source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34397 | 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 running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train? | [
"150",
"781",
"767",
"277"
] | A | Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter
Answer: A |
AQUA-RAT | AQUA-RAT-34398 | # 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.
Price of a book increases 15% successively (2times) what is the new price of the book more compared to that of the old price: | [
"32.25%",
"23.34%",
"36%",
"39%"
] | A | New price is 1.15*1.15*old price = 1.3225 *old price
Increase in price = 0.3225 * old price
% increase = 100* 0.3225 * old price /old price = 32.25 %
ANSWER:A |
AQUA-RAT | AQUA-RAT-34399 | 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.
A is 16th position and 29th from down of the people who passed. 5 failed , 6 didn't give . how many boys are there in the class? | [
"52",
"53",
"54",
"55"
] | D | A is 16th position, means = 15 + A
29th from down, means = A + 28
Total passed = A + 15 + 28 = 44
Total failed = 5
Didn't give = 6
Therefore, total boys = 44 + 5 + 6 = 55
ANSWER:D |
AQUA-RAT | AQUA-RAT-34400 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is: | [
"2 : 3",
"4 : 3",
"6 : 7",
"9 : 16"
] | B | Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
Answer B |
AQUA-RAT | AQUA-RAT-34401 | 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.
9 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio | [
"1:2",
"1:3",
"2:1",
"2:5"
] | D | Explanation:
Let 1 man 1 day work = x
1 boy 1 day work = y
then 9x + 2y = 4(x+y)
=> 5x = 2y
=> x/y = 2/5
=> x:y = 2:5
Option D |
AQUA-RAT | AQUA-RAT-34402 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 12, without any of the digits repeating? | [
"100",
"98",
"81",
"96"
] | D | 0,1,2,3,4,5
One digit will have to remain out for all 5 digit numbers;
if 0 is out; Leftover digits will be 1,2,3,4,5 = Sum(1,2,3,4,5)=15. Ignore
if 1 is out; Leftover digits will be 0,2,3,4,5 = Sum(0,2,3,4,5)=14. Ignore
if 3 is out; Leftover digits will be 0,1,2,4,5 = Sum(0,1,2,4,5)=12.
4*4! = 4*24 = 96
if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. Ignore
if 5 is out; Leftover digits will be 0,1,2,3,4 = Sum(0,1,2,3,4)=10. Ignore
Total count of numbers divisible by 12 = 96
Ans:D |
AQUA-RAT | AQUA-RAT-34403 | ## anonymous 5 years ago 100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!
1. anonymous
you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..
2. anonymous
$\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950$
3. anonymous
Could you solve it using factorials, please?
4. anonymous
hmm i found the answer as 9900
5. amistre64
if there are 3 people: 1,2 ; 1,3 ; 2,3 .... is that right?
6. anonymous
Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...
7. anonymous
yea but its harder to calculate this like that... so i think C(100,2) better way of solve
8. amistre64
5054 if we do that ..... add all the numbers from 1 to 100 :)
9. anonymous
No, all the numbers from 1 to 99.
10. amistre64
n(n+1) ------ :) 2
11. amistre64
ack .... 99 then lol
12. amistre64
wouldnt the last guy have noone to shake hands with?
13. anonymous
mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,
14. anonymous
Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.
15. anonymous
yea you right
16. amistre64
The following is multiple choice question (with options) to answer.
6 people meet for a business lunch. Each person shakes hands once with each other person present. How many handshakes take place? | [
"30",
"21",
"18",
"15"
] | D | the formula to count handshakes is n(n−1)2n(n−1)2
Where n is the number of people
=> 6(6-1)/2 = 6*5/2 = 30/2 = 15
=> the answer is D(15) |
AQUA-RAT | AQUA-RAT-34404 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? | [
"10",
"5",
"15",
"20"
] | D | Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Answer is D. |
AQUA-RAT | AQUA-RAT-34405 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
Which of the following correctly lists the data sets in order of greatest to least standard deviation?
I. 4, 12, 15, 17, 19, 25
II. 42, 42, 42, 42, 42, 42
III. 57, 58, 59, 50, 51, 52 | [
"II, I, III",
"I, III, II",
"I, II, III",
"II, III, I"
] | C | out of three given sets,
II has all numbers same, so it has zero standard deviation hence it is Least
only answer choices C and D have II in least position.
Out of I ans III
III has consecutive numbers , hence all are closedly spaced hence std deviation will be less
while in I,numbers are spread widely hence std deviation should be more than III
hence order should be II, III, I
IMO C |
AQUA-RAT | AQUA-RAT-34406 | ### Show Tags
09 Oct 2015, 14:30
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
I found an almost identical question with the same question stem, but different statements.
Here it is: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Manager
Joined: 12 Sep 2015
Posts: 80
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
08 Apr 2016, 21:31
MrSobe17 wrote:
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
z can take any value in that case. Think of a case in which z = 12.
$$5^2 * 3 * 2^2 = 3 * x^y$$
Here x = 10
_________________
The following is multiple choice question (with options) to answer.
X, Y, and Z are consecutive numbers and X > Y > Z. Also, 2X + 3Y + 3Z = 5Y + 11. What is the value of Z? | [
"2",
"3",
"4",
"5"
] | B | If X, Y, and Z are consecutive numbers and X > Y > Z, then Y = Z+1 and X = Z+2.
2X + 3Y + 3Z = 5Y + 11
2Z+4+3Z+3+3Z = 5Z+5+11
3Z = 9
Z = 3
The answer is B. |
AQUA-RAT | AQUA-RAT-34407 | # How to calculate the area covered by any spherical rectangle?
Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a sphere with a radius $R$?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
• What do you mean by "rectangle"? Do you just want a quadrilateral, or some stronger condition? (Note that you cannot have four right angles in a quadrilateral on the sphere.) – Nick Matteo Mar 25 '15 at 13:16
• Yes, a quadrilateral having equal opposite sides (but not parallel) each as a great circle arc on a sphere. – Harish Chandra Rajpoot Mar 25 '15 at 13:21
• What do you mean by length and width? As the sides are not parallel, we don't have the usual idea of width and length. Do you simply mean the lengths of the sides? – robjohn Apr 4 '15 at 14:12
• Yes, length & width are simply the sides of the rectangle as great circles arcs on the spherical surface. – Harish Chandra Rajpoot May 28 '15 at 23:56
Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides $$w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2$$
The following is multiple choice question (with options) to answer.
The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1600 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units? | [
"140",
"99",
"88",
"160"
] | D | Given that the area of the square = 1600 sq.units
=> Side of square = √1600 = 40 units
The radius of the circle = side of the square = 40 units Length of the rectangle = 2/5 * 40 = 16 units
Given that breadth = 10 units
Area of the rectangle = lb = 16 * 10 = 160 sq.units
Answer: D |
AQUA-RAT | AQUA-RAT-34408 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
Senior Manager
Joined: 04 Aug 2010
Posts: 322
Schools: Dartmouth College
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
The following is multiple choice question (with options) to answer.
If a tap could fill entire tank in 20 hrs due to leakage, then in how much time
tank can be emptied by leakage if tap can fill entire tank in 16 hrs without leakage | [
"12hrs",
"24hrs",
"36hrs",
"80hrs"
] | D | time take to fill[withleakage] =20hrs so workdone in 1hr=1/20
time tkae to fill[without leakage]=16hrs so workdone in 1hr=1/16
if u subtract both u'll get time taken by leakage to empty...
1/20-1/16=1/80 so 80hrs
ANSWER:D |
AQUA-RAT | AQUA-RAT-34409 | Related Rate Prob - Ships sailing in different directions
• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?
Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$
$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$
$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$
$2D*\frac{dx}{dt}= 1250t-2400$
$\frac{dx}{dt}=\frac{1250t-2400}{2D}$
$D=\sqrt{5^2+45^2}=5\sqrt{82}$
$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$
a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h
b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h
The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet
The following is multiple choice question (with options) to answer.
A man can row a distance of 5 km in 60 min with the help of the tide. The direction of the tide reverses with the same speed. Now he travels a further 20 km in 10 hours. How much time he would have saved if the direction of tide has not changed? | [
"2",
"8",
"1",
"6"
] | D | Explanation:
He covered 5 km in 1 hour , so he might cover 20 km in 4 hours.
But he took 10 hours.
He would have saved 10 – 4 = 6 hours.
Answer: D |
AQUA-RAT | AQUA-RAT-34410 | 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500
501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
The following is multiple choice question (with options) to answer.
Look at this series: 544, 509, 474, 439, 404, 369 ... What number should come next? | [
"334",
"408",
"306",
"507"
] | A | 334
This is a simple subtraction series. Each number is 35 less than the previous number.
A |
AQUA-RAT | AQUA-RAT-34411 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
The cost of type 1 rice is Rs 15 per Kg and type 2 rice
is Rs 20 per Kg. If both type1 and type 2 are mixed in the
ratio of 2:3,then the price per Kg of the mixed variety
of rice is? | [
"15",
"17",
"19",
"18"
] | D | Let the price of the mixed variety be Rs x per Kg.
Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2 rice Rs 20
Mean Price Rs x
20-x x-15
(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.
