source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34197 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
Robert's salary was decreased by 30% and subsequently increased by 30%. how much percentage does he lose? | [
"10%",
"9%",
"25%",
"30%"
] | B | let original salary be $100
Salary after decreasing 30% = 100 - 100 x 30/100 = $70
Salary after increasing 30% on $70 = 70 + 70 x 30/100 = $91
Percentage of loss = 100 - 91 = 9%
Answer : B |
AQUA-RAT | AQUA-RAT-34198 | meteorology, atmosphere, temperature
So, the temperature is falling over night after the sunset, but rises again after the sunrise. Thus, the temperature is at its lowest point in the morning.
Appendix for all you loving calculations:
Note: Simplified to toy model, no atmosphere
The cooling rate of the Earth is approximatelly given by the Stefan-Boltzmann equation:
$$j_E=\sigma\cdot T^4=5.670 \cdot 10^{-8} \frac{W}{m^2 K^4} \cdot (288.15 K)^4 = 390 \frac{W}{m^2}$$
The maximum heating rate of Sun in the zenith is $j_{\text{S max}}=1361 \frac{W}{m^2}$. So, the heating rate of Sun at altitude $\alpha$ is:
$$j_S=j_{\text{S max}}\cdot \sin{\alpha}=1361 \frac{W}{m^2}\cdot \sin{\alpha}$$
When is the heating rate equal to zero?
$$0=j_S-j_E=1361 \frac{W}{m^2}\cdot \sin{\alpha} - 390 \frac{W}{m^2}$$
$$1361 \frac{W}{m^2}\cdot \sin{\alpha} = 390 \frac{W}{m^2}$$
$$\alpha = 17 °$$
So, with our calculations, the minimum temperature is at the time when the altitude is equal to 17°.
The following is multiple choice question (with options) to answer.
The average of temperatures at noontime from Monday to Friday is 45; the lowest one is 42, what is the possible maximum range of the temperatures? | [
"15",
"25",
"40",
"45"
] | A | there are 5 days so the sum of temperature can be 45*5=225
lowest is 42. to find the maximum range we can say the temperature was the lowest for 4 of the 5 days
so 4*42=168.
on the fifth day it is 225-168=57
range is therefore 57-42=15
answer A |
AQUA-RAT | AQUA-RAT-34199 | python, python-3.x, calculator
Title: Taxi fare calculator, trip recorder, and fuel calculator I solved this practice problem in preparation for school exams.
The owner of a taxi company wants a system that calculates how much money his taxis take in one day.
Write and test a program for the owner.
Your program must include appropriate prompts for the entry of data.
Error messages and other output need to be set out clearly and understandably.
All variables, constants and other identifiers must have meaningful names.
You will need to complete these three tasks. Each task must be fully tested.
TASK 1 – calculate the money owed for one trip
The cost of a trip in a taxi is calculated based on the numbers of KMs traveled and the type of taxi that you are traveling in. The taxi company has three different types of taxi available:
a saloon car, that seats up to 4,
a people carrier, that seats up 8,
a mini-van, that seats up to 12.
For a trip in a saloon car the base amount is RM2.50 and then a charge of RM1.00 per KM. For a trip in a people carrier the base amount is RM4.00 and a charge of RM1.25 per KM. For a trip in a mini- van the base amount is RM5.00 and a charge of RM1.50 per KM. The minimum trip length is 3KM and the maximum trip length is 50KM. Once the trip is complete a 6% service tax is added.
TASK 2 – record what happens in a day
The owner of the taxi company wants to keep a record of the journeys done by each one of his taxis in a day. For each one of his three taxis record the length of each trip and the number of people carried. Your program should store a maximum of 24 trips or 350km worth of trips, whichever comes first. Your program should be able to output a list of the jobs done by each one of the three taxis.
TASK 3 – calculate the money taken for all the taxis at the end of the day.
At the end of the day use the data stored for each taxi to calculate the total amount of money taken and the total number of people carried by each one of the three taxis. Using the average price of RM2.79 per litre use the information in the table below to calculate the fuel cost for each taxi:
The following is multiple choice question (with options) to answer.
A certain taxi company charges $2.50 for the first 1/5 of a mile plus $0.40 for each additional 1/5 of a mile. What would this company charge for a taxi ride that was 8 miles long? | [
"15.6",
"18.1",
"17.5",
"18.7"
] | B | A certain taxi company charges $2.50 for the first 1/5 of a mile plus $0.40 for each additional 1/5 of a mile. What would this company charge for a taxi ride that was 8 miles long?
A. 15.60
B. 16.00
C. 17.50
D. 18.70
E. 19.10
1/5 miles = 0.2 miles.
The cost of 8 miles long ride would be $2.50 for the first 0.2 miles plus (8-0.2)/0.2*0.4 = $2.50 + $15.6 = $18.1.
Answer: B. |
AQUA-RAT | AQUA-RAT-34200 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A man misses a bus by 40 minutes if he travels at 30 kmph. If he travels at 40 kmph, then also he misses the bus by 10 minutes. What is the minimum speed required to catch the bus on time? | [
"43 kmph",
"45 kmph",
"47 kmph",
"49 kmph"
] | B | B
Let the distance to be travelled to catch the bus be x km
x/30 - x/40 = 30/60 => (4x - 3x)/120 = 1/2 => x = 60 km
By traavelling 30 kmph time taken = 60/30 = 2 hours
By taking 2 hours, he is late by 40 min. So, he has to cover 60 km in at most speed = 60/(4/3) = 45 kmph. |
AQUA-RAT | AQUA-RAT-34201 | +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 be the product of (25-1)*(25-2)*(25-3)*..............*(25-49)*(25-50)? | [
"0",
"1",
"625",
"less than -100,000"
] | A | One of the terms is (25-25) so the product is 0.
The answer is A. |
AQUA-RAT | AQUA-RAT-34202 | Taking 1 and subtracting all parts of the equation from it gives:
$1-p_{n} > 1-0.9999... \ge 0$
Then, we observe that:
$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}$
and hence
$\frac{1}{10^n} > 1-0.9999... \ge 0$.
But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that
$1-0.9999... = 0$
and therefore that
$0.9999... = 1$.
What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1.
Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.
This entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.
### 94 Responses to Q: Is 0.9999… repeating really equal to 1?
1. mathman says:
lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”
The following is multiple choice question (with options) to answer.
0.9999+0.1111=? | [
"1",
"1.0001",
"1.0021",
"1.111"
] | D | 0.9999 + 0.1111
= 0.9999 + 0.111 + 0.0001
= (0.9999 + 0.0001) + 0.111
= 1 + 0.111
= 1.111
D |
AQUA-RAT | AQUA-RAT-34203 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
Why have we done 4,5 and 6 selection?What i can discern from the question is different pairings form different teams. So if AB is paired once whether in row 1 or 2, its pairing is done.
Please explain what is wrong in my interpretation of the question?
Math Expert
Joined: 02 Aug 2009
Posts: 7755
Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
### Show Tags
07 Oct 2017, 06:13
1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
Hi..
let's do the Q step by step and understand what is the TRICK in it
STEP I;-
make teams of 4
choosing 4 out of 8 is 8C4=$$\frac{8!}{4!4!}$$
TIME saver- dont simplify as more steps are involved
STEP II :-
Make different pairs in 4
choosing 2 out of 4 = 4C2 = $$\frac{4!}{2!2!}$$
The TRICKY part - when we choose two out of 4, the other 2 are already there as SECOND pair. BUT 4C2 counts them as 2 separate ways
so divide by 2!
so $$\frac{4!}{2!2!*2}$$
Also it is also valid for the second set of 4..
The following is multiple choice question (with options) to answer.
A pet store holds cats and dogs. If the difference between the number of cats and the number of dogs is 3. What could be the ratio of Cats to dogs in the pet store? | [
"12:15",
"1:4",
"1:5",
"2:5"
] | A | Say theratio of cats to dogs is a/b. Then the numberof cats would be ax and the number of dogs bx, for some positive integer x.
We are told that ax - bx = 3 --> x(a - b) = 3. Since 3 is a prime number it could be broken into the product of two positive multiples only in one way: x(a - b) = 1*13.
The above implies that either x = 1 and (a - b) = 3 or x = 3 and (a - b )= 1.
Therefore the correct answer should have the difference between numerator and denominator equal to 1 or 13.
For the original question only option which fits is E, 4:5. Cats = 3*4 = 12 and dogs = 3*5 = 15.
Answer: A. |
AQUA-RAT | AQUA-RAT-34204 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
How many ways could five people sit at a table with eight seats in which three of the eight seats will remain empty? | [
"6720",
"6520",
"56",
"120"
] | A | ways in which 5 seats out 8 can be selected = 8C5
ways in which 5 people can be arranged in 5 seats = 5!
Total ways of arrangement = 8C5*5!
=(8!/(5!*3!))*5!
=56*120
=6720
Ans = A |
AQUA-RAT | AQUA-RAT-34205 | javascript
Title: More efficient version of an ID calculator in JavaScript The following function takes two numbers that are linked with a "user" and calculates an ID number based on that. I have been trying to make this as clean as possible, and would like some advice on how to make this more efficient. an example of the input would be "12195491" for the num and "3120" for the ts, which would output "8511"
function getidnumber(num, ts) {
num = num.substr(4, 4);
ts = ((ts == undefined) ? "3452" : (ts));
var _local5 = "";
var _local1 = 0;
while (_local1 < num.length) {
var _local4 = Number(num.substr(_local1, 1));
var _local3 = Number(ts.substr(_local1, 1));
var _local2 = String(_local4 + _local3);
_local5 = _local5 + _local2.substr(_local2.length - 1);
_local1++;
}
return("@user" + _local5);
}; Here's a better implementation.
