I had to rep). – mikeTheLiar Jun 5 '13 at 19:55
@mikeTheLiar I don't think my answer was wrong completely... if you know the three angles of a triangle, you can never know the size, but you know the shape, which is sufficient to answer the question. Let's change it to: 'Given three data, you can find any triangle given you have one length among these data, otherwise you get a similar triangle with unknown size" – Óscar Gómez Alcañiz Jun 8 '13 at 17:00
The following is multiple choice question (with options) to answer.
An isosceles triangle has a perimetre of 37 centimetres and its base has a length of 9 centimetres. The other two sides will each have a length of
? | [
"14 cm",
"15 cm",
"16 cm",
"17 cm"
] | A | Solution:
The isosceles triangle has two equal sides. The sum of the equal sides will be 37 - 9 = 28 cm . Thus, one of them will be 28 ÷ 2 = 14 cm
Answer A |
AQUA-RAT | AQUA-RAT-36005 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
Of the employees in a company, 30 % are female who have a master degree. If 20 % of the female employees do not have a master degree, what percent of the employees in the company are female? | [
"25 %",
"30 %",
"32.5 %",
"37.5 %"
] | D | No. of female employees with Masters = 0.3 x E (From the question)
No. of female employees without masters = 0.2 x F (From the question)
Therefore No. of female employees with masters = F - 0.2 F = 0.8 F
The 2 expressions equal each other therefore 0.8F = 0.3E; F/E = 0.3/0.8 = 37.5%
Ans: D |
AQUA-RAT | AQUA-RAT-36006 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land Tails up exectly twice in 3 consecutive flips ? | [
"0.375",
"0.25",
"0.325",
"0.5"
] | A | Total number of ways in which H or T can appear in 3 tosses of coin is
= 2 * 2 * 2 = 8 ways
For 2 T and 1 TH
Thus probability is
= P(HTT) + P(TTH) + P(THT)
= 1/8 + 1/8 + 1/8
= 3/8
= .375
Answer :A |
AQUA-RAT | AQUA-RAT-36007 | # Making Friends around a Circular Table
I have $n$ people seated around a circular table, initially in arbitrary order. At each step, I choose two people and switch their seats. What is the minimum number of steps required such that every person has sat either to the right or to the left of everyone else?
To be specific, we consider two different cases:
1. You can only switch people who are sitting next to each other.
2. You can switch any two people, no matter where they are on the table.
The small cases are relatively simple: if we denote the answer in case 1 and 2 for a given value of $n$ as $f(n)$ and $g(n)$ respectively, then we have $f(x)=g(x)=0$ for $x=1, 2, 3$, $f(4)=g(4)=1$. I’m not sure how I would generalize to larger values, though.
(I initially claimed that $f(5)=g(5)=2$, but corrected it based on @Ryan’s comment).
If you’re interested, this question came up in a conversation with my friends when we were trying to figure out the best way for a large party of people during dinner to all get to know each other.
Edit: The table below compares the current best known value for case 2, $g(n)$, to the theoretical lower bound $\lceil{\frac{1}{8}n(n-3)}\rceil$ for a range of values of $n$. Solutions up to $n=14$ are known to be optimal, in large part due to the work of Andrew Szymczak and PeterKošinár.
The moves corresponding to the current best value are found below. Each ordered pair $(i, j)$ indicates that we switch the people in seats $(i, j)$ with each other, with the seats being labeled from $1 \ldots n$ consecutively around the table.
The following is multiple choice question (with options) to answer.
Find the no.of ways of arranging the boy and 4 guests at a circular table so that the boy always sits in a particular seat? | [
"7!",
"4!",
"9!",
"10!"
] | B | Ans.(B)
Sol. Total number of persons = 5 Host can sit in a particular seat in one way. Now, remaining positions are defined relative to the host. Hence, the remaining can sit in 4 places in 4P4 = 4! Ways ... The number of required arrangements = 4! x 1= 4! = 4! ways |
AQUA-RAT | AQUA-RAT-36008 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference of 2 digit number & the number obtained by interchanging the digits is 36. What is the difference the sum and the number if the ratio between the digits of the number is 1:2 ? | [
"4",
"8",
"9",
"10"
] | B | Let the number be xy.
Given xy – yx = 36.
This means the number is greater is than the number got on reversing the digits.
This shows that the ten’s digit x > unit digit y.
Also given ratio between digits is 1 : 2 => x = 2y
(10x + y) – (10y +x) = 36 => x – y = 4 => 2y – y =4.
Hence, (x + y) – (x – y) = 3y – y = 2y = 8
B |
AQUA-RAT | AQUA-RAT-36009 | • #### 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.
The length of the rectangular field is double its width. Inside the field there is square shaped pond 5m long. If the area of the pond is 1/8 of the area of the field. What is the length of the field? | [
"54",
"32",
"75",
"20"
] | D | A/8 = 5 * 5 => A = 5 * 5 * 8
x * 2x = 5 * 5 * 8
x = 10 => 2x = 20
Answer:D |
AQUA-RAT | AQUA-RAT-36010 | Alternatively, the lcm of 54 and 72 can be found using the prime factorization of 54 and 72:
• The prime factorization of 54 is: 2 x 3 x 3 x 3
• The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(54,54) = 216
In any case, the easiest way to compute the lcm of two numbers like 54 and 72 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 54,72. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 54 and 72
What is the least common multiple of 54 and 72 used for? Answer: It is helpful for adding and subtracting fractions like 1/54 and 1/72. Just multiply the dividends and divisors by 4 and 3, respectively, such that the divisors have the value of 216, the lcm of 54 and 72.
$\frac{1}{54} + \frac{1}{72} = \frac{4}{216} + \frac{3}{216} = \frac{7}{216}$. $\hspace{30px}\frac{1}{54} – \frac{1}{72} = \frac{4}{216} – \frac{3}{216} = \frac{1}{216}$.
## Properties of LCM of 54 and 72
The most important properties of the lcm(54,72) are:
• Commutative property: lcm(54,72) = lcm(72,54)
• Associative property: lcm(54,72,n) = lcm(lcm(72,54),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The following is multiple choice question (with options) to answer.
Given 2 numbers; 8 and q, the LCM and HCF of both numbers is 72 and 1 respectively. Find q? | [
"3",
"5",
"6",
"9"
] | D | HCF x LCM = Product of Numbers
1 x 72 = 8 x q
q = (1 x 72) / 8
other number = 9
ANSWER : D |
AQUA-RAT | AQUA-RAT-36011 | ### Show Tags
09 Jan 2010, 05:34
24
28
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?
A.54
B.432
C.2160
D.2916
E.148,824
i ve got it right.but this problem is very time consuming.can anyone suggest shorter method
We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.
Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).
9C1*6C*8C1*5C1=2160.
Answer: C.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]
### Show Tags
25 Jan 2013, 13:03
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Slightly different way of thinking:
On the 9x6 grid of possibilities, I can imagine a bunch of rectangles (with sides parallel to x and y axes). Each of these rectangles contains 4 triangles that fit the description of the question stem.
therefore:
Answer = ( # of Rectangles I can make on the grid) x 4
To create the rectangle, I need to pick 2 points on the x direction, and 2 points on the y direction. Therefore:
The following is multiple choice question (with options) to answer.
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and Y coordinates of P,Q and R are to be integers that satisfy the inequalitites -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed? | [
"110",
"1100",
"9900",
"10000"
] | C | total values for x=10;y=11
x1,y1=10*11.......................coordinates of 1st pnt
x2,y2=9*1(y2=y1)............... coordinates of 2nd pnt y coordinates will be same as that of 1st pnt bcoz it is parallel to x axis
x3,y3=1*10(x2=x3)..........coordinates of 3rd pt. x coordinates will be same as that of 2nd point bcoz to make a right angle it has to be parallel to y axis
tot ways=10*11*9*1*1*10=9900
ANS:C |
AQUA-RAT | AQUA-RAT-36012 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
What is the dividend. divisor 16, the quotient is 6 and the remainder is 2 | [
"86",
"87",
"88",
"89"
] | C | C = d * Q + R
C = 16 * 6 + 2
C = 86 + 2
C = 88 |
AQUA-RAT | AQUA-RAT-36013 | # 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.
If a trader sold two cars each at Rs. 325475 and gains 12% on the first and loses 12% on the second, then his profit or loss percent on the whole is? | [
"1.44%",
"1.74%",
"1.84%",
"1.47%"
] | A | SP of each car is Rs. 325475, he gains 12% on first car and losses 12% on second car.
In this case, there will be loss and percentage of loss is given by
= [(profit%)(loss%)]/100
= (12)(12)/100 % = 1.44%
Answer:A |
AQUA-RAT | AQUA-RAT-36014 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
Three hundred students at College Q study a foreign language. Of these, 120 of those students study French, and 170 study Spanish. If at least 90 students who study a foreign language at College Q study neither French nor Spanish, then the number of students who study Spanish but not French could be any number from | [
"10 to 40",
"50 to 90",
"60 to 100",
"60 to 110"
] | B | 120 students study French
180 students do not study French
170 students study Spanish
130 students do not study Spanish
90 students study neither French nor Spanish
180-130=50
180-90=90
B. 50 to 90 |
AQUA-RAT | AQUA-RAT-36015 | (1) $$a^2+b^2>16$$
Doesn't tell us about the value of a. a could be 2 or 10 or many other values.
(2) a=|b|+5
a is 5 or greater since |b| is at least 0. This tells us that 'a' does not lie between -4 and 4. Hence this statement is sufficient to answer the question.
_________________
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Veritas Prep | GMAT Instructor
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The following is multiple choice question (with options) to answer.
If a - b = 5 and a2 + b2 = 34, find the value of ab. | [
"A)4.5",
"B)12",
"C)15",
"D)18"
] | A | Explanation:
2ab = (a2 + b2) - (a - b)2
= 34 - 25 = 9
ab = 4.5.
Answer: A |
AQUA-RAT | AQUA-RAT-36016 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Dawson got a weekly raise of $160. If he gets paid every other week, write an integer describing how the raise will affect his paycheck. | [
"150",
"160",
"170",
"180"
] | B | Let the 1st paycheck be x (integer).
Dawson got a weekly raise of $ 160.
So after completing the 1st week she will get $ (x+160).
Similarly after completing the 2nd week she will get $ (x + 160) + $ 160.