ANSWER : D 18 |
AQUA-RAT | AQUA-RAT-34412 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
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The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
The salary of a worker is first increased by 30% and afterwards reduced by 30%. What is net change in his salary? | [
"3% decrease",
"5% decrease",
"9% decrease",
"4% decrease"
] | C | (30 * 30)/100 = 9% decrease
Answer:C: |
AQUA-RAT | AQUA-RAT-34413 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
A money lender lent Rs. 1000 at 4% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years. | [
"3.2",
"3.75",
"4",
"4.25"
] | A | (1000xtx4/100) + (1400xtx5/100) = 350 → t =3.2 answer A |
AQUA-RAT | AQUA-RAT-34414 | $\implies \text{Var} \left(X\right) = \frac{1}{n} \cdot \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{1}{n} \cdot \frac{n \left(n + 1\right)}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{\left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{n + 1}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \left[\frac{2 n + 1}{3} - \frac{n + 1}{2}\right]$
$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \cdot \frac{n - 1}{6}$
$\implies \text{Var} \left(X\right) = \frac{{n}^{2} - 1}{12}$
So, Standard deviation of $\left\{1 , 2 , 3 , \ldots . , n\right\}$ is ${\left[\text{Var} \left(X\right)\right]}^{\frac{1}{2}} = {\left[\frac{{n}^{2} - 1}{12}\right]}^{\frac{1}{2}}$
In particular, your case the standard deviation of $\left\{1 , 2 , 3 , 4 , 5\right\}$
$= {\left[\frac{{5}^{2} - 1}{12}\right]}^{\frac{1}{2}} = \sqrt{2}$.
The following is multiple choice question (with options) to answer.
If d is the standard deviation x, y, and z, what is the standard deviation of x + 4, y + 4, z + 4 ? | [
"4d",
"d",
"5d",
"d+4"
] | B | CONCEPT: Standard Deviation is Defined as Average Deviation of Terms in the set from the Mean value of the set. i.e.
1) It depends on the separation between the successive terms of the set
2) If a Constant Value is Added/Subtracted in every terms of set then the Separation between successive terms does NOT change Hence S.D. remains Constant
e.g.{1, 2, 3, 4, 5} will have same standard Deviation as {1+10, 2+10, 3+10, 4+10, 5+10}
3) If a Constant Value is Multiplied in every terms then the Separation between succesive terms gets multiplied by the constant Hence S.D. remains gets multiplied by same Number
e.g. {0.7, 1.4, 2.1, 2.8, 3.5} will have Standard Deviation = 0.7* Standard deviation of set {1, 2, 3, 4, 5}
When 4 is added in each term of set {x, y, z} then the new set {x+4, y+4, z+4} will remain same as the previous standard deviation i.e. d
Answer: B |
AQUA-RAT | AQUA-RAT-34415 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman makes a score of 100 runs in the 11th inning and thus increases his average by 5. Find his average after 11th inning. | [
"40",
"50",
"62",
"45"
] | B | Let the average after 11th inning = x
Then, average after 10th inning = x-5
10(x-5)+100 = 11x
x = 100-50 = 50
Answer is B |
AQUA-RAT | AQUA-RAT-34416 | # 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 simple interest on a certain sum of money for 8 years at 4% per annum is same as the simple interest on Rs. 560 for 8 years at the rate of 12% per annum then the sum of money is | [
"Rs.1820",
"Rs.1040",
"Rs.1120",
"Rs.1680"
] | D | Explanation:
Let the sum of money be x
then
(x × 4 × 8)/100 = (560 × 12 × 8)/100
x × 4 × 8 = 560 × 12 × 8
x × 4 = 560 × 12x = 560 × 3 = 1680
Answer: Option D |
AQUA-RAT | AQUA-RAT-34417 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
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.
Kudos [?]: 24 [2], given: 0
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
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 persons going for tour is 16 years. Twenty new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years. The number of persons initially going for tour is: | [
"5",
"10",
"20",
"30"
] | C | Explanation :
Let the initial number of persons be x. Then,
=> 16x + 20 x 15 -15.5 (x + 20)
=> 0.5x = 10
=> x = 20
Answer : C |
AQUA-RAT | AQUA-RAT-34418 | 2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)
3.
Amount of A = Concentration of A * Volume of mixture
Amount = C*V
( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt)
When you add water to this solution, the amount of milk does not change.
So Initial Conc * Initial Volume = Final Conc * Final Volume
$$C_i * V_i = C_f * V_f$$
$$C_f = C_i * (V_i/V_f)$$
In the question above, we find the final concentration of wine. Initial concentration $$C_i$$ = 1 (because it is pure wine)
When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt.
So $$C_f = 1 * 2/4$$
Since it is done 19 times, $$C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})$$
The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has $$V_f$$ as 22, you get amount of wine = 1/77 * 22 = 2/7 lt
Kudos +1 Karishma
Is there an fast way to compute the result of the multiplacation series like we have for $$Cf$$? I actually did the long way .
Re: A container has 3L of pure wine. 1L from the container is taken out an [#permalink] 06 Feb 2013, 22:54
Go to page 1 2 Next [ 30 posts ]
# A container has 3L of pure wine. 1L from the container is taken out an
The following is multiple choice question (with options) to answer.
A mixture of 70 liters of wine and water contains 10% water. How much water must be added to make water 12 ½% of the total mixture? | [
"8",
"7",
"5",
"2"
] | D | 70 * (10/100) = 7
Wine Water
87 1/2% 12 1/2%
87 1/2% ------- 63
12 1/2% -------? => 9-7=2
Answer: D |
AQUA-RAT | AQUA-RAT-34419 | # Complex root of equation 1 [closed]
let $$a,b$$ & $$c$$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $$\frac a{a-2},\frac b{b-2}$$ and $$\frac c{c-2}$$
hence or otherwise find value of $$(a-2)(b-2)(c-2)$$
• Since $p(x)=(x-a)(x-b)(x-c)$ you simply have $(a-2)(b-2)(c-2)=-p(2)=3$. – Jack D'Aurizio Nov 20 '18 at 18:56
• – lab bhattacharjee Nov 20 '18 at 19:24
HINT
We have that $$(x-a)(x-b)(x-c)=x^3-3x^2+1$$ and then
• $$a+b+c=3$$
• $$ab+bc+ca=0$$
• $$abc=-1$$
(see figure below)
Invert the given relationship $$y=\dfrac{r}{r-2} \ \ \ \ (1)$$ into $$r=\dfrac{2y}{y-1} \ \ \ \ (2)$$
As $$r$$ is a root of the given cubic, we have $$r^3-3r^2+1=0 \ \ \ \ (3)$$
It suffices then to plug relationship (2) into (3) to get on the LHS a rational expression ; equating its numerator to zero gives :
$$3y^3 - 9y^2 - 3y + 1=0 \ \ \ \ (4)$$
The following is multiple choice question (with options) to answer.
Which of the following are roots of an equation (x^-2)-(x^-1)-30=0 | [
"1/5 and -1/6",
"-1/5 and 1/6",
"1/5 and 1/6",
"-1/5 and -1/6"
] | B | Given: (x^-2) - (x^-1) - 30 = 0
Rewrite as: 1/(x²) - 1/x - 30 = 0
Remove fractions by multiplying both sides by x² to get: 1 - 1x - 30x² = 0
Rearrange to get: 30x² + x - 1 = 0
Factor to get: (5x + 1)(6x - 1) = 0
So, EITHER 5x + 1 OR 6x - 1 = 0
If 5x + 1 = 0, then x = -1/5
If 6x - 1 = 0, then x = 1/6
So, the roots (solutions) are -1/5 and 1/6
The answer is B. |
AQUA-RAT | AQUA-RAT-34420 | So by my equation I get that they are of equal purity.
Kalyan.
4. ## Re: Purity
Hello, PASCALfan!
Two vessels contains equal amount of water and beer.
1 spoon beer is added to the water and mixed well.
From this mixture 1 spoon is added to the beer.
Which of the liquids is more pure?
This is the classic "Wine and Water" problem.
As long as the amount transferred each time remains the same,
. . the final concentrations are equal.
This is the basis of a stunning card trick.
You are seated at a small table across from your volunteer.
You show him a deck of cards, fanning them
. . so he can see they are all facing one way.
Both of you place your hands under table.
You hand him the deck.
Instruct him to select a secret number from 1 to 20,
. . and keep it a secret. .Call it n.
Have him count off the top n cards,
. . turn them over and replace them on the deck.
Have him shuffle the deck thoroughly.
Then count off the top n cards
. . and hand them to you under the table.
Now you remind him:
. . you don't know his secret number,
. . he doesn't know how many face-up cards he has,
. . you don't know how many face-up cards you have.
Despite this, you will try to make your number of face-up cards
. . equal his number of face-up cards.
He hears the sound of cards being counted and moved.
After a few moment, you bring your cards to the top of the table
. . and count the face-up cards.
Instruct him to do the same . . . and the numbers match!
[Acknowledge the thunderous applause ... modestly, of course.]
Spoiler:
Secret: When he hands you the packet, turn it over.
5. ## Re: Purity
In fact, the answer does not change regardless of how well the liquids were mixed in the process. Indeed, the volume of the liquid in each vessel after the two operations is the same as it was initially. Therefore, in the end the volume of beer in water has to equal the volume of water in beer.
This is one of my favorite math problems because it shows that there is more to math than handling large formulas.
6. ## Re: Purity
Thanks for the card trick..
Regards
The following is multiple choice question (with options) to answer.
A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required? | [
"35",
"36",
"37",
"38"
] | C | If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.
So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.