Used the unary + operator for number conversion.
Used null instead of undefined since it's shorter and produces the same result because of type coersion.
Avoid performing a substring operation by starting to iterate from index 4.
Cached num.length into len so that we save on property lookups when the condition is evaluated for every loop iteration.
Removed uneeded parenthesis.
Took advantage of the += operator.
Made sure that every variables were locally scoped. The _local1 variable in the selected answer isin't properly scoped.
Used a single var statement; it's a better practice to declare variables at the top of the function for readability.
Stole the % 10 idea from the other answer since I thought it was great ;)
function getidnumber (num, ts) {
var i = 4,
len = num.length,
res = '@user';
ts = ts == null? '3452' : ts;
The following is multiple choice question (with options) to answer.
What is the difference between local value & face value of 7 in the numeral 65793? | [
"693",
"656",
"691",
"9890"
] | A | (Local value of 7) - (Face value of 7)
= (700 - 7) = 693
A |
AQUA-RAT | AQUA-RAT-34206 | # how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
The following is multiple choice question (with options) to answer.
A cube of volume 1000 cm 3 is divided into small cubes of 1 cm 3 ..& all the outer surfaces are painted how many cubes will be there with (at least one side painted or no side painted) | [
"477",
"488",
"456",
"425"
] | B | Number of cubes with No side painted = 8*8*8 = 512
Number of cubes with atleast one side painted = 1000-512= 488
ANSWER:B |
AQUA-RAT | AQUA-RAT-34207 | Alice speaks the truth with probability 3/4 and Bob speaks the truth with probability 2/3. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is 3 while Bob says (to Carl) the number is 1. Find the probability that the number is actually 1.
UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ({1,2,⋯,6} - {The number that actually showed up}). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.
My attempt:
Case 1: Number 1 showed up.
Chance of all this happening = $\frac{1}{6}\cdot \frac{2}{3}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)=\frac{1}{180}$
Case 2: Number 3 showed up
Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{120}$
Case 3: Other number showed up
Chance of all this happening = $\frac{4}{6}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{450}$
So, total = $\overline{)\frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}$
Is my attempt correct? If not, how to do this problem?
You can still ask an expert for help
## Want to know more about Probability?
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
The following is multiple choice question (with options) to answer.
If A speaks the truth 60% of the times, B speaks the truth 50% of the times. What is the probability that at least one will tell the truth | [
"0.8",
"0.9",
"1.0",
"1.2"
] | A | probability of A speaks truth p(A)=6/10;false=4/10
probability of B speaks truth p(B)=5/10;false=5/10.For given qtn
Ans=1-(neither of them tell truth).Because A & B are independent events
=1-[(4/10)*(5/10)]=4/5=0.8
ANSWER:A |
AQUA-RAT | AQUA-RAT-34208 | astrophotography
if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
as if light years are a measure of time. A light year is a measure of distance, the distance light travels in a year in a vacuum.
The following is multiple choice question (with options) to answer.
The distance that light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100000 years is: | [
"111 × 1827 miles",
"999× 1238 miles",
"346 × 1012 miles",
"587 x 10^15 miles"
] | D | The distance of the light travels in one years is:
5,870,000,000,000 = 587 * 10 ^10
The distance of the light travels in 100000 years is:
= 587 * 10^10 x 10 ^5 = 587 x 10^15 miles
Answer :D |
AQUA-RAT | AQUA-RAT-34209 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
_________________
Please press +1 Kudos if this post helps.
Intern
Joined: 16 May 2017
Posts: 14
GPA: 3.8
WE: Medicine and Health (Health Care)
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 151 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A? | [
"1289",
"1198",
"281",
"1208"
] | D | Explanation:
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 151 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 151 = 3k + 151 => 2k = 302 => k = 151
The initial quantity of water in A = 8k = 8 * 151 = 1208 liters.
Answer: Option D |
AQUA-RAT | AQUA-RAT-34210 | ### 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.
John bought 2 shares and sold them for $24 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had | [
"a profit of $10",
"a profit of $2",
"a loss of $2",
"a loss of $10"
] | C | Loss% = (%age Profit or loss / 10)^2 = (20/10)^2 = 4% loss
Total Selling Price = 24*2 = $48
Total Cost Price = 48/(0.96) = $50
Loss = 50-48 = $2
Answer: Option C |
AQUA-RAT | AQUA-RAT-34211 | # Exponential Distribution and Expected Times
Suppose two people A,B are assigned to do an individual task and then a group task.
Person A completes his individual task on average around 30 minutes. Person B completes his individual task on average around 40 minutes. After a person completes his individual task, he will move on to the group task. The group task can be completed alone by either A or B on average around 1 hour. The group task can be completed together on average around 30 minutes.
What is the expected time to finish all tasks?
My attempt: I know that it should be a max time it takes for the individual tasks to be done + additional time to finish group task.
I would have
E(max(A,B)) + E(T) where A is time it takes A for his own task and B is time it takes for his own task. The trouble I am having is conditioning on T, which is additional time to do the group task after all individual tasks have been finished.
I know that if can be broken up into two cases and further into two mini-cases:
Case 1: A>B Our time will be A if A > B + T. Our time will be A + T if A < B + T.
Case 2: B>A Our time will be B if A > B + T. Our time will be B + T if A < B + T.
I am not sure how to represent E(T) from here. Intuitively, seems like E(T) could be (0 + Average Time it Takes to Complete Group Task Together) divided by 2, so E(T) = 15 minutes.
Hence my answer would be E(max(A,B)) + E(T) = E(A+B-min(A,B)) + E(T) = 30 + 40 - 120/7 + 15 = 475/7 minutes.
Is my reasoning correct?
It depends on the distribution of times. You have used the words "Exponential distribution" in the title, so let's suppose that the times are distributed that way and use the memoryless property.
Person A works at a rate of $2$ per hour and person B at a rate of $1.5$ per hour on their individual tasks. Individually they work on the group task at a rate of $1$ per hour and together at a rate of $2$ per hour.
The following is multiple choice question (with options) to answer.
Dan can do a job alone in 12 hours. Annie, working alone, can do the same job in just 9 hours. If Dan works alone for 8 hours and then stops, how many hours will it take Annie, working alone, to complete the job? | [
"2",
"3",
"4",
"5"
] | B | Dan can complete 1/12 of the job per hour.
In 8 hours, Dan completes 8(1/12) = 2/3 of the job.
Annie can complete 1/9 of the job per hour.
To complete the job, Annie will take 1/3 / 1/9 = 3 hours.
The answer is B. |
AQUA-RAT | AQUA-RAT-34212 | # Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$
Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $$\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$$ where $$r-\frac{s}{2}$$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $$\frac{\pi r^2}. {2}$$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($$2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$$). In conclusion the resultant equation is (with $$r=x$$ and $$s=y$$):
$$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$$
The following is multiple choice question (with options) to answer.
The ratio of the radius of two circles is 1: 5, and then the ratio of their areas is? | [
"1:7",
"2:9",
"1:25",
"3:7"
] | C | r1: r2 = 1: 5
Î r1^2: Î r2^2
r1^2: r2^2 = 1:25
Answer: C |
AQUA-RAT | AQUA-RAT-34213 | Ex. 1.2 | Q 20 | Page 28
105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?
Ex. 1.2 | Q 21 | Page 28
The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Ex. 1.2 | Q 22 | Page 28
Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
## RD Sharma solutions for Class 10th Board Exam Mathematics chapter 1 - Real Numbers
RD Sharma solutions for Class 10th Board Exam Maths chapter 1 (Real Numbers) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE 10 Mathematics solutions in a manner that help students grasp basic concepts better and faster.
Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10th Board Exam Mathematics chapter 1 Real Numbers are Revisiting Irrational Numbers, Euclid’s Division Lemma, Fundamental Theorem of Arithmetic, Fundamental Theorem of Arithmetic Motivating Through Examples, Proofs of Irrationality, Revisiting Rational Numbers and Their Decimal Expansions, Introduction of Real Numbers, Real Numbers Examples and Solutions.
Using RD Sharma Class 10th Board Exam solutions Real Numbers exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10th Board Exam prefer RD Sharma Textbook Solutions to score more in exam.
The following is multiple choice question (with options) to answer.
10 camels cost as much as 24 horses, 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants. If the cost of 10 elephants is Rs.120000, find the cost of a camel? | [
"s.9800",
"s.3800",
"s.9800",
"s.4800"
] | D | Cost of the camel = P
10 camels = 24 horses
16 horses = 4 oxen
6 oxen = 4 elephants
10 elephants = Rs.120000
P = Rs.[(24 * 4 * 4 * 120000)/(10 * 16 * 6 * 10)]
P = Rs.(46080000/9600) => P = Rs.4800
Answer:D |
AQUA-RAT | AQUA-RAT-34214 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
A certain car uses 12 gallons of gasoline in traveling 360 miles. In order for the car to travel the same distance using 10 gallons of gasoline, by how many miles per gallon must the car’s gas mileage be increased? | [
"2",
"4",
"6",
"8"
] | C | 360/10 = 36. The difference is 36 - 30 = 6.
Answer C |
AQUA-RAT | AQUA-RAT-34215 | evolution, zoology, anatomy, species
Title: Examples of animals with 12-28 legs? Many commonly known animals' limbs usually number between 0 and 10. For example, a non-exhaustive list:
snakes have 0
Members of Bipedidae have 2 legs. Birds and humans have 2 legs (but 4 limbs)
Most mammals, reptiles, amphibians have 4 legs
Echinoderms (e.g., sea stars) typically have 5 legs.
