source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39597 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
One week, a certain truck rental lot had a total of 26 trucks, all of which were on the lot Monday morning. If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 14 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week? | [
"18",
"16",
"12",
"24"
] | D | N - Not rented trucks; R - Rented trucks
N + R = 26
N + R/2 = 14
R = 24
D |
AQUA-RAT | AQUA-RAT-39598 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling an article for $225 a person gains $75. What is the gain %? | [
"25%",
"30%",
"50%",
"20%"
] | C | S.P. = $225
Gain = $75
C.P. = 225-75 = 150
Gain% = 75/150 * 100 = 50%
Answer is C |
AQUA-RAT | AQUA-RAT-39599 | sides of equal length. Q2. Isosceles Right Triangle has one of the angles exactly 90 degrees and two sides which is equal to each other. Correct Answer: D. Step 1: An isosceles right triangle has two sides equal and measure of one of its angles is 90°. The isosceles right triangle has a right angle and two acute angles with a measure of 45° each, thus, two sides of the triangle are equal and the other is different. Also iSOSceles has two equal "Sides" joined by an "Odd" side. In an isosceles right triangle the length of two sides of the triangle are equal. a) The right angled triangles which are having two equal sides are called as isosceles right angled triangles. Can an isosceles triangle be the right angle or scalene triangle? The term "right" triangle may mislead you to think "left" or "wrong" triangles exist; they do not. Sorry!, This page is not available for now to bookmark. Alphabetically they go 3, 2, none: 1. Scalene: means "uneven" or "odd", so no equal sides. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Therefore, AB = AC and AB ≅ AC. So, the two triangles formed from the original isosceles triangle are congruent because all three sides of one are congruent to all three sides of the other. This is an interesting one. Two isosceles triangles are always similar. Vedantu Mark the vertices of the triangles as $$\text{A}$$, $$\text{B}$$, and $$\text{C}$$. If the triangle breaks upward, it is a bullish sign, but if it breaks downward, it is a bearish sign. A triangle comprises three sides which make three angles with each other. isosceles triangle (plural isosceles triangles) A triangle having at least two sides equalUsage notes []. The sides and angles of the triangle could vary. Isosceles right triangle: In this triangle, one interior angle measures 90° , and the other two angles measure 45° each. In defining the types of triangles, our class was stumped by a question asked by one of the student. Repeaters, Vedantu The different types of triangle are: In a Isosceles
The following is multiple choice question (with options) to answer.
Triangle XYZ is an isosceles right triangle. If side XY is longer than side YZ, and the area of the triangle is 49, what is the measure of side XY? | [
"4",
"4√2",
"14",
"8√2"
] | C | ans C ..14..
xy being larger means it is the hyp..
area =(1/2)*(yz)^2=4 or yz=7*\sqrt{2}..
therefore hyp=xy=14 |
AQUA-RAT | AQUA-RAT-39600 | (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.
The regular price per can of a certain brand of soda is $0.40. If the regular price per can is discounted 15 percent when the soda is purchased in 24-can cases, what is the price of 100 cans of this brand of soda purchased in 24-can cases? | [
"$16.32",
"$18.00",
"$21.60",
"$34"
] | D | The discounted price of one can of soda is (0.85)($0.40), or $0.34. Therefore, the price of 72 cans of soda at the discounted price would be (100)($0.34)= 34
Answer: D. |
AQUA-RAT | AQUA-RAT-39601 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
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26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
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The following is multiple choice question (with options) to answer.
Out of 40 applicants to a law school, 15 majored in political science, 20 had a grade point average higher than 3.0, and 10 did not major in political science and had a GPA equal to or lower than 3.0. How many T applicants majored in political science and had a GPA higher than 3.0? | [
"5",
"10",
"15",
"25"
] | A | Total applicants = 40
Political science = 15 and Non political science = 40 - 15 = 25
GPA > 3.0 = 20 and GPA <= 3.0 = 20
10 Non political science students had GPA <= 3.0 --> 15 Non political science students had GPA > 3.0
GPA > 3.0 in political science = Total - (GPA > 3.0 in non political science) T= 20 - 15 = 5
Answer: A |
AQUA-RAT | AQUA-RAT-39602 | autocad
Title: What is the suitable multiplier for all units? I create Auto CAD Drawing using Primary units decimal (mm). I need to add alternative unit as Architectural. What is the suitable multiplier for all units?
After adding new value So
1 Millimeter = 0.0393701 Inches
You can get this from many places, some being listed as holding the national standards, while others are just repeating the information. This one has some of the history and explanation which you might find useful.
from:
https://www.squareyards.com/blog/millimeter-to-inch-cnvart
The following is multiple choice question (with options) to answer.
How many inches are in 2000 millimeters? (round your answer to the nearest hundredth of of an inch). | [
"98.74 inches",
"78.74 inches",
"88.74 inches",
"58.74 inches"
] | B | One inch is the same as 25.4 mm. Let x inches be the same as 1000 mm
x = 1 inch * 2000 mm / 25.4 mm = 78.74 inches
correct answer B |
AQUA-RAT | AQUA-RAT-39603 | Veritas Prep Reviews
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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]
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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]
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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.
If it takes a machine 3⁄5 minute to produce one item, how many items will it produce in 2 hours? | [
"1⁄3",
"4⁄3",
"80",
"200"
] | D | 1 item takes 3/5 min so it takes 120 min to produce x
3x/5=120 the x=200
Answer:D |
AQUA-RAT | AQUA-RAT-39604 | # 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.
A word is defined as a sequence of five dots arranged in a row. Each dot is colored either red, yellow, or blue. How many distinct words can be formed? | [
"124",
"150",
"212",
"243"
] | D | 3^5 = 243
The answer is D. |
AQUA-RAT | AQUA-RAT-39605 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train? | [
"297m",
"276m",
"267m",
"180m"
] | B | Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
Answer: B |
AQUA-RAT | AQUA-RAT-39606 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
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26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
Based on this year's costs, a company budgets P dollars for hiring N new employees next year. If the average cost of hiring each employee were to decrease 20% from this year's cost, then the greatest number of employees could hire next year using P dollars would be: | [
"Equal to N",
"25% greater than N",
"20% greater than N",
"25% less than N"
] | B | Correct answer is (B).
This year, the price of an employee is price1=P/N.
If this price decreases by 20% it becomes Price2=P/N*0.8
Then with P dollars, you can hire P/Price2 employees i.e. P/(P/N*0.8) i.e. N/1.1 i.e. 1.25*N
Which is 25% greater than N. |
AQUA-RAT | AQUA-RAT-39607 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
There are a total of 15 numbers with an average of 11. The average of first seven numbers is 14 and the average of last seven numbers is 6. What is the middle number? | [
"31",
"24",
"29",
"25"
] | D | The total of 15 numbers = 15X11 = 165
The total of first 7 and last 7 numbers is = 7 X 14+7 X 6 = 140
So, the middle number is (165 - 140 ) = 25
D |
AQUA-RAT | AQUA-RAT-39608 | Let $P(X_1, \ldots, X_n)$ a polynomial. Substitute $X_i\mapsto X_i + a_i$ and get $$P(X_1+ a_1, \ldots, X_n + a_n) = \sum c_{\alpha} X_1^{\alpha_1} \ldots X_n^{\alpha_n}$$ and so
$$P(X_1, \ldots, X_n) = \sum c_{\alpha} (X_1-a_1)^{\alpha_1} \ldots (X_n-a_n)^{\alpha_n}$$
Note that $c_{(0,\ldots, 0)} = P(a_1, \ldots, a_n)$. Moreover, $$P(X_1, \ldots, X_n) = c_{(0,\ldots, 0)}+ \sum (X_i - a_i) g_i(X_1, \ldots, X_n)$$ as all the other terms $c_{\alpha}(X-a)^{\alpha}$ are divisible by some $(X_i - a_i)$. Therefore $$P(X_1, \ldots, X_n) - P(a_1, \ldots, a_n) \in (X_1-a_1, \ldots, X_n - a_n)$$
The following is multiple choice question (with options) to answer.
(X^n - A^n) is completely divisible by (X - A), when | [
"n is any natural number",
"n is an even natural number",
"n is an odd natural number",
"n is prime"
] | A | For every natural number n, (X^n - A^n) is always divisible by (X - A)
Answer A. |
AQUA-RAT | AQUA-RAT-39609 | ### Show Tags
19 Aug 2015, 01:34
2
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BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
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19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of a group of n people is 15 years old. One more person aged 35 joins the group and the new average is 17 years old. What is the value of n? | [
"8",
"9",
"10",
"11"
] | B | 15n + 35 = 17(n+1)
2n = 18
n = 9
The answer is B. |
AQUA-RAT | AQUA-RAT-39610 | # Polynomial Challenge III
#### anemone
##### MHB POTW Director
Staff member
Let $f(x)=x^4+px^3+qx^2+rx+s$, where $p,\,q,\,r,\,s$ are real constants. Suppose $f(3)=2481$, $f(2)=1654$, $f(1)=827$.
Determine the value of $\dfrac{f(-5)+f(9)}{4}$.
#### mente oscura
##### Well-known member
Let $f(x)=x^4+px^3+qx^2+rx+s$, where $p,\,q,\,r,\,s$ are real constants. Suppose $f(3)=2481$, $f(2)=1654$, $f(1)=827$.
Determine the value of $\dfrac{f(-5)+f(9)}{4}$.
Hello.
$$f(1)=827=827*1$$
$$f(2)=1654=827*2$$
$$f(3)=2481=827*3$$
Therefore:
$$\dfrac{f(-5)+f(9)}{4}=\dfrac{827*(-5)+827*(9)}{4}=\dfrac{827(-5+9)}{4}=827$$
Regards
#### anemone
##### MHB POTW Director
Staff member
Hello.
$$f(1)=827=827*1$$
$$f(2)=1654=827*2$$
$$f(3)=2481=827*3$$
Therefore:
The following is multiple choice question (with options) to answer.