Now, number of cans of Maaza = 80/16 = 5
Number of cans of Pepsi = 144/16 = 9
Number of cans of Sprite = 368/16 = 23
Thus, the total number of cans required = 5 + 9 + 23 = 37 Answer: C |
AQUA-RAT | AQUA-RAT-34421 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains move in the same direction at speeds 50 kmph and 32 kmph respectively. A man in the slower train observes that 15 seconds elapse before the faster train completely passes by him. What is the length of faster train? | [
"100m",
"75m",
"120m",
"50m"
] | B | since both trains move in same direction so:
average speed=50-32=18kmph=5mps
speeed=length of train /time
length of train=5*15=75m
ANSWER:B |
AQUA-RAT | AQUA-RAT-34422 | 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.
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are: | [
"39, 30",
"41, 32",
"42, 33",
"43, 34"
] | C | Let their marks be (x + 9) and x.
Then, x + 9 = 56/100(x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
answer :C |
AQUA-RAT | AQUA-RAT-34423 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 60 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one? | [
"21",
"88",
"40",
"99"
] | C | Relative speed = 60 + 30 = 90 km/hr.
90 * 5/18 = 25 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 *1/25 = 40 sec
Answer: C |
AQUA-RAT | AQUA-RAT-34424 | • Choose $5$ members (combinations). Make sure you do not have exactly $0$ or $1$ mathematicians (subtractions of other combinations) – Henry May 15 '18 at 17:05
• I get $251$ as the solution. There are $462$ combinations in total, one of them contains only computer scientists, and $210$ ($7!/4!$) contains 1 mathematician. $462-1-210=252$. – Hanlon May 15 '18 at 17:14
• $7+5=12$ so there are ${12 \choose 5}$ not ${11 \choose 5}$ combinations in total. More than one of them only involves computer scientists ... – Henry May 15 '18 at 17:25
• I meant 4 not 5 mathematicians. Sorry. And how can there be more than one combination that involves only computer scientists? – Hanlon May 15 '18 at 17:27
• Actually, it should be $426$. We place 7 computer scientists in 4 boxes, we have $7!/(7-4)!$ permutations, which is $840$, and then we divide by $4!$ because there are 4 boxes and we get $35$ combinations. $462-1-35=426$ – Hanlon May 15 '18 at 18:27
The following is multiple choice question (with options) to answer.
There are n members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 55 possible combinations in which Michael is not selected, what is the value of n? | [
"11",
"12",
"15",
"18"
] | B | Combinations of two persons, in which Michael was not selected = 55
number of ways two persons can be selected from m people = m*(m-1)/2
Let m be the number of people excluding Michael, then m*(m-1) = 110 => m=11
Thus, n = m + 1(Michael) = 12
option B |
AQUA-RAT | AQUA-RAT-34425 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Jim is now twice as old as Fred, who is nine years older than Sam. Six years ago, Jim was six times as old as Sam. How old is Jim now? | [
"30",
"34",
"38",
"42"
] | D | J = 2F = 2(S+9) = 2S+18
J-6 = 6*(S-6)
(2S+18)-6 = 6S-36
S = 12 and so J = 42
The answer is D. |
AQUA-RAT | AQUA-RAT-34426 | # Linear Algebra Money Question
bleedblue1234
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # $10 bills A+B+C = 32 1A+5B+10C = 100 ## The Attempt at a Solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. ## Answers and Replies E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. azizlwl You can narrow the selection.$1 can only be in a group of 5.
Homework Helper
This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.
Last edited by a moderator:
Homework Helper
Dearly Missed
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # \$10 bills
A+B+C = 32
1A+5B+10C = 100
## The Attempt at a Solution
The following is multiple choice question (with options) to answer.
A man has Rs. 80 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? | [
"15",
"60",
"75",
"90"
] | A | Answer : A
Let number of notes of each denomination be x.
Then, x + 5x + 10x = 80
16x = 80
x =5.
Hence, total number of notes = 3x = 15. |
AQUA-RAT | AQUA-RAT-34427 | & = \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.
(128.5 x 32) + (13.8 x 30) = ? x 25 | [
"524.48",
"556.02",
"574.36",
"181.04"
] | D | Explanation :
? = (128.5 x 32) + (13.8 x 30)/25
= 4112 + 414/25 = 181.04
Answer : Option D |
AQUA-RAT | AQUA-RAT-34428 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes respectively. Both the pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after | [
"6 minutes",
"9 minutes",
"45 minutes",
"3 minutes"
] | B | If pipe B is turned off after x mins, then
(2*30)/75 + x/45 =1
x/45 = 1-60/75 = 1/5
x= 45/5= 9 mins
ANSWER:B |
AQUA-RAT | AQUA-RAT-34429 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If a(a - 6) = 27 and b(b - 6) = 27, where a ≠ b, then a + b =? | [
"−48",
"6",
"−2",
"46"
] | B | i.e. if a =-3 then b = 9
or if a = 9 then b =-3
But in each case a+b = -3+9 = 6
Answer: option B |
AQUA-RAT | AQUA-RAT-34430 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
What is the greater of the two numbers whose product is 2048, given that the sum of the two numbers exceeds their difference by 64? | [
"A)90",
"B)32",
"C)64",
"D)70"
] | C | Let the greater and the smaller number be g and s respectively.
gs = 2048
g + s exceeds g - s by 64 i.e., g + s - (g - s) = 64
i.e., 2s = 64 => s = 32.
g = 2048/s = 64.
ANSWER:C |
AQUA-RAT | AQUA-RAT-34431 | ## Extra 01
How many arrangements can be made using the letters from the word COURAGE? What if the arrangements must contain a vowel in the beginning?
• $$4 \times 6!$$
## Extra Problem 02
How many arrangements are possible using the words
• EYE
• CARAVAN?
## 3(a)
There are (p+q) items, of which p items are homogeneous and q items are heterogeneous. How many arrangements are possible?
## 2(j)
There are 10 letters, of which some are homogeneous while others are heterogeneous. The letters can be arranged in 30240 ways. How many homogeneous letters are there?
Let, $$m = \text{number of homogeneous items}$$
• n(arrangements) = 30240 = $$\frac {10!}{m!}$$
• $$m! = \frac{10!}{30240}=120$$
• m = 5
## 2(k)
A library has 8 copies of one book, 3 copies of another two books each, 5 copies of another two books each and single copy of 10 books. In how many ways can they be arranged?
Total books = $$1 \times 8+3 \times 2+5 \times 2 + 8 \times 1 + 10$$ = 42
• n(arrangements) = $$\frac{42}{8!(3!)^2(5!)^2}$$
## 2(l)
A man has one white, two red, and three green flags; how many different signals can he produce, each containing five flags and one above another?
Flags: W = 2, R = 2, G = 3, Total = 7
Answer
## 2 (m)
A man has one white, two red, and three green flags. How many different signals can he make, if he uses five flags, one above another?
## 3(a)
How many different arragnements can be made using the letters of the word ENGINEERING? In how many of them do the three E’s stand together? In how many do the E’s stand first?
i
ii
iii
## 3(b)
In how many ways can the letters of the word CHITTAGONG be arranged, so that all vowels are together?
The following is multiple choice question (with options) to answer.
For an upcoming charity event, a male vocalist has agreed to sing 5 out of 8 “old songs” and 3 out of 7 “new songs.” How many ways can the singer make his selection? | [
"25",
"50",
"150",
"217.78"
] | D | =8C5*7C3
=8*7/3*7*5/3
=217.78
Ans = D |
AQUA-RAT | AQUA-RAT-34432 | java, xml, swing
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/12 00:00:00</date>
<riddle>ipso lorem 7</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/19 00:00:00</date>
<riddle>ipso lorem 8</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/26 00:00:00</date>
<riddle>ipso lorem 9</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/02 00:00:00</date>
<riddle>ipso lorem 10</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/09 00:00:00</date>
<riddle>ipso lorem 11</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/16 00:00:00</date>
<riddle>ipso lorem 12</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/23 00:00:00</date>
<riddle>ipso lorem 13</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/30 00:00:00</date>
<riddle>ipso lorem 14</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
</weeks>
The following is multiple choice question (with options) to answer.
On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006? | [
"Saturday",
"Friday",
"Monday",
"Tuesday"
] | B | Explanation:
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday
S0, 8th Dec, 2006 is Friday.
Answer: B) Friday |
AQUA-RAT | AQUA-RAT-34433 | & = \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.
[( 3.242 x 10 ) / 100] = ? | [
"0.045388",
"4.5388",
"0.03242",
"473.88"
] | C | Answer
Multiplying 3.242 x 10=3.242
Now divide 3.242 by 100
So, 3.242 ÷ 100= 0.03242
∴ Shift the decimal two places to the left as 100
Correct Option: C |
AQUA-RAT | AQUA-RAT-34434 | # 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.
Profit selling 10 candles equals selling price of 3 bulbs. While loss on selling 10 bulbs equal selling price of 4 candles. Also profit percentage equals to the loss percentage and cost of a candle is half of the cost of a bulb. What is the ratio of selling price of candles to the selling price of a bulb? | [
"5:4",
"3:2",
"4:5",
"3:4"
] | B | Solution:
Price Candle Bulb
SP A(let) C(let)
CP B D
and, C = 2A
Profit = 10 (B-A) = 3D
Loss = 10 (C-D) = 4B
Profit % = (3D*100)/10A
Loss % = (4B*100)/10C
Now, By the questions,
(3D*100)/10A = (4B*100)/10C
B/D = 3/2.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34435 | ## 1 Answer
You also cannot succeed with $$8$$ tiles. Each of the tiles can only cover one of the squares marked with an $$\times$$: $$\begin{array}{|c|c|c|c|c|c|c|} \hline \times &\;\,&\;\,&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline \end{array}$$
Here is a solution with $$9$$ tiles:
• Wow that's so nice... and I am blind. – abc... Feb 23 at 3:53
• Very nice. For a slightly prettier solution with 9 tiles, you can start with a rotationally symmetric arrangement of 8 tiles covering all but the center square. Namely, put tiles on b1&c2, d2&e1, g2&f3, f4&g5, f7&e6, d6&c7, a6&b5, b4&a3; every square but d4 is covered. – bof Feb 23 at 5:40
The following is multiple choice question (with options) to answer.