Insects typically have 6 legs
Octopi and arachnids have 8 legs
decapods (e.g., crabs) have 10 legs
....But I can't really think of many examples of animals containing more legs until you reach 30+ legs in centipedes and millipedes. Some millipedes even have as many as 750 legs! The lone example I am aware of, the sunflower sea star, typically has 16-24 (though up to 40) limbs.
So my question is: what are some examples of animals with 12-28 legs? As a couple of counterexamples, species in the classes Symphyla (Pseudocentipedes) and Pauropoda within Myriapoda have 8-11 and 12 leg pairs respectively, so between 16 to 24 legs (sometimes with one or two leg pair stronlgy reduced in size).
(species in Symphyla, from wikipedia)
Another common and species-rich group with 14 walking legs (7 leg pairs) is Isopoda.
(Isopod, picture from wikipedia)
You also need to define 'legs' for the discussion to be meaningful. As you say, decapods have 10 legs on their thoracic segments (thoracic appendages), but they can also have appendages on their abdomens (Pleopods/swimming legs), which will place many decapods in the 10-20 leg range.
(Decapod abdominal appendages/legs in yellow, from wikipedia)
So overall, in Arthropoda, having 12-28 legs doesn't seem all that uncommon. There are probably other Arthropod groups besides those mentioned here that also have leg counts in this range.
However, for a general account, the most likely answer (if there is indeed a relative lack of 12-28 legged animals) is probably evolutionary contingencies and strongly conservative body plans within organism groups.
The following is multiple choice question (with options) to answer.
In a group of ducks and cows, the total number of legs are 26 more than twice the no. of heads. Find the total no.of buffaloes. | [
"11",
"12",
"13",
"16"
] | C | Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 26
=> 2x = 26=> x = 13
C |
AQUA-RAT | AQUA-RAT-34216 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10
eg
unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number
To get the value we want we need to floor the log first
unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct
or
unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000
Get digit at position of positive integer
To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.
const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);
or
const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;
// Note brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;
digitAt(567, 0); // 7
digitAt(567, 1); // 6
digitAt(567, 2); // 5
The following is multiple choice question (with options) to answer.
(((13!)^16)-((13!)^8))/(((13!)^8)+((13!)^4))=a
then what is the units digit for a/((13!)^4)= | [
"7",
"8",
"9",
"6"
] | C | (((13!)^16)-((13!)^8))/(((13!)^8)+((13!)^4))=a
solving it , we get
a/((13!)^4)= ((13!)^4-1)
last digit of (13!)^4 will be 0.
so last digit of ((13!)^4-1) will be 9.
ANSWER:C |
AQUA-RAT | AQUA-RAT-34217 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
Suggest Corrections
2
Similar questions
View More
People also searched for
View More
The following is multiple choice question (with options) to answer.
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is: | [
"1520 m2",
"2420 m2",
"2480 m2",
"2520 m2"
] | D | Explanation:
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) m2 = 2520 m2.
Answer: D |
AQUA-RAT | AQUA-RAT-34218 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
John and Ingrid pay 30% and 40% tax annually, respectively. If John makes $58000 and Ingrid makes $72000, what is their combined tax rate? | [
"32%",
"34.4%",
"35%",
"35.6%"
] | D | (1) When 30 and 40 has equal weight or weight = 1/2, the answer would be 35.
(2) When 40 has larger weight than 30, the answer would be in between 35 and 40. Unfortunately, we have 2 answer choices D and E that fit that condition so we need to narrow down our range.
(3) Get 72000/130000 = 36/65. 36/65 is a little above 36/72 = 1/2. Thus, our answer is just a little above 35.
Answer: D |
AQUA-RAT | AQUA-RAT-34219 | Now we calculate $B$. 5 is prime, so $XYZ$ is divisible by 5 if and only if one of $X$, $Y$, or $Z$ is divisible by 5. We can use inclusion-exclusion: $B$ is the sum of the cases where (at least) $X$, $Y$, or $Z$ is divisible by 5, minus the cases where (at least) two are divisible by 5, plus the cases where all three are divisible by 5. That is, $$\begin{eqnarray}B&=&D_x + D_{y} + D_z \\ &&- D_{xy} - D_{xz} - D_{yz} \\ &&+ D_{xyz}\end{eqnarray}$$
Where $D_{xy}$ denotes the number of choices of $(X,Y,Z)$ where $5\mid X$ and $5\mid Y$, and similarly for the others.
By symmetry, $D_x = D_y = D_z$, and $D_{xy} = D_{xz} = D_{yz}$. Also, it is impossible to have two of $(X,Y,Z)$ divisible by 5 without the third also being divisible by 5, so $D_{xy} = D_{xyz}$. So the previous equation reduces to:
$$B = 3D_x - 2D_{xyz}$$
We calculate $D_x$: $X$ will be a multiple of 5 whenever $X\in\{0,5,10,15,20\}$, so we want $$\begin{eqnarray} &&\sum_{5\mid X} C(X) \\ &=& \sum_{X\in\{0,5,10,15,20\}} (21-X) \\ &=& 21\cdot5 - (0+5+10+15+20) \\ &=& 105 - 50 \\ &=& 55 \end{eqnarray}$$
The following is multiple choice question (with options) to answer.
If x is divisible by 6 and 5, which of the following must divide evenly into x? I. 64 II. 32 III. 15 | [
"I only",
"II only",
"III only",
"II,III only"
] | C | If x is divisible by 6,5 means it will be = or > 30, 60,90 etc...
That are not divisible by 32, 64.
So, the answer is C |
AQUA-RAT | AQUA-RAT-34220 | 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 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is? | [
"65 km/hr",
"50 km/hr",
"21 km/hr",
"25 km/hr"
] | B | Speed of the train relative to man
= (125/10) m/sec
= (25/2) m/sec. [(25/2) * (18/5)] km/hr
= 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed
= (x - 5) km/hr. x - 5 = 45 ==> x
= 50 km/hr.
Answer: B |
AQUA-RAT | AQUA-RAT-34221 | 2. ### math
list all the 3 digit numbers that fit these clues. the hundreds digits is less than 3. the tens digit is less than 2. the ones digit is greater than 7
3. ### Math
Think of a five-digit number composed of odd numbers. The thousands digit is two less than the ten thousands digit but two more than the hundreds digit. The tens digit is two more than ones digit which is four less than the
4. ### Math (Confused)
How many international direct-dialing numbers are possible if each number consists of a four-digit area code (the first digit of which must be nonzero) and a five-digit telephone numbers (the first digit must be nonzero)? a.
1. ### math
how many positive 4-digit numbers are there with an even digit in the hundreds position and an odd digit in the tens position? a. 10,000 b. 5,040 c. 2,500 d. 2,250
2. ### math
Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3-digit integer to the original 4-digit
3. ### Math
7 digit number no repetition of numbers digit 5 in thousands place, greatest digit in the millions place, digit in the hundred thousands place is twice the digit in the hundreds place, digit in the hundreds place is twice the
4. ### math
the hundred thousands digit of a six-digit even numbers is 3 more than the thousand digit,which is twice the ones digit.give at least four numbers that satisfy the given condition.
The following is multiple choice question (with options) to answer.
How many numbers between 400 and 1000 can be made with the digits 2,3,4,5,6 and 0? | [
"30",
"60",
"90",
"120"
] | B | Required number = 3 * 5P2 = 3*20 = 60
Answer is B |
AQUA-RAT | AQUA-RAT-34222 | 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.
The length of a train and that of a platform are equal. If with a speed of 144 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is? | [
"757",
"758",
"718",
"1200"
] | D | Speed = [144 * 5/18] m/sec = 40 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 40 è x = 40 * 60 / 2
=1200
Answer:D |
AQUA-RAT | AQUA-RAT-34223 | 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 15 sec. What is the length of the train? | [
"118",
"150",
"277",
"250"
] | D | Speed = 60 * 5/18 = 50/3 m/sec
Length of the train = speed * time = 50/3 * 15 = 250 m
Answer: D |
AQUA-RAT | AQUA-RAT-34224 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
The salary of a typist was at first raised by 10% and then the same was reduced by 5%. If he presently draws Rs.6270.What was his original salary? | [
"6000",
"2999",
"1000",
"2651"
] | A | X * (110/100) * (95/100) = 6270
X * (11/10) * (1/100) = 66
X = 6000
Answer: A |
AQUA-RAT | AQUA-RAT-34225 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is: | [
"1%",
"2%",
"3%",
"5%"
] | D | D
5%
Increase in 10 years = (262500 - 175000) = 87500.
Increase% = (87500/175000 x 100)% = 50%.
Required average = (50/10)% = 5%. |
AQUA-RAT | AQUA-RAT-34226 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
Find the cost of fencing around a circular field of diameter 42 m at the rate of Rs.3 a meter? | [
"138",
"390",
"393",
"279"
] | C | 2 * 22/7 * 21 = 131
131 * 3
= Rs.393
Answer: C |
AQUA-RAT | AQUA-RAT-34227 | Within a class of$M=30$children, knowing that the year counts$A=365$days... The probability that at least$n=2$children have their birthday the same day is: $$P(365,30,2)\simeq 70,6\%$$ The probability that at least$n=3$children have their birthday the same day is: $$P(365,30,3)\simeq 2,85\%$$ The probability that at least$n=4$children have their birthday the same day is: $$P(365,30,4)\simeq 0,0532\%$$ Nicolas Just like to point out that Trazom's answer is incorrect for the general case - the sets being counted in the outer sum overlap. I don't have enough reputation to comment. I wrote a blog post about the general case here : https://swarbrickjones.wordpress.com/2016/05/08/the-birthday-problem-ii-three-people-or-more/ Anyone looking for generalized birthday problem i.e. How many people are required such that M of them share same birthday with certain probability. This link explain various method for calculating probability of generalized birthday problem. http://mathworld.wolfram.com/BirthdayProblem.html Also this paper talk more about various kind of coincidences we face ib life, interesting read. https://www.stat.berkeley.edu/~aldous/157/Papers/diaconis_mosteller.pdf • "Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." – Shraddheya Shendre Sep 10 '17 at 5:35 I am looking at this question and the complicated answers and it's confusing me. Supposing I want to solve in a group of 100 people. what is the probability that at least 3 people share a birthday. So I start from very basic - if there are 3 people, the probability of them sharing a birthday is $$\frac{1}{365}
The following is multiple choice question (with options) to answer.