= $ (x + 160 + 160)
= $ (x + 320)
So in this way end of every week her salary will increase by (B)$ 160. |
AQUA-RAT | AQUA-RAT-36017 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
$500 will become $1000 in 6 years find the rate of interest? | [
"A)16.67%",
"B)20%",
"C)25%",
"D)30%"
] | A | SI = simple interest = A-P = 1000-500 = $500
R = 100SI/PT = 100*500/500*6 =16.67%
Answer is A |
AQUA-RAT | AQUA-RAT-36018 | combinatorics, sets, arithmetic, binary
100000 -> (>2 consecutive 0s OR 1s) -> remove from subset
100001 -> (>2 consecutive 0s OR 1s) -> remove from subset
100010 -> (>2 consecutive 0s OR 1s) -> remove from subset
100011 -> (>2 consecutive 0s OR 1s) -> remove from subset
100100 -> (<= 2 consecutive 0s AND 1s) -> order [14]
100101 -> (<= 2 consecutive 0s AND 1s) -> order [15]
100110 -> (<= 2 consecutive 0s AND 1s) -> order [16]
100111 -> (>2 consecutive 0s OR 1s) -> remove from subset
101000 -> (>2 consecutive 0s OR 1s) -> remove from subset
101001 -> (<= 2 consecutive 0s AND 1s) -> order [17]
101010 -> (<= 2 consecutive 0s AND 1s) -> order [18]
101011 -> (<= 2 consecutive 0s AND 1s) -> order [19]
101100 -> (<= 2 consecutive 0s AND 1s) -> order [20]
101101 -> (<= 2 consecutive 0s AND 1s) -> order [21]
101110 -> (>2 consecutive 0s OR 1s) -> remove from subset
101111 -> (>2 consecutive 0s OR 1s) -> remove from subset
The following is multiple choice question (with options) to answer.
Which one of the following numbers can be removed from the set S = {0, 2, 4, 5, 9} without changing the average of set S? | [
"0",
"2",
"4",
"5"
] | C | Solution:
The average of the elements in the original set S is: (0+2+4+5+9) /5 =20 /5 =4.
If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}.
The average of the elements is,
(0+2+5+9) /4=16 /4 =4.
Answer: Option C |
AQUA-RAT | AQUA-RAT-36019 | 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 swims downstream 72 km and upstream 45 km taking 12 hours each time; what is the speed of the current? | [
"1.125",
"1.128",
"1.126",
"1.123"
] | A | 72 --- 12 DS = 6
? ---- 1
45 ---- 12 US = 3.75
? ---- 1 S = ?
S = (6 - 3.75)/2 = 1.125
Answer:A |
AQUA-RAT | AQUA-RAT-36020 | Consider seperate cases of $$m$$, $$m+1$$ even.
Notice that all but one of the terms in the summation are even.
Notes
• Having read the official solution, I'm amazed this (natural to me) approach worked out so quickly but they didn't use it.
• Defining $$y$$ is pretty natural given $$x$$, and it's a useful trick in such situations. (EG It's also in the official solution.)
• Bonus: $$y^n = \sqrt{ k_n + 1 } -\sqrt{k_n}$$
• $$k_1 = m, k_2 = 4m^2 + 4m, k_3 = 16m^3 + 24m^2 + 9m$$
• I'm slightly amazed that there is a closed form expression for $$k_n$$. I looked at initial $$k_n$$ and couldn't spot a generalizable pattern.
• I very much appreciate your efforts to guide users with observations and hints, and not always straight out answers. For many students, this is more valuable that doing it for them! Thanks! Jan 3 at 22:08
• (+1) Great solution! Jan 3 at 22:08
This is a USAMTS problem. The official solution can be found here. You are actually on the right track.
Below is my solution. Notice that I used the same notation as in the original problem.
My proof:
We prove by induction that: If $$m$$ is odd then $$x^m = a_m\sqrt n + b_m \sqrt{n+1}, na_m^2+1 = (n+1) b_m^2$$ If $$m$$ is even then $$x^m = c_m + d_m \sqrt{n(n+1)}, c_m^2 = n(n+1) d_m^2 +1$$
When $$m=1$$ it's true.
If it's true for $$m$$, we prove it's true for $$m+1$$.
The following is multiple choice question (with options) to answer.
If y is a positive even integer, and n and m are consecutive integers, then (n - m)^y/(m - n)^y = | [
"-2",
"-1",
"0",
"1"
] | D | y= +ve even integer
Since,n and m are consecutive integers , their difference will be 1
((n-m)^y)/((m-n)^y)= ((n-m)/(m-n))^y = (-1)^y
Since we are raising the difference of n and m to power y , which is even , the answer will be 1 .
Answer D |
AQUA-RAT | AQUA-RAT-36021 | 1. ## Average Speed
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?
A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5
(30 + 40)/2 = 35
So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.
2. ## Re: Average Speed
Originally Posted by harpazo
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?
A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5
(30 + 40)/2 = 35
So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.
The idea is this: $r\cdot t=D$ The distance $D$ is constant.
Going $30\cdot t_1=D$ and coming back we have $40\cdot t_2=D$
Thus $t_1=\dfrac{D}{30}~\&~t_2=\dfrac{D}{40}$
You want $Av\cdot(t_1+t_2)=2D$
Can you solve for $Av~?$
3. ## Re: Average Speed
Originally Posted by harpazo
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?
A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5
(30 + 40)/2 = 35
The following is multiple choice question (with options) to answer.
A car traveled 75% of the way from town A to town B at an average speed of 60 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 40 miles per hour. What is S ? | [
"10",
"20",
"25",
"30"
] | B | Total distance = 100 miles (easier to work with %)
75% of the distance = 75 miles
25% of the distance = 25 miles
1st part of the trip → 75/60 = 1.25
2nd part of the trip → 25/S = t
Total trip → (75+25)/40 = 1.25 + t » 100/40 = 1.25 + t » 2.5 = 1.25 + t » t = 1.25
Back to 2nd part of the trip formula: 25/S = 1.25 » S = 20
Ans B |
AQUA-RAT | AQUA-RAT-36022 | # Calculating probability for forming a triangle
I am having trouble coming up with a solution for this problem:
There is a stick of unit length. We break it into two parts. Now, we pick the bigger one and break it into two parts. I want to calculate the probability that the three pieces form a triangle.
The problem is from "Introduction to Probability, Charles M. Grinstead", Chapter 2.2, Exercise 13
• What have you attempted? – ncmathsadist Aug 3 '15 at 20:51
• You will need to add constraints on the likelihood of the pieces being certain sizes... Are the two breaks both uniform on $(0,1)$? Do you pick a break first uniformly then uniformly break the largest of the subsequent pieces...? Not enough information. – jameselmore Aug 3 '15 at 20:53
Let $\lambda$ be the length of the bigger piece and we split it into two smaller pieces $\lambda\mu$ and $\lambda(1-\mu)$. It is clear $\lambda$ and $\mu$ are uniform random variables $\sim \mathcal{U}(\frac12,1)$ and $\mathcal{U}(0,1)$ respectively.
In order for the three pieces with lengths $\;1-\lambda, \lambda\mu, \lambda(1-\mu)\;$ to form a triangle, the necessary and sufficient conditions are the fulfillment of following three triangular inequalities: $$\begin{cases} \lambda \mu + \lambda (1-\mu) &\ge 1-\lambda\\ \lambda \mu + (1 - \lambda) &\ge \lambda (1-\mu)\\ \lambda (1-\mu) + (1-\lambda) &\ge \lambda \mu \end{cases} \quad\iff\quad \begin{cases} \lambda \ge \frac12\\ \mu \ge 1 - \frac{1}{2\lambda}\\ \frac{1}{2\lambda} \ge \mu \end{cases}$$ The first inequality is trivially satisfied because we are told to break the bigger piece.
The probability we seek is given by:
The following is multiple choice question (with options) to answer.
Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle? | [
"440/455",
"412/455",
"434/455",
"449/455"
] | B | Total ways of selecting 3 dots out of 15 is 15C3 = 455 If 3 dots are collinear then triangle may not be formed. Now look at the above diagram. If we select any 3 dots from the red lines they may not form a triangle. They are 3 x 5C3 = 30. If we select the three letters from blue lines, they may not form a triangle. They are in total 5 ways. Also there are 6 others lines which don't form a triangle. Also another two orange lines. Total = 30 + 5 + 6 + 2 = 43. So we can form a triangle in 455 - 43 = 412. So answer could be 412/455.
ANSWER:B |
AQUA-RAT | AQUA-RAT-36023 | • Finally, suppose $$f(n) = r = 7$$. Here, the possible $$n$$ are $$23, 26, 29, 31, 32$$, $$34, 35, 37, 38$$, $$40, 41, 43, 46, 49$$. For each $$n$$, again we want pairs $$(a,b)$$ such that $$3a + 8b = 49 - n$$. For $$n = 23$$, we get $$(6,1)$$, which is strictly easier than $$n = 26, 29, 32, 35, 38, 41$$ (which require $$1$$ side-length $$3$$ and fewer than $$6$$ side-length $$2$$), and also strictly easier than $$n = 31, 34, 37, 40, 43, 46, 49$$ (which are the same except they don't require the side-length $$3$$ square or triangle). That's all the possible $$n$$, so we just have to show $$n = 23$$ is achievable. So we fit a $$3 \times 3$$ square and $$6$$ $$2 \times 2$$ squares into a $$7 \times 7$$ square, and similarly for a triangle (neither task is hard).
The following is multiple choice question (with options) to answer.
If f(2, 7) = 57 and f(1, 6) = 37, what is the value of f(3, 10)? | [
"107",
"127",
"147",
"None of these"
] | B | Solution:
The function f(a, b) = a3 + b2
f(2, 7) therefore is = 23 + 72 = 8 + 49 = 57 and
f(1, 6) = 13 + 62 = 1 + 36 = 37.
Therefore, f(3, 10) = 33 + 102 = 27 + 100 = 127.
Answer B |
AQUA-RAT | AQUA-RAT-36024 | thermodynamics
Say you own a theme park where, every hour, each adult has to pay ${\rm d}\mu_a=\$7$ (money=energy) and each kid has to pay ${\rm d}\mu_k=\$3$.
We assign positive signs to when they put money out of their pockets.
If there are $n_a=10$ adults and $n_k=5$ kids in the park, and if this number of people stays fixed (${\rm d}n_\text{totoal}=0$ over some time), then they together have to spend
$$n_a\cdot {\rm d}\mu_a + n_k\cdot {\rm d}\mu_k = \$70 + \$15 = \$85$$
Now say you, the park owner, are not allowed to actually make any money (${\rm d}G=0$) and you relax the condition that the kids have to pay any money at all. That is, now just the adult have to pay ... and since the money has to go somewhere, the kids are on the receiving end. Then the ten parents still spend $\$70$ per hour and now the five kids will split that money among each other. Each kids receives
$$-{\rm d}\mu_k = \frac{1}{n_k}\cdot n_a\cdot {\rm d}\mu_a = \frac{1}{5}\cdot \$70 = \$14$$
The comparison with "money per time" lacks in that here people don't bring in money just from coming to the park (as it's the case with particles coming into the system and energy). But I hope this clear up the meaning of factor $\frac{1}{n_k}$ clear.