Can you please walk me through how to best approach this problem? Thanks
If #p# = ap^3+ bp – 1 where a and b are constants, and #-4# = 2, what is the value of #4#? | [
"-4",
"0",
"-2",
"-3"
] | A | #p# = ap^3 + bp - 1
#-4# = 2
putting p = -4 in above equation
-64a -(4b +1) = 2 or
#-4# = (64a+4b+1) = -2
therefore 64a+4b = -3 .....(1
now putting p = 4
#4# = 64 a+4b - 1
using equation 1(64a+4b = -3)
#4# = -3-1 = -4
hence A |
AQUA-RAT | AQUA-RAT-39611 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invested Rs.6600, Rs.4400 and Rs.11000 respectively, in a partnership business. Find the share of A in profit of Rs.12100 after a year? | [
"Rs.3300",
"Rs.3100",
"Rs.3500",
"Rs.4200"
] | A | 6600:4400:11000
3:2:5
3/10 * 11000
= Rs.3300
AnswerA |
AQUA-RAT | AQUA-RAT-39612 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election between two candidates, 20% of votes are were declares invalid. First candidate got 480 votes which were 60% of the total valid votes . The total number of votes enrolled in that election was: | [
"2000 votes",
"2500 votes",
"3000 votes",
"4000 votes"
] | C | Answer: Option 'B'
(100%-20%=80%
48%-32%=16%
16% = 16×30 = 480
100% =100×30 =3000 votes)
C |
AQUA-RAT | AQUA-RAT-39613 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train is 460 meter long is running at a speed of 45 km/hour. In what time will it pass a bridge of 140 meter length | [
"20 seconds",
"30 seconds",
"40 seconds",
"48 seconds"
] | D | Explanation:
Speed = 45 Km/hr = 45*(5/18) m/sec
= 25/2 m/sec
Total distance = 460+140 = 600 meter
Time = Distance/speed
=600∗2/25=48seconds
Option D |
AQUA-RAT | AQUA-RAT-39614 | Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 121\cdot 7,$ and $77\cdot 11$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2x$. Thus the possibilities for $x$ are $423$, $57$, and $33$.
Now plug in these values into the equation $n = \frac{-1 + x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $\boxed{\textbf{(A)}\ 255}$.
~Edits by BakedPotato66
## Solution 2
As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$. Clearly $m > n$, so let $m = n + a$, where $a$ is some positive integer. We have:
$(n+a)^2 = n(n+1) + 212$. Expanding, grouping like terms and factoring, we get: $n = \frac{(212 - a^2)}{(2a - 1)}$.
We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ ($14^2 = 196 < 212 < 15^2 = 225$). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$. These gives $n$ values of $211, 28,$ and $16$, respectively. $211 + 28 + 16 = 255$. Hence, the answer is $\boxed{\textbf{(A)}\ 255}$.
## Solution 3
The following is multiple choice question (with options) to answer.
(x + 6) is a factor in x^2 - mx - 42. What is the value of m? | [
"2",
"2.2",
"1",
"4"
] | C | I solved the second degree equation and found it like this:
x^2 - mx - 42 = 0
(x-7)(x+6) = 0
x=7 or x= -6
Substituting both values for x in the equation we find:
x^2 - mx -42
=> (-6)^2 - m(-6) = 42
=> 36+ 6m= 42
=> 6m = 42-36=6
=> m = 1
And with 7, using a similar process we end up with:
(7)^2 - m(7) =42
-7m = 42-49= -7
m=-1
Ao,ANS C |
AQUA-RAT | AQUA-RAT-39615 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invested Rs.6300, Rs.4200 and Rs.10500 respectively, in a partnership business. Find the share of A in profit of Rs.12100 after a year? | [
"3630",
"2890",
"2707",
"2700"
] | A | 6300:4200:10500
3:2:5
3/10 * 12100
= 3630
Answer:A |
AQUA-RAT | AQUA-RAT-39616 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
How many 1 1/4 foot pieces of cable can be cut out of a piece that is 30 feet long? | [
"a.10",
"b.16",
"c.17",
"d.24"
] | D | Pretty Simple..the answer is choice (D) 24.
All that one has to do is :
1: Convert 1 1/4 into an improper fraction (5/4) or a decimal (1.25)
2: Divide 30 by 5/4 (or 1.25)
Answer =24 (D) |
AQUA-RAT | AQUA-RAT-39617 | # How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
The following is multiple choice question (with options) to answer.
In how many ways can 4 black and 5 white chess pieces be arranged in a row such that they occupy alternate places? Assume that the pieces are distinct. | [
"20",
"1440",
"120",
"2880"
] | D | 4 black chess pieces can be arranged in 4! ways and 5 white pieces can be arranged in 5 ! ways.
W_W_W_W
Number of ways = 4!*5! = 24 * 120 = 2880
Answer D |
AQUA-RAT | AQUA-RAT-39618 | Difficult Probability Solved QuestionAptitude Discussion
Q. If the integers $m$ and $n$ are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 equals:
✔ A. $\dfrac{1}{4}$ ✖ B. $\dfrac{1}{7}$ ✖ C. $\dfrac{1}{8}$ ✖ D. $\dfrac{1}{49}$
Solution:
Option(A) is correct
Table below can be scrolled horizontally
Form of the exponent
$m$ $4x +1$ $4x+3$ $4x+2$ $4x$
$n$ $4y+3$ $4y+1$ $4y$ $4y+2$
last digit of
$7^m+7^n$
$0$ $0$ $0$ $0$
Number of
selections
$25 \times 25$ $25 \times 25$ $25 \times 25$ $25 \times 25$
If a number ends in a 0 then the number must be divisible by 5.
Hence required probability is,
$=\dfrac{625 \times 4}{100^2}$
$=\dfrac{1}{4}$
Edit: Thank you, Barry, for the very good explanation in the comments.
Edit 2: Thank you Vaibhav, corrected the typo now it's $25 \times 25$ and not $26 \times 25$.
Edit 3: For yet another approach of solving this question, check comment by Murugan.
(7) Comment(s)
Murugan
()
This sum is very simple. Power cycles of 7 are 7, 9, 3, 1
So totally 4 possibilities.
Total possibilities are $^4P_1 \times ^4P_1=16$
For selecting a number from 1 to 100 which are divisible by 5,
$m=9$, $n=1$ or $m=1$, $n=9$ or $m=7$, $n=3$ or $m=3$, $n=7$
i.e. 4 chances.
The following is multiple choice question (with options) to answer.
If an integer x is to be selected at random from 1 to 200, inclusive, what is probability x(x+1) will be divisible by 4? | [
"0.5",
"1",
"0.25",
"0.75"
] | A | because x(x+1) is always an even product of even*odd or odd*even factors,
there is a probability of 1 that that it will be divisible by 2,
and, thus, a probability of 1/2 that it will be divisible by 4
1*1/2=1/2 =0.5
Answer : A |
AQUA-RAT | AQUA-RAT-39619 | python, beginner, parsing, csv
Full text: On the other hand, we denounce with righteous indignation and dislike men who are so beguiled and demoralized by the charms of pleasure of the moment, so blinded by desire, that they cannot foresee the pain and trouble that are bound to ensue; and equal blame belongs to those who fail in their duty through weakness of will, which is the same as saying through shrinking from toil and pain. These cases are perfectly simple and easy to distinguish. In a free hour, when our power of choice is untrammelled and when nothing prevents our being able to do what we like best, every pleasure is to be welcomed and every pain avoided. But in certain circumstances and owing to the claims of duty or the obligations of business it will frequently occur that pleasures have to be repudiated and annoyances accepted. The wise man therefore always holds in these matters to this principle of selection: he rejects pleasures to secure other greater pleasures, or else he endures pains to avoid worse pains
On the other hand, we denounce with righteous indignation and dislike men who are so beguiled and demoralized by the charms of pleasure of the moment, so blinded by desire, that they cannot foresee the pain and trouble that are bound to ensue; and equal blame belongs to those who fail in their duty through weakness of will, which is the same as saying through shrinking from toil and pain. These cases are perfectly simple and easy to distinguish. In a free hour, when our power of choice is untrammelled and when nothing prevents our being able to do what we like best, every pleasure is to be welcomed and every pain avoided. But in certain circumstances and owing to the claims of duty or the obligations of business it will frequently occur that pleasures have to be repudiated and annoyances accepted. The wise man therefore always holds in these matters to this principle of selection: he rejects pleasures to secure other greater pleasures, or else he endures pains to avoid worse pains
The following is multiple choice question (with options) to answer.
To be undaunted is to be impervious, undisturbed and ‘fearless’. | [
"2",
"HH",
"28",
"28J"
] | D | Answer:D |
AQUA-RAT | AQUA-RAT-39620 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
In a company, 50 percent of the employees are men. If 60 percent of the employees are unionized and 70 percent of these are men, what percent of the non-union employees are women? | [
"85%",
"80%",
"75%",
"70%"
] | B | The percent of employees who are unionized and men is 0.7*0.6 = 42%
The percent of employees who are unionized and women is 60 - 42 = 18%
50% of all employees are women, so non-union women are 50%-18% = 32%
40% of all employees are non-union.
The percent of non-union employees who are women is 32%/40% = 80%
The answer is B. |
AQUA-RAT | AQUA-RAT-39621 | in usage again whenever we have to find the angle of elevation or the height of a pole or a mountain. For example, an orientation represented as 225° is equivalent to -135° (360° - 225°). It is also known as a 45-90-45 triangle. It is half of a complete angle. Observe the right triangle angles. This is the only right triangle that is an isosceles triangle. Pythagorean triplet is a set of three whole numbers $$\text{a, b and c}$$ that satisfy Pythagorean theorem. A triangle, one with equal sides ∆ also demonstrates acute angles. Have that clean right angle congruence theorem example neat right Angle, Hypotenuse, Side ) applies only to triangles! Examples of right angle in a Sentence A square has four right angles. Step by step guide to finding missing sides and angles of a Right Triangle. A right angle has a value of 90 degrees ($90^\circ$). For the right triangle $$\triangle\,ABC$$ shown on the right, find the values of all six trigonometric functions of the acute angles $$A$$ and $$B$$. Check your understanding of right angles and their mathematical uses with this practice quiz and worksheet. Example: like an X that crosses in the middle and it does not have a right angle. No, a triangle cannot have two right angles because the sum of the three angles in a triangle is 180°. The rays first, let 's explore a right triangle and their mathematical uses with this practice quiz worksheet! Most house rooftops, as the vertex the angle of 90° this triangle, interior! Recognize right angles, arc lengths, and the other rhombus and a horizontal line make most right. Added every day corresponding to the right triangle looks how to draw a segment..., 5 ) arranged at right angles because the sum of an acute angle 0°, which also defines lines! A congruent triangle are congruent 10 feet directly above the base and of. Triangle pattern made of stars up to n lines royalty-free stock photos, illustrations and vectors the..., which also defines perpendicular lines straight angle, this one 's going from the left to addition! Chorion was breached at a right triangle that is an isosceles triangle a...... see full
The following is multiple choice question (with options) to answer.
a right triangle has three sides (F, G, H). If F=5, G=8.66, and H=10, which angle is 90 degrees | [
"FG",
"HG",
"HF",
"n/a"
] | A | A) the angle FG is a right angle. HF, and HG are acute. |
AQUA-RAT | AQUA-RAT-39622 | I am also in High School, and it is my understanding that if there is a number squared and then square rooted then it is (by definition) just the original number. Taking your example of -5, plugging it in you would get sqrt (-5)^2 = -5. (-5)^2 is 25 and square rooting it is 5 which is not equal to -5. To avoid this, you must specify that if X is negative then sqrt X^2 = -x and if X is positve then sqrt X^2 = X. This would be expressed mathematically using inequalites as seen in your question.