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child? | [
"4 years",
"6 years",
"8 years",
"10 years"
] | A | Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34228 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A, B and C rents a pasture for Rs.841. A put in 12 horses for 8 months, B 16 horses for 9 months and 18 horses for 6 months. How much should C pay? | [
"270",
"279",
"276",
"261"
] | D | 12*8 :16*9 = 18*6
8: 12: 9
9/29 * 841 = 261
Answer: D |
AQUA-RAT | AQUA-RAT-34229 | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
The following is multiple choice question (with options) to answer.
Find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 21 ? | [
"9, 6, 11",
"9, 6, 17",
"9, 6, 19",
"9, 6, 12"
] | D | Let the numbers be a, b and c.
Given that a, b and c are in the ratio 3 : 2 : 4.
let, a = 3x, b = 2x and c = 4x
Given, (a+b) + (c - b) = 21
= > a + b + c - b = 21 = > a + c = 21
= > 3x + 4x = 21 = >7x = 21 = > x = 3
a , b , c are 3x, 2x, 4x.
a, b, c are 9 , 6 , 12.
Answer: D |
AQUA-RAT | AQUA-RAT-34230 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The owner of a furniture shop charges his customer 100% more than the cost price. If a customer paid Rs. 1000 for a computer table, then what was the cost price of the computer table? | [
"800",
"650",
"500",
"600"
] | C | CP = SP * (100/(100 + profit%))
= 1000(100/200) = Rs. 500.
Answer: C |
AQUA-RAT | AQUA-RAT-34231 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10
eg
unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number
To get the value we want we need to floor the log first
unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct
or
unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000
Get digit at position of positive integer
To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.
const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);
or
const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;
// Note brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;
digitAt(567, 0); // 7
digitAt(567, 1); // 6
digitAt(567, 2); // 5
The following is multiple choice question (with options) to answer.
Find the unit's digit in 264^102+264^103 | [
"0",
"2",
"3",
"5"
] | A | Required unit's digit = unit's digit in 4102+4103.4102+4103.
Now, 4242 gives unit digit 6.
⇒ 41024102 gives unit digit 6.
⇒ 41034103 gives unit digit of the product 6×46×4 i.e., 4.
Hence, unit's digit in 264102+264103264102+264103
= unit's digit in (6+4)=0
A |
AQUA-RAT | AQUA-RAT-34232 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 5:3 and B:C = 5:2. If the total runs scored by all of them are 60, the runs scored by B are? | [
"20.23",
"20.13",
"30.93",
"19.56"
] | D | A:B = 5:3
B:C = 5:2
A:B:C = 25:15:6
15/46 * 60 = 19.56
ANSWER:D |
AQUA-RAT | AQUA-RAT-34233 | # 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 boy incurs 5% loss by selling a book for Rs. 1000. At what price should the book be sold to earn 5 % profit? | [
"Rs. 1105.26",
"Rs. 1251.50",
"Rs. 1085.13",
"Rs. 1885.13"
] | A | Explanation:
Let the new selling price be Rs. x
(100 – loss %) /1000 = (100 + gain %) / x
(100 – 5) / 1000 = (100 + 5) / x
x = (105 × 1000) / 95 = 105000 / 95
x = 1105.26.
ANSWER A |
AQUA-RAT | AQUA-RAT-34234 | # 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.
In a store, the profit is 320% of the cost. If the cost increases by 25% but the SP remains constant, approximately what % of the sellingprice is the profit? | [
"50%",
"70%",
"90%",
"100%"
] | B | Let C.P. = Rs. 100. Then, profit = Rs. 320, S.P. = Rs. 420
New C.P. = 125% of Rs. 100 = Rs. 125.
New S.P. = Rs. 420
Profit = 420 - 125 = Rs. 295
Required percentage = 295/420 * 100 = 1475/21 = 70%
B |
AQUA-RAT | AQUA-RAT-34235 | In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.
The following is multiple choice question (with options) to answer.
The average of ten numbers is 0. In those 10, at the least, how many may be greater than or equal to zero? | [
"0",
"1",
"2",
"3"
] | B | Explanation :
Average of 10 numbers = 0
Sum of 10 numbers =(0 x 10) =0.
It is quite possible that 9 of these numbers may be negative and if there sum is -a, then 10th number is (a).
Answer : B |
AQUA-RAT | AQUA-RAT-34236 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 57 kmph, find the speed of the stream? | [
"16 kmph",
"19 kmph",
"14 kmph",
"11 kmph"
] | B | The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream
= (2+1)/(2-1) = 3/1
= 3:1
Speed of the stream
= 57/3 = 19 kmph.
Answer: B |
AQUA-RAT | AQUA-RAT-34237 | 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.
In the county of Veenapaniville, there are a total of 50 high schools, of three kinds: 25 public schools, 16 parochial schools, and 9 private independent schools. These 50 schools are divided between three districts: A, B, and C. District A has 18 high schools total. District B has 17 high schools total, and only one of those are private independent schools. If District C has an equal number of each of the three kinds of schools, how many private independent schools are there in District A? | [
"2",
"3",
"4",
"5"
] | B | Total Private Schools = 9
Dist A: High Schools = 18 ==> Private Schools = ?
Dist B: High Schools = 17 ==> Private Schools = 1
Dist C: High Schools = 15 ==> Private Schools = 5
Therefore, 9 - 1 - 5 ==> 3
Answer B) |
AQUA-RAT | AQUA-RAT-34238 | In both ratios, it is Length-to-Width.
. . That is, .the longer side : the shorter side.
If you don't "line them up" correctly, you're asking for trouble.
Code:
* - - - - - - - - - *
| | * - - *
| | | |
40 | | 3 | |
| | | |
* - - - - - - - - - * * - - *
60 2
These two rectangles are similar; the side are proportional.
But not because: . $\frac{60}{40} \:=\:\frac{2}{3}$ . . . . which is not true.
Get the idea?
8. Originally Posted by Soroban
We can't order a proportion randomly.
In both ratios, it is Length-to-Width.
. . That is, .the longer side : the shorter side.
If you don't "line them up" correctly, you're asking for trouble.
Code:
* - - - - - - - - - *
| | * - - *
| | | |
40 | | 3 | |
| | | |
* - - - - - - - - - * * - - *
60 2
These two rectangles are similar; the side are proportional.
But not because: . $\frac{60}{40} \:=\:\frac{2}{3}$ . . . . which is not true.
Get the idea?
Yes. So in your example it should be 60/40 : 3/2
In the original example it's 1 over the second time because that's the longer side.
Thank you!
1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2
Sorry, what I meant to ask is how do I get from
1 - 1 / sq.rt 2
to
(sq.rt 2 - 1) / sq.rt 2
.........
1 - 1 / sq.rt 2 = (sq.rt 2 - 1) / sq.rt 2
Sorry, what I meant to ask is how do I get from
1 - 1 / sq.rt 2
to
(sq.rt 2 - 1) / sq.rt 2
1 - 1/sqrt(2)
The following is multiple choice question (with options) to answer.
An order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2. Subsequently, the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter. Find the ratio of the areas of the carpets in both the cases. | [
"8 : 6",
"8 : 7",
"8 : 8",
"8 : 9"
] | B | Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.
Let the dimensions of the carpet in the second case be 7y, 3y units respectively.
From the data,.
2(3x + 2x) = 2(7y + 3y)
=> 5x = 10y
=> x = 2y
Required ratio of the areas of the carpet in both the cases
= 3x * 2x : 7y : 3y
= 6x2 : 21y2
= 6 * (2y)2 : 21y2
= 6 * 4y2 : 21y2
= 8 : 7
Answer: Option B |
AQUA-RAT | AQUA-RAT-34239 | # 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}}$$
\end{document}
The following is multiple choice question (with options) to answer.
In one hour a boat goes 18 km long the stream and 6 km against the stream.The speed of the boat in still water is? | [
"12",
"14",
"16",
"17"
] | A | Speed in still water A = ½ ( 18+6) km/hr A= 12 kmph
Answer: A |
AQUA-RAT | AQUA-RAT-34240 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 3:4:5. If the total value of all the coins in the bag is Rs.400, find the number of 50 paise coins ? | [
"498",
"488",
"428",
"528"
] | C | 50*3k + 20*4k + 10*5k = 40000
280k = 40000 => k = 142.85
50p coins = 3k = 3*142.85 = 428
20p coins = 4k = 4*142.85 = 571
10p coins = 5k = 5*142.85 = 714
ANSWER:C |
AQUA-RAT | AQUA-RAT-34241 | 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.