A room 3 m 21 cm long and 7m 77 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor. | [
"27636",
"27640",
"27647",
"27713"
] | D | Explanation:
Area of the room = (321 x 777) cm2.
Size of largest square tile = H.C.F. of 321 cm and 777 cm = 3 cm.
Area of 1 tile = (3 x 3) cm2.
Number of tiles required =(321×777)/(3×3)=27713
Answer: Option D |
AQUA-RAT | AQUA-RAT-34436 | You want to end up with exactly 6 gallons in the unmarked container.
(YES: the unmarked container > 6 gallons!)
Your mission is to do it in exactly 9 moves, and using the MINIMUM number of gallons of water
from the water supply.
A "move" is any pouring, filling or emptying.
For example, filling the 9-gal from the water supply,
then pouring it to fill the 4-gal container,
then emptying it is 3 moves: fill - pour - empty.
How do you do it?
Spoiler:
(1) Fill the "9".
Code:
* *
|/////|
|/////| * *
|/////| | |
|/////| | |
* * |//9//| | 0 |
| | |/////| | |
| 0 | |/////| | |
*-----* *-----* *-----*
4 9 U
(2) Pour "9" into "4".
Code:
* *
| |
| | * *
| | | |
|/////| | |
* * |/////| | 0 |
|/////| |//5//| | |
|//4//| |/////| | |
*-----* *-----* *-----*
4 9 U
(3) Empty "4",
Code:
* *
| |
| | * *
| | | |
|/////| | |
* * |/////| | 0 |
| | |//5//| | |
| 0 | |/////| | |
*-----* *-----* *-----*
4 9 U
(4) Pour "9" into "4".
Code:
* *
| |
| | * *
| | | |
| | | |
* * | | | 0 |
|/////| | | | |
|//4//| |//1//| | |
* - - * * - - * * - - *
4 9 U
(5) Pour "9" into "U".
The following is multiple choice question (with options) to answer.
Jill has 42 gallons of water stored in quart, half-gallon, and one gallon jars. She has equal numbers of each size jar holding the liquid. What is the total number of water filled jars? | [
"3",
"6",
"9",
"72"
] | D | Let the number of each size of jar = wthen 1/4w + 1/2w + w = 42 1 3/4w = 42w=24The total number of jars = 3w =72Answer: D |
AQUA-RAT | AQUA-RAT-34437 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is? | [
"21",
"25",
"20",
"29"
] | C | M = 15
S = 5
DS = 15 + 5 = 20
Answer: C |
AQUA-RAT | AQUA-RAT-34438 | ### 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,C invested the capitals in a business. If A invested Rs. 30,000 more thanC and B invested Rs.10,000 more than A, then how should a profit of Rs.15,000 be divided among A,B and C? Given : Capital of C is Rs.10,000 | [
"6000, 4500, 4500",
"6200, 5800, 3000",
"6000, 7500, 1500",
"6500, 5500, 3000"
] | C | A==40,000
B==50,000
C==10,000
so, ration are 4:5:1
profit 15000
so,, 15000/10==1500
so,, A==6000, B==7500,, C==1500
ANSWER:C |
AQUA-RAT | AQUA-RAT-34439 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P? | [
"3 : 6",
"3 : 9",
"3 : 3",
"3 : 4"
] | D | Since S # 1/t
S1 : S2 = 1/t1 : 1/t2 = 1/8 : 1/6 = 3 : 4
Answer:D |
AQUA-RAT | AQUA-RAT-34440 | 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.
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is: | [
"5 : 2",
"7 : 3",
"9 : 2",
"13 : 4"
] | B | Explanation:
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
3x + 20 = 2x + 40
x = 20.
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
Answer: B |
AQUA-RAT | AQUA-RAT-34441 | Now, let's handle potential factors of $$\3\$$ in $$\n\$$. We know that $$\10^{\varphi(n)}-1\$$ is a multiple of $$\n\$$, but $$\\frac{10^{\varphi(n)}-1}9\$$ might not be. But, $$\\frac{10^{9\varphi(n)}-1}9\$$ is a multiple of $$\9\$$ because it consists of $$\9\varphi(n)\$$ ones, so the sum of its digits a multiple of $$\9\$$. And we've noted that multiplying the exponent $$\k\$$ by a constant preserves the divisibility.
Now, if $$\n\$$ has factors of $$\2\$$'s and $$\5\$$'s, we need to add zeroes to end of the output. It way more than suffices to use $$\n\$$ zeroes (in fact $$\\log_2(n)\$$ would do). So, if our input $$\n\$$ is split as $$\n = 2^a \times 5^b \times m\$$, it suffices to have $$\9\varphi(m)\$$ ones to be a multiple of $$\n\$$, multiplied by $$\10^n\$$ to be a multiple of $$\2^a \times 5^b\$$. And, since $$\n\$$ is a multiple of $$\m\$$, it suffices to use $$\9\varphi(n)\$$ ones. So, it works to have $$\9\varphi(n)\$$ ones followed by $$\n\$$ zeroes.
The following is multiple choice question (with options) to answer.
What is the greatest positive integer T such that 3^T is a factor of 9^10? | [
"5",
"T=9",
"T=10",
"T=20"
] | D | What is the greatest positive integer T such that 3^T is a factor of 9^10?
9^10 = (3^2)^10 = 3^20
D. 20 |
AQUA-RAT | AQUA-RAT-34442 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A farmer had d animals, all of which the farmer bought for the same price. The farmer decided to sell the animals. 20 of the animals were sold at a profit of 10% each while the rest were sold at a loss of 20% each. Overall, the farmer made neither a profit nor a loss. Which of the following is equal to d? | [
"24",
"26",
"28",
"30"
] | D | Let P be the price the farmer paid for each animal.
The total profit on the 20 animals is 20*0.1*P.
The total loss on the other (d-20) animals is (d-20)*0.2*P.
(d-20)*0.2*P = 20*0.1*P
0.2*d = 2 + 4
d = 30
The answer is D. |
AQUA-RAT | AQUA-RAT-34443 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A batsman in his 12th innings makes a score of 65 and thereby increases his average by 3 runs. What is his average after the 12th innings if he had never been ‘not out’? | [
"32",
"43",
"44",
"45"
] | A | Let ‘x’ be the average score after 12 th innings
⇒ 12x = 11 × (x – 3) + 65
∴ x = 32
Answer A |
AQUA-RAT | AQUA-RAT-34444 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 282 runs? | [
"6.25",
"6.22",
"6.29",
"6.39"
] | A | Required run rate
= [282 - (3.2 * 10)]/40
= 250/40
= 6.25
Answer: A |
AQUA-RAT | AQUA-RAT-34445 | ### Show Tags
23 Oct 2009, 23:42
14
KUDOS
Expert's post
12
This post was
BOOKMARKED
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.
In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.
Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).
Consider:
$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:
$$ttttt|||$$
Means that first nephew will get all the tickets,
$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1
And so on.
How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.
The following is multiple choice question (with options) to answer.
Visitors to show were charged Rs.15 each on the first day. Rs.7.50 on the second day, Rs.2.50 on the third day and total attendance on the three days were in ratio 2:5:13 respectively. The average charge per person for the whole show is? | [
"2",
"7",
"5",
"6"
] | C | 2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5
Answer: C |
AQUA-RAT | AQUA-RAT-34446 | 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.
There were two candidates in an election. Winner candidate received 55% of votes and won the election by 100 votes. Find the number of votes casted to the winning candidate? | [
"550",
"744",
"255",
"199"
] | A | W = 55% L = 45%
55% - 45% = 10%
10% -------- 100
55% -------- ? => 550
Answer:A |
AQUA-RAT | AQUA-RAT-34447 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is: | [
"200",
"600",
"800",
"1200"
] | D | Let the parts be x, y and [2600 – (x + y)].Then,
x×4×1/100=y×6×1/100=[2600−(x+y)]×8×1/100
∴ y⁄x = 4⁄6 = 2⁄3 or y = 2⁄3x.
So, x×4×1100=[2600−53x]×8100x×4×1100=[2600−53x]×8100
⇒ 4x = (7800−5x)×8/3⇒52x=(7800×8)
⇒ x = (7800×8/52)=1200.
∴ Money invested at 4% = 1200.
Answer D |
AQUA-RAT | AQUA-RAT-34448 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 23 students in a class. In how many different ways can a committee of 3 students be formed? | [
"1254",
"1482",
"1771",
"1875"
] | C | 23C3 = 23*22*21 / 6 = 1771
The answer is C. |
AQUA-RAT | AQUA-RAT-34449 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
What will come in place of the x in the following Number series? 2, 4, 8, x , 22 | [
"33",
"16",
"12",
"14"
] | D | (D)
The pattern is + 2, + 4, + 6, +8 ………..