• "Square Rooting" does NOT by definition mean just take the original number. This is because for $\sqrt{}$ to behave like a function from $\Bbb{R} \to \Bbb{R}$ (which we want for a lot of reasons), it must take in a single number. This makes $\sqrt{x^2} = \sqrt{(-x)^2}$ because before you can take the root, you need to simplify the inside, deleting its "past". If we wanted to keep the "past", it would need to come from $\Bbb{R}^n, n > 1$. To do so, it takes only one branch, and mathematicians decided for a variety of reaasons that it would be the positive branch. Thus, $\sqrt{x^2} = |x|$. Sep 25 at 5:58
The following is multiple choice question (with options) to answer.
A positive number x is multiplied by 5, and this product is then divided by 3. If the positive square root of the result of these two operations equals x, what is the value of x ? | [
" 9/4",
" 3/2",
" 4/3",
" 5/3"
] | D | sq rt(5x/3) = x
=> 5x/3 = x^2
=> x = 5/3
Ans - D |
AQUA-RAT | AQUA-RAT-39623 | 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.
Students in Class I and II of a school are in the ratio of 3 : 5. Had 15 students leave the school from each class, the ratio would have become 1 : 2. How many total students were there in the beginning? | [
"120",
"64",
"96",
"80"
] | A | class 1 have 3x student and class 2 have 5x student.
3x-15/5x-15=1/2,6x-30=5x-15,x=15
45+75=120 student
answer A |
AQUA-RAT | AQUA-RAT-39624 | It helps to try out the strategy with a smaller number of students. With only five students you teacher's method gives $$4^5-\binom{4}{1}\cdot3^5+\binom{4}{2}\cdot2^5-\binom{4}{3}\cdot1^5=1024-972+192-4=240$$ ways. You can check that this is right by observing that some book must be distributed to two students and each of the other books to only one. There are four choices for which book goes to two students, and $$\binom{5}{2}$$ choices of the two students, and $$3!$$ ways to distribute the other three books to the remaining students: $$4\cdot10\cdot6=240$$.
Now try your method, which gives $$\binom{5}{4}\cdot4!\cdot4^1=480$$ ways. Why is your number too big? It's because distributing different books to students $$1$$, $$2$$, $$3$$, $$4$$ and any book to student $$5$$ includes some of the same configurations as distributing different books to students $$1$$, $$3$$, $$4$$, $$5$$ and any book to student $$2$$. For example, if students $$1$$, $$2$$, $$3$$, $$4$$ get $$H$$, $$R$$, $$L$$, $$E$$ in that order and student $$5$$ gets $$R$$, that's the same as students $$1$$, $$3$$, $$4$$, $$5$$ getting $$H$$, $$L$$, $$E$$, $$R$$ in that order and student $$2$$ getting $$R$$.
The following is multiple choice question (with options) to answer.
The maximum number of students among them 848 pens and 630 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils? | [
"10",
"4",
"2",
"14"
] | C | number of pens = 848
number of pencils = 630
Required number of students = H.C.F. of 848 and 630 = 2
Answer is C |
AQUA-RAT | AQUA-RAT-39625 | => 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.
The average monthly salary of 8 workers and one supervisor in a factory was 430.@SSWhen@SSthe@SSsupervisor@CC@SSwhose@SSsalary@SSwas@SS430.@SSWhen@SSthe@SSsupervisor@CC@SSwhose@SSsalary@SSwas@SS430.Whenthesupervisor,whosesalarywas430. When the supervisor, whose salary was 870 per month, retired, a new person was appointed and then the average salary of 9 people was $$390 per month. The salary of the new supervisor is: | [
"233",
"600",
"510",
"771"
] | C | Explanation:
Total salary of 8 workers and supervisor together = 9 × 430 = 3870
Now total salary of 8 workers = 3870 − 870 = 3000
Total salary of 9 workers including the new supervisor = 9 × 390 = 3510
Salary of the new supervisor = 3510 − 3000 = 510
Answer: C |
AQUA-RAT | AQUA-RAT-39626 | • #### 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.
How long will a boy take to run round a square field of side 35 meters, If he runs at the rate of 9 km/hr? | [
"22",
"38",
"37",
"56"
] | D | Answer: D) 56 sec |
AQUA-RAT | AQUA-RAT-39627 | What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$
The following is multiple choice question (with options) to answer.
12 spheres of the same size are made from melting a solid cylinder of 8 cm diameter and 1 cm height. What is the diameter of each sphere? | [
"2 cm",
"4 cm",
"8 cm",
"10 cm"
] | A | Volume of cylinder = pi*r^2*h
Volume of a sphere = 4*pi*R^3 / 3
12*4*pi*R^3 / 3 = pi*r^2*h
R^3 = r^2*h / 16 = 1 cm^3
R = 1 cm
D = 2 cm
The answer is A. |
AQUA-RAT | AQUA-RAT-39628 | This is an essential point. A mathematical concept typically can be modeled in a number of different ways. We must not confuse the concept with one model. This is another aspect of the abstractness of mathematics.
I continued:
Now let's look back at your problem.
You are trying to answer this question:
Each time Sue rents a bicycle, she pays a fixed base
cost plus an hourly rate for the time the bicycle is
rented. Last Saturday she paid $12.00 to rent a bicycle for 6 hours, and yesterday she paid$9.50 to
rent a bicycle for 4 hours. Which of the following
equations shows the total cost C, in dollars, for
Sue to rent a bicycle for n hours?
That problem is not about a graph, but about rental costs. Your approach to solving it was to represent it as a problem about finding the equation of a line by looking for the slope and intercept. (There are other ways to model the problem, but you chose this because it is familiar. A standard method of problem solving is to model a new problem as one that is familiar, so you can use tools you already have.)
So Sam has shown that he can choose a representation for a problem. Good!
You also recognized that "a fixed base cost plus an hourly rate" means that the cost is a linear function. Suppose that function is
C = mn + b
or, using the variables you are more familiar with,
y = mx + b
Again, you think of it this way just because you want to turn the problem into one you know how to solve. All these variable names -- x, y, m, and b -- are things you chose to introduce, along with the idea of graphing. (By the way, the variables m and b here are parameters, which means they are considered fixed for the sake of the problem, namely graphing a line, whereas x and y are considered to actually vary. But really they are all just variables -- letters representing numbers you don't know. They just play different roles.)
Sam also recognized that the letters n and C played the roles commonly taken by x and y. The concept of a parameter may be the hardest here.
## Parameters transformed
The following is multiple choice question (with options) to answer.
A small company is planning to rent either computer A or computer B to print customer mailing lists. Both computer A and computer B must be rented on an hourly basis. The rental fee is based only on the amount of time the computer is turned on. It will cost 40 percent more per hour to rent computer A than to rent computer B. Computer B would, however, require 20 hours more than computer A to do the job. If either computer A, or computer B were rented the total cost to rent the computer would be $850.00. What would be the approximate hourly charge to rent computer B? | [
"$9.40",
"$11.30",
"$12.14",
"$17.80"
] | C | Pa = price of A
Pb = price of B
Ta = time for A to complete the job
Tb = time for B to complete the job
Given
Pa = 1.4 Pb
Ta+20 = Tb
Pa*Ta = Pb*Tb=850
1.4Pb * (Tb-20) = Pb*Tb
1.4 Pb Tb - Pb Tb = 1.4 Pb * 20
0.4PbTb = 28Pb
Tb = 28/0.4 = 70
Pb=850/70 ~ 12.14
C |
AQUA-RAT | AQUA-RAT-39629 | # 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 owner of a furniture shop charges his customer 60% more than the cost price. If a customer paid Rs. 2000 for a computer table, then what was the cost price of the computer table? | [
"2789",
"2772",
"1250",
"6725"
] | C | CP = SP * (100/(100 + profit%))
= 1250(100/160) = Rs. 1250.
Answer: C |
AQUA-RAT | AQUA-RAT-39630 | 2. Estimate the probability that in 18 independent measurements of this tower, the average of the measurements is between 199 and 201, inclusive.
Exercise $$\PageIndex{8}$$
For Example $$\PageIndex{9}$$ estimate $$P(S_{30} = 0)$$. That is, estimate the probability that the errors cancel out and the student’s grade point average is correct.
Exercise $$\PageIndex{9}$$
Prove the Law of Large Numbers using the Central Limit Theorem.
Exercise $$\PageIndex{10}$$
Peter and Paul match pennies 10,000 times. Describe briefly what each of the following theorems tells you about Peter’s fortune.
1. The Law of Large Numbers.
2. The Central Limit Theorem.
Exercise $$\PageIndex{11}$$
A tourist in Las Vegas was attracted by a certain gambling game in which the customer stakes 1 dollar on each play; a win then pays the customer 2 dollars plus the return of her stake, although a loss costs her only her stake. Las Vegas insiders, and alert students of probability theory, know that the probability of winning at this game is 1/4. When driven from the tables by hunger, the tourist had played this game 240 times. Assuming that no near miracles happened, about how much poorer was the tourist upon leaving the casino? What is the probability that she lost no money?
Exercise $$\PageIndex{12}$$
We have seen that, in playing roulette at Monte Carlo (Example [exam 6.7]), betting 1 dollar on red or 1 dollar on 17 amounts to choosing between the distributions $m_X = \pmatrix{ -1 & -1/2 & 1 \cr 18/37 & 1/37 & 18/37\cr }$ or $m_X = \pmatrix{ -1 & 35 \cr 36/37 & 1/37 \cr }$ You plan to choose one of these methods and use it to make 100 1-dollar bets using the method chosen. Using the Central Limit Theorem, estimate the probability of winning any money for each of the two games. Compare your estimates with the actual probabilities, which can be shown, from exact calculations, to equal .437 and .509 to three decimal places.
The following is multiple choice question (with options) to answer.
Tom spent six lucky days in Las Vegas. On his first day he won a net amount of only $20, but on each of the following days, the daily net amount he won grew by d dollars. If Tom won a total net amount of $1620 during his stay in Las Vegas, how much did he win on the last day? | [
"330",
"500",
"520",
"540"
] | C | I drew a diagram:
1 - $20
2 -
3 -
4 - $320
5 -
6 -
Total: $1,620
Between 1 and 6 are 5 days where he won $1,600. This means he averaged $320 per day (1600/5). You can put $320 by 4 because it's the middle number. Now you just find the two points betwenn $20 and $320 (320-20 = 300 / 3 = 100). So each day, he earned $100 more. This means on day 6, tom earned$520. Answer choice C. |
AQUA-RAT | AQUA-RAT-39631 | PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.
PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.
IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
The following is multiple choice question (with options) to answer.
The average of first nine prime numbers which are odd is? | [
"16.1",
"15.1",
"14.1",
"17.1"
] | C | Explanation:
Sum of first 9 prime no. which are odd = 158
Average = 127/9 = 14.1
Answer:C |
AQUA-RAT | AQUA-RAT-39632 | You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $\frac 12\cdot2.5\cdot2.5\tan(60)=\frac{25\sqrt3}{8}$. Multiplying this area by $12$ (the amount of triangles) gives us $\frac{75\sqrt3}{2}\approx64.9519$
The following is multiple choice question (with options) to answer.
If two sides of a triangle are 12 and 6, which of the following could be the area of triangle?
1. 35
2. 48
3. 56 | [
"a) 1 only",
"b) 1 and 2 only",
"c) 1 and 3 only",
"d) 2 and 3 only"
] | D | when two sides of a triangle are known , the maximum area occurs when the angle between those two sides is 90.
lets say base = 12 , height =8 (angle = 90)
maximum area = (1/2)*12*8 = 48 ( this is the same maximum area even when base is 8 and height is 12).
if we fix the base and keep lowering the angle, it will result in a lower height . hence the resulting area will always be < 48.
C,D and E are ruled out.
3 and 2 are possible areas as their areas are less than 48.
Hence the answer is D. |
AQUA-RAT | AQUA-RAT-39633 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? | [
"4.05%",
"4.02%",
"4%",
"3%"
] | A | Percentage error in calculated area
=(2+2+(2×2)/100)%
=4.04%
ANSWER:A |
AQUA-RAT | AQUA-RAT-39634 | (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.
Joan and Karl each bought a bicycle and the sum of their purchases was $800. If twice of what Joan paid was $40 more than what Karl paid, what did Joan pay for her bicycle? | [
"$280",
"$300",
"$320",
"$340"
] | A | J+K=800 so K=800-J
2J=K+40
2J=(800-J)+40
3J=840
J=280
The answer is A. |
AQUA-RAT | AQUA-RAT-39635 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer k such that k+n is a multiple of 21? | [
"2",
"4",
"6",
"8"
] | A | n = 3p + 1 = 7q + 5
n+2 = 3p + 3 = 7q + 7
n+2 is a multiple of 3 and 7, so it is a multiple of 21.
The answer is A. |
AQUA-RAT | AQUA-RAT-39636 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was : | [
"2500",
"2600",
"2700",
"2900"
] | C | Total number of votes = 7500
Given that 20% of Percentage votes were invalid
Valid votes = 80%
Total valid votes = 7500x (80/100)
1st candidate got 55% of the total valid votes.
Hence the 2nd candidate should have got 45% of the total valid votes
Valid votes that 2nd candidate got = total valid votes x 45/100
7500x (80/100)x45/100= 2700
answer :C |
AQUA-RAT | AQUA-RAT-39637 | & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}
The following is multiple choice question (with options) to answer.
Which of the following equivalent to (1/125)^2? | [
"(0.03)^5",
"(0.04)^3",
"(0.05)^6",
"(0.06)^3"
] | B | (1/125)^2 = (1/5^3)^2 = (1/5)^6 = (1/25)^3 = (0.04)^3
Answer: Option B |
AQUA-RAT | AQUA-RAT-39638 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
There are 7 magazines lying on a table; 3 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazines will be selected? | [
"1/2",
"7/8",
"32/35",
"11/12"
] | B | answer isB
Total probability=7C3=48
4C3 +4C2*4C1+4C1*4C2=4+24+24=42
therefore the probability that at least one of the fashion magazines will be selected= 42/48=7/8
B |
AQUA-RAT | AQUA-RAT-39639 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
if x/4-x-3/6=1,then find the value of x. | [
"6",
"4",
"5",
"1"
] | A | (x /4)-((x-3)/6)=1=> (3x-2(x-3) )/12 = 1 => 3x-2x+6=12 => x=6.
Answer is A |
AQUA-RAT | AQUA-RAT-39640 | Ans.: A
Director
Joined: 12 Nov 2016
Posts: 776
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
### Show Tags
19 Apr 2017, 23:24
dynamo wrote:
Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?
A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8
We can approach this problem using conversion factors
If the total distance is 2 miles and Dan runs at a rate of 10 miles per hour then he will cover in 1/5 of an hour or 1 mile in 6 minutes
1 mile/6 minutes = 2 miles/ 20 minutes
cross multiply add and then divide by 2
(16 miles /2) / 60 minutes
(16 miles/2)/ 1 minute
8 miles/ 1 minute
1 minute/ 8 miles
Intern
Joined: 09 Oct 2016
Posts: 34
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
### Show Tags
01 Aug 2017, 02:38
Dienekes wrote:
If distance traveled is same but at varying rates, the average rate can be found through the shortcut $$\frac{2ab}{a+b}$$ where a and b are individual rates.
In this problem $$a=10$$ and $$b=6$$ therefore average speed = $$\frac{15}{2}$$ miles per hour, i.e. $$\frac{1}{8}$$ miles per minute.
P.S. Added tags as saw this question in an MGMAT CAT.
Thank you for the explanation. I picked answer choice "B" because I did not pay attention to the unit (minutes) in the question.
SVP
Joined: 12 Sep 2015
Posts: 2298
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 8 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute? | [
"1/9",
"2/15",
"3/15",
"1/4"
] | A | Dan travels 4 miles round trip.
Running part: (2/8 = 1/4*60 = 15 minutes)
Swimming Part: (2/6 = 1/3*60 = 20 minutes)
4 miles in (15+20) minutes
4/35 = 1/9 mile per minute
Answer: 1/9 mile per minute |
AQUA-RAT | AQUA-RAT-39641 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
A man has only 20-paise and 25-paise coins in a bag. If he has 50 coins in all totaling to Rs.11.25, then the number of 20-paise coins is | [
"28",
"27",
"26",
"25"
] | D | x=20p;y=25p;
x+y=50;
0.20x+0.25y=11.25;
25x+25y=1250;
20x+25y=1125;
=>x=25;
ANSWER:D |
AQUA-RAT | AQUA-RAT-39642 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A takes twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2days. B can do the work alone in? | [
"8",
"10",
"12",
"15"
] | C | Suppose A,B,C takes x,x/2,x/3 hours to finish the work
Then, 1/x + 2/x + 3/x = 1/2
6/x = 1/2
x = 12
Answer is C |
AQUA-RAT | AQUA-RAT-39643 | (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.
The cost of 2 books and 2 magazines is $26. The cost of 1 book and 3 magazines is $27. How much does 1 magazine cost? | [
"8",
"5",
"7",
"4"
] | C | Let the cost of 1 book = x, let the cost of 1 magazine = y
2x + 2y = 26
x = (26 - 2y)/2
x = 13 - y
Again,
(13 - y) + 3y = 27
13 + 2y = 27
2y = 14
y = 7
Answer: C |
AQUA-RAT | AQUA-RAT-39644 | Just to check each case: \begin{align*} f(S_T)=\begin{cases} 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + 0 + 0 = 3 & \text{if }S_T\leq 30, \\ 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + (30 - S_T) + 0 = 33- S_T & \text{if }30
I know @KeSchn already answered but hope this helps since this is how I usually do these. Of course, you can do this multiple ways but this gets to a correct answer relatively quickly.
Edit: Gordon's answer is definitely the way to go if you're comfortable with indicator functions. It does everything the graphical methods do without requiring any visualizing etc
The following is multiple choice question (with options) to answer.
If 45-[28-{37-(15-*)}]= 57, then * is equal to: | [
"-29",
"-19",
"18",
"29"
] | C | 45-[28-{37-(15-*)}]= 57 => 45-[28-{37-15+*}]=57
45-[28-37+15-*]=57 => 45[43-37-*]=57
45-[6-*]=57 => 45-6+*=57
39+*=57 => *=57-39
= 18
ANSWER:C |
AQUA-RAT | AQUA-RAT-39645 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Five different books (A, B, C, D and E) are to be arranged on a shelf. Books C,D and E are to be arranged first and second starting from the right of the shelf. The number of different orders in which books A, B and may be arranged is | [
"1!",
"2!",
"5!",
"15!"
] | B | Solution
Since books C,D and E are arranged first and second, only books A and B will change order. Therefore it an arrangement problem involving 2 items and the number of different order is given by
2!
Answer B |
AQUA-RAT | AQUA-RAT-39646 | Points where a line intersects the function $(x-1)^{\frac 1 3}$
I'm trying to do the following exercise:
Find the equation of the line tangent to the graph of $f(x)= (x-1)^{1/3}$ at the point $x=2$. Find also if that line intersects the graph of f(x) at any other point. If it does, then find its coordinates.
Well, I begin by differentiating f(x) to get the slope:
$f(x) = (x-1)^{1/3}$
$f'(x) = (1/3)(x-1)^{-2/3}$
$f'(2) = (1/3)(2-1)^{-2/3}$
$f'(2) = (1/3)1^{-2/3}$
$f'(2) = (1/3)1$
$f'(2) = 1/3$
So the equation of the tangent line at 2 should be $y = (1/3)x$ + constant
To find the constant I use the value of f(2) which is 1.
$f(2) = (2-1)^{1/3}$
$f(2) = 1^{1/3}$
$f(2) = 1$
$f(2) =$ tangent line at 2
$1 = (1/3)2 + constant$
$1 = 2/3 + constant$
$1 - 2/3 = constant$
$1/3 = constant$
Therefore the equation of the tangent line at 2 is $y=(1/3)(x+1)$
Now I try to find out whether there are other points where this line intersects the function.
$(x-1)^{1/3} = (1/3)(x+1)$
$(x-1) = (1/27)(x+1)^3$
$\frac{x-1}{(x+1)^3} = (1/27)$
And here I have no idea of how to continue.
The following is multiple choice question (with options) to answer.
A line has a slope of 3/4 and intersects the point R (-12, -39). At which point does this line intersect the x-axis? | [
"(40,0)",
"(30,0)",
"(0,40)",
"(40,30)"
] | A | Assume that the equation of the line is y=mx+c, where m and c are the slope and y-intercept.
You are also given that the line crosses the point (-12,-39), this means that this point will also lie on the line above.
Thus you get -39=m*(-12)+c , with m=3/4 as the slope is given to be 3/4.