A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1200. How many workers were part of the latter production run? | [
"A)6",
"B)14",
"C)11",
"D)15"
] | B | The daily average production is given by 5n + 20- given
In the first k days, 500 units are produced
= (5n+20)K =500
k = 500/5n+20...................................1
5 workers were added = 5(n+5)+20 = 5n +45
cumulative is 1200 .. thus for the current period = 1200 -500 = 700
(5n+45)K= 700
k = 700/5n+45........................................2
equate 1 and 2
500/5n+20 = 700 /5n+45
500(5n+45) = 700(5n+20)
25n + 225 = 35n + 140
-10n = -85
n = 8.5
thus n+5 = 14
hence B |
AQUA-RAT | AQUA-RAT-34242 | another explanation that works. They are 2, 4, 6, 8,10, 12,14, 16 and so on. For these examples we’ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number). The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: ∑ = = (+). First Name. 2, 7, 12, .,to 10 term We know that Sum of AP = /2 (2a + (n 1) d) Here n = 10, a = 2, & d = 7 2 = 2 Putting these in formula , Sum = /2 (2a + (n 1) d) = 10/2 (2 2 + (10 1) Here, we are implementing a C program that will be used to find the sum of all numbers from 0 to N without using loop. C Program To Print Sum of Series 1+1/2+1/3+1/4+…+1/n Here, we have listed How To Print The Sum of Series of Numbers in the format 1 + 1/2 + 1/3 + … +1/n in C Programming Language. Calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using pen and paper. How does this summation calculator work? sum for i = 1 to n-1 of i * i * (i-1) / 2. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. To find sum of even numbers we need to iterate through even numbers from 1 to n. Initialize a loop from 2 to N and increment 2 on each iteration. View PDF. sum = n(n+1)/2 We want to add 1 bean to 2 beans to 3 beans… all the way up to 5 beans. In mathematics, the infinite series 1 − 1 + 1 − 1 + ⋯, also written ∑ = ∞ (−) is sometimes called Grandi's series, after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703.It is
The following is multiple choice question (with options) to answer.
a series of natural number 1+2+3+4+5+6+7+8+9+10 is divisibel by 11 so we has to find how many such series is possible upto n where n is less than 1000. | [
"90",
"99",
"180",
"96"
] | C | the total multiple of 11 less then 1000 are 90 .
and for every multiple there are two such series which are divisible by 11.
. .
. sum of series = n(n+1)/2
for multiple 11 we can take n=10 and 11
for 22 '' '' ''' n=21 and 22
and so on and for 990 multiple of 11 we can take n=989 and 990
so total series formed are 90*2= 180
ANSWER:C |
AQUA-RAT | AQUA-RAT-34243 | statistics, logistic-regression, sas
Unless I misunderstood you, it seems you want to have an input parameter that indicates how reflexively a respondent will be to just respond with 'No'. The problem is, as I see it, you don't know the answer to this. You have no way to measure this directly. No?
If you have more data about the respondents, maybe you can back into this. Huge assumption here, but if you could get a count of the number of recommendations made vs the number of times they have said 'No' you could come up with some factor:
$$f_{(No)} = {{Number\ of\ "No"\ Responses}\over{Number\ of\ Recommendations}}$$
This may get you what you are looking for, but it assumes you can tie back the data to the respondent and that the respondent is, in fact, the Branch Manger or Sales Representative, not the customer. The problem with this approach is that it assumes that all the recommendations were equally competitive.
HTH
The following is multiple choice question (with options) to answer.
A company conducted a survey about its two brands, A and B. x percent of respondents liked product A, (x – 20) percent liked product B, 23 percent liked both products, and 23 percent liked neither product. What is the minimum number G of people surveyed by the company? | [
"46",
"G=80",
"G=90",
"G=100"
] | D | 100 = x + x - 20 + 23 - 23
x = 60,
So, Product A = 60%, Product B = 40%, Both = 23%, Neither = 23%
23% of the total no. of people should be an integer. So, A,BC are out.
60% of D and 40% of D are both integers. So, D satisfies all conditions.
So, answer is D. |
AQUA-RAT | AQUA-RAT-34244 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department,what is the greatest number of distinct teams to which the project could be assigned? | [
"4^3",
"4^4",
"4^5",
"6(4^4)"
] | B | First member: 4*4=16 choices;
Second member: 4*3=12 choices (as one department already provided with a member);
Third member: 4*2=8 choices (as two departments already provided with members);
So, we have: 16*12*8.
Since the order of the members in a team does not matter (we don't actually have 1st, 2nd and 3rd members) then we should divide above by 3! to get rid of duplication: 16*12*8/3!=4^4.
ANSWER:B |
AQUA-RAT | AQUA-RAT-34245 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Working at their respective constant rates, Paul, Abdul and Adam alone can finish a certain work in 3, 4, and 5 hours respectively. If all three work together to finish the work, what fraction M of the work will be done by Adam? | [
" 1/4",
" 12/47",
" 1/3",
" 5/12"
] | B | Let the total work be 60 units. Pual does 60/3 =20 units of work per hr. abdul does 15 units per hr and adam does 12 units per hr. If all work together they do (20 + 15 + 12) units per hr = 47 units per hr. So the time taken to finish the work = 60/47 hrs.
adam will do 60/47 * 12 units of work in 60/47 hr . fraction of work adam does =work done by adam / total work
M>( 60/47 *12)/60 =12/47
Answer B |
AQUA-RAT | AQUA-RAT-34246 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A car during its journey travels 30 minutes at a speed of 60 kmph, another 45 minutes at a speed of 60 kmph, and 2 hours at a speed of 80 kmph.The average speed of the car is | [
"63.07 kmph",
"64 kmph",
"72.30 kmph",
"64.02 kmph"
] | C | First car travels 30 min at speed of 60 kmph
distance = 60 x 1/2 = 30 m
Then car travels 45 min at a speed of 60 kmph
distance = 45 min at speed of 60 kmph
distance = 60 x 3/4 = 45 m
at last it travels 2 hours at speed of 70 kmph
distance = 80 x 2 = 160 m
Total distance = 30 + 45 + 160 = 235
Total time= 1/2 + 3/4 + 2 = 3.25
Average speed of the car = 235/3.25 = 72.30
ANSWER:C |
AQUA-RAT | AQUA-RAT-34247 | # Largest consecutive integer using basic operations and optimal digits?
If you are first time reading this, you may want to read the summary section last.
## Solution summary and questions
Sequence values
If the allowed operations are $$(+,-,\times,\div)$$ and parentheses $$(\space)$$, then the $$f(n)$$ is the number of consecutive integers we can make starting at $$1$$, given $$n$$ "digits" that each needs to be used exactly once. Equivalently, it is the largest integer we can make such that all smaller integers $$\in\mathbb N$$ can be made as well.
The $$d$$ is (optimal?) digit set used, where $$d=\{d_1,\dots,d_n\}=(d_1,\dots,d_n)\in\mathbb N^n$$.
So far I have the following data given $$n=1,\dots,5$$ digits to use (best so far):
$$\begin{array}{ccccccc} n & 1&2&3&4&5&6\\ f(n)& 1 & 3 & 10 & 52 & 351 & ? \\ d & \{1\} & \{1, 2\} & \{1, 2, 4\} & \{2, 3, 4, 22\} & \{3, 6, 8, 12, 37\} & ? \end{array}$$
The $$1,3,10$$ are optimal. The $$52$$ entry is probably optimal, but I'm not sure if this can be proven.
The $$f(5)\ge 351$$ can be obtained using $$\{3, 6, 8, 12, 37\}$$. I don't think it can be better.
Notice the weird similarity to the $$A052446$$ sequence. Pure coincidence?
$$Q_1$$: Is there any reason to suspect $$f(4)=52,f(5)=351$$ aren't the optimal solutions?
The following is multiple choice question (with options) to answer.
What is the greatest of 3 consecutive integers whose sum is 21 ? | [
"6",
"7",
"8",
"9"
] | C | The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3
If the sum is 24, we need to solve the equation 3n + 3 = 21;
=> 3n = 18;
=> n = 6
The greatest of the three numbers is therefore 6 + 2 = 8 Answer: C |
AQUA-RAT | AQUA-RAT-34248 | The first few values of $m$ and $n$ are
$(m,n) \in \{(11, 3), (15, 4), (18, 5), (22, 6), (26, 7), (29, 8), (33, 9), (36, 10)\}$
You can see that $m=11$ and $n=3$ will be the smallest values of $m$ and $n$.
Note also that there is no smallest value of $\dfrac mn$ since $\displaystyle \lim_{n \to \infty} \dfrac mn = \log_2 12$ and $\log_2 12$ is irrational.
From $2^{\frac{m}{n}}<13$ follows $\dfrac{m}{n}<\log_2{13}$
Developing $\log_2{13}$ in continued fraction we get $\{3,1,2,2,1,\ldots\}$
Using $3+\dfrac{1}{1+\dfrac{1}{2}}$ we get $\dfrac{11}{3}$ which gives $m=11;\;n=3$
Going on we find $3+\dfrac{1}{1+\frac{1}{2+\frac{1}{2}}}=\dfrac{26}{7}$ which gives $m=26;\;n=7$ and so on
$$3.5849...=\log_{2}12<\frac{m}{n}<\log_{2}13=3.70...$$ Thus, for $\frac{m}{n}=3\frac{2}{3}$ it occurs.
If we want to make $3$ in the denominator be smaller than we'll get $n=2$ and it's impossible.
Thus, $3\frac{2}{3}=\frac{11}{3}$ is our answer.
The following is multiple choice question (with options) to answer.
Which of the following fraction is the smallest? | [
"12/14",
"13/19",
"17/21",
"7/8"
] | B | Explanation:
12/14 = 0.857, 13/19 = 0.684, 17/21 = 0.8095 and 7/8 = 0.875
Since 0.684 is the smallest, so 13/19 is the smallest fraction.