The following is multiple choice question (with options) to answer.
Usually the room tariff in this hotel is higher. At present, it is low because of the ______ season.
a. peak | [
"PINK",
"bkue",
"orancge",
"off"
] | D | Explanation:
Because of the Off season.
Answer:D |
AQUA-RAT | AQUA-RAT-36025 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color? | [
"5/6",
"5/9",
"5/4",
"5/2"
] | B | Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9
Answer: B |
AQUA-RAT | AQUA-RAT-36026 | 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 650 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 48 km/hr? | [
"25",
"45",
"54",
"44"
] | C | Speed of train relative to man = 48- 3 = 45 km/hr.
= 45 * 5/18 = 25/2 m/sec.
Time taken to pass the man = 650 * 2/25 = 54 sec.
Answer: C |
AQUA-RAT | AQUA-RAT-36027 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
What will be the compound interest on Rs. 25000 after 3 years at the rate of 12 % per annum | [
"Rs 10123.20",
"Rs 10123.30",
"Rs 10123.40",
"Rs 10123.50"
] | A | Explanation:
(25000×(1+12/100)3)
=>25000×28/25×28/25×28/25
=>35123.20
So Compound interest will be 35123.20 - 25000
= Rs 10123.20
Option A |
AQUA-RAT | AQUA-RAT-36028 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes? | [
"192 km",
"162 km",
"172 km",
"168 km"
] | B | 12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km.
Answer:B |
AQUA-RAT | AQUA-RAT-36029 | Difference between revisions of "2014 AMC 10A Problems/Problem 17"
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$
Solution 1 (Clean Counting)
First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.
--happiface
Solution 2 (Bashy Casework)
Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.
The following is multiple choice question (with options) to answer.
Two 6 faced dice are thrown together. The probability that all the three show the same number on them is? | [
"1/32",
"1/6",
"1/33",
"1/38"
] | B | The three dice can fall in 6 * 6 = 36 ways.
Hence the probability is 6/36
= 1/6
Answer: B |
AQUA-RAT | AQUA-RAT-36030 | P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2 P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2
(1)
Show whether or not the pair {A,B}{A,B} is independent.
### Solution
P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc)P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc), and P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc)P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc).
PA = 0.4*0.7 + 0.3*0.3
PA = 0.3700
PB = 0.6*0.7 + 0.2*0.3
PB = 0.4800
PA*PB
ans = 0.1776
PAB = 0.4*0.6*0.7 + 0.3*0.2*0.3
PAB = 0.1860 % PAB not equal PA*PB; not independent
## Exercise 2
Suppose {A1,A2,A3} ci |C{A1,A2,A3} ci |C and ci |Cc ci |Cc, with P(C)=0.4P(C)=0.4, and
The following is multiple choice question (with options) to answer.
12 is 6% of a, and 6 is 12% of b. c equals b/a. What is the value of c? | [
"1 / 2",
"1 / 3",
"1 / 4",
"1 /6"
] | C | 6a/100 = 12
a = 200
12b/100 = 6
b = 50
c = b/a = 50 / 200 = 1/4
The answer is C. |
AQUA-RAT | AQUA-RAT-36031 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
If 6 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is? | [
"20:15:18",
"20:15:16",
"20:15:12",
"20:15:11"
] | C | 6A = 8B = 10 C
A:B:C = 1/6:1/8:1/10
= 20:15:12
Answer: C |
AQUA-RAT | AQUA-RAT-36032 | ## Extra 01
How many arrangements can be made using the letters from the word COURAGE? What if the arrangements must contain a vowel in the beginning?
• $$4 \times 6!$$
## Extra Problem 02
How many arrangements are possible using the words
• EYE
• CARAVAN?
## 3(a)
There are (p+q) items, of which p items are homogeneous and q items are heterogeneous. How many arrangements are possible?
## 2(j)
There are 10 letters, of which some are homogeneous while others are heterogeneous. The letters can be arranged in 30240 ways. How many homogeneous letters are there?
Let, $$m = \text{number of homogeneous items}$$
• n(arrangements) = 30240 = $$\frac {10!}{m!}$$
• $$m! = \frac{10!}{30240}=120$$
• m = 5
## 2(k)
A library has 8 copies of one book, 3 copies of another two books each, 5 copies of another two books each and single copy of 10 books. In how many ways can they be arranged?
Total books = $$1 \times 8+3 \times 2+5 \times 2 + 8 \times 1 + 10$$ = 42
• n(arrangements) = $$\frac{42}{8!(3!)^2(5!)^2}$$
## 2(l)
A man has one white, two red, and three green flags; how many different signals can he produce, each containing five flags and one above another?
Flags: W = 2, R = 2, G = 3, Total = 7
Answer
## 2 (m)
A man has one white, two red, and three green flags. How many different signals can he make, if he uses five flags, one above another?
## 3(a)
How many different arragnements can be made using the letters of the word ENGINEERING? In how many of them do the three E’s stand together? In how many do the E’s stand first?
i
ii
iii
## 3(b)
In how many ways can the letters of the word CHITTAGONG be arranged, so that all vowels are together?
The following is multiple choice question (with options) to answer.
In how many way the letter of the word "APPLE" can be arranged | [
"20",
"40",
"60",
"80"
] | C | Explanation:
Friends the main point to note in this question is letter "P" is written twice in the word.
Easy way to solve this type of permutation question is as,
So word APPLE contains 1A, 2P, 1L and 1E
Required number =
=5!/1!∗2!∗1!∗1!=5∗4∗3∗2!/2!=60
Option C |
AQUA-RAT | AQUA-RAT-36033 | Is this approach okay?
Math Expert
Joined: 02 Aug 2009
Posts: 7334
Re: A certain jar contains only b black marbles, w white marbles [#permalink]
### Show Tags
23 Jan 2019, 22:46
1
blitzkriegxX wrote:
anilnandyala wrote:
A certain jar contains only b black marbles, w white marbles and r red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater then the probability that the marble chosen will be white?
(1) r/(b+w) > w/(b+r)
(2) b-w > r
Hi chetan2u
Is it really necessary to do the long simplifications for st 1 ?
$$\frac{r}{(b+w)} > \frac{w}{(b+r)}$$
Since we know "b" is just some constant, we can just start by taking the above as -
$$\frac{r}{(w)} > \frac{w}{(r)}$$
After which it simply is -
$$r^2 > w^2$$
or
$$r > w$$
Is this approach okay?
Yes, this is much simpler and better..
You can do this as all variables are positive and the question finally boils down to which is more r or w..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html
GMAT Expert
The following is multiple choice question (with options) to answer.
Jar A has 16% more marbles than Jar B . What percent of marbles from Jar A need to be moved into Jar B so that both jars have equal marbles? | [
"5%",
"7%",
"9%",
"11%"
] | B | An easy way to solve this question is by number plugging. Assume there are 100 marbles in Jar B then in Jar A there will be 116 marbles. Now, for both jars to have equal marbles we should move 8 marbles from A to B, which is 8/116=~7% of A.
Answer: B. |
AQUA-RAT | AQUA-RAT-36034 | visible-light
The color we call 'red' is the color with the longest wavelength.
Prisms and water droplets are not the only things that affect the propagation direction of light. As light travels through the atmosphere it has a small probability of being scattered. The short wavelengths (the colors we perceive as shades of blue) are more prone to being scattered.
In the evening and in the morning the Sun is close to the horizon and the light has to travel through a lot of atmosphere to get to us. In general the composition of the light from the Sun is such that we will perceive that as white light. But with the atmosphere scattering the blue end of the spectrum, we often see the Sun having a red color when the Sun is close to the horizon.
The Sun gives us warmth, so naturally all us humans will associate the color we call 'Red' with warmth. Conversely, we will all associate absence of any hint of red (the blue colors) as comparatively cold.
We perceive Yellow as a particularly bright color, because we perceive incoming light as yellow when it has such a composition that it triggers two of our three light-sensitive molecules in about equal measure. Put differently: we perceive a mix of different colors of light as Yellow when it has the Red portion and The green portion of the spectrum in it, but little blue portion.
To get from light entering the eye to the perception of color by the visual cortex is a long way; it involves stages of processing information. It is remarkable that we experience different colors so vividly different, given that the perception of color is very much a constructed perception.
I suppose that our brain makes the different colors so vividly different by having specific strong associations. For instance, it seems to me that we will naturally associate the spectrum of greens that we perceive with the things around us that are predominantly green. It seems to me that these associations, not all of them conscious associations, allow us to feel a lot of distinction between the colors we perceive.
Your question, 'why are the colors ordered that way'.
The following is multiple choice question (with options) to answer.
If green means red, red means white, white means blue, blue means orange and orange means green, what is the colour of snow ? | [
"Blue",
"Green",
"Yellow",
"Red"
] | A | Explanation:
The colour of snow is 'blue' and as given 'White' means Blue'.
so, the snow is 'blue'.
Answer: A |
AQUA-RAT | AQUA-RAT-36035 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum? | [
"288",
"327",
"585",
"972"
] | C | 90 = (450*4*R)/100
R = 5%
I = (450*6*5)/100 = 135
450 + 135 = 585
Answer: C |
AQUA-RAT | AQUA-RAT-36036 | # 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 fetched total simple interest of 4016.25 at the rate of 13 p.c.p.a. in 5 years. What is the sum? | [
"6178.85",
"8032.5",
"4462.5",
"8900"
] | A | Let the sums be P.
Now, 65% of P = 4016.25
or, P = 6178.85
Answer A |
AQUA-RAT | AQUA-RAT-36037 | 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 100 freshmen at a particular college, all of whom must take at least one of the three core classes: Art, Biology, and Calculus. Of these freshmen, 21 take only Biology, 10 take only Calculus, 5 take all three classes, and 20 take Art and exactly one of the other two core classes. If the number of freshmen who take only Art is 3 times the number of freshmen who take every core class except Art, how many freshmen take Art? | [
"25",
"32",
"36",
"58"
] | D | Make a venn diagram to get a clear picture. Look at the diagram:
Each letter represents only one color. b represents the people who take only Art. d represents people who take only Art and Bio etc.
d + f = 20 (People who take Art and one other class)
b = 3e (people who take only Art is 3 times the people who take Bio and Calculus)
21 + 10 + 5 + b + d + e + f = 100 (Total people)
b + b/3 = 44
b = 33
Number of freshmen who take Art = 33 + 20 + 5 = 58
Answer D |
AQUA-RAT | AQUA-RAT-36038 | Question
(a) The “lead” in pencils is a graphite composition with a Young's modulus of about $1 \times 10^9 \textrm{ N/m}^2$. Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils?