After substituting the above values, you get c = -30.
Thus the equation of the line is y=0.75*x-30 and the point where it will intersect the x-axis will be with y coordinate = 0.
Put y=0 in the above equation of the line and you will get, x=40.
Thus, the point R of intersection is (40,0).
A is the correct answer. |
AQUA-RAT | AQUA-RAT-39647 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B enter into partnership with capital as 7:9. At the end of 8 months, A withdraws. If they receive the profits in the ratio of 8:9 find how long B's capital was used? | [
"8",
"4",
"5",
"7"
] | D | 7 * 8 : 9 * x = 8:9 => x= 7
Answer:D |
AQUA-RAT | AQUA-RAT-39648 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A pipe can fill a cistern in 20 minutes whereas the cistern when fill can be emptied by a leak in 28 minutes. When both pipes are opened, find when the cistern will be full? | [
"27",
"76",
"70",
"21"
] | C | 1/20 - 1/28 = 1/70
70 minutes.Answer: C |
AQUA-RAT | AQUA-RAT-39649 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
A and B start a business with Rs.2000 and Rs.8000 respectively. How should they share their profits at the end of one year? | [
"3:5",
"1:4",
"3:1",
"3:2"
] | B | They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
2000 : 8000 => 1:4
Answer:B |
AQUA-RAT | AQUA-RAT-39650 | ## Output
The sum of series till 7 is 1260
This approach is not efficient as it takes time complexity of the order of N.
Another approach is using the mathematical formula for the sum of series. As we have discussed in the problem description, the series can be said the summation of (n)*(n+1)*(n+2).
Using this information let’s create a general formula for the sum.
$Sum =\sum_{\square=1}^\square\blacksquare((\square)\ast(\square+1)\ast(\square+2))$
$=\sum\lbrace{(n^2+n)(n+2)}\rbrace$
$=\sum\lbrace{n^3 + n^2 + 2n^2 + 2n}\rbrace$
$=\sum\lbrace{n^3 +3n^2 + 2n}\rbrace$
$=\sum_{\square=1}^\square\blacksquare\square^3+3\sum_{\square=1}^\square\blacksquare\square^2+2\sum_{\square=1}^\square\blacksquare\square^\blacksquare$
Now, using the general formula for the sums,
$\sum_{\square=1}^\square\blacksquare\square^3=\frac{(\square\ast(\square+1))^2}{2}$
$\sum_{\square=1}^\square\blacksquare\square^2=\frac{(\square\ast(\square+1)\ast(2\square+1))^\blacksquare}{6}$
$\sum_{\square=1}^\square\blacksquare\square^\blacksquare=\frac{(\square\ast(\square+1)^\blacksquare}{2}$
Adding all these to sum formula makes,
The following is multiple choice question (with options) to answer.
Find the sum to 200 terms of the series 1 + 4 + 6 + 5 + 11 + 6 + .... | [
"30,200",
"29,800",
"30,400",
"40,400"
] | A | Spot the above series is a combination of two APs.
The 1st AP is (1 + 6 + 11 + ....) and the 2nd AP is (4 + 5 + 6 + ...)
Since the terms of the two series alternate
S =(1 + 6 + 11 + .... to 100 terms) + (4 + 5 + 6 + .... to 100 terms)
=> 100[2 x 1 + 99 x 5]/2 + 100[2 x 4 + 99 x 1]/2 --->(Using the formula for the sum of an AP)
=>50[497 + 107] =50[604] = 30200
Alternatively, we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(1 + 4) + (6 + 5) + (11 + 6).... -> 5, 11, 17,....
Now, a= 5, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 5 + 99 x 6]
=> 50[604]
=> 30200
ANSWER:A |
AQUA-RAT | AQUA-RAT-39651 | 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.
P is five times as old as Q, and Q is seven years older than R. If R is z years old, what is P’s age in terms of z? | [
"5z + 7",
"5z − 35",
"5z + 35",
"(z + 7)/5"
] | C | P = 5Q
Q = R+7 = z+7
P = 5(z+7) = 5z + 35
The answer is C. |
AQUA-RAT | AQUA-RAT-39652 | 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.
A certain roller coaster ride has between 29 and 150 people waiting in line to board. If riders are let on only in groups of 5 there will be 2 riders that do not get on. If the riders are let on only in groups of 6 all riders will be able to get on. Which of the following is the sum of the greatest possible number of people in the line and the least possible number of people in the line? | [
"269",
"184",
"174",
"169"
] | C | Let x be the number of people
x=5m+2
x=6n
x=30p+12
Least = 30+12= 42
Highest = 30*4+12 = 132
Total = 174
ANSWER:C |
AQUA-RAT | AQUA-RAT-39653 | # 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.
If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 54/r years. If Joe's parents invested $5,000 in a long-term bond that pays 6 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Joe is ready for college? | [
" $20000",
" $15000",
" $12000",
" $10000"
] | D | Since investment doubles in 54/r years, then for r=6 it'll double in 54/6=~9 years (we are not asked about the exact amount so such an approximation will do). Thus after 18 years investment will become $5,000*2=$10,000 .
Answer: D. |
AQUA-RAT | AQUA-RAT-39654 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A pair of articles was bought for $50 at a discount of 60%. What must be the marked price of each of the article? | [
"$25",
"$10",
"$29.65",
"$35.95"
] | B | S.P. of each of the article = 50/2 = $25
Let M.P = $x
40% of x = 25
x = 25*.4 =$10
Answer is B |
AQUA-RAT | AQUA-RAT-39655 | • is 107
• is 560
• is 840
• cannot be determined from the given information
### Key Concepts
Integers
Arrangement
Algebra
AIME I, 1990, Question 8
Combinatorics by Brualdi
## Try with Hints
Let the columns be labelled A,B and C such that first three targets are A, A and A the next three being B, B and B and the next being C and C in which we consider the string AAABBBCC.
Since the arrangement of the strings is one-one correspondence and onto to the order of shooting for example first A is shot first, second A is shot second, third A is shot third, first B is shot fourth, second B is shot fifth, third B is shot sixth, first C is shot seventh, second C is shot eighth,
or, here arrangement of the strings is bijective to the order of the shots taken
the required answer is the number of ways to arrange the letters which is $\frac{8!}{3!3!2!}$=560.
Categories
## Digits and Order | AIME I, 1992 | Question 2
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Order.
## Digits and order – AIME I, 1992
A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. Find number of ascending positive integers are there.
• is 107
• is 502
• is 840
• cannot be determined from the given information
### Key Concepts
Integers
Digits
Order
AIME I, 1992, Question 2
Elementary Number Theory by David Burton
## Try with Hints
There are nine digits that we use 1,2,3,4,5,6,7,8,9.
Here each digit may or may not be present.
$\Rightarrow 2^{9}$=512 potential ascending numbers, one for subset of {1,2,3,4,5,6,7,8,9}
Subtracting empty set and single digit set
=512-10
=502.
Categories
## Remainders and Functions | AIME I, 1994 | Question 7
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.
## Remainders and Functions – AIME I, 1994
The following is multiple choice question (with options) to answer.
If APPLE is coded as 0 and FLOWER is coded as 63, then BASKET is coded as | [
"2",
"19",
"35",
"40"
] | D | APPLE = 1+16+16+12+5=50 i.e 5*0=0
FLOWER = 6+12+15+23+5+18=79 i.e 7*9=63
BASKET =2+1+19+11+5+20=58 5*8=40
ANSWER:D |
AQUA-RAT | AQUA-RAT-39656 | 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$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Annie will be half as old as Larry in 3 years. Annie will also be one-third as old as Jerome in 5 years. If Jerome is 12 years older than Larry, how old is Annie? | [
"5",
"8",
"19",
"26"
] | A | J = L+12
(1) 2(A+3) = L+3
(2) 3(A+5) = J+5 = L+17
Let's subtract equation (1) from equation (2)
A+9 = 14
A = 5
The answer is A. |
AQUA-RAT | AQUA-RAT-39657 | ## Circular Combination
1. Arrange 8 dancers in circular fashion.
2. Use 8 pearls in a band to make a necklace.
3. Arrange 8 science and 7 arts students circularly so no two arts students are together.
• 7!
• $$\frac{(8-1)!}{2}$$
• $$^8P_7 \times \space 7!$$
# Combination
## Concept
A, B, C
In how many ways can we select 2?
• AB, AC, BC
• When making teams $$AB \equiv BA$$
## Formulae And Notation
Formulae (Don’t Memorize!)
• $$^nC_r = {n! \over r!(n-r)!}$$
• $$^nC_r \times ^rP_r = ^nP_r = ^nP_r \times r!$$ (Permutation-combination relationship)
• $$^{n+1}C_r = ^nC_r + ^nC_{r-1}$$
• $$^nC_r=?$$ (from above, expanding twice up to n-2)
• $$^nC_r = ^nC_{n-r}$$
Notation
• $$^nC_r \equiv$$ $$n \choose r$$
• Select 5 people from 6 $$\rightarrow ^6C_5 =$$ $$6 \choose 5$$
## Expaniosn of $$^nC_r$$
The following is multiple choice question (with options) to answer.
Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants? | [
"41",
"66",
"36",
"70"
] | B | 8 shirts
9 pants
we can combine 2 shirts with (9-3) pants
2*6=12
we can combine the other shirts (6) with any pants (9)
6*9=54
Thus total :12+54=66 ANS:B |
AQUA-RAT | AQUA-RAT-39658 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
4,25,49,121,169, | [
"149",
"169",
"289",
"209"
] | C | 17^2= 289 because follow sequence of square of the prime numbers
ANSWER:C |
AQUA-RAT | AQUA-RAT-39659 | to solve for the other. Logic To Calculate Percentage Difference Between 2 Numbers. Sale Discount Calculator - Percent Off Mortgage Loan Calculator - Finance Fraction Calculator - Simplify Reduce Engine Motor Horsepower Calculator Earned Value Project Management Present Worth Calculator - Finance Constant Acceleration Motion Physics Statistics Equations Formulas Weight Loss Diet Calculator Body Mass Index BMI Calculator Light. The commonly used way is to go from right to left, which gives us a positive number. Posted: (1 week ago) Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Just subtract the past value from the current value. Click the calculate button 3. Calculate percentage difference between two columns I have a input text file in this format: ITEM1 10. The percentage difference between the two values is calculated by dividing the value of the difference between the two numbers by the average of the two numbers. First, you need Excel to subtract the first number from the second number to find the difference between them. Listing Results about Calculate Variance Between Two Numbers Real Estate. To write an increase or decrease as a percentage, use the formula actual increase or decrease original cost × 100%. The growth rate can be listed for real or nominal GDP. This difference needs to be divided between the first number (the one that doesn't change). We then append the percent sign. Follow 52 views (last 30 days) Show older comments. Enter an old number in cell A1 and a new number in cell B1. The percent difference formula or the percent difference equation of two numbers a and b is: ((a - b) / (a+b)/2) × 100, where a > b Calculate the percentage difference between the numbers 35 and 65. The first step to the equation is simple enough. Step 1: Calculate the difference (subtract one value from the other) ignore any negative sign. With a Difference From, Percent Difference From, or Percent From calculation, there are always two values to consider: the current value, and the value from which the difference should be calculated. A percentage variance, aka percent change, describes a proportional change between two numbers, an original value and a new value. It has more of an impact when you say, "There was a 50 percent increase in attendance at the concert compared to last year," versus when you say, "There were 100 more people at the concert this year than
The following is multiple choice question (with options) to answer.