ANSWER: B |
AQUA-RAT | AQUA-RAT-34249 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.20000? | [
"6620",
"3277",
"2688",
"2998"
] | A | A = 20000(11/10)3
= 26620
= 20000
----------
6620
Answer:A |
AQUA-RAT | AQUA-RAT-34250 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A man covers a certain distance by car driving at 30 km/hr and he returns back to the starting point riding on a scooter at 10 km/hr. Find his average speed for the whole journey. | [
"8 Km/hr",
"17 Km/hr",
"15 Km/hr",
"12 km/hr"
] | C | Explanation:
= (2 x a x b)/(a + b)
= (2 x 30 x 10)/(30 + 10)
= (2 x 30 x 10)/40
= 15 km/hr
Answer: Option C |
AQUA-RAT | AQUA-RAT-34251 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
The Course Used By GMAT Club Moderators To Earn 750+
souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4234
Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
Four car rental agencies A, B, C and D rented a plot for parking their cars during the night. A parked 15 cars for 12 days, B parked 12 cars for 20 days, C parked 18 cars for 18 days and D parked 16 cars for 15 days. If A paid Rs. 1125 as rent for parking his cars, what is the total rent paid by all the four agencies? | [
"2388",
"2778",
"1279",
"6150"
] | D | The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1125
45k = 1125 => k = 25
246k = 246(25) = Rs. 6150
Thus the total rent paid by all the four agencies is Rs. 6150.
Answer: D |
AQUA-RAT | AQUA-RAT-34252 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A sum of money at simple interest amounts to Rs. 830 in 3 years and to Rs. 854 in 4 years. The sum is: | [
"647",
"698",
"758",
"847"
] | C | S.I. for 1 year = Rs. (854 - 830) = Rs. 24.
S.I. for 3 years = Rs.(24 x 3) = Rs. 72.
Principal = Rs. (830 - 72) = Rs. 758.
Answer:C |
AQUA-RAT | AQUA-RAT-34253 | 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.
Thirty percent of the members of a swim club have passed the lifesaving test. Among the members who havenotpassed the test, 5 have taken the preparatory course and 30 have not taken the course. How many members are there in the swim club? | [
" 60",
" 80",
" 50",
" 120"
] | C | 30% of the members have passed the test, thus 70% have not passed the test.
We also know that 30+5=42 members have not passed the test, thus 0.7*Total=35 --> Total=50.
Answer: C. |
AQUA-RAT | AQUA-RAT-34254 | Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
Hence answer will be (E) $693 The wage difference is not with respect to time , correct approach will be with respect to work done.. ( Errorr is highlighted in red ) Further explanation as provide to you via PM Wages/Salary can be of 2 types ( Something You will certainly admit ) - 1. Hourly Wage payment System ( Most White collar Jobs ) 2. Unit of Work Done ( Most Blue collar Jobs ) Here it is given the part of job completed by A and B , both have diff efficiency.. Suppose total Work is 18 units Efficiency of A = 11 and Efficiency of B = 7 Now, tell me , whom will you give more money A or B ( According to the amount of work completed ) Certainly A !!! Continuing with the same example - Time Taken by A = 18/11 = 1.63 Hours Time Taken by B = 18/7 = 2.57 Hours Now, tell me , whom will you give more money A or B ??? B is inefficient , straightay, he takes more time to complete less amount of work than A... Hence Hourly wage approach is not suited here, hope this helps.. _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 110 Location: United States GPA: 3.37 Re: Elana was working to code protocols for computer processing. She did [#permalink] ### Show Tags 02 Nov 2016, 11:44 Elena did 11/18 which means that Andrew did 7/18. They both earn the same hourly rate. The difference between the work they did is 4/18. We know that 4/18 equals$154 because that is the difference in pay. Divide $154 by 4 so we can see how much 1/11th of the job is worth.$154/4=\$38.5
The following is multiple choice question (with options) to answer.
A company pays project contractors a rate of a dollars for the first hour and b dollars for each additional hour after the first, where a > b.
In a given month, a contractor worked on two different projects that lasted 6 and 8 hours, respectively. The company has the option to pay for each project individually or for all the projects at the end of the month. Which arrangement would be cheaper for the company and how much would the company save? | [
"Per month, with savings of $(5a + 5b)",
"Per month, with savings of $(5a - 5b)",
"The two options would cost an equal amount.",
"Per project, with savings of $(5a + 5b)"
] | B | Per Project, company will pay as follows:
For 6 hours work = a+5b
For 8 hours work = a+7b
Total = 2a+12b
Per Month, company will pay for 14 hours work = a+13b
Total per contract - total per month
6a+8b - (a+13b)
5a-5b
Since a>b Amount 6a+8b(per contract amount) > a+13b (per project amount) by 5a-5b.
Hence per month payment will be cheaper by 5a-5b .
OA B |
AQUA-RAT | AQUA-RAT-34255 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular field is to be fenced on three sides leaving a side of 30 feet uncovered. If the area of the field is 720 sq. feet, how many feet of fencing will be required? | [
"78",
"40",
"68",
"88"
] | A | EXPLANATION
We have: l = 30 ft and lb = 720 sq. ft.
So, b = 24 ft.
Length of fencing = (l + 2b) = (30 + 48) ft = 78 ft.
Answer A |
AQUA-RAT | AQUA-RAT-34256 | # Constant acceleration problem with a car
1. Sep 8, 2016
### Aman Abraha
1. Question to problem
A car moving with constant acceleration covered the distance between two points 57.9 m apart in 5.02 s. Its speed as it passes the second point was 14.4 m/s. (a) What was the speed at the first point? (b) What was the acceleration? (c) At what prior distance from the first point was the car at rest?
I'm using wileyplus.com to submit answers, and the only one that I'm stuck on is problem (c). A and B are correct.
2. Relevant equations
(a)V average= V final + V initial/ 2
(b)V final= V initial + at
(c)V final^2 = V initial^2 +2ad
3. Attempt at solution
(a) 11.5 m/s= 14.4 m/s + V initial / 2
V initial= 8.60 m/s
(b) 14.4 m/s = 8.60 m/s + a(5.02 s)
a= 1.16 m/s^2
(c) (8.60 m/s)^2 = (0 m/s)^2 + 2d(1.16 m/s^2)
d=31.9 m
For (c) it says on wiley plus that it is incorrect.
Any suggestion on what I did wrong?
Last edited by a moderator: Sep 8, 2016
2. Sep 8, 2016
### RUber
I am not familiar with the equation you used for (c).
In part (a), you found the velocity at the first point. In part (b) you found the constant acceleration.
How long did it take the car to reach the velocity from (a)?
How much distance would a car cover from rest in that amount of time at the acceleration you found in (b)?
3. Sep 8, 2016
### kuruman
The following is multiple choice question (with options) to answer.
Car Z travels 45 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour, but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour. How many miles does car Z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour? | [
"320",
"360",
"400",
"408.3"
] | B | The question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour. We therefore first calculate the fuel efficiency at that speed.
The stem tells us that at 45 miles/hour, the car will run 45 miles/gallon and at 60 miles/hour, that distance decreases by 20%. We can therefore conclude that the car will travel 36 miles/gallon at a constant speed of 60 miles/gallon. With 10 gallons of fuel, the car can therefore travel 36 miles/gallon * 10 gallons = 360 miles.
Answer B. |
AQUA-RAT | AQUA-RAT-34257 | 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 cyclist start on a circular track from a given point but in opposite direction with speeds of 7m/s and 8m/s. If the circumference of the circle is 180meters, after what time will they meet at the starting point? | [
"20sec",
"12sec",
"30sec",
"50sec"
] | B | They meet every 180/7+8 = 12sec
Answer is B |
AQUA-RAT | AQUA-RAT-34258 | 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.
5 skilled workers can build a wall in 20days; 8
semi-skilled workers can build a wall in 25 days; 10
unskilled workers can build a wall in 30days. If a te
am has 2 skilled, 6 semi-skilled and 5 unskilled
workers, how long will it take to build the wall? | [
"15days",
"16days",
"14days",
"18days"
] | A | For Skilled
5 workers --------20 days
5 workers 1 day work= 1/20
1 worker's 1 day work= 1/(5*20)
Similarly,
For Semi-Skilled------
1 worker's 1 day work= 1/(8*25)
For Unskilled------
1 worker's 1 day work= 1/(10*30)
For 2 skilled,6 semi-skilled and 5 unskilled
workers
One day work= 2*[1/(5*20)] +6*[1/(8*25)] + 5*[ 1/(10*30)] = 1/15
Therefore no. of days taken= 15
ANSWER:A |
AQUA-RAT | AQUA-RAT-34259 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
Twenty coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head? | [
"2^19",
"2^10",
"3 * 2^8",
"3 * 2^9"
] | A | Fix the third coin as H. The remaining 19 coins have 2^19 outcomes.
A |
AQUA-RAT | AQUA-RAT-34260 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The manufacturing cost of a shoe is Rs.180 and the transportation lost is Rs.500 for 100 shoes. What will be the selling price if it is sold at 20% gains | [
"Rs 222",
"Rs 216",
"Rs 220",
"Rs 210"
] | A | Explanation :
Total cost of a watch = 180 + (500/100) = 185.
Gain = 20% => SP = 1.2CP = 1.2 X 185 = 222
Answer : A |
AQUA-RAT | AQUA-RAT-34261 | # 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.