$1.2 \textrm{ mm}$
Solution Video
# OpenStax College Physics Solution, Chapter 5, Problem 31 (Problems & Exercises) (1:30)
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## Calculator Screenshots
Video Transcript
Submitted by shelby.keith on Mon, 02/25/2019 - 20:55
Good solution
Submitted by C-Mak Breezy on Wed, 11/27/2019 - 08:05
How come when I put the equation '0.50x10^-3/2' it gives me 2.5x10^4? Your solution is showing the result as 0.25x10^3. Am I missing something?
Submitted by ShaunDychko on Wed, 11/27/2019 - 10:07
I agree with your calculator. The two figures are the same quantities, expressed differently. If you try subtracting one from the other you'll find the answer is zero, as expected for numbers that are the same quantities. The 0.25x10^3 is something I just did in my head rather than using a calculator.
In reply to by C-Mak Breezy
The following is multiple choice question (with options) to answer.
If 1/2 of a pencil is blue, 1/8 of the remaining is red and the remaining 3/8 is white, what is the total length of the pencil? | [
"5",
"3",
"1",
"4"
] | C | Blue is 1/2 but can also be written as 4/8
Red is 1/8
White is 3/8
4/8+1/8+3/8= 8/8=1
Answer is C) 1 |
AQUA-RAT | AQUA-RAT-36039 | In the odd cases, write the even numbers in ascending order and the odd numbers in descending order, repeat this series $\frac{1}{2}(n-1)$ times, follow with the even numbers in ascending order (omitting $n-1$), the odd numbers in descending order (omitting 1), and conclude with all the numbers (odd and even) in their natural order (omitting 1 and n). Thus for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73 moves.
(source)
which I had tried to solve but could not find the reasoning behind the moves used in the solution could anybody tell how to handle this type of puzzles without hit and trial?
The following is multiple choice question (with options) to answer.
Look at this series: 7, 10, 8, 11, 9, 12, 10, 13, 11, 14, ... What number should come next? | [
"10",
"11",
"12",
"13"
] | C | This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.
The answer is C. |
AQUA-RAT | AQUA-RAT-36040 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
In a party attended by 13 persons, each clinch their glass with every other. How many glass clinches? | [
"51",
"52",
"78",
"54"
] | C | Total no. of person = 11
Total no. of glass clinches = n(n-1)/2
=13*12/2
= 78
ANSWER:C |
AQUA-RAT | AQUA-RAT-36041 | 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 persons start running simultaneously around a circular track of length 200 m from the same point at speeds of 6 km/hr and 12 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions? | [
"28 sec",
"37 sec",
"17 sec",
"60 sec"
] | D | Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 200 / (6+ 12)5/18 = 300* 18 / 18 * 5 = 60 seconds.
Answer : D |
AQUA-RAT | AQUA-RAT-36042 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
4 shepherds were watching over the flocks and they were commenting on how many sheep they each had. If ram had 3 more sheep than he would have one less than rahul. wheras akar has the same number as the other 3 shepherds put togeher. If john had 3 less sheep he would have exactly trile the number of ram. If they were evenly distributed if they would each have 11 seep how many sheep did ram have? | [
"2",
"3",
"4",
"5"
] | B | akar has=ram+rahul+john
after evenly distribution each has 11. so, total no. is 44
so, akar has=22 & ram+rahul+john=22 also ram=rahul-4 & john-3=3*ram solving these we get the sol.
ANSWER:B |
AQUA-RAT | AQUA-RAT-36043 | c#, uwp
await Task.Delay(300);
}
private async void DecreaseMonth()
{
ScrollViewer listScrollViewer = GetScrollViewer(GridDays);
FirstDay = (Convert.ToUInt16(Math.Truncate(listScrollViewer.VerticalOffset / 50)) * 7);
int numberOfDates;
bool firstdayfound = false;
DateTime indexSucceedingMonth;
double avanzo = (listScrollViewer.VerticalOffset / 50) - (Math.Truncate(listScrollViewer.VerticalOffset / 50));
if (avanzo == 0)
{
numberOfDates = 42;
}
else
{
numberOfDates = 49;
}
int nDaysMonthFirstDate = FindNumberOfDays(ViewModel.listDateDay[FirstDay].Month);
int firstDayPresent = ViewModel.listDateDay[FirstDay].Day;
int nDaysPresentFirstDate = nDaysMonthFirstDate - (firstDayPresent - 1);
int nDaysPresentSecondDate = numberOfDates - (nDaysPresentFirstDate);
if (nDaysPresentFirstDate > nDaysPresentSecondDate)
{
DateTime IncreasedDate = ViewModel.listDateDay[FirstDay].AddMonths(-1);
indexSucceedingMonth = new DateTime(IncreasedDate.Year, IncreasedDate.Month, 1, 0, 0, 0);
while (firstdayfound == false)
{
if (ViewModel.listDateDay[FirstDay] == indexSucceedingMonth)
{
firstdayfound = true;
}
FirstDay -= 1;
}
}
else
{
DateTime IncreasedDate = ViewModel.listDateDay[FirstDay];
indexSucceedingMonth = new DateTime(IncreasedDate.Year, IncreasedDate.Month, 1, 0, 0, 0);
while (firstdayfound == false)
{
if (ViewModel.listDateDay[FirstDay] == indexSucceedingMonth)
{
firstdayfound = true;
}
FirstDay -= 1;
}
}
The following is multiple choice question (with options) to answer.
Reduce
368/575 to the lowest terms. | [
"30/25",
"28/29",
"27/29",
"16/25"
] | D | Explanation:
We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25
Option D |
AQUA-RAT | AQUA-RAT-36044 | # 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 30 km downstream and 20 km upstream in 4 hours. He can row 45 km downstream and 40 km upstream in 7 hours. Find the speed of man in still water?
A. 15 kmph B. 10 kmph | [
"12.9 kmph",
"28.8 kmph",
"28.9 kmph",
"12.5 kmph"
] | D | Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a - b) = 4 and 45/(a + b) + 40/(a - b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph.
Answer:D |
AQUA-RAT | AQUA-RAT-36045 | 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 man started driving at a constant speed, from the site of a blast, the moment he heard the blast. He heard a second blast after a time of 30 mins and 15 seconds. If the second blast occurred exactly 30 mins after the first, how many meters was he from the site when he heard the second blast? ( speed of sound = 330 m/s) | [
"1650",
"2750",
"3850",
"4950"
] | D | The distance the sound traveled to the man is 15*330 = 4950 meters
The answer is D. |
AQUA-RAT | AQUA-RAT-36046 | java, console, calculator, finance
taxableIncome <= hohFiler[5].maxSalary //$432,200
)
{
HeadOfHouseFilers.TierFive();
}
//TierSix()
else if (
taxableIncome >= hohFiler[5].minSalary && //$405,101
taxableIncome <= hohFiler[5].maxSalary && //$432,200
taxableIncome > hohFiler[6].minSalary //$432,201
)
{
HeadOfHouseFilers.TierSix();
}
//TierSeven()
else if (
taxableIncome >= hohFiler[6].minSalary && //$432,201
taxableIncome <= hohFiler[6].maxSalary //Double.MAX_VALUE
)
{
HeadOfHouseFilers.TierSeven();
}
break;
The following is multiple choice question (with options) to answer.
If the personal income tax rate is lowered from 42% to 32%, what is the differential savings for a tax payer having an annual income before tax to the tune of $42400? | [
"$3500",
"$5000",
"$4240",
"$7000"
] | C | Saving = (42-32)% of 42400 = 4240.
Answer:C |
AQUA-RAT | AQUA-RAT-36047 | that can be factored with ease!
Use the fact that $73 \times 137=10001 = 10^4+1$.
Now, mark off the number in groups of four digits starting from the right, and add the four-digit groups together with alternating signs.
Applying the above rule, $1000027$ in groups of $4$ is $\underbrace{0100}$ $\underbrace{0027}$. Adding the groups with alternate signs gives $73$. Therefore, $1000027$ is divisble by $73$ and gives $13699$ as quotient.
For $13699$ apply the divisbility test by $7$ by marking of the digits in groups of $3s$ by starting from the right and adding together with alternate signs. Therefore, adding the groups $\underbrace{013}$ $\underbrace{699}$ with alternate signs gives $686$ which is divsible by $7$ and hence $13699$ is disvisble by $7$
$13699$ when divided by $7$ gives $1957$.
Now, $1957$ is $1900+57$ and hence is divisible by $19$ giving the quotient as $103$.
Combining them all gives $1000027 = 7\times19\times73\times103$
To begin with note that
$1000027 = 100^3 + 3^3 = (100+3)(100^2−3⋅100+32) = 103 \cdot 9709 = 103 \cdot (1001 \cdot 9 + 700) = 103 \cdot 7 \cdot (13 \cdot 11 \cdot 9+100) = 103 \cdot 7 \cdot 1387$
where we have used the well-known fact that $1001 = 7 \cdot 11 \cdot 13$
The following is multiple choice question (with options) to answer.
The number 523hbc is divisible by 7,8,9. Then what is the value of h*b*c | [
"504",
"532",
"210",
"180"
] | D | LCM of 7, 8 and 9 is 504, thus 523hbc must be divisible by 504.
523hbc=523000+hbc
523000 divided by 504 gives a remainder of 352.
Hence, 352hbc=k*504.
k=1 hbc=152 --> h*b*c=10
k=2 hbc=656 --> h*b*c=180
As hbc is three digit number k can not be more than 2.
Two answers? Well only one is listed in answer choices, so D.
Answer: D. |
AQUA-RAT | AQUA-RAT-36048 | (Quad with diagonals that bisect each other. Complex Quadrilaterals. Perimeter is the _____ around a closed plane figure. A quadrilateral is a trapezoid or a trapezium if 2 of its sides parallel to each other. • its diagonals bisect each other. If a quadrilateral is a parallelogram, then it's ____ bisect each other. Parallelograms are special types of quadrilaterals with opposite sides parallel. It is a quadrilateral where both pairs of opposite sides are parallel. (iii) A quadrilateral is a parallelogram if it has one pair of opposite sides parallel and equal. 4) Theorem 6.2B states: If both pairs of opposite _____ of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Parallelogram. One special kind of polygons is called a parallelogram. The sum of all the angles of a quadrilateral is equal to: a. Get better grades with tutoring from top-rated private tutors. This lesson introduces four methods of identifying parallelograms. The area of a composite figure can be found by adding or _____ the areas of the shapes it contains. Prove that MO¯ bisects ∠NMP and ∠NOP, and that NP¯ bisects ∠MNO and ∠MPO. 15) Find the area of region 1 of the composite figure to the nearest tenth: 16) Find the area of region 2 of the composite figure to the nearest tenth: 17) Find the area of region 3 of the composite figure to the nearest tenth: 18) Find the area of region 4 of the composite figure to the nearest tenth: 19) Find the total area of the composite figure to the nearest tenth: 20) Find the area of region 1 of the composite figure: 21) Find the area of region 2 of the composite figure: 22) Find the total area of the composite figure: The center of a regular polygon is a point _____ from each of its vertices. = 8 cm, find the area of a quardiletral is 360° matter which you! Something else, like a parallelogram supplementary ( a ) square ( B ) rhombus ( ). Are ___________ parallel, which means two pairs of opposite
The following is multiple choice question (with options) to answer.