When 1/10 percent of 4,000 is subtracted from 1/10 of 4,000, the difference is | [
"396",
"36",
"3996",
"40"
] | A | (1/10)*4000 - (1/10)%*4000 =
400 - (1/1000)*4000 =
400 - 4 = 396
The answer is A. |
AQUA-RAT | AQUA-RAT-39660 | homework-and-exercises, fluid-statics, bernoulli-equation
You ignore the area of the hole.
Rate of water going in: $r'= \frac{0,2cm^3}{s}$
Rate of water going out: $r=1mm^2*\sqrt{2gh}$
Set the two rates equal and solve for $h$
$\frac{0,2cm^3}{s} = 1mm^2*\sqrt{2gh}$
$\frac{0,2cm^3}{s*0,01cm^2} = \sqrt{2gh}$
$\frac{20cm}{s} = \sqrt{2gh}$
The units work out correctly
I will leave the rest to you
Hint square both sides
Solving for the $t$ in $h(t)$ would be a more complex
The following is multiple choice question (with options) to answer.
A river 3m deep and 36 m wide is flowing at the rate of 2 kmph the amount of water that runs into the sea per minute is? | [
"4500",
"2678",
"3600",
"3400"
] | C | (2000 * 3 * 36)/60
= 3600 M3
Answer:C |
AQUA-RAT | AQUA-RAT-39661 | (D)
Manager
Joined: 03 Aug 2017
Posts: 64
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink]
### Show Tags
05 Oct 2019, 22:16
Bunuel wrote:
If 20 typists can type 48 letters in 20 minutes, then how many letters will 30 typists working at the same rate complete in 1 hour?
A. 63
B. 72
C. 144
D. 216
E. 400
This i how i solved...
20 Typist can type 48 letters / 20 Min therofre in 1 Min 20 Typist can type = 48/20 = 2.4 Letters per Min
In 60 Min 20 Typist can type = 2.4 *60 =144 Words per min...
We are also told now there are 30 Typist so if 20 typist can type 144 words so 30 Typist can type X words per min.....
Solve for X
20/144 = 30 / x .....
x = 72*3 = 216 Words per min
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink] 05 Oct 2019, 22:16
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Rates for having a manuscript typed at a certain typing service are $6 per page for the first time a page is typed and $4 per page each time a page is revised. If a certain manuscript has 100 pages, of which 35 were revised only once, 15 were revised twice, and the rest required no revisions, what was the total cost of having the manuscript typed? | [
"$860",
"$850",
"$840",
"$830"
] | A | 50 pages typed 1x
35 pages typed 2x (original + one revision)
15 pages typed 3x (original + two revisions)
50(6)+35(6+4)+15(6+4+4)=300+350+210=860
Answer - A |
AQUA-RAT | AQUA-RAT-39662 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 20 days and B can do it in 15 days and C can do it 20 days. They started the work together and A leaves after 2 days and B leaves after 4 days from the beginning. How long will work lost? | [
"10 2/5",
"10 2/0",
"12 2/3",
"10 2/1"
] | C | 2/20 + 4/15 + x/20 = 1
x = 38/3 = 12 2/3 Answer: C |
AQUA-RAT | AQUA-RAT-39663 | evolution, mathematical-models, theoretical-biology, population-dynamics, population-genetics
Title: Effective population size when the population sizes varies from season to season Let's think of a species which has four generations per year and which population size changes from season to season so that the population size is 100 in summer, 200 in spring, 50 in autumn and 20 in winter for example. In such case, the effective population size $N_e$ can be calculated by:
$$N_e = \frac{n}{\sum_{i=1}^n\frac{1}{N_i}}$$
where $n$ is the number of generation per year (4 in my example) and each $N_i$ correspond to the population size in one season.
My question
Can you please provide an explanation of why this formula (based on the harmonic mean) holds true to define the effective population size?
The following is multiple choice question (with options) to answer.
At the start of an experiment, a certain population consisted of 3 animals. At the end of each month after the start of the experiment, the population size was double its size at the beginning of that month. Which of the following represents the population size at the end of 6 months? | [
" 2^3",
" 3^2",
" 2(3^10)",
" 3(2^6)"
] | D | 3*2^n where n is the #of month --> 3*2^6 Answer (D) |
AQUA-RAT | AQUA-RAT-39664 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.42, the number of 25 paisa coins is? | [
"16",
"42",
"16",
"84"
] | B | 5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42
Answer: B |
AQUA-RAT | AQUA-RAT-39665 | python, programming-challenge, primes, factors
Title: Calculate the first largest factor of 600851475143 ٍProblem description:
The prime factors of 13195 are 5, 7, 13 and 29. Largest prime factor 13195 is 29.
What is the largest prime factor of the number 600851475143 ?
Prime Factor:
any of the prime numbers that can be multiplied to give the original number.
Example:
Find the prime factors of 100:
100 ÷ 2 = 50; save 2
50 ÷ 2 = 25; save 2
25 ÷ 2 = 12.5, not evenly so divide by next highest number, 3
25 ÷ 3 = 8.333, not evenly so divide by next highest number, 4
But, 4 is a multiple of 2 so it has already been checked, so divide by next highest number, 5
25 ÷ 5 = 5; save 5
5 ÷ 5 = 1; save 5
List the resulting prime factors as a sequence of multiples, 2 x 2 x 5 x 5 or as factors with exponents, 2^2 x 5^2.
My Solution
This is my solution for problem 3 of Project Euler using Python:
def FLPF(n):
'''Find Largest Prime Factor
'''
PrimeFactor = 1
i = 2
while i <= n / i:
if n % i == 0:
PrimeFactor = i
n /= i
else:
i += 1
if PrimeFactor < n: PrimeFactor = int(n)
return PrimeFactor Couple of points:
The following is multiple choice question (with options) to answer.
The HCF of two numbers is 42 and the other two factors of their LCM are 11 and 12. What is the largest number. | [
"462",
"450",
"480",
"504"
] | D | Explanation:
HCF of the two numbers = 42
HCF will be always a factor of LCM
42 is factor of LCM
other two factors are 11 & 12
Then the numbers are (42 11) and (42 x 12)
= 462 and 504
Answer: Option D |
AQUA-RAT | AQUA-RAT-39666 | Thanks
• Your "40% overlap" would mean taht the surface overlaping is 40% of the surface of which rectangle ? – Evargalo Sep 28 '17 at 16:13
• I want to know if, say, 40% of the other rectangle is contained inside the self rectangle – John Lexus Sep 28 '17 at 16:14
For the equations, I will let the left of the first rectangle be $l_0$, the right be $r_0$, the top $t_0$ and the bottom $b_0$. The second rectangle is $l_1, r_1,$ etc. Their areas will be $A_0$ and $A_1$.
If the boxes don't overlap, obviously the percentage overlap is $0$.
If your boxes are found to be colliding, simply use this formula to calculate the area that is overlapping:
$$A_{overlap} = (\max(l_0, l_1)-\min(r_0, r_1))\cdot(\max(t_0, t_1)-\min(b_0, b_1)).$$
Now there are two ways to calculate a percentage error that could make sense for your explanation. If we're just checking the percentage within the first rectangle (or "self" in your program), the percent overlap is simple: $$P_{overlap} = \frac{A_{overlap}}{A_{self}}.$$
If you want the percentage to be equal whether it's calculated from either rectangle, the equation you're looking for is: $$\frac{A_{overlap}}{A_0+A_1-A_{overlap}}.$$
The following is multiple choice question (with options) to answer.
What will be the percentage increase in the area of the cube's surface if each of the cube's edges grows by 20%? | [
"33%",
"44%",
"55%",
"56%"
] | B | The question is very easy. My logic is the following:
A surface = 6*a^2
After 20% increase A surface = 6* ((1.2a)^2) = 6*1.44*a^2
The increase in the surface area = (6*1.44*a^2 - 6*a^2)/6*a^2 = (6*a^2(1.44-1))/(6*a^2) = 1.44-1=0.44 = 44%
ANSWER:B |
AQUA-RAT | AQUA-RAT-39667 | This leaves ten admissible configurations, and for each there are $(3!)^2=36$ ways of permuting the boys and girls, for a total of 360 ways.
• ok, but where exatcly am i going wrong? – Anvit Dec 10 '17 at 11:21
• @AnvitGarg Note that you have considered the case that boys and girls should sit in alternate positions. But, the question says that a boy can be adjacent to atleast one girl. – Rohan Dec 10 '17 at 11:22
• @AnvitGarg You overlooked permutations such as $b_1g_1g_2b_2b_3g_3$ in which two or more girls are consecutive. – N. F. Taussig Dec 10 '17 at 11:22
The following is multiple choice question (with options) to answer.
there are 3 girls and 6 boys. how many kids are there? | [
"10",
"19",
"20",
"9"
] | D | D |
AQUA-RAT | AQUA-RAT-39668 | ### Show Tags
21 Jul 2010, 07:18
4
for venn diagrams in case of 3 cases:
Total = m(a) + m(b) + m(c) + m(a&b) + m(b&c) + m(c&a) - 2*m(a&b&c)
So, 70 = 40 + 30 + 35 - m(a&b) - m(b&c) - m(c&a) - 2*15
=> -35 = - [ m(a&b) + m(b&c) + m(c&a) ] - 30
Therefore, m(a&b) + m(b&c) + m(c&a) = 5
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Re: There are 70 students in Math or English or German. Exactly 40 are in [#permalink]
### Show Tags
05 Feb 2020, 04:51
2
ramana wrote:
There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.
OA given is 5
please explain your answers.
what is the significance of 'exactly' (Exactly 40 are in Math,
30 in German, 35 in English ) in the stem?