So the missing term is = 8 + 6= 14 |
AQUA-RAT | AQUA-RAT-34450 | ### 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 20%. What must be the marked price of each of the article? | [
"$25",
"$31.25",
"$29.65",
"$35.95"
] | B | S.P. of each of the article = 50/2 = $25
Let M.P = $x
80% of x = 25
x = 25*100/80 =$31.25
Answer is B |
AQUA-RAT | AQUA-RAT-34451 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum? | [
"227",
"887",
"585",
"2679"
] | C | 90 = (450*4*R)/100
R = 5%
I = (450*6*5)/100 = 135
450 + 135 = 585
Answer:C |
AQUA-RAT | AQUA-RAT-34452 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water? | [
"2",
"8",
"9",
"6"
] | B | 30 --- 3 DS = 10
? ---- 1
18 ---- 3 US = 6
? ---- 1 M = ?
M = (10 + 6)/2 = 8
Answer: B |
AQUA-RAT | AQUA-RAT-34453 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream? | [
"30/45 hours",
"30/14 hours",
"37/13 hours",
"30/13 hours"
] | D | Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26
= 30/13 hours.
Answer: D |
AQUA-RAT | AQUA-RAT-34454 | 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 300 meter long train crosses a platform in 30 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform? | [
"287m",
"278m",
"350m",
"200m"
] | D | Speed = [300 / 18] m/sec = 50/3 m/sec.
Let the length of the platform be x meters.
Then, x + 300 / 30
= 50/3
3(x + 300)
= 1500 è x
= 200m.
Answer:D |
AQUA-RAT | AQUA-RAT-34455 | Substitute 0.085 for interest rate, $3000 for the amount deposited and$6000 for the amount after t years in the formula, “A=Pert” as,
$6000=$3000e0.085t
Divide both sides by $3000 as,$6000$3000=$3000e0.085t$3000e0.085t=2 Take natural logarithm on both sides as, ln(e0.085t)=ln2 Now, apply the inverse property of lnex as, 0.085t=ln2 Divide both sides by 0.085 as, 0.085t0.085=ln20.085t0.6930.0858.15 years So, the deposited amount will be doubled in approximately 8.15 years when the interest rate is 8.5%. Now, as the amount will become quadruple of the deposited amount after t years, then the amount after t years is 4P, that is, A=4P=4($3000)=\$12000
Substitute 0
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The following is multiple choice question (with options) to answer.
Find the least number of complete years in which a sum of money put out at 15% compound interest will be more than double of itself? | [
"5",
"6",
"7",
"8"
] | A | 5 years
Answer: A |
AQUA-RAT | AQUA-RAT-34456 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
A baker filled with a measuring cup with 2/5 cup water. He poured 1/2 of the water into the batter, and then spilled 1/8 of the water on the floor. How much water will the baker needed to add what is left in the cup to have 50% more than what he started with? | [
"1/8 cup",
"3/8 cup",
"21/40 cup",
"1/2 cup"
] | C | 2/5 is the original water in cup .half in batter.So left is 1/5 out which 1/8 is spilled out.So again left with 3/40.
so 50% more than what he started was =2/5+1/2*(2/5)=3/5
Amount of water needed to add = 3/5 - 3/40=21/40
C |
AQUA-RAT | AQUA-RAT-34457 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
A man walks at a speed of 3 km/hr and runs at a speed of 6 km/hr. How much time will the man require to cover a distance of 10 1/2 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running? | [
"2.62",
"2.56",
"2.5",
"2.8"
] | A | Required time = (5 1/4)/3 + (5 1/4)/6 = 2.625 hours.
Answer:A |
AQUA-RAT | AQUA-RAT-34458 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
do you know how many 2's are there between the terms 112 to 375? | [
"A)156",
"B)157",
"C)158",
"D)159"
] | A | Let us calculate total 2's in the units place. (122, 132, 142 ... 192), (201, 212, 222, ... 292), (302, 312, ... 372) = 8 + 10 + 8 = 26
Total 2's in tenth's place, (120, 121, 122, ..., 129) + (220, 221, ..., 229) + (320, 321, ..., 329) = 30
Total 2's in hundred's place = (200, 201, ... 299) = 100.
Total 2's between 112 and 375 = 26 + 30 + 100 = 156
Answer:A |
AQUA-RAT | AQUA-RAT-34459 | 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.
Today Jim is twice as old as Fred, and Sam is 2 years younger than Fred. Five years ago Jim was 5 times as old as Sam. How old is Jim now? | [
"8",
"12",
"16",
"20"
] | D | We're asked how old Jim is NOW. We're given three facts to work with:
1) Today, Jim is TWICE as old as Fred
2) Today, Sam is 2 years younger than Fred
3) Five years ago, Jim was 5 times as old as Sam.
let's TEST Answer D: 20
IF....Jim is currently 20 years old....
Fred is 10 years old
Sam is 8 years old
5 years ago, Jim was 15 and Sam was 3, so Jim WAS 5 times Sam's age. This is an exact MATCH for what we were told, so this MUST be the answer.
D |
AQUA-RAT | AQUA-RAT-34460 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article offering a discount of 10% and earned a profit of 25%. What would have been the percentage of profit earned if no discount was offered? | [
"39%",
"52%",
"15%",
"21%"
] | A | Let C.P. be $100.
Then, S.P. = $ 125
Let marked price be $ x. Then, 90/100 * x = 125
x = 12500/90 = $ 139
Now, S.P. = $ 139, C.P. = Rs. 100
Profit % = 39%.
Answer: A |
AQUA-RAT | AQUA-RAT-34461 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man purchased some fruits for Rs. 1000. He sold fruits worth 400 at 10% profit. At what profit per cent, must he sell the rest in order to gain 20% on the whole? | [
"26(2/3)%",
"25%",
"30%",
"33(1/3)%"
] | A | Solution: To get 20 % profit on whole,
1000(CP)==20%(gain)==>1200(SP).
Total Profit = 1200 - 1000 = Rs. 200.
400 ==10%(gain)==>440.
He gets Rs. 40 profit on 400 hundred. Rest Profit = 200 - 40 = 160.
Then he must get Rs. 160 as profit on Rs. 600;
Hence, % profit = 160*100/600 = 26.66%.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34462 | solid cylinder has radius of base 14 cm and height 15 cm. cm and height 7 cm. Two cylinder surface area calculators based on radius or diameter. 14 * 25 * 10 = 314. The lateral area of a cylinder is the area of its curved surface, excluding the area of its bases. Formulas for Area and Volume 1) The lateral surface of the right cylinder is the surface without the top and bottom surfaces. 52 cm 2 + 94. Surface of Cylinder: 2π · r · (r + h) Surface of Sphere: 4π · r² Surface of Cone: π · r( r + √(h²+r²) ) Volume of Cuboids, Rectangular Prisms and Cubes. The area of a circular disk is π r 2 so each disk has an area of π × 5 2 = 78. Enjoy the calculator! Buy a comprehensive geometric formulas ebook. The cardboard costs 0. The surface area SA is the area of the ends (which are just circles), plus the area of the side, which is a circle's circumference times the height h of the cylinder: SA cyl = 2π r 2 + 2π rh Depending on the class you're taking, you might also need to know the formula for the volume V of a cone with base radius r and height h :. You'll gain access to interventions, extensions, task implementation guides, and more for this instructional video. The total surface area (TSA) includes the area of the circular top and base, as well as the curved surface area (CSA). Solve advanced problems in Physics, Mathematics and Engineering. Just like ordinary area, the units are the squares of the units of length (that is, if a shape's sides are measured in meters, then the shape's area is measured in square meters). The volume of a cylinder of height 8 cm is 1232 cm³. Find the surface area of the red. Higher ability full lesson (5. The height of the prism is 15 centimeters. The surface area of a capsule can be determined by combining the surface area equations for a sphere and the lateral surface area of a cylinder. 14 x diameter 2 / 2. _ ____ 2) Volume = _27 cubic cm. Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. To find the surface area of
The following is multiple choice question (with options) to answer.
Find the volume , curved surface area and the total surface area of a cylinder with diameter of base 7 cm and height 40 cm. | [
"952cm^2",
"957cm^2",
"954cm^2",
"958cm^2"
] | B | Volume = ∏r2 h = ((22/7)x(7/2)x(7/2)x40) = 1540 cm^3. .
Curved surface area = 2∏rh = (2x(22/7)x(7/2)x40)= 880 cm^2 .
Total surface area = 2∏rh + 2∏r2 = 2∏r (h + r)
= (2 x (22/7) x (7/2) x (40+3.5)) cm2
= 957 cm^2
Answer is B |
AQUA-RAT | AQUA-RAT-34463 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief is noticed by a policeman from a distance of 250 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes? | [
"100",
"277",
"2987",
"150"
] | D | Relative speed of the thief and policeman = 11 - 10 = 1 km/hr.
Distance covered in 6 minutes = 1/60 * 6 = 1/10 km = 100 m.
Distance between the thief and policeman = 250 - 100 = 150 m.
Answer: D |
AQUA-RAT | AQUA-RAT-34464 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Abhishek....
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
Working simultaneously and independently at an identical constant rate, 20 machines of a certain type can produce a total of x units of product P in 4 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 8 days? | [
"30",
"32",
"35",
"38"
] | A | The rate of 20 machines is rate=job/time=x/4 units per day --> the rate of 1 machine 1/20*(x/4)=x/80 units per day;
Now, again as {time}*{combined rate}={job done}
then 8*(m*x/80)=3x --> m=30.