Find the area of the quadrilateral of one of its diagonals is 30 cm and its off sets 9 cm and 6 cm? | [
"225 cm2",
"150 cm2",
"127 cm2",
"177 cm2"
] | A | 1/2 * 30(9 + 6)
= 225 cm2
Answer: A |
AQUA-RAT | AQUA-RAT-36049 | An imaginary circle is one in which the radius is the square root of a negative number—i.e., imaginary.
• Sir, can we call a point is Cartesian coordinates "a point circle" ? Jul 19, 2019 at 14:44
• Sir, we are newbies.Can you please explain what can a circle look like if it's radius is zero? Jul 19, 2019 at 15:05
• Who is "we"? And surely, you needn't call me "sir." Anyway, a circle is a point if its radius is zero. This seems to violate intuition regarding what a circle is, but keep in mind that a circle is merely the set of points that are a fixed distance away from the center. If that fixed distance is zero, then that set of points is simply the center point itself. In the same way, a line segment of length zero is a single point, a square of side length zero is a single point, a sphere of radius zero is a single point, and so on. Jul 19, 2019 at 16:54
A circle can be described by the equation $$r^2 = (x-h)^2 + (y-k)^2$$ where $$r$$ is the radius of the circle and $$(h,k)$$ is the center of the circle.
1. If $$r^2 = 0$$, then this equation has only one solution, the point $$(h,k)$$. Thus the equation describes a "point circle".
2. If $$r^2 > 0$$, then this equation has many solutions of the form $$(x,y)$$, where both $$x$$ and $$y$$ are real numbers. The solutions form a circle in the real plane, thus the equation describes a "real circle".
3. If $$r^2 < 0$$, then this equation has many solutions of the form $$(x,y)$$, but these solutions will be complex (not real). Because the equation is that for a circle, but the solutions are not real, we could describe this as an "imaginary circle" or an "unreal circle".
## Explanation
The following is multiple choice question (with options) to answer.
In the xy-plane, the point (3, 1) is the center of a circle. The point (3, -2) lies inside the circle and the point (-2, 1) lies outside the circle. If the radius r of the circle is an integer, then r = | [
"2",
"3",
"4",
"5"
] | C | An easy way to solve this question will be just to mark the points on the coordinate plane.
You'll see that the distance between the center (3, 1) and the point inside the circle (3, -2) is 3 units
(both points are on x=3 line so the distance will simply be 1-(-2)=3)
So the radius must be more than 3 units.
The distance between the center (3,1) and the point outside the circle (-2,1) is 5 units
(both points are on y=1 line so the distance will simply be 3-(-2)=5)
So the radius must be less than 5 units, which implies 3 < r < 5, thus as r is an integer then r=4.
Answer: C. |
AQUA-RAT | AQUA-RAT-36050 | 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 long does a train 165 meters long running at the rate of 90 kmph take to cross a bridge 660 meters in length? | [
"28",
"27",
"33",
"18"
] | C | T = (660 + 165)/90 * 18/5
T = 33
Answer: C |
AQUA-RAT | AQUA-RAT-36051 | 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 much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction? | [
"50",
"26",
"28",
"69"
] | A | The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-36052 | So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
VP
Joined: 07 Dec 2014
Posts: 1128
Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A reduction of 44% in the price of bananas would enable a man to obtain 64 more for Rs.40, what is reduced price per dozen? | [
"2.3",
"8.3",
"7.3",
"3.3"
] | D | Explanation:
40*(44/100) = 17.6 --- 64
? --- 12 => Rs.3.30
Answer: D |
AQUA-RAT | AQUA-RAT-36053 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
X and Y are integers, X is even and negative, Y is odd and positive. Which of the following could be false?
1. (X + Y) is an even number.
2. Y^(X + Y) is not an integer.
3. X^Y is a positive number. | [
"2 only.",
"3 only.",
"1 and 3 only.",
"2 and 3 only."
] | A | Statement I is not TRUE
Statement II is TRUE
Statement III is not TRUE
Answer : A |
AQUA-RAT | AQUA-RAT-36054 | There will be one solution: name it $x_0$.
Find $y$ at $x_0$, and call it $y_0$.
Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$
The following is multiple choice question (with options) to answer.
fill in ht blank:
(a) the point with coordinates (0,0) is called ____ of a rectangular coordinate system,
(b) to find the y-intercept of a line, we let ___ equal 0 and solve for ___; to find x- intercept , we let ____ equal 0 and solve for___ | [
"(a) the point with coordinates (0,0) is called axis of a rectangular coordinate system, (b) to find the y-intercept of a line, we let x equal 0 and solve for y ; to find x- intercept , we let y equal 0 and solve for y",
"(a) the point with coordinates (0,0) is called end of a rectangular coordinate system, (b) t... | C | (a) the point with coordinates (0,0) is called origin of a rectangular coordinate system, (b) to find the y-intercept of a line, we let x equal 0 and solve for y ; to find x- intercept , we let y equal 0 and solve for x
correct answer (C) |
AQUA-RAT | AQUA-RAT-36055 | # 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 three letter words are formed using the letters of the word TIME? | [
"28",
"21",
"26",
"24"
] | D | The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃
= 4 * 3 * 2 = 24.
Answer: D |
AQUA-RAT | AQUA-RAT-36056 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
Calculate the largest 6 digit number which is exactly divisible by 55? | [
"999955",
"999900",
"999845",
"999790"
] | A | Largest 4 digit number is 999999
After doing 999999 ÷ 55 we get remainder 44
Hence largest 4 digit number exactly divisible by 88 = 999999 - 44 = 999955
A |
AQUA-RAT | AQUA-RAT-36057 | $$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games
The following is multiple choice question (with options) to answer.
A baseball team won 45 percent of the first 80 games it played. How many of the remaining 82 games will the team have to win in order to have won exactly 50 percent of all the games it played? | [
"36",
"45",
"50",
"55"
] | B | Number of games won by baseball team in first 80 games = .45* 80 = 36
Total number of games = 80 + 82 = 162
% of wins required in 162 = 50
Number of wins required in 162 = 81
Number of wins required in remaining 82 games = 81 - 36 = 45
Answer B |
AQUA-RAT | AQUA-RAT-36058 | 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 persons start running simultaneously around a circular track of length 400 m from the same point at speeds of 15 kmph and 25 kmph. When will they meet for the first time any where on the track if they are moving in the opposite direction ? | [
"144",
"36",
"124",
"32"
] | B | Time taken to meet the first time = length of track/relative speed
= 400/ (15 +25) (5/18)
= 400/40 *(18/5) = 36 sec.
ANSWER:B |
AQUA-RAT | AQUA-RAT-36059 | 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 takes 3hrs less than a truck to cover the distance frm Bombay to pune(192 km). The speed of the truck is 16km/hr slower than car.Find the speed of the car? | [
"40.38",
"38.38",
"41.3",
"48"
] | A | Let the speed of car = x km/hrs
then that of truck = (x-16) km/hrs
According to the given condition
[192/(x-16)]- [192/x] = 3
on solving the equation we have x = 8[(sq-root of 17) + 1] =40.38
ANSWER:A |
AQUA-RAT | AQUA-RAT-36060 | The pattern emerges and the final answer could be written as
x = (1/39)[4(10^d) - 16]
where is an integer and d > 0, which represents the number of digits in x.
For d = 5 we obtain our answer. You still have to try d for 1 through 4, but it's still easier to do it this way.
It literally took me 30 seconds to derive this expression and had the answer in no time.
What if the question was asking for the next smallest number that satisfies this condition? Yeah... try doing it the other way. For that situation d = 11, so you'd have to do it many time. That number is 10,256,410,256.
See a pattern?
The following is multiple choice question (with options) to answer.
If d=1/(2^3*5^11) is expressed as a terminating decimal, how many nonzero digits will d have? | [
"One",
"Two",
"Three",
"Seven"
] | C | Another way to do it is :
We know x^a*y^a=(X*Y)^a
given = 1/(2^3*5^11)
= Multiply and divide by 2^8
=2^8/(2^3*2^8*5^11)
=2^8/10^11
=> non zero digits are 256=> Ans C |
AQUA-RAT | AQUA-RAT-36061 | # How many $2$'s are needed?
There is a positive integer $N$. $N$ is made up of only two distinct digits- $2$ and $3$. $N+18$ is divisible by $37$. What is the minunum amount of times the number $2$ can appear in $N$? I'm pretty sure the answer is only one time. But how can I prove this?
• are there any better tags I could use? – Joao Dec 3 '14 at 22:22
• I'm afraid I don't share your optimism; the smallest count of twos I managed to come up was three. – Peter Košinár Dec 3 '14 at 22:33
• @PeterKošinár but is it possible to prove that the smallest amount of two's is 3? – Joao Dec 3 '14 at 22:39
• Note that $1000\equiv 1 \mod 37$ so you can reduce the problem to analysing the sum of digits (with some cases to check). – Mark Bennet Dec 3 '14 at 23:01
• @MarkBennet Thanks. That should be an answer. – Joao Dec 3 '14 at 23:08
## 3 Answers
Let's start with a simple observation:
The number $37$ plays very nicely with powers of $10$... The value $10^m$ can only have three possible remainders modulo $37$: $1$, $10$ and $26$. If $N+18$ is divisible by $37$, $N$ must be congruent to $19$ modulo $37$. This observation will be quite useful in subsequent thoughts.
The number $N=2323323$ satisfies the given conditions and contains three twos. In order to show that we cannot satisfy the conditions given with fewer twos, we need to consider three cases:
The following is multiple choice question (with options) to answer.
What least number should be added to 1101, so that the sum is completely divisible by 24 | [
"A)4",
"B)1",
"C)2",
"D)3"
] | D | Explanation:
(1056 / 24) gives remainder 21
21 + 3 = 24, So we need to add 3.
Answer: Option D |
AQUA-RAT | AQUA-RAT-36062 | $9a^2 - 2a^3 = 18a^2 - 4a^3$
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?
$$\int _{ a }^{ 2a }{ \left( 3x-{ x }^{ 2 } \right) dx } =2\int _{ 0 }^{ a }{ \left( 3x-{ x }^{ 2 } \right) dx } \\ \frac { 36{ a }^{ 2 }-16{ a }^{ 3 } }{ 6 } -\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ \frac { 27{ a }^{ 2 }-14{ a }^{ 3 } }{ 6 } =\frac { 9{ a }^{ 2 }-2{ a }^{ 3 } }{ 3 } \\ 27{ a }^{ 2 }-14{ a }^{ 3 }=18{ a }^{ 2 }-4{ a }^{ 3 }\\ 9{ a }^{ 2 }-10{ a }^{ 3 }=0\\$$ clearly $a\neq 0$ so the answer is $$\color{red}{a=\frac { 9 }{ 10 }}$$
Since $a\neq 0$ (as this would be absurd), we can solve your equation as follows
\begin{align}9a^2−2a^3&=18a^2−4a^3\\ 9-2a&=18-4a\tag{divide through by $a$}\\ 2a&=9\\ a&=\frac 92\end{align}
However, as noted in other answers, this is incorrect.