Letting M = math, E = English, and G = German, we can use the formula for a 3-category scenario:
Total = n(M) + n(E) + n(G) - n(M and E) - n(M and G) - n(E and G) + n(all 3) - n(none)
70 = 40 + 35 + 30 - n(M and E) - n(M and G) - n(E and G) + 15 - 0
70 = 120 - n(M and E) - n(M and G) - n(E and G)
50 = n(M and E) + n(M and G) + n(E and G)
The following is multiple choice question (with options) to answer.
In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German? | [
"30",
"10",
"18",
"28"
] | C | Answer
Let A be the set of students who have enrolled for English and B be the set of students who have enrolled for German.
Then, (A U B) is the set of students who have enrolled at least one of the two subjects. As the students of the class have enrolled for at least one of the two subjects, A U B = 40
We know A U B = A + B - (A n B)
i.e, 40 = A + 22 - 12
or A = 30 which is the set of students who have enrolled for English and includes those who have enrolled for both the subjects.
However, we need to find out the number of students who have enrolled for only English = Students enrolled for English - Students enrolled for both German and English
= 30 - 12 = 18.
choice is (C) |
AQUA-RAT | AQUA-RAT-39669 | I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?
6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
The following is multiple choice question (with options) to answer.
There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices. | [
"102",
"116",
"103",
"512"
] | B | The number of triangle can be formed by 10 points = 10C3.
Similarly, the number of triangle can be formed by 4 points when no one is collinear=4C3.
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
= 10C3-4C3 = 120-4 = 116.
Ans: B |
AQUA-RAT | AQUA-RAT-39670 | In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.
The following is multiple choice question (with options) to answer.
If 99<x<199 and 10<y<100, then the product xy CANNOT be equal to: | [
"19,104",
"19,903",
"19,356.732",
"19,502"
] | B | Correct Answer: (B)
Determine the range of xy by multiplying the two extremes of each individual range together. The smallest value of xy must be greater than 99 * 10. The largest value must be less than 199 * 100. This means that 990 < xy < 19,900. (B) is outside of this range, so it is not a possible product of xy. |
AQUA-RAT | AQUA-RAT-39671 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 4 : 5 : 6 and their average is 25. The largest number is : | [
"30",
"98",
"27",
"21"
] | A | Explanation:
Let the numbers be 4x, 5x and 6x, Then, (4x + 5x + 6x ) / 3 = 25
=> 5x = 25
=> x = 5.
Largest number 6x = 30.'
Answer: A) 30 |
AQUA-RAT | AQUA-RAT-39672 | # If both integers $x$ and $y$ can be represented as $a^2 + b^2 + 4ab$, prove that $xy$ can also be represented like this …
There is a set $Q$ which contains all integral values that can be represented by $$a^2 + b^2 + 4ab$$, where $a$ and $b$ are also integers. If some integers $x$ and $y$ exist in this set, prove that $xy$ does too.
I really have no idea how I can go about solving this. I tried simple multiplication of the two assuming one to be $(a^2 + 4ab + b^2)$ and other as $(c^2 + 4cd + d^2)$ but ultimately it leads to a long equation I can make no tail of :/
Any help whatsoever would be greatly appreciated
-
I have updated your post to LaTeX. Please see that the updates are correct. – Jeel Shah Jan 20 '14 at 12:48
@hardmath Fixed! Thanks for the catch! – Jeel Shah Jan 20 '14 at 13:01
## 2 Answers
Since $a^2+b^2+4ab=(a+2b)^2-3b^2$, your numbers are exactly the numbers of the form $x^2-3y^2$. Now $x^2-3y^2$ is the norm of the algrebraic number $x+y\sqrt{3}$, so you have the identity
$$(x^2-3y^2)(u^2-3v^2)=(xu+3yv)^2-3(xv+yu)^2$$
(multiplicativity of norms).
-
Thank you so much, now I can finally sleep with this homework done. – skatter Jan 20 '14 at 12:54
To make the resulting identity explicit in terms of $a, b, c, d$, if $f(x,y) = x^2 + 4xy + y^2$, then $$f(ac-bd,ad+4bd+bc) = f(a,b) f(c,d).$$ – heropup Jan 20 '14 at 13:13
Generalization:
The following is multiple choice question (with options) to answer.
If x and y are integers such that x^2= 2y and xy = 256, then x – y = ? | [
"-30",
"-24",
"-5",
"5"
] | B | here x and y are integers .
x^2=2y,xy =256.
substitute (x^2)/2=y in xy => x^3=256*2=>x^3=512.
here x^3 is positive ,
x is also positive.
x=8 then y=32.
x-y=-24
so option B is correct |
AQUA-RAT | AQUA-RAT-39673 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
The entire contents of a full sack of flour can be used to make 15 cupcakes and 8 pizzas. The same full sack of flour can be used to make 5 cupcakes and 14 pizzas. If a full sack of flour is used to make only pizzas, how many pizzas can be made? | [
"18 1/4",
"18 1/2",
"18 3/4",
"17"
] | D | 15x+8y=5x+14y
x=3/5y. Answer D |
AQUA-RAT | AQUA-RAT-39674 | # 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 an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price? | [
"Rs.1520",
"Rs.1570",
"Rs.1500",
"Rs.1550"
] | C | Let the cost price of an article be Rs. x.
(19% of x) - (12% of x) = 105
19x/100 - 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs.1500
Answer:C |
AQUA-RAT | AQUA-RAT-39675 | # probability question on a customer
A very frustrated customer is trying to find an electronic receipt on their phone so they can return an item. The trouble is that the customer has three email accounts and can't remember which, one, account it was sent to. The customer assumes that there is an equal probability for each account. To add complexity the phone only has enough battery power to search one of the three email accounts
Unfortunately, the store says that this is the last day it will accept the return, and it is only 3 minutes till close, so there will not be any chance to get to a charger. The customer randomly decides to search one of the emails, without any bias.
Suppose, due to the organization and the various spam filters of the accounts, the chance of finding the receipt even if they were to search in the correct account is not a guarantee. The probability of finding it in account 1, assuming it was sent there is 62%, 54% if in account 2 and 56% for account 3.
Part (a) What is the probability that if the receipt is in account 2 that the customer will find it?
I think this one is (1/3)*(0.54) The third is for choosing the account and then times 0.54 is the prob for finding from the question. but the question seems like a conditional probability so I'm not sure.
Part (b) Calculate the probability the receipt was in account 2, if the search in account 2 is unsuccessful.
I think this one is ((1/3)(0.16))/((1/3)(0.31)+(1/3)(0.16)+(1/3)(0.49))
Part(c) What is the probability this person finds the email?
(1/3)(0.69)+(1/3)(0.84)+(1/3)(0.51) because we can either find it in account 1 or 2 or 3
Any help would be appreciated!
The following is multiple choice question (with options) to answer.
A manufacturer is using glass as the surface for the multi-touch screen of its smartphone. The glass on the manufactured phone has a 7% probability of not passing quality control tests. The quality control manager bundles the smartphone in groups of 10. If that bundle has any smartphone that does not pass the quality control test, the entire bundle of 10 is rejected.
What is the probability that a smartphone bundle that will be rejected by quality control? | [
" 0.25",
" 1-0.93^10",
" 1-0.95^10",
" 1-0.05^10"
] | B | Find the probability of the opposite event and subtract from 1.
The opposite event is that bundle will NOT be rejected by quality control, which will happen if all 10 phones pass the test, so P(all 10 phones pass test)=0.93^10.
P(at least one phone do not pass the test)=1- P(all 10 phones pass test)=1-0.93^10.
Answer: B. |
AQUA-RAT | AQUA-RAT-39676 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
If there are half as many women as men in a group and an equal number of men and women do not own cars - a group that is 30% of the total. What fraction of the total is men who own cars? | [
"3⁄20",
"31⁄60",
"9⁄40",
"1⁄3"
] | B | Consider a group of 100 women and 200 men, a total of 300 people. 30% of them, which is 90, form a group of people who don't own a car.
Half of them are men, and the other half are women, more precisely 45.
It means that there are 200 - 45 = 155 men who own a car, and this represents 155/300 = 31/60 of the total.
Answer B |
AQUA-RAT | AQUA-RAT-39677 | # Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$
Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $$\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$$ where $$r-\frac{s}{2}$$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $$\frac{\pi r^2}. {2}$$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($$2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$$). In conclusion the resultant equation is (with $$r=x$$ and $$s=y$$):
$$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$$
The following is multiple choice question (with options) to answer.
The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second? | [
"74%",
"44%",
"68%",
"64%"
] | D | r = 3 πr2 = 9
r = 5 πr2 = 25
25 π – 16 π
100 ---- ? => 64%
Answer: D |
AQUA-RAT | AQUA-RAT-39678 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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The following is multiple choice question (with options) to answer.
Running at the same constant rate, 6 identical machines can produce a total of 390 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes? | [
"648",
"1800",
"2600",
"10800"
] | C | Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)
Machines 6 : 10 :: 390 : x
Time (in minutes) 1 : 4
6 x 1 x x = 10 x 4 x 390
x = (10 x 4 x 390)/(6)
x = 2600.
ANSWER:C |
AQUA-RAT | AQUA-RAT-39679 | 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}$$
The following is multiple choice question (with options) to answer.
Each child has 2 pencils and 13 Skittles. If there are 11 children, how many pencils are there in total? | [
"16",
"12",
"18",
"22"
] | D | 2*11=22.Answer is D. |
AQUA-RAT | AQUA-RAT-39680 | 2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
### Show Tags
30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
My 2 cents.
It is important to realize from the Question stem that the 3 equations intersect as (p,r).
For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.
For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.
So D.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Sep 2017, 19:04
1
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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The following is multiple choice question (with options) to answer.
Line R has the equation 3x + y = 7. Which of the following lines is perpendicular to Line R? | [
" y = 3x + 4",
" y = –3x – 6",
" y = (1/3)x – 1",
" y = (–1/3)x + 2"
] | C | I first rewrote the equation in the standard y=mx+b form. Therefore, Line R as presented, 3x + y = 7, can be rewritten as follows: y=-3x+7.
Thought process next is what line would be perpendicular to Line R? Any line with a reciprocal of the slope but in the opposite direction. The reciprocal of any fraction/integer is 1 over that number/integer. Therefore the reciprocal of -3 is -1/3 - need to drop the negative sign because the line would kinda run parallel and we want perpendicular. Scan the answers choices and notice C as the only one. |
AQUA-RAT | AQUA-RAT-39681 | ### Show Tags
19 Aug 2015, 01:34
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Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
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19 Aug 2015, 01:55
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Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
In a village the average age of n people is 42 years. But after the verification it was found that the age of a person had been considered 20 years less than the actual age, so the new average, after the correction,increased by 1. The value of n is? | [
"33",
"20",
"88",
"37"
] | B | It is the same as a person with 20 years more age replaces an existing person of the group ( or village)
Since the total age of the village having n persons, is being increased by 20 years and the average age of village is being increased by 1 year, hence there are total 20 people in the village.