Answer: A. |
AQUA-RAT | AQUA-RAT-34465 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
(6+√3+√5)^2-(√3+√5)^2=? | [
"1+2√3+2√5",
"36+12√3+12√5",
"1-√3+√5",
"1+√3-√5"
] | B | (6+√3+√5)^2-(√3+√5)^2
(6+√3)^2 + 5 +2 (6+√3) (√5) - ( 3 +5 + 2√3√5)
36+ 3+ 12√3 +5 +12 √5 + 2 √15 - 3-5 -2√15
36+12√3+12√5
B is the answer |
AQUA-RAT | AQUA-RAT-34466 | Q 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:
If same pair of coins is tossed at random, find the probability of getting:
SOLUTION:
Number of trials = 500
Number of outcomes of two heads (HH) = 105
Number of outcomes of one head (HT or TH) = 275
Number of outcomes of no head (TT) = 120
(i) Probability of getting two heads = $\frac{Frequency\;of\;getting\;2\;heads}{Total\;No.\;of\;trails}$ = $\frac{105}{500}$ = $\frac{21}{100}$
(ii) Probability of getting one head = $\frac{Frequency\;of\;getting\;1\;heads}{Total\;No.\;of\;trails}$ = $\frac{275}{500}$ = $\frac{11}{20}$
(iii) Probability of getting no head = $\frac{Frequency\;of\;getting\;no\;heads}{Total\;No.\;of\;trails}$ = $\frac{120}{500}$ = $\frac{6}{25}$
#### Practise This Question
The lines shown below never meet at any point. So, these lines are not parallel. Say true or false.
The following is multiple choice question (with options) to answer.
Two unbiased coins are tossed. Find the probability of getting two heads? | [
"1/2",
"1/3",
"1/4",
"2/3"
] | C | S = {HH,HT,TH,TT}
E = Event of getting two heads
E = {TT}
P(E) = 1/4
Answer is C |
AQUA-RAT | AQUA-RAT-34467 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 742, 743, 633, 853, 871, 990, 532 | [
"532",
"990",
"633",
"742"
] | D | Explanation :
In all numbers except 742, the difference of third and first digit is the middle digit.
Answer : Option D |
AQUA-RAT | AQUA-RAT-34468 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
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.
Kudos [?]: 24 [2], given: 0
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
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 monthly salary of 18 employees in an organisation is Rs. 2000. If the manager's salary is added, then the average salary increases by Rs. 200. What is the manager's monthly salary? | [
"Rs.5800",
"Rs.3618",
"Rs.3600",
"Rs.3619"
] | A | Manager's monthly salary
= Rs. (2200 * 19 - 2000 * 18)
=Rs.5800 Answer:A |
AQUA-RAT | AQUA-RAT-34469 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
The area of a square is equal to nine times the area of a rectangle of dimensions 121 cm * 144 cm. What is the perimeter of the square? | [
"1500 cm",
"1800 cm",
"1584 cm",
"1718 cm"
] | C | Area of the square = s * s = 9(121 * 144)
=> s = 3* 11 * 12 = 396 cm
Perimeter of the square = 4 * 396 = 1584 cm.
Answer: C |
AQUA-RAT | AQUA-RAT-34470 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
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• 7 minutes ago
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The following is multiple choice question (with options) to answer.
(.24 x .35 ) / (.14 x .15 x .02) is equal to ? | [
"2",
"20",
"200",
"2000"
] | C | Answer
(.24 x .35) / (.14 x .15 x .02)
= (24 x 35 x 100 ) / (14 x 15 x 2)
=200
Correct Option: C |
AQUA-RAT | AQUA-RAT-34471 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 50 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one? | [
"21",
"88",
"45",
"99"
] | C | Relative speed = 50 + 30 = 80 km/hr.
80 * 5/18 = 200/9 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 9/200 = 45 sec
Answer: C |
AQUA-RAT | AQUA-RAT-34472 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers? | [
"488",
"540",
"552",
"568"
] | C | Sum of n even consecutive integers = n(n+1)
Sum of 8 consecutive even integers that start at some point after n = (n+8)(n+9)
Given => (n+8)(n+9) - n(n+1) = 424
16n +72 = 424 - General equation for sum of consecutive 8 digits where n is where counting starts
Solving we get n = 22 (Where the count starts)
Sum of 8 consecutive integers after n = 22 will be count starting at n1 = 22+8 = 30
So Answer = 16*30 +72 = 552
ANSWER:C |
AQUA-RAT | AQUA-RAT-34473 | 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, candidate A got 70% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid vote polled in favor of candidate. | [
"330000",
"340000",
"347000",
"333200"
] | D | Total number of invalid votes = 15 % of 560000
= 15/100 × 560000
= 8400000/100
= 84000
Total number of valid votes 560000 – 84000 = 476000
Percentage of votes polled in favour of candidate A = 70 %
Therefore, the number of valid votes polled in favour of candidate A = 70 % of 476000
= 70/100 × 476000
= 33320000/100
= 333200
D) |
AQUA-RAT | AQUA-RAT-34474 | Now all you have to do is to choose k so that –17+55k lies between 100 and 200.
Originally Posted by barhin
Question 2
Let X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 194 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)
For this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X – 12 = (multiple of 194). Then add suitable multiples of 194 to find the two smallest positive values for X.
3. Hello, barhin!
$\text{(1) Of all integer pairs }(x,y)\text{ that satisfy the equation: }\: 42x+55y\:=\:1$
. . $\text{only one such pair has }100
. . $\text{What is this value of }x\,?$
The problem can be solved without Euclid's algorithm.
. . But, of course, this solution takes much longer.
We have: . $42x + 55y \:=\:1$
$\text{Then: }\:x \:=\:\frac{1-55y}{42} \:=\:\frac{-42y + 1 - 13y}{42} \quad\Rightarrow\quad x \:=\:-y + \frac{1-13y}{42}$ .[1]
Since $x$ is an integer, $1-13y$ must be a multiple of 42.
. . $1-13y \:=\:42a \quad\Rightarrow\quad y \:=\:\frac{1-42a}{13} \quad\Rightarrow\quad y \:=\:-3a + \frac{1-3a}{13}$ .[2]
Since $y$ in an integer, $1-3a$ must be a multiple of 13.
The following is multiple choice question (with options) to answer.
For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 2835 what is the least possible value of k? | [
"5",
"7",
"9",
"11"
] | C | 2835 = 3*3*3*3*5*7
Thus k must include numbers at least up to the number 9 so that there are at least four appearances of 3 (that is: 3, 6, and 9=3*3).
The answer is C. |
AQUA-RAT | AQUA-RAT-34475 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
On July 1 of last year, total employees at company E was decreased by 10 percent. Without any change in the salaries of the remaining employees, the average (arithmetic mean) employee salary was 10 percent more after the decrease in the number of employees than before the decrease. The total of the combined salaries of all the employees at Company E after July 1 last year was what percent E of thatbeforeJuly 1 last year? | [
"90%",
"99%",
"100%",
"101%"
] | B | the total number of employees = n
the average salary = x
total salary to all emplyoees = xn
after
the total number of employees = n - 0.1n = 0.9n
the average salary = x + 10% of x = 1.1x
total salary to all emplyoees = 0.9n (1.1x)
total salary after as a % of total salary before E = [0.9n (1.1x)]/xn= 0.99 or 99%=B |
AQUA-RAT | AQUA-RAT-34476 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
An exam paper consists of 5 problems, each problem having 3 internal choices. In how many ways can a candidate attempt one or more problems? | [
"1020",
"1023",
"1030",
"1054"
] | B | Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
B |
AQUA-RAT | AQUA-RAT-34477 | 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 300 m long train crosses a platform in 42 sec while it crosses a signal pole in 18 sec. What is the length of the platform? | [
"286 m",
"350 m",
"277 m",
"400 m"
] | D | Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/42 = 50/3
x = 400 m.
Answer:D |
AQUA-RAT | AQUA-RAT-34478 | (A) 9
(B) 12
(C) 18
(D) 24
(E) 27
Weight of 2nd piece = 4 pound
Since the weight is directly proportional to the square of its length., we may write
$$\frac{16}{36^2}$$ = $$\frac{4}{x^2}$$
Solving above, we get x = 18
_________________
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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the [#permalink]
### Show Tags
30 Jun 2017, 04:02
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?
(A) 9
(B) 12
(C) 18
(D) 24
(E) 27
$$20$$ pounds wire is cut into two pieces = $$16$$ pounds $$+$$ $$4$$ pounds
The piece weighs $$16$$ pounds is $$36$$ feet long.
Ratio of weight and length of $$16$$ pounds piece $$= \frac{16}{36^2} = \frac{4 * 4}{36 * 36} = \frac{1}{81}$$
Therefore required ratio of other part would also be $$\frac{1}{81}$$
Ratio $$= \frac{4}{x^2} =$$ $$\frac{1}{81}$$
$$x^2 = 81*4$$ $$=> x = \sqrt{81*4}$$
$$x = 9 * 2 = 18$$
Hence length of $$4$$ pounds wire $$= 18$$
The following is multiple choice question (with options) to answer.