Your error is somewhere in your calculation of the area of $Q$, you should get $$\int_a^{2a}(3x-x^2)\text dx =\frac{27a^2-14a^3}{6}$$
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of 2a+16, 3a-8 is 94, what is the value of a? | [
"25",
"30",
"28",
"36"
] | D | AM of 2a+16, 3a-8=2a+16+ 3a-8 /2= 5a+8/2
Given that 5a+8 /2 = 94
a= 36
Answer is D |
AQUA-RAT | AQUA-RAT-36063 | (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.
Following an increase in prices, the price of a candy box was 20 pounds and the price of a can of soda was 6 pounds. If the price of a candy box was raised by 25%, and the price of a can of soda was raised by 50%. What was the price of a box of candy plus a can of soda before prices were raised? | [
"11.",
"20.",
"13.",
"14."
] | B | Price of candy before price increase= 20/1.25= 16
Price of soda before price increase= 6/1.5= 4
Total price = 16+4= 20
B is the answer |
AQUA-RAT | AQUA-RAT-36064 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
The rate of Interest on a sum of money is 8% p.a. for the first 3 years, 4% p.a. for the next 4 years, and 5% for the period beyond 7 years. If the S.I, Occured on the sum for the total period of 8 years is Rs. 540/-, the sum is | [
"1,200",
"2,000",
"2,100",
"2,250"
] | A | Explanation:
I1 = (P x 3 x 8)/100 = 6P/25
I2 = (P x 4 x 4)/100 = 4P/25
I3 = (P x 1 x 5)/100 = P/20
6P/25 + 4P/25 + P/20 = 540
9P/20 = 540
P = 1200
Answer: Option A |
AQUA-RAT | AQUA-RAT-36065 | Veritas Prep Reviews
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### Show Tags
05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
A mail handler needs 3 hours to sort the mail. His assistant takes 6 hours to sort the mail. How many hours will it take for both of them working together to sort the mail? | [
"3",
"2",
"1",
"2 1/2"
] | B | Work hrs=AB/(A+B)= 18/9 =2
Answer is B |
AQUA-RAT | AQUA-RAT-36066 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Three pipes, A, B, & C are attached to a tank. A & B can fill it in 20 & 30 minutes respectively while C can empty it in 15 minutes. If A, B & C are kept open successively for 0.5 minute each, how soon will the tank be filled? | [
"1 hour 30 min",
"4 hours",
"3 hours",
"5 hours"
] | A | in three minute 1/20+1/30-1/15=1/60 part is filled
1.5 min--------1/60 parts
x min--------- 1 part(full)
x=90 min =1 hour 30 min
ANSWER:A |
AQUA-RAT | AQUA-RAT-36067 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
If the value of a piece of property decreases by 10% while the new tax rate on the property represent 110% of the original tax rate, what is the effect on the taxes? | [
"taxes increase by 10%",
"taxes increase by 1%",
"there is no change in taxes",
"taxes decrease by 1%"
] | D | Consider the value of property = 100, Original tax rate 10%, hence tax value = 100*10% = 10.
Decreased value of property = 90, New tax rate = 110% of 10 = 11%, hence tax value = 90*11% = 9.9
10-9.9 = decrease by 1%.
Answer D. |
AQUA-RAT | AQUA-RAT-36068 | of sides of the parallelogram ( - 2, 3 ), (,... D, the fourth vertex or one of the fourth vertex of the fourth?... Of line segment joining the points and is given points are vertices of a.... The parallelogram whose three if three vertices of a parallelogram are of a parallelogram are ( -2, -1 ), B ( 8,1 ) B. Points P, Q, R are 3 vertices of parallelogram Worksheet 8., and ( 9, 8 ) the distance formula to find out the measure sides! For three vertices of a parallelogram 4th vertex two points or vertices can be used to draw of! Class 11 Science Class 11 C are the three given points are vertices of parallelogram ABCD are (... Let ’ s call a, B, C are the possible locations the! Measure of sides of the fourth vertex using the distance formula to find out the of... Where students can interact with teachers/experts/students to get solutions to their queries vertices are a... ’ s call a, B ( 6, 7 ) and (,. Formula to find out the measure of sides of the parallelogram consecutive vertices a. Find the coordinates of D, the fourth vertex R are 3 vertices of a parallelogram are (,. Three digit numbers divisible by 6, R are 3 vertices of a parallelogram are (,. In the figure interact with teachers/experts/students to get solutions to their queries =... The parallelogram to their queries the 4th vertex line segment joining the points and is using distance! Welcome to Sarthaks eConnect: a unique platform where students can interact with teachers/experts/students to get solutions their. Sides of the parallelogram of sides of the fourth vertex given points vertices. Verify the given points point D, the fourth vertex of the parallelogram 3 of! Shown in the figure are ; a ( 0,2 ), B, C the... Points are vertices of a parallelogram are ( -2, -1 ), and 9. Call a, B ( 6, 7 ) and ( 4,3 ) D the..., B ( 8,1 ), ( 3, 8 ) also an opposite vertices of a parallelogram shown... Midpoint of line segment joining
The following is multiple choice question (with options) to answer.
ABCD is a parallelogram on xy rectangular coordinate plane, three of the four vertices of parallelogram are (5,15) , (5,55) and (-5,15). Which can be the coordinate of fourth vertex? | [
"(15,55)",
"(5,15)",
"(-5,-25)",
"(-5,10)"
] | A | A(-5, 15)
B(5, 15)
C(5, 55)
X(-5, 55): Parallelogram=Square=AXCB, where AX||CB, AX=CB, XC||AB, XC=AB
Y(-5, 5): Parallelogram=AYBC, where AY||BC, AY=BC, AC||YB, AC=YB
Z(15, 55): Parallelogram=ACZB, where AC||ZB, AC=ZB, CZ||AB, CZ=AB
Ans:A |
AQUA-RAT | AQUA-RAT-36069 | inorganic-chemistry, aqueous-solution, solubility
Title: Solutions of several salts What happens when you have a solution of a salt, and then try to dissolve another salt in that solution? I know about the common ion-effect etc, but how do you solve problems like this:
How much AgCl can you solve per L in a 3.8g/L NaCl solution? To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,
$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$
This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.
If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.
$$\begin{array}{lcl}
x&=&[\ce{Ag+}]=[\ce{Cl-}]\\
x^2 &=& 1.8\times 10^{-10} \\
x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5}
\end{array}$$
We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.
The following is multiple choice question (with options) to answer.
Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? | [
"6:4",
"6:14",
"4:4",
"4:6"
] | C | let:
x = ounces taken from solution A (20% salt)
y = ounces taken from solution B (80% salt)
to prepare 50 ounce 50% salt.
first equation is simple:
x + y = 50
to get another equation so as to be able to solve, compute salt contents.
20% of x + 80% of y = 50% of 50 or
x/5 + 4/5 * y = 25 or
x+4y = 125
solve two equations to get:
x = 25
y = 25
so solutions has to mix in
1:1 oops 4:4
ANSWER:C |
AQUA-RAT | AQUA-RAT-36070 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Three years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now? | [
"20 years",
"24 years",
"27 years",
"30 years"
] | B | Sol.
Present age of ( A + B) = ( 18 × 2 + 3 × 2) years = 42 years.
Present age of ( A + B + C) = (22 × 3) years = 66 years.
Therefore C’s age = (66 – 42) years
= 24 years.
Answer B |
AQUA-RAT | AQUA-RAT-36071 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief steals at a car at 2.30 p.m. and drives it at 60 km/hr. The theft is discovered at 3 p.m. and the owner sets off in another car at 75 km/hr. When will he overtake the thief? | [
"2",
"78",
"5",
"6"
] | C | Suppose the thief is overtaken x hrs after 2.30 p.m.
Then, distance covered by the owner in (x - 1/2) hrs.
60x = 75(x - 1/2) => x = 5/2 hrs.
So, the thief is overtaken at 5 p.m.
Answer: C |
AQUA-RAT | AQUA-RAT-36072 | • #### 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.
The ratio between the length and the breadth of a rectangular park is 3:2. If a man cycling along theboundary of the park at the speed of 12 km/hr completes one round in 8 min, then the area of the park is? | [
"163600 m",
"178600 m",
"153600 m",
"163600 m"
] | C | Perimeter = Distance covered in 8 min. = 12000 x 8 m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m2 = 153600 m
C |
AQUA-RAT | AQUA-RAT-36073 | optimization
Title: Algorithm for campaign optimization (Digital Advertising) Suppose i am running an Ad thru an Ad exchange A, and i have a set of campaigns running on it. I have
The spend of the campaign.
The budget allocated to it.
The number of hours it took to exhaust it's budget.
Total installs.
Total Revenue.
I want to write an algorithm to divide the total budget x to n campaigns, allocating higher budgets to a campaign delivering higher revenue. sounds like a great place for a weighted average and a little algebra.
budget_n = (revenue_n / total_revenue) * total_budget
This is a simple solution you may have already tried, but in principle, the revenue fraction can be as complex as you need it to be as all the fractions across all campaigns add up to 1.
The following is multiple choice question (with options) to answer.
If a company allocates 10 percent of its budget to advertising, 10 percent to capital improvements, and 55 percent to salaries, what fraction of its budget remains for other allocations? | [
"4/5",
"1/4",
"3/10",
"1/5"
] | B | 10 + 10 + 55 = 75% 100 - 75 = 25% to all others
B |
AQUA-RAT | AQUA-RAT-36074 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
If the price of sugar rises from Rs. 10 per kg to Rs. 11 per kg, a person, to have no increase in the expenditure on sugar, will have to reduce his consumption of sugar by | [
"9%",
"20%",
"25%",
"30%"
] | A | Sol.
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 10 = x × 11 = x = 91 kg.
∴ Reduction in consumption = 9%.
Answer A |
AQUA-RAT | AQUA-RAT-36075 | 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 said to his son, "I was as old as you are at present at the time of your birth." If the father's age is 40 years now, the son's age five years back was? | [
"16 years",
"15 years",
"18 years",
"19 years"
] | B | Let the son's present age be x years.
Then, (40 - x) = x
2x = 40 => x = 20
Son's age 5 years back = (20 - 5)
= 15 years.