Alternatively : ( n x 42 ) + 20 = ( n x 43 )
=> n=20
Answer: B |
AQUA-RAT | AQUA-RAT-39682 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of the even integers from 0 to 80 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 40 inclusive? | [
"20",
"24",
"28",
"36"
] | A | The sum of even numbers from 0 to N is 2 + 4 +...+ N
= 2(1 + 2 + ...+ N/2)
= 2(N/2)(N/2+1)/2 = (N/2)(N/2+1)
The average is (N/2)(N/2+1)/(N/2+1) = N/2
The average of the even numbers from 0 to 80 is 80/2 = 40
The average of the even numbers from 0 to 40 is 40/2 = 20
The answer is A. |
AQUA-RAT | AQUA-RAT-39683 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train, 130130 metres long travels at a speed of 4545 km/hr crosses a bridge in 3030 seconds. The length of the bridge is | [
"270 metres",
"245 metres",
"235 metres",
"220 metres"
] | B | Explanation:
Solution 1Speed =45 km/hr=45×518 m/s=12.5 m/s=45 km/hr=45×518 m/s=12.5 m/s
Time =30 s=30 s
Distance travelled =12.5×30=375 m=12.5×30=375 m
Length of the bridge =375−130=245 m
ANSWER IS B |
AQUA-RAT | AQUA-RAT-39684 | 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.
At what rate percent per annum will the simple interest on a sum of money be 1/5 of the amount in 10 years? | [
"2%",
"7%",
"9%",
"3%"
] | A | Let sum = x. Then, S.I. = x/5, Time
= 10 years.
Rate = (100 * x) / (x * 5 * 10)
= 2%
Answer: A |
AQUA-RAT | AQUA-RAT-39685 | But $45 < \frac{\alpha}{2} < 90$. Therefore $\cos \left( \frac{\alpha}{2}\right) = \sqrt{0.02} = 0.1414$.
The following is multiple choice question (with options) to answer.
45 x ? = 35% of 900 | [
"16.2",
"7",
"5",
"500"
] | B | Answer
Let 45 x A = (35 x 900)/100
∴ A = (35 x 9)/45 = 7
Correct Option: B |
AQUA-RAT | AQUA-RAT-39686 | "How does this help me?" you say
I'm very glad you asked
Let's think about $8$ as our 100% and we want to find 2 parts of 8, so using what we just learned we say that "2 parts of 8 is the same as $\frac{2}{8}$." To turn $\frac{2}{8}$ into a percentage we just need the denominator to equal 100 since earlier we said that 8 is our 100%
We could use an equation for this, so let's use an equation for this :D
We know that $8 \cdot \left(\text{some number}\right) = 100$ so;
$8 x = 100$
$\frac{1}{8} \cdot 8 x = 100 \cdot \frac{1}{8} = \frac{100}{8} = \frac{25}{2} = 12.5$
Now we know that if we multiply the denominator by 12.5 we will get 100 and if we multiply the numerator by 12.5 we will get a ratio with 100 as the denominator, so our percentage must be the numerator!
2/8*12.5/12.5 = 25/100 = 25%
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The following is multiple choice question (with options) to answer.
If a% of b is the same as 8/9 of 450, then the value of ab is? | [
"40000",
"45120",
"65200",
"32560"
] | A | a/100 * b = 8/9 * 450
xy = 400*100 = 40000
Answer is A |
AQUA-RAT | AQUA-RAT-39687 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
How many four-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number. | [
"480",
"475",
"485",
"490"
] | B | N=(4*5-1)*5*5=475
where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
B |
AQUA-RAT | AQUA-RAT-39688 | 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.
The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each? | [
"Rs. 4800",
"Rs. 3600",
"Rs. 3560",
"Rs. 4530"
] | D | Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530
ANSWER:D |
AQUA-RAT | AQUA-RAT-39689 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a loss of 15%. If it was sold for Rs.400 more, there would have been a gain of 5%. What is the cost price? | [
"1600",
"2998",
"2778",
"2788"
] | A | 85%
120%
--------
25% ---- 400
100% ---- ? => Rs.1600
Answer:A |
AQUA-RAT | AQUA-RAT-39690 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
Look at this series: 22, 21, 23, 22, 24, 23, ... What number should come next? | [
"22",
"24",
"25",
"26"
] | C | Explanation: In this simple alternating subtraction and addition series; 1 is subtracted, then 2 is added, and so on.
Answer: Option C |
AQUA-RAT | AQUA-RAT-39691 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Posts: 12145
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
If the Average of 10 consecutive integers is 22.5 then the 10th integer is :- | [
"15",
"20",
"23",
"27"
] | D | The average falls between the 5th and 6th integers,
integer 5 = 22, integer 6 = 23.
Counting up to the tenth integer
we get 27.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39692 | c++, beginner, calculator, reference
There are two problems with it. First, the movie name I gave was not Journey to Mars so that's a basic problem that comes from not passing that value into printReport. The second problem is more subtle. Notice that the gross amount is $954.50, but the sum of the donated amount and the net sale is only $954.49. That missing $0.01 is going to drive an accountant insane! This is a fundamental problem with using double (or any floating-point representation) for money values. An alternative is to keep a number of cents as an integer value internally. For more depth about floating point issues, I'd recommend the excellent article "What every computer scientist should know about floating-point arithmetic" by David Goldberg.
Don't abuse using namespace std
Putting using namespace std at the top of every program is a bad habit that you'd do well to avoid.
Use objects
As mentioned in the other review, using one or more objects would help this code a lot. Here's a start. First you could have a class named Movie:
class Movie
{
public:
void getInput();
void calculate();
void printReport() const;
private:
std::string movieName;
double adultTicketPrice;
double childTicketPrice;
int noOfAdultTicketSold;
int noOfChildTicketSold;
double percentageDonation;
// calculated values
double grossAmount;
double amountDonated;
double netSale;
};
Now all of your functions are easily changed into member functions. For example, this:
void getInput(string& movieName, double& adultTicketPrice,
double& childTicketPrice, int& noOfAdultTicketSold,
int& noOfChildTicketSold, double& percentageDonation) {
becomes this:
void Movie::getInput() {
And your movieSalesReport() is considerably simplified:
void movieSalesReport() {
Movie movie;
movie.getInput();
movie.calculate();
movie.printReport();
}
It's not ideal, but it should get you started.
The following is multiple choice question (with options) to answer.
The entrance ticket at the Imax theatre in Chennai is worth Rs 250. The sale of tickets increased by 50% when the price of the ticket was lowered. However, the collections recorded a decrease of 17.5%. The price of the ticket is reduced by how much rupees? | [
"Rs150",
"Rs112.50",
"Rs105",
"Rs120"
] | B | let the no. of tickets sold before = x
then the no. of tickets sold after = 1.5x
Collection Before = x*250
Collection after = 1.5x* = (250x*82.5)100
=137.5
change in price =250 - 137.5
=112.5
ANSWER:B |
AQUA-RAT | AQUA-RAT-39693 | (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.
Harkamal purchased 8kg of grapes at the rate of 90 per kg and 9 kg of mangoes at the rate of 55 per kg. How much amount did he pay to the shopkeeper? | [
"A)1055",
"B)1215",
"C)1065",
"D)1070"
] | B | Cost of 8 kg grapes = 90 × 8 = 720.
Cost of 9 kg of mangoes = 55 × 9 = 495.
Total cost he has to pay = 720 + 495 = 1215.
B) |
AQUA-RAT | AQUA-RAT-39694 | # Combination Question
#### Swazination
##### New member
How many combinations can I make if I have 4 letters A,G,C,T and I want to make groups of 5. Yes, you can repeat the letters.
#### Denis
##### Senior Member
From AAAAA to TGCAA
or (if replaced by digits 1 to 4):
from 1111 to 43211
Yes?
#### ksdhart2
##### Full Member
Combinations/Probability problems of this sort almost always come down to pattern recognition. In fact, learning how to find a pattern and extrapolate what comes next will help you greatly in nearly every aspect of mathematics. So let's try using those skills and see what we can come up with for this problem. Let's start with a much simpler version and build up as we go along. Suppose we had two letters (A and G) and one slot. How many combinations would there be? Well, obviously the answer is two. That's not very interesting nor does it seem helpful just yet, but we'll keep in mind all the same.
Now suppose we had two letters and two slots. How many combinations would there be then? Thinking about it, we can force the first slot to be A, then there's two possibilities for the second slot. Similarly, if we force the first slot to be G, we have two additional combinations. So we have two possibilities for the first slot, and for each of those two possibilities, we have two possibilities for the second slot. That gives us a total of 2 * 2 = 4 = 22 combinations.
Now suppose we had two letters and three slots. How many combinations would there be then? I see that we can arbitrarily fix the first two slots and then let the third one vary. Since we already know that there's four possible ways to arrange the first two slots, and for each of those four possibilities, there's two choices for the third slot, that gives us a total of 4 * 2 = 8 = 23 combinations. Are you seeing a pattern? How many combinations would there be with four slots? Five slots? n slots?
The following is multiple choice question (with options) to answer.
A store sells eight different kinds of shirts. If we want to buy two different kinds of shirts, how many possible combinations of two shirts can we buy? | [
"22",
"24",
"26",
"28"
] | D | 8C2=28.
The answer is D. |
AQUA-RAT | AQUA-RAT-39695 | (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.
In a barrel of juice there is 40 liters; in a barrel of beer there are 80 liters. If the price ratio between barrels of juice to a barrel of beer is 3:4, what is the price ratio between one liter of juice and one liter of beer? | [
"3:2.",
"2:1.",
"3:1.",
"4:3."
] | A | Price of 40 L juice= 3x
1L= 3x/40
Price of 80 L beer= 4x
1L= 4x/80
Ratio of 1 L price = 3x/40/4x/80= 3:2
A is the answer |
AQUA-RAT | AQUA-RAT-39696 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A individuals pledged to pay equal contributions so that a charity’s goal of $x could be reached. If d of the contributors failed to pay their share, which of the following represents the additional number of dollars that each of the remaining individuals would have to pay in order to allow the charity to reach its goal? | [
"dx/A(A - d)",
"dx",
"a",
"A-d"
] | A | Number of individuals = A
Amount paid by each individual = n
Total expected amount = Charity's goal = nA = x
n = x/A
Number of individuals who fail to pay = d
Contribution from individuals who would fail to pay = dx/A --> Additional amount
Number of individuals who are paying = A - d
Additional amount has to be divided among the (A - d) individuals --> dx/A(A - d)
A |
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