If a 9 ton whale is divided into 2 equal large pieces and 3 equal small pieces, how much does a small piece weigh if all of the small pieces put together weigh as much as one of the large pieces? | [
"1 ton",
"1 1/2 tons",
"2 tons",
"2 1/2 tons"
] | A | A. Each large piece is a third of the whale. Each small piece is a third of the remaining third of the whale. One third of one third is one ninth. One ninth of nine tons is one ton, so the answer is A |
AQUA-RAT | AQUA-RAT-34479 | rd, eg, qg, l1, 80, pj, kk, bv, hk, mc, sw, jh, pn, ty, ck, an, tf, rf, rm, n7, s5, yf, fk, ai, lg, ih, eo, 32, 2x, hh, em,
The following is multiple choice question (with options) to answer.
W, S, Q, M, ? | [
"M",
"K",
"H",
"Y"
] | B | Explanation:
No explanation is available for this question!
ANSWER: B |
AQUA-RAT | AQUA-RAT-34480 | # Find the least next N-digit number with the same sum of digits.
Given a number of N-digits A, I want to find the next least N-digit number B having the same sum of digits as A, if such a number exists. The original number A can start with a 0. For ex: A-> 111 then B-> 120, A->09999 B-> 18999, A->999 then B-> doesn't exist.
You get the required number by adding $9$, $90$, $900$ etc to $A$, depending on the digits of $A$.
First Case If $A$ does not end in a row of $9$s find the first (starting at the units end) non-zero digit. Write a $9$ under that digit and $0$s under all digits to the right of it. Add the two and you get $B$.
Example: $A=3450$. The first non-zero digit is the 5 so we write a $9$ under that and a $0$ to its right and add:
\begin{align} 3450\\ 90\\ \hline 3540 \end{align}
There is a problem if the digit to the left of the chosen non-zero digit is a $9$. In this case we write a $9$ under that $9$ and $0$s to its right. And if there are several $9$ we put our $9$ under the highest one.
Example: $A=3950$. The first non-zero digit is the 5 but there is a $9$ to its left. We write a $9$ under that $9$ instead and $0$s to its right and add:
\begin{align} 3950\\ 900\\ \hline 4850 \end{align}
Second case If $A$ does end in a row of $9$s write a $9$ under the highest of the row of $9$s and $0$s under all digits to the right of it. Add the two and you get $B$. As you say, if $A$ is entirely $9$s there is no solution.
The following is multiple choice question (with options) to answer.
The sum of digits of a two digit number is 12,The difference between the digits is 6. Find the number | [
"85",
"93",
"83",
"72"
] | B | Description :
=> x+y=12, x-y=6
Adding these 2x =18
=> x=9, y=3.
Thus the number is 93
Answer B |
AQUA-RAT | AQUA-RAT-34481 | (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.
ax ± b = c. All problems like the following lead eventually to an equation in that simple form.
Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse. How much was the blouse? | [
"28",
"26",
"24",
"22"
] | A | Every word problem has an unknown number. In this problem, it is the price of the blouse. Always let x represent the unknown number. That is, let x answer the question.
Let x, then, be how much she spent for the blouse. The problem states that "This" -- that is, $42 -- was $14 less than two times x.
Here is the equation:
2x − 14 = 42.
2x=42+14
=56.
x=56/2
= 28
The blouse cost $28.
Answer is A. |
AQUA-RAT | AQUA-RAT-34482 | energy, temperature, dimensional-analysis, si-units, metrology
Title: Why is there a separate SI unit for temperature? Assuming I got the below facts correct:
There are 7 SI units - mass (kg), length (m), time (s), current (A), temperature (K), luminous intensity (cd), amount of substance (mol).
Thermal energy is essentially nothing but kinetic energy.
Temperature is a measure of thermal energy (hence measure of kinetic energy).
Kinetic energy can be expressed in $[M^1 L^2 T^{-2}]$ (a function of mass, length and time).
Per these facts, is kelvin (SI unit for temperature) also a function of mass, length and time. If so, why do we need a separate SI unit specially for temperature? Am I missing something? Good question. In any system of units the proportionality factor between the unit of temperature and the unit of energy is the Boltzmann constant. If we set the Boltzmann constant to $1$ - as it is in the system of Planck units - then in effect we are measuring temperature and energy in the same units.
However, using Planck units in everyday life would create practical difficulties, because the Planck temperature is so very large - the temperature of a cup of tea is approximately $2 \times 10^{-30}$ Planck units. You can think of this as being a measure of how small atoms are - an everyday object contains a large number of atoms, so its average thermal energy per atom is very small compared to its macroscopic kinetic energy, potential energy etc. So to produce a set of units in which both typical macroscopic energies and temperatures have reasonable values (not too large, not too small) you need to set the Boltzmann constant to a very small value.
The following is multiple choice question (with options) to answer.
The daily high temperatures were recorded at an Antarctic weather station. If a temperature of -38.2 degrees Celsius is 2 units of standard deviation below the mean temperature, and a temperature of -22.6 degrees Celsius is 4 units of standard deviation above the mean, which of the following temperatures is 5 unit of standard deviation above the mean? | [
"-35.6 degrees Celsius",
"-20.0 degrees Celsius",
"-30.4 degrees Celsius",
"-27.8 degrees Celsius"
] | B | Let us suppose mean = M and SD = d.
So, we are given M - 2d = -38.2
and M + 4d = -22.6
solving both the equations, we will get M = -33 an d = 2.6
So, we need to find out M + 5d = -33 +5* 2.6 = -20. Hence, B |
AQUA-RAT | AQUA-RAT-34483 | Case 4: One digit is used four times, while two other digits are used once each.
Subcase 1: The leading digit is repeated.
We have nine ways of choosing the leading digit and $\binom{5}{3}$ ways of choosing the other three positions in which it appears. We have nine choices for the leftmost open position and eight choices for the remaining position. $$9 \cdot \binom{5}{3} \cdot 9 \cdot 8 = 6480$$
Subcase 2: The leading digit is not repeated.
We have nine ways of choosing the leading digit. We have nine ways of choosing the repeated digit and $\binom{5}{4}$ ways of selecting four of the five open positions in which to place it. We have eight ways of filling the remaining open position. $$9 \cdot 9 \cdot \binom{5}{4} \cdot 8 = 3240$$
That gives a total of $$9 + 405 + 81 + 810 + 405 + 6480 + 3240 = 11,430$$ excluded cases.
Hence, there are $$900,000 - 11,430 = 888,570$$ six-digit positive integers in which no digit appears more than three times.
This is neatly handled using exponential generating functions. Assuming first that we are allowed to have 0 as the first digit (e.g. we're talking about license plates or lock combinations): each of $10$ digits can occur up to $3$ times, and the order of the symbols matters. The answer is $$\left[\frac{x^6}{6!}\right] \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^{10} = 987,300.$$
The following is multiple choice question (with options) to answer.
A license plate in the country Kerrania consists of four digits followed by two letters. The letters A, B, and C are used only by government vehicles while the letters D through Z are used by non-government vehicles. Kerrania's intelligence agency has recently captured a message from the country Gonzalia indicating that an electronic transmitter has been installed in a Kerrania government vehicle with a license plate starting with 79. If it takes the police 15 minutes to inspect each vehicle, what is the probability that the police will find the transmitter within three hours? | [
"18/79",
"1/75",
"1/25",
"1/50"
] | B | If it takes 15 minutes to inspect one vehicle, the # of vehicles that can be inspected in 3 hours (180 minutes) = 180/15 = 12. Hence, for calculating the probability that the police will find the transmitter within three hours, the favorable cases = 12. Now, we need to figure out the total # of cases.
The total # of cases = Total # of such cars possible. The details given about the car is that it starts with 79, which leaves 2 more digits, both of which can be filled by all 10 numbers (0-9). In addition, we have 3 letters, each of which can be filled by any from the set {A,B,C}. Hence the total # of such cars possible = 10*10*3*3 = 900
So, the probability that the police will find the transmitter within three hours = 12/900 = 1/75. Option B |
AQUA-RAT | AQUA-RAT-34484 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
Let y = 2g + x^2 and g = 4x + 1.
If g^2 = 289, then which of the following can be a value of 2y + 3g? | [
"100",
"130",
"141",
"151"
] | D | 800score Official Solution:
First note that g can either be 17 or -17. We are going to have to use both to find all solutions before we can choose a final answer. We begin by substituting g = 17 and then g = -17. If g = 17, then x = 4. If g = -17, then x = -18/4. Since all of our solutions are whole numbers, we can dismiss -18/4. We use g = 17 and x = 4 to determine the value of y: y = 2(17) + 16 =50. Finally, we substitute the values for y and g into the last expression to determine its value : 2(50) + 3(17) = 100 + 51 = 151;the correct answer is (D). |
AQUA-RAT | AQUA-RAT-34485 | ### Show Tags
04 Aug 2019, 03:54
generis wrote:
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?
A. 4
B. 1
C. 0
D. -1
E. -4
Don't remove brackets and solve. The question asks for the logic behind absolute value.
$$y$$ is a positive integer
$$|x| < 5 − y$$
LHS is nonnegative - it is positive or 0.
Least value for x is ZERO
Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.
The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0
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Why can't x be a negative integer, in that case it should be -1 i.e. option D.
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
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08 Aug 2019, 18:54
I don't understand why x can't be negative
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
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The following is multiple choice question (with options) to answer.
For any number y, y* is defined as the greatest positive even integer less than or equal to y. What is the value of 6.2 – 6.2*? | [
"0.2",
"1.2",
"1.8",
"2.2"
] | D | Since y* is defined as the greatest positive even integer less than or equal to y, then 6.2* = 4 (the greatest positive even integer less than or equal to 6.2 is 4).