Answer: B |
AQUA-RAT | AQUA-RAT-36076 | Sort by:
1)If r is an odd prime, let it be of the form 2n+1 where n is a positive integer. If we put this value of r in our equation, we get 2(n+1)q-2np=(2n+1)^2. As L.H.S. is even and R.H.S is odd we won't have any solutions for p,q,r in this case. So r=2. We get 3q-p=4 which means q=(p+4)/3. We can see that q can be an integer when p=2,5,11,17,23,29,41,47,53,59,71,..... After which p+q will exceed 111. So when p is 71 then q is not prime, when p is 59 q is 21 which is again not a prime. But when p is 53, q is 19 which is indeed a prime. So our largest solution set for p,q is 53 and 19 and r=2. So largest possible value of pqr is 53×19×2=2014.
- 4 years, 6 months ago
Great solution! You would get 7 out of 7. Only 1 improvement is to make your proof look neater (paragraphing, spacing, etc.)
- 4 years, 6 months ago
$\Large{{ 5 }^{ a }+{ b }^{ 2 }={ 3 }^{ c }\\ \Longrightarrow \frac { { 5 }^{ a }+{ b }^{ 2 } }{ 3 } ={ 3 }^{ c-1 }}$
Since $3^{c-1}$ is an integer. We have
${ 5 }^{ a }+{ b }^{ 2 }\equiv 0\quad \left( \mod 3 \right)$
The following is multiple choice question (with options) to answer.
If q and r are both odd numbers N, which of the following must also be odd? | [
"q – r",
"(q + r)^2",
"q(q + r)",
"(qr)^2"
] | D | We're told that q and r are both ODD numbers N. We're asked which of the following must also be odd.
IF...
q = 1
r = 3
Answer A) q – r = 1-3 = -2 NOT odd
Answer B) (q + r)^2 = (1+3)^2 = 16 NOT odd
Answer C) q(q + r) = (1)(4) = 4 NOT odd
Answer D) (qr)^2 = (3)^2 = 9 This IS ODD
Answer E) q/r = 1/3 NOT odd
Final Answer:
D |
AQUA-RAT | AQUA-RAT-36077 | # 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 boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water? | [
"40 minutes",
"1 hour",
"1 hour 15 min",
"1 hour 30 min"
] | C | Solution
Rate downstream = (1 / 10 x 60)km/hr
= 6 km / hr.
Rate upstream = 2 km/hr.
Speed in still water = 1 / 2(6 + 2)km/hr
= 4 km.
Required time = (5 / 4) km/hr
= 1x 1/4 km/hr
= 1 hr 15 min.
Answer C |
AQUA-RAT | AQUA-RAT-36078 | I'll have a go and answer this the maths-lite way (though there are a number of answers with more mathematic rigor and .. dare I say it vigor posted here already).
Note that there is:
• 1 result with a face value 1
• 3 results with a face value 2,
• 5 results with a face value 3,
• 7 results with a face value 4,
• 9 results with a face value 5, and
• 11 results with a face value 6
The Average is defined to be: $$\text{Average} = \frac{\text{Sum of the Results}}{\text{Total number of Results}}$$
The Sum of the Results is: $$\begin{eqnarray} \text{Sum} &=& (1 \times 1) + (3 \times 2) + (5 \times 3) + (7 \times 4) + (9 \times 5) + (11 \times 6) \nonumber \\ &=& 1 + 6 + 15 + 28 + 45 + 66 \nonumber \\ &=& 161 \nonumber \end{eqnarray}$$
The Total number of Results is: $6 \times 6 = 36$
So the Average is: $$\text{Average} = \frac{161}{36} \approx 4.472$$
-
This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is
$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$
and we can write out this sum explicitly. In particular, we can expand to get
$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is
$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$
or after some rearrangement
The following is multiple choice question (with options) to answer.
The average of 60 results is 40 and the average of other 40 results is 60 . what is the average of all the results? | [
"24",
"25",
"48",
"50"
] | C | Answer
Sum of 100 result = sum of 60 result + sum of 40 result.
= 60 x 40 + 40 x 60
= 4800/100
Correct Option: C |
AQUA-RAT | AQUA-RAT-36079 | # 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 list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is? | [
"8%",
"4%",
"6%",
"3%"
] | B | 65*(90/100)*((100-x)/100) = 56.16
x = 4%
Answer: B |
AQUA-RAT | AQUA-RAT-36080 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Word Problems.
A salesman's income consists of commission and base salary. His weekly income totals over the past 5 weeks have been $406, $413, $420, $436 and $395. What must his average (arithmetic mean) income over the next two weeks be to decrease his average weekly income to $400 over the 7-week period? | [
"$350",
"$400",
"$365",
"$385"
] | C | OFFICIAL SOLUTION:
(C) First, we need to add up the wages over the past 5 weeks: $406 + $413 + $420 + $436 + $395 = $2070.
To average $400 over 7 weeks, the salesman would need to earn: $400 × 7 = $2800.
Subtract $2070 from $2800 to determine how much he would need to earn, in total, over the next 2 weeks to average $400 for the 7 weeks: $2800 – $2070 = $730.
Dividing $730 by 2 will give us the amount he needs to earn on average over the next 2 weeks: $730/2 = $365.
The correct answer is choice (C). |
AQUA-RAT | AQUA-RAT-36081 | # Minimum tickets required for specified probability of winning lottery
In a lottery, 1/10 of the 50 000 000 tickets give a prize.
What is the minimum amount of tickets one should buy to have at least a 50% chance to win?
Would be very glad if you could explain your methodology when resolving this. Please consider this as homework if you will.
## 1 Answer
I should really ask what your thoughts are so far on this. This problem is very closely related to the "birthday problem". The easiest way to do these problems is to count the possible ways of either winning or losing. Usually one is much easier to do than the other, so the key is to find the right one. Before we get into the actual calculation, let's start with some heuristics.
Heuristics: Let $n$ be the total number of tickets and $m$ be the number of winning tickets. In this case $n = 50\,000\,000$ and $m = 5\,000\,000$. When $n$ is very large then purchasing multiple distinguishable tickets is almost the same as sampling with replacement from the population of tickets.
Let's suppose that, instead of having to purchase $k$ separate tickets, we purchased a ticket, looked to see if it was a winner and then returned it to the lottery. We then repeat this procedure where each such draw is independent from all of the previous ones. Then the probability of winning after purchasing $k$ tickets is just $$\Pr( \text{we won} \mid \text{purchased k tickets} ) = 1 - \left(\frac{n-m}{n}\right)^k .$$
For our case then, the right-hand side is $1 - (9/10)^k$ and so we set this equal to $1/2$ and solve for $k$ in order to get the number of tickets.
But, we're actually sampling without replacement. Below, we'll go through the development, with the point being that the heuristics above are more than good enough for the present problem and many similar ones.
There are $50\,000\,000$ tickets. Of these $5\,000\,000$ are winning ones and $45\,000\,000$ are losing ones. We seek
The following is multiple choice question (with options) to answer.
The probability of a lottery ticket being a prized ticket is 0.2. When 4 tickets are purchased, the probability of winning a prize on atleast one ticket is? | [
"0.5904",
"0.5914",
"0.5912",
"0.5916"
] | A | Explanation:
P(winning prize atleast on one ticket)
= 1 - P("Losing on all tickets")
= 1 - (0.8)4 = (1 + (0.8)2)(1 - (0.8)2)
= (1.64)(0.36) = 0.5904
A) |
AQUA-RAT | AQUA-RAT-36082 | java, console, calculator, finance
taxableIncome <= hohFiler[5].maxSalary //$432,200
)
{
HeadOfHouseFilers.TierFive();
}
//TierSix()
else if (
taxableIncome >= hohFiler[5].minSalary && //$405,101
taxableIncome <= hohFiler[5].maxSalary && //$432,200
taxableIncome > hohFiler[6].minSalary //$432,201
)
{
HeadOfHouseFilers.TierSix();
}
//TierSeven()
else if (
taxableIncome >= hohFiler[6].minSalary && //$432,201
taxableIncome <= hohFiler[6].maxSalary //Double.MAX_VALUE
)
{
HeadOfHouseFilers.TierSeven();
}
break;
The following is multiple choice question (with options) to answer.
Country C imposes a two-tiered tax on imported cars: the first tier imposes a tax of 12% of the car's price up to a certain price level. If the car's price is higher than the first tier's level, the tax on the portion of the price that exceeds this value is 8%. If Ron imported a $12,000 imported car and ended up paying $1440 in taxes, what is the first tier's price level? | [
"$1600",
"$6000",
"$6050",
"$1200"
] | D | Let T be the tier price, P be total price = 12000
Per the given conditions:
0.12T + 0.08(P-T) = 1440 ----> T= 12000. D is the correct answer. |
AQUA-RAT | AQUA-RAT-36083 | $x+4=-(y+3)$ or $x+4=y+3$
$x+y=-7$ or $x-y=-1$
Solving $x+y=-7$ and $2x+y=-5$ simultaneously, we get $x=2$ which contradicts our assumption that $1-x\geq0$.
Solving $x-y=-1$ and $2x+y=-5$ simultaneously, we get $x=-2$, $y=-1$ and this satisfies all of the constraints and thus, this is a pair of valid solutions.
Case B (i)+$y\geq1$:
$x+y<0$, $1-x\geq0$ and $y\geq1$
From $|x+y|+|1-x|=6$, we get $2x+y=-5$.
From $|x+y+1|+|1-y|=4$:
$|x+y+1|-(1-y)=4$
$|x-5-2x+1|=5-y$
$|-x-4|=5-y$
$|x+4|=5-y$
$x+4=-(5-y)$ or $x+4=5-y$
$x-y=-9$ or $x+y=1$
Solving $x-y=-9$ and $2x+y=-5$ simultaneously, we get $\displaystyle x=-\frac{14}{3}$, $\displaystyle y=\frac{13}{3}$ and this satisfies all of the constraints and thus, this is another pair of valid solutions.
Since $x+y=1$ contradicts the assumption that $x+y<0$, we'll just ignore this case.
Case B (ii)
+$y<1$ :$x+y<0$, $1-x<0$ and $y<1$
From $|x+y|+|1-x|=6$:
The following is multiple choice question (with options) to answer.
If x : y = 3 : 4, find (4x + 5y) : (5x - 2y). | [
"32/7",
"13/7",
"2/7",
"32/17"
] | A | . X/Y=3/4 (4x+5y)/(5x+2y)= (4( x/y)+5)/(5 (x/y)-2) =(4(3/4)+5)/(5(3/4)-2)
=(3+5)/(7/4)=32/7
ANSWER A 32/7 |
AQUA-RAT | AQUA-RAT-36084 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
In how many ways can 10 people, A, B, C, D, E, F, G, H, I, J be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other? | [
"384",
"396",
"576",
"3628704"
] | D | Number of total arrangements = 10!