Shopkeeper rise price by 33% and gives successive discount of 10% and 15%. What is overall % gain or loss? | [
"1.745%",
"4.745%",
"3.745%",
"6.745%"
] | A | Let d initial price be 100
33 % rise
now price = 133/100*100 = 133
10% discount
Then price = 133 * 90/100 = 119.7
15 % discount
Then price = 119.7 * 85/100 = 101.745
So Gain = 101.745 - 100 = 1.745
Gain % = Gain * 100 /CP
==> 1.745 * 100 /100 = 1.745%
ANSWER:A |
AQUA-RAT | AQUA-RAT-34262 | 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 relay race covers 1 1/2 miles,and each runner on a team will run 1/4 of a mile. How many runners are needed for a team? | [
"9",
"6",
"3",
"10"
] | B | Number of runners=total distance/distance each runner runs
=1 1/2 ÷ 1/4
=3/2 ÷ 1/4 = 3/2*4/1
=12/2
= 6.
Answer is B. |
AQUA-RAT | AQUA-RAT-34263 | 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.
Today Jennifer, who is 30 years old, and her daughter, who is 6 years old, celebrate their birthdays. How many years will pass before Jennifer’s age is twice her daughter’s age? | [
"18",
"20",
"22",
"24"
] | A | Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
After x years passes Jennifer’s age will be (30+x) years old, and her daughter’s age will be (6+x) years old. Since the Jennifer’s age is twice her daughter’s age (30+x)= 2 * (6+x) ---> 30+x=12+2x ---> x= 18.
The answer is (A). |
AQUA-RAT | AQUA-RAT-34264 | Including leading zeros you can see this as follows.
Let $d$ be a digit from $0,\ldots , 9$
• there is exactly 1 3-digit number $ddd$ $\Rightarrow 3$ occurrences
• there are $\binom{3}{2}\cdot 9$ 3-digit numbers, where $d$ occurs exactly twice $\Rightarrow 3\cdot 9 \cdot 2 = 54$ occurrences
• there are $\binom{3}{1}\cdot 9^2$ 3-digit numbers, where $d$ occurs exactly once $\Rightarrow 3\cdot 9^2 = 243$ occurrences
All together $300$ occurrences.
Here is another shorter argument:
If you write down all 3-digit codes with digits $0,\ldots , 9$, you write all together $3000$ digits. Each digit occurs equally often. So, each digit occurs $$\frac{3000}{10}= 300 \mbox{ times}$$
The following is multiple choice question (with options) to answer.
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? | [
"72",
"144",
"216",
"283"
] | C | here is the answer in pure probability terms.
let the 3 digit number be XYZ
now none of them contain zero --> they can vary from 1 - 9
so lets take the value of X, X can be from 1 - 9, hence X can be selected in 9 ways
since Y =X, after selecting X you can select Y in 1 way only
Z can be an selected from 1 -9 (minus 1 because it cannot be the same as X) in exactly 8 ways
hence total number of ways of selecting the number is 9 * 8 = 72.
BUT we are asked for all possible numbers that can be formed.
So XYZ ( or XXZ in this case) can be moved around in 3 ways
XXZ
XZX
ZXX
hence the total number of numbers is 72 *3 = 216
ANSWER:C |
AQUA-RAT | AQUA-RAT-34265 | (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.
There were 35 students in a hostel. Due to the admission of 7 new students, ;he expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. Wbat was the original expenditure of the mess? | [
"320",
"120",
"400",
"420"
] | D | Sol. Let the original average expenditure be Rs. x. Then,
42 (x - 1) - 35x=42  7x= 84  x =12.
Original expenditure = Rs. (35 x 12) =Rs. 420. .
ANSWER D |
AQUA-RAT | AQUA-RAT-34266 | homework-and-exercises, kinematics
Title: Train problem question in kinematics engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last
compartment with velocity v. The middle point of the train passes past the same pole with a velocity of.
My thinking:
Q1 Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
Q2 Shouldn’t the middle part also cover it with u velocity.?
Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
You are right in saying that two ends of a same rigid body can't have different values of speed but you should note that your line is correct only if you are talking about the speed of engine and the last compartment at a given point of time.
In your question , the engine passes the pole with speed $u$ and note that since the train can't elongate or compress , the speed of the middle part as well as the last compartment at that instant is $u$ .
When the last compartment reaches the pole, in that time interval the train has accelerated (first line of your question). So, at that instant, the speed of the engine , the middle part and the last compartment is $v$.
So your first question is just a mere confusion.
For the center of the train to pass the pole , the train has to travel for some time and since it is an accelerated motion , the speed with which it passes the pole is different than $u$.
Note: The value of speed with which the center passes the pole can be calculated using the three equation of motion involving constant acceleration :
$$ v = u + at$$
$$s = ut + \frac{1}{2} a t^2$$
$$v^2 = u^2 + 2as$$
Hope it helps .
The following is multiple choice question (with options) to answer.
A train of length L is traveling at a constant velocity and passes a pole in t seconds. If the same train travelling at the same velocity passes a platform in 3.5t seconds, then what is the length of the platform? | [
"0.5L",
"1.5L",
"2.5L",
"3.5L"
] | C | The train passes a pole in t seconds, so velocity v = L/t
(L+P)/v = 3.5t
(L+P) / (L/t) = 3.5t
P = 2.5L
The answer is C. |
AQUA-RAT | AQUA-RAT-34267 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
In how many days does B alone complete the work
1. B and C together can complete the work in 8 days
2. A and B together can complete the work in 12 days | [
"Statement 1 alone is sufficient",
"Statement 2 alone is sufficient",
"Statement 1 and 2are sufficient",
"Statement 1 and 2are not sufficient"
] | D | Explanation:
We cannot find the work done by B.
If C+A work is given we can find it by
B’s work = (A+B+C) work - (A+C) work
Answer: Option D |
AQUA-RAT | AQUA-RAT-34268 | => 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.
A project scheduled to be carried out over a single fiscal year has a budget of $12,600, divided into12 equal monthly allocations. At the end of the fourth month of that fiscal year, the total amount actually spent on the project was $5,580. By how much was the project over its budget? | [
" $380",
" $540",
" $1,050",
" $1,380"
] | D | Difficulty level: 600
Each month's budget = 12600/12 = 1050
Budget for 4 months = 4*1050 = 4200
Actual amount spent = 5580
Amount spent over the budget = 5580 - 4200 = 1380
Answer (D),
Regards, |
AQUA-RAT | AQUA-RAT-34269 | 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 student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by 1/2. What is the number of students in the class? | [
"20",
"30",
"10",
"40"
] | D | Let the total number of students = x
The average marks increased by 1/2 due to an increase of 83 - 63 = 20 marks.
But total increase in the marks = (1/2)x=x/2
Hence we can write as
x/2=20
⇒x=20×2=40
Answer is D. |
AQUA-RAT | AQUA-RAT-34270 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
Suggest Corrections
2
Similar questions
View More
People also searched for
View More
The following is multiple choice question (with options) to answer.
The length of a rectangle is halved, while its breadth is tripled. Wat isthe % change in area? | [
"20%",
"40%",
"50%",
"60%"
] | C | Let original length = x and original breadth = y.
Original area = xy.
New length = x .
2
New breadth = 3y.
New area = x x 3y = 3 xy.
2 2
Increase % = 1 xy x 1 x 100 % = 50%.
2 xy
C |
AQUA-RAT | AQUA-RAT-34271 | # 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.
In 1998 the profits of company N were 10 percent of revenues. In 1999, the revenues of company N fell by 20 percent, but profits were 10 percent of revenues. The profits in 1999 were what percent of the profits in 1998? | [
"80%",
"105%",
"120%",
"124.2%"
] | A | 0,08R = x/100*0.1R
Answer A |
AQUA-RAT | AQUA-RAT-34272 | + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 14+3x \\ 8. & 7+2x &-2 a_1 + 2 b& 3 a_1 -b & 13+3x \end{array}$$
The following is multiple choice question (with options) to answer.
Can you solve it?
2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=?? | [
"68",
"78",
"88",
"98"
] | D | 2+3=2*[3+(2-1)]=8
3+7=3*[7+(3-1)]=27
4+5=4*[5+(4-1)]=32
5+8=5*[8+(5-1)]=60
6+7=6*[7+(6-1)]=72
therefore
7+8=7*[8+(7-1)]=98
x+y=x[y+(x-1)]=x^2+xy-x
correct answer is D)98 |
AQUA-RAT | AQUA-RAT-34273 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
If x and y are positive integers, and 3x^2=3y, then which of the following must be a multiple of 9?
I. x^2
II. y^2
III. xy | [
"I only",
"II only",
"III only",
"I and II only"
] | A | 4x^2 = 3y
since x,y are positive integers, x^2 = x*x is divisible by 3 -> x is divisible by 3 and y is divisible by x^2
-> x^2 and y is divisible by 9 -> y^2 is divisible by 9
(1),(2), and (3) must be true
Answer is A |
AQUA-RAT | AQUA-RAT-34274 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10
eg
unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number
To get the value we want we need to floor the log first
unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct
or
unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000
Get digit at position of positive integer
To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.
const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);
or
const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;
// Note brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;
digitAt(567, 0); // 7
digitAt(567, 1); // 6
digitAt(567, 2); // 5
The following is multiple choice question (with options) to answer.
Find the unit's digit in (264)^102 + (264)^103 | [
"4",
"1",
"5",
"0"
] | D | Required unit's digit = unit's digit in (4)^102 + (4)^103.
Now, 4^2 gives unit digit 6.
Therefore, (4)^102 gives unit digit 6.
Therefore, (4)^103 gives unit digit of the product (6 x 4) i.e., 4.
Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.