Hence, 6.2 – 6.2* = 6.2 - 4 = 2.2
Answer: D. |
AQUA-RAT | AQUA-RAT-34486 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population P ; grows as a linear function of time t, where t is the number of years after 2003.
a) How many students will be enrolled in the school in 2010? | [
"1800",
"1900",
"1700",
"1500"
] | B | a) The given information may be written as ordered pairs (t , P). The year 2003 correspond to t = 0 and the year 2006 corresponds to t = 3, hence the 2 ordered pairs
(0, 1200) and (3, 1500)
Since the population grows linearly with the time t, we use the two ordered pairs to find the slope m of the graph of P as follows
m = (1500 - 1200) / (6 - 3) = 100 students / year
The slope m = 100 means that the students population grows by 100 students every year. From 2003 to 2010 there are 7 years and the students population in 2010 will be
P(2010) = P(2003) + 7 * 100 = 1200 + 700 = 1900 students
Answer B |
AQUA-RAT | AQUA-RAT-34487 | # 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 selling price is doubled, the profit triples. Find the profit percent. | [
"200/3",
"100",
"316/3",
"120"
] | B | Explanation :
Let the C.P. be x and the S.P. be y
So the profit is (y - x) ---------------------------- (1)
Now, the S.P. is doubled. So the new S.P. = 2y
The new profit = (2y - x)
Given that when S.P. is doubled, profit increases 3 times
=> New profit = 3 * old profit
=> (2y - x) = 3(y - x)
=> y = 2x
So, the profit = (y - x) = (2x - x) = x
% profit = (x/x∗100)% = 100%
Answer : B |
AQUA-RAT | AQUA-RAT-34488 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 20 boys and 15 girls are present in a class. If 5 student will chosen for a competition an at least two students must be girl? | [
"309128",
"320000",
"251230",
"301000"
] | A | 15 girls and 20 boys. Five students needed
It is always better to subtract the total from none in case of at least one problems.
Total ways = 35C5 = 324632
Total ways without any girl = 20C5 = 15504
Hence ways in which at least one women will be present = 324632 - 15504 = 309128
Correct option: A |
AQUA-RAT | AQUA-RAT-34489 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
Find avrg speed if a man travels at speed of 32km/hr up and 48km/hr dawn at an altitude of 230m. | [
"25.8",
"26.8",
"38.4",
"34.4"
] | C | avg speed=2*x*y/(x+y)
=2*32*48(32+48)=38.4
ANSWER:C |
AQUA-RAT | AQUA-RAT-34490 | homework-and-exercises, electric-circuits, electric-current, electrical-resistance, batteries
Title: What happens when the length of the uniform wire is more than or less than 54 cm? I understand how to solve this (when $V_{BC}$ is equal to the emf of the second cell then the current through the second cell is zero). However, I'm curious what happens when the length BC is more than or less than 54 cm? Would current flow through the branch? In what direction?
The following is multiple choice question (with options) to answer.
22 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be : | [
"84 m",
"28 m",
"120 m",
"137 m"
] | B | Sol.
Let the length of the wire b h.
Radius = 1/2 mm = 1/20 cm. Then,
22/7 * 1/20 * 1/20 * h = 22 ⇔ = [22 * 20 * 20 * 7 / 22] = 2800cm = 28 m.
Answer B |
AQUA-RAT | AQUA-RAT-34491 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
A farmer has three different crops on his farm; cabbage, carrots, and parsnips, where 10 percent of the crops are cabbage and 35 percent are carrots. If the farm produces 650 pounds of parsnips, and the weight of all crops are evenly distributed, how many pounds of crops are produced in total? | [
" 6,500 pounds",
" 2,000 pounds",
" 10 pounds",
" 1,000 pounds"
] | D | Let the total pounds of crops be x
% of cabbage = 10%
% of carrots = 35%
% of parsnips = 100% - 45% = 65%
But this number is given as 500 pounds
so 65% of x = 650 and x = (650)(100/65)
x = 1,000
Therefore there a total of 1,000 pounds of crops.
Correct answer - D |
AQUA-RAT | AQUA-RAT-34492 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
How many times are the hands of a clock at right angle in 2 weeks? | [
"616",
"611",
"661",
"116"
] | A | In 1 day, they are at right angles 44 times.
In 14 days, they are at right angles 616 times. Answer: Option A |
AQUA-RAT | AQUA-RAT-34493 | java, algorithm, knapsack-problem
Title: Stampcalculator - Given a set of stamps, what combinations are there to reach a certain amount? Background
My mother has a hobby of buying and reselling books via online trading sites. After a price is agreed on, the books have to be put into an envelope for mailing. On this envelope, stamps have to be applied to pay for postage (mailing fee). My mother buys stamps in bulk at various prices, varying from 75 to 90% of face value. As the postage prices change every year, new stamp combinations have to be calculated that will satisfy the postage fee with a minimum amount of hassle.
My mother is not all that good with calculations. And having to manually recalculate what stamp combinations are best to get to the postage required for a 250 - 500 gram package takes a lot of time. Therefore, she calculates a few options via trial and error - sometimes overshooting the postage amount by a few cents - and just uses those.
Some stamps are preferred to be used, whereas others aren't, due to the cost not always being the face value of a stamp. Asking my mother about the stamp prices individually would have taken too much time, so I just made a program that generates a list of options. It requires recompilation when you need to add different stamps or want to get to a different limit, but that's okay. The whole point of this was to not spend too much time on it (the goal is to save time!) - which is why I wrote it in the total time of 1 hour (from idea to solution).
Problem description
Given a target amount, a set of stamp denominations, and a maximum amount of usable stamps, print a list of stamp combinations that will get to the target amount.
Programmed in 1 hour. I have placed everything in one file to ensure that it can run in Ideone, so that she could theoretically make the changes herself, if needed.
How it works
Using a list of stamps, make partial combinations that are at or below the target amount. For each combination, make a new combinations via duplicating the current combination and adding a stamp with a value equal to or below the last added stamp. Combinations that go past the goal are discarded. Combinations that meet the goal are added to a solutions list.
The following is multiple choice question (with options) to answer.
The number of stamps that P and Q had were in the ratio of 7:2 respectively. After P gave Q 15 stamps, the ratio of the number of P's stamps to the number of Q's stamps was 2:1. As a result of the gift, P had how many more stamps than Q? | [
"25",
"35",
"45",
"55"
] | C | P started with 7k stamps and Q started with 2k stamps.
(7k-15)/(2k+15) = 2/1
3k = 45
k = 15
P has 7(15) - 15 = 90 stamps and Q has 2(15)+15 = 45 stamps.
The answer is C. |
AQUA-RAT | AQUA-RAT-34494 | The area of the square in both the cases $= 1^{2} = 1$
Now, we can find the areas of unshaded regions.
• $\text{Case M:}$
• The area of unshaded regions $= 1-4\pi\text{R}^{2} = 1 – 4\pi \left(\frac{1}{4}\right)^{2} = 1 – \frac{\pi}{4}$
• $\text{Case N:}$
• The area of unshaded regions $= 1-9\pi\text{r}^{2} = 1-9\pi \left(\frac{1}{6}\right)^{2} = 1 – \frac{\pi}{4}$
$\therefore$ The ratio of the areas of unshaded regions of $\text{case M}$ to that of $\text{case N}$ is $1:1.$
Correct Answer $:\text{B}$
${\color{Magenta}{\textbf{PS:}}}$
• ${\color{Green}{\text{The area of square} = \text{(side)}^{2}}}$
• ${\color{Lime}{\text{The area of circle} = \pi \times (\text{radius})^{2}}}$
18.0k points 4 7 16
4 7 16
The following is multiple choice question (with options) to answer.
If the radii of umbra and penumbra cast by an object on a wall are of the ratio 2:6, what is the area of the penumbra ring around the umbra of the latter’s radius is 40 cms? | [
"40288.57cm^2",
"40388.57cm^2",
"40488.57cm^2",
"40588.57cm^2"
] | A | et the radius of umbra and penumbra are 2k and 6k.
Then as given radius of umbra=40cm
so 2k=40
k=20
radius of penumbra=20*6=120
area of penumbra ring around the umbra=area of penumbra-area of umbra
22/7*[(120)^2-(40)^2]
=40288.57cm^2
ANSWER:A |
AQUA-RAT | AQUA-RAT-34495 | ### 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 start a business each investing 20,000. After 5 months A withdrew 5000, B withdrew 4000 and C invests 6000 more. At the end of the year, a total profit of 69,900 was recorded. Find the share of B. | [
"20,000",
"21,200",
"28,200",
"20,500"
] | C | Ratio of the capitals of A, B and C
= 20000 × 5 + 15000 × 7 : 20000 × 5 + 16000 × 7 : 20000 × 5 + 26000 × 7
= 205000 : 212000 : 282000 = 205 : 212 : 282.
B’s share = (69900 × 212⁄699) = 21200;
Answer C |
AQUA-RAT | AQUA-RAT-34496 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
There are 36 carpenters in a crew. On a certain day, 29 were present. What percent showed up for work? (round to the nearest tenth) | [
"50.6%",
"60.6%",
"70.6%",
"80.6%"
] | D | x/100=29/36
Multiply the opposites:
29 x 100 = 2900
Divide by the remaining number:
80.55/2900.00
80.6%
correct answer D |
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