Restriction 1= ABCD not next to each other --> let say AB and CD are considered as one unit, respectively
Restriction 2= AB is not the same as BA + CD is not the same as DC --> the number will increase by 2*2
Total number of arrangements - Number out of restrictions = Result
10! - (4!*2*2) = 3628800 - (24*2*2) = 3628704
Answer D |
AQUA-RAT | AQUA-RAT-36085 | Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.
Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.
Thus $$P=x+y+10=\sqrt{200}+10$$.
Sufficient.
Hi Bunuel,
Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to "square" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above.
Thanks!
It should come with practice...
We know the values of xy and x^2+y^2, while need to get the value of x+y. Now, if you add twice xy to x^2+y^2 you get the square of x+y, hence squaring x+y is quite natural thing to do.
Similar questions to practice:
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-166186.html
what-is-the-perimeter-of-rectangle-r-96381.html
Hope this helps.
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
### Show Tags
02 Aug 2014, 20:14
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?
The following is multiple choice question (with options) to answer.
The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 − 20x + 25. If the sum of the perimeters of both squares is 36, what is the value of x? | [
"3",
"5",
"7",
"9"
] | A | The areas are (x+5)^2 and (2x-5)^2.
The lengths of the sides are x+5 and 2x-5.
If we add the two perimeters:
4(x+5) + 4(2x-5) = 36
12x = 36
x = 3
The answer is A. |
AQUA-RAT | AQUA-RAT-36086 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
What is the rate percent when the simple interest on Rs.810 amount to Rs.155 in 4 Years? | [
"4.78",
"4.68",
"4.58",
"4.48"
] | A | 155 = (810*4*R)/100
R = 4.78%
Answer: A |
AQUA-RAT | AQUA-RAT-36087 | # 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 5 different balls can be arranged in to 3 different boxes so that no box remains empty? | [
"125",
"150",
"175",
"200"
] | B | 1. The total number of possibilities including empty boxes: 35=24335=243
2. Two of the boxes are empty: C32=3C23=3
3. One but not two of the boxes is empty: 3∗(25−2)=903∗(25−2)=90
4. the total number of possibilities excluding empty boxes: 243 - 3 - 90 = 150
ANS:B |
AQUA-RAT | AQUA-RAT-36088 | # 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 percentage profit earned by selling an article for $1920 is equal to the percentage loss incurred by selling the same article for $1280. At what price should the article be sold to make 25% profit? | [
"1000",
"3000",
"2000",
"4000"
] | C | B
2000
Let C.P. be $x.
Then, (1920 - x)/x * 100 = (x - 1280)/x * 100
1920 - x = x - 1280
2x = 3200 => x = 1600
Required S.P. = 125 % of $1600 = 125/100 * 1600 = $2000. |
AQUA-RAT | AQUA-RAT-36089 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10
eg
unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number
To get the value we want we need to floor the log first
unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct
or
unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000
Get digit at position of positive integer
To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.
const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);
or
const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;
// Note brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;
digitAt(567, 0); // 7
digitAt(567, 1); // 6
digitAt(567, 2); // 5
The following is multiple choice question (with options) to answer.
Mr. Smith calculated the average of 10three digit numbers. But due to a mistake he reversed the digits of a number and thus his average increased by 39.6. The difference between the unit digit and hundreds digit of that number is : | [
"a) 4",
"b) 3",
"c) 2",
"d) 1"
] | A | Since the average increased by 39.6 and there were a total of 10 numbers, it means the incorrect number was 396 greater than the correct number.
Say, the correct number was abc (where a, b and c are the digits of the 3 digit number)
Then the incorrect number was cba.
100c + 10b + a - (100a + 10b + c) = 396
99c - 99a = 99(c - a) = 396
396 = 99*4 = 99(c - a)
So c - a = 4
Answer (A |
AQUA-RAT | AQUA-RAT-36090 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
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A total of 22 men and 26 women were at a party, and the average [#permalink]
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04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
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Abhishek....
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
In a University the average age in an arts class is 21 and average age in technical class is 18. What is the age average of University if there are 8 arts classes and 5 technical classes? | [
"18.75",
"19.85",
"20.32",
"20.9"
] | B | Average age of University= ((8x21)+(5x18))/13
=19.85
Answer: B |
AQUA-RAT | AQUA-RAT-36091 | So we get $\mathbf{1} \cdot \mathbf{s} \leq \sqrt{N}\sqrt{M + \mathbf{1}\cdot \mathbf{s}}$.
So we have $A \leq \sqrt{N} \sqrt{M+A}$. Squaring we get $A^2 \leq N (M+A)$. So we have $(A - N)A \leq NM$. Thus $A-N \leq \sqrt{NM}$. Thus $A \leq \sqrt{NM}+N$.
Thus $\sum_k s_k \leq \sqrt{N} \sqrt{N(N-1)} + N \approx N \sqrt{N}$.
Some amusement:
The kids card game Spot It! has 55 cards, each with 8 animals (and 57 different animals in total). All the cards are different, and any pair of cards has exactly one pair of animals in common. This is $\mathbb{PF}_7$ with two missing lines (probably for printing reasons).
There's also Spot It Jr.! with 31 cards and 6 animals per card (and 31 different animals in total). This is $\mathbb{PF}_5$.
Yes it is possible to improve on $2n$.
The following is multiple choice question (with options) to answer.
In a certain deck of cards, each card has a positive integer written on it, in a multiplication game a child draws a card and multiplies the integer on the card with the next large integer. If the each possible product is between 11 and 200, then the least and greatest integer on the card would be | [
"3 and 15",
"3 and 20",
"4 and 13",
"4 and 14"
] | A | Given: 11<x(x+1)<200.
Now, it's better to test the answer choices here rather than to solve:
If x=3 then x(x+1)=12>11 --> so, the least value is 4. Test for the largest value: if x=15 then
x(x+1)=15*16=240>200 --> discard B.
Answer: A. |
AQUA-RAT | AQUA-RAT-36092 | # Real Roots and Differentiation
Prove that the equation $$x^5 − 1102x^4 − 2015x = 0$$ has at least three real roots.
So do I sub in values of negative and positive values of $$x$$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $$x$$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.
• one root is $x=0$. So your problem is reduced to $x^4-1102x^3-2015=0$. Then, try to find two pairs $x_1, x_2$ where $x^4-1102x^3-2015$ changes sign. You can then conclude from the Intermediate value theorem, that there are at least 2 additional roots. – MrYouMath Sep 17 '15 at 18:10
• let $$f(x)=x^5-1102x^4-2015x$$ then calculate $$f(-2)$$ and $$f(-1)$$ and $$f(1000)$$ and $$f(2000)$$ – Dr. Sonnhard Graubner Sep 17 '15 at 18:13
It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done.
Observe, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.
Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$
The following is multiple choice question (with options) to answer.
How many real roots does the equation x^2y+15xy+64y=0 have if y < 0? | [
"0",
"1",
"2",
"3"
] | C | x^2y+15xy+64y=0
=> y ( x^2 + 15x + 64) = 0
=> y (x+8)^2 = 0
if y<0 , then x=-7
So although there are 2 factors , they are the same x=-7 .
The equations has 2 distinct real root .
Answer C |
AQUA-RAT | AQUA-RAT-36093 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
A fruit-salad mixture consists of apples, peaches, and grapes in the ratio 6:5:4, respectively, by weight. If 60 pounds of the mixture is prepared, the mixture includes how many more pounds of apples than grapes? | [
"15",
"12",
"8",
"6"
] | C | We can first set up our ratio using variable multipliers. We are given that a fruit-salad mixture consists of apples, peaches, and grapes, in the ratio of 6:5:4, respectively, by weight. Thus, we can say:
apples : peaches : grapes = 6x : 5x : 4x
We are given that 60 pounds of the mixture is prepared so we can set up the following question and determine a value for x:
6x + 5x + 4x = 60
15x = 60
x = 4
Now we can determine the number of pounds of apples and of grapes.
pounds of grapes = (4)(4) = 16
pounds of apples = (6)(4) = 24
Thus we know that there are 24 - 16 = 8 more pounds of apples than grapes.
Answer is C. |
AQUA-RAT | AQUA-RAT-36094 | java, console, calculator, finance
taxableIncome <= hohFiler[5].maxSalary //$432,200
)
{
HeadOfHouseFilers.TierFive();
}
//TierSix()
else if (
taxableIncome >= hohFiler[5].minSalary && //$405,101
taxableIncome <= hohFiler[5].maxSalary && //$432,200
taxableIncome > hohFiler[6].minSalary //$432,201
)
{
HeadOfHouseFilers.TierSix();
}
//TierSeven()
else if (
taxableIncome >= hohFiler[6].minSalary && //$432,201
taxableIncome <= hohFiler[6].maxSalary //Double.MAX_VALUE
)
{
HeadOfHouseFilers.TierSeven();
}
break;
The following is multiple choice question (with options) to answer.
If the personal income tax rate is lowered from 40% to 33%, what is the differential savings for a tax payer having an annual income before tax to the tune of $45000? | [
"$3500",
"$5000",
"$3150",
"$7000"
] | C | Saving = (40-33)% of 45000 = 3150.
Answer:C |
AQUA-RAT | AQUA-RAT-36095 | # Find three integers $x$ so that $271x \equiv 272\pmod{2015}$
I know that $\forall{a,n}\in\mathbb{Z}:\Bigl[\gcd(a,n)=1\Bigr]\implies\Bigl[\exists{k}\in\mathbb{Z}:ak\equiv1\pmod{n}\Bigr]$
In other words, for every pair of co-prime integers $a$ and $n$, there is an integer $k$ which is the inverse of $a\pmod{n}$.
Using Euclidean algorithm, I have found:
• The inverse of $271\pmod{2015}$ is $461$
• The inverse of $2015\pmod{271}$ is $209$
I do not understand how to apply this in the context of the question above.
I tried simplifying it in the same way but I ended up stuck and unable to find an integer $x$ such that $271x\equiv272\pmod{2015}$.
I can find many different inverses I think, but none of them are integers.
The following is multiple choice question (with options) to answer.
E is the set of all positive integers x such that x^2 is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer x in E?
I. 9
II. 15
III. 27 | [
"I only",
"II only",
"I and II only",
"I and III only"
] | C | Ans: C
E->{x} where x^2 is a multiple of both 27 and 375 means 3^3 and (5^3)*3
means x must contain 3^2 and 5^2
so with these conditions we know that 9=3^2 and 15=3*5 both have required factors for the divisibility of lowest int for x which is 9*25
but 27 is not a divisor because it can't divide 9*25 fully.
so Ans : C |
AQUA-RAT | AQUA-RAT-36096 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
The first three terms of a proportion are 2, 10 and 20. The fourth term is? | [
"105",
"102",
"190",
"100"
] | D | (10*20)/2 = 100
Answer:D |
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