Answer is D. |
AQUA-RAT | AQUA-RAT-34275 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The current of a stream runs at the rate of 4 kmph. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of the boat in still water? | [
"7",
"2",
"8",
"6"
] | C | S = 4
M = x
DS = x + 4
US = x - 4
6/(x + 4) + 6/(x - 4) = 2
x = 8
Answer:C |
AQUA-RAT | AQUA-RAT-34276 | Next, let $a \in A$ be any self-adjoint element. As $(\star)$ implies that $f(a) g(a) = - g(a) f(a)$, we get $$f(a^{2}) g(a^{2}) = f(a) f(a) g(a) g(a) = - f(a) g(a) f(a) g(a) = f(a) g(a) g(a) f(a),$$ and similarly, $$g(a^{2}) f(a^{2}) = g(a) g(a) f(a) f(a) = - g(a) f(a) g(a) f(a) = f(a) g(a) g(a) f(a).$$ We also know from $(\star)$ that $f(a^{2}) g(a^{2}) + g(a^{2}) f(a^{2}) = 0_{B}$, so $f(a) g(a) g(a) f(a) = 0_{B}$. Hence, by the self-adjointness of $a$, we have $$[f(a) g(a)] [f(a) g(a)]^{*} = f(a) g(a) g(a) f(a) = 0_{B}.$$ Therefore, $f(a) g(a) = 0_{B}$, and by interchanging $f$ and $g$, we also obtain $g(a) f(a) = 0_{B}$. As our choice of $a$ was arbitrary, the discussion in this paragraph applies to all self-adjoint elements of $A$.
The following is multiple choice question (with options) to answer.
For which of the following does g(a)−g(b)=g(a−b) for all values of a and b? | [
"g(x)=x^2",
"g(x)=x/2",
"g(x)=x+5",
"g(x)=2x−1"
] | B | To solve this easiest way is just put the value and see that if it equals or not.
with option 1. g(a) = a^2 and g(b) = b^2
so L.H.S = a^2 - b^2
and R.H.S = (a-b)^2 ==> a^2 + b^2 -2ab.
so L.H.S not equal to R.H.S
with option 2. g(a) = a/2 and g(b) = b/2
L.H.S = a/2 - b/2 ==> 1/2(a-b)
R.H.S = (a-b)/2
so L.H.S = R.H.S which is the correct answer.
answer:B |
AQUA-RAT | AQUA-RAT-34277 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Find the simple interest on Rs.500 for 9 months at 6 paisa per month? | [
"Rs.345",
"Rs.270",
"Rs.275",
"Rs.324"
] | B | Explanation:
I = (500*9*6)/100 = 270
Answer IS B |
AQUA-RAT | AQUA-RAT-34278 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of 13 articles is equal to the selling price of 11 articles. Find the profit percent? | [
"18 2/17",
"18 2/11",
"18 2/15",
"18 2/19"
] | B | 13 CP = 11 SP
11 --- 2 CP
100 --- ?
=>18 2/11%
Answer:B |
AQUA-RAT | AQUA-RAT-34279 | #### romsek
MHF Helper
Ah. I see. The book doesn't say that it equals 0.0333, as I mentioned in my OP, it just asks to find the decimal version to the fourth decimal. So instead of it repeating endlessly, it just stops at 0.0333. So the book's answers are "1/30; 0.0333."
What I didn't get, was how to make sense of the fact that 333/10000 and 1/30 both equal 0.0333, even though they aren't equivalent fractions, and if 333/10000 would be an incorrect answer for the fractional part of the question.
$\dfrac 1{30} \neq 0.0333$
$\dfrac 1{30} = 0.033333333333333333333333333333333333 \dots$ with 3's going on forever
1 person
The following is multiple choice question (with options) to answer.
3+33+333+3.33=? | [
"362.33",
"372.33",
"702.33",
"702"
] | B | 3
33
333
3.33
----------
372.33
----------
Answer is B |
AQUA-RAT | AQUA-RAT-34280 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
How long will it take for a sum of money to grow from Rs.1250 to Rs.10,000, if it is invested at 12.5% p.a simple interest? | [
"38",
"56",
"22",
"088"
] | B | Explanation:
Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum
In this case, Rs. 1250 has become Rs.10,000.
Therefore, the interest earned = 10,000 – 1250 = 8750.
8750 = [(1250 x n x 12.5)/100]
=> n = 700 / 12.5 = 56 years.
Answer: B) 56years |
AQUA-RAT | AQUA-RAT-34281 | So:
W_2(L,F) = 0, If 2F+1 > L
W_2(L,F) = L - 2F, If 2F+1 <= L
We can compute W_2(L), for all valid F, by summing W_2(L,F) for F ranging from 1 to (L-1)/2:
...skipping the algebra...
If L is odd, W_2(L) = (L^2 - 2L + 1) / 4
If L is even, W_2(L) = (L^2 - 2L) / 4
Now, we can use this result in calculating the three-number problem:
If we have a first number F, then our second number, S, and third number, T, must satisfy: F < S < T F + S + T <= L
Assuming we have chosen an F small enough that there are S and T that can satisfy F + S + T <= L, then
W_3(L,F) = W_2(L - 3F)
W_3(L,F) = ((L-3F)^2-2*(L-3F)+1)/4, if 3F-L is odd
W_3(L,F) = ((L-3F)^2-2*(L-3F))/4, if 3F-L is even
Our upper limit for F is F + (F + 1) + (F + 2) <= L or F <= (L - 3) / 3
Finally, we can compute W_3(L) for all valid F, by summing W_3(L,F) for F ranging from 1 to I = Int((L-3)/3), where INT() rounds down to the nearest integer.
...skipping even more algebra...
If (L-3)/3 is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I)/8
The following is multiple choice question (with options) to answer.
If the operation Ø is defined for all positive integers x and w by x Ø w = (2^x)/(2^w) then (3 Ø 1) Ø 2 = ? | [
"2",
"4",
"8",
"16"
] | B | 3 Ø 1 = 2^3/2^1 = 4
4 Ø 2 = 2^4/2^2 = 4
The answer is B. |
AQUA-RAT | AQUA-RAT-34282 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The profit obtained by selling an article for Rs. 66 is the same as the loss obtained by selling it for Rs. 52. What is the cost price of the article? | [
"Rs. 40",
"Rs. 50",
"Rs. 49",
"Rs. 59"
] | D | S.P 1- C.P = C.P – S.P 2
66 - C.P = C.P - 52
2 C.P = 66 + 52;
C.P = 118/2 = 59
ANSWER:D |
AQUA-RAT | AQUA-RAT-34283 | > w2:=-(1/2)*r^2+3*t^2+(27/8)*s^2; 1 2 1 2 8 2 - - s + - r + - t 2 3 3 5 2 5 2 2 2 - - t + - s + - r 3 2 3 1 2 2 27 2 - - r + 3 t + -- s 2 8 > H2:=expand(-1/16*u2^2-1/16*v2^2-1/16*w2^2+1/8*u2*v2+1/8*u2*w2+1/8*v2*w2); 49 2 2 49 2 2 49 2 2 49 4 49 4 441 4 -- t r + -- t s + --- r s - -- t - --- r - ---- s 72 32 128 36 576 1024 > simplify(H1/H2); 576 --- 49 >
The following is multiple choice question (with options) to answer.
If 13 = 13w/(1-w) ,then (2w)2 = | [
"1/4",
"1/2",
"1",
"2"
] | D | 13-13w=13w
26w=13
w=1/2
2w=1
2w*2=1*2=2
ANSWER:D |
AQUA-RAT | AQUA-RAT-34284 | \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain
The following is multiple choice question (with options) to answer.
If a*b*c=130, b*c*d = 65, c*d*e=750 and d*e*f=250 the (a*f)/(c*d) = ? | [
"1/2",
"2/3",
"3/4",
"4/3"
] | B | Explanation :
a∗b∗c/b∗c∗d= 130/65 => a/d = 2
d∗e∗f/c∗d∗e= 250/750 => f/c = 1/3
a/d* f/c = 2 * 1/3 = 2/3
Answer : B |
AQUA-RAT | AQUA-RAT-34285 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 7 days of 9 hours each and B alone can do it in 6 days of 7 hours each.how long will they take it to do working together
8 2/5 hours a day? | [
"3 days",
"4 days",
"5 days",
"6 days"
] | A | A's work per hour=1/63
B's work per hour=1/42
A & B's work per hour together=(1/63)+(1/42)=5/126
so a & B together complete the work in 126/5 hours...
if they work 8 2/5=42/5 hours a day,
it will take (126/5)/(42/5) days=(126/5)*(5/42)=3 days...
ANSWER:A |
AQUA-RAT | AQUA-RAT-34286 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Find value of x from given equation 2x + 1 = -17 | [
"-8",
"-9",
"-7",
"7"
] | B | 1. Subtract 1 from both sides:
2x + 1 - 1 = -17 - 1
2. Simplify both sides:
2x = -18
3. Divide both sides by 2:
4. Simplify both sides:
x = -9
B |
AQUA-RAT | AQUA-RAT-34287 | Here is an idea which should work. Let $n >0$ be a positive integer. Define $k_0=n$, and recursively, as long as $k_j \neq 0$ define $$k_{j} =2^{k_{j+1}} m_{j+1} \, \mbox { with } m_{j+1} \mbox{ odd }$$ This process ends after $t$ steps, when $k_t=0$, or equivalently $k_{t-1}$ is odd. Now define f(n) =( \frac{m_1-1}{2}, \frac{m_2-1}{2},.., ... 2 If something is not in B \cup C, then it is in neither B nor C. Because if it was in B, then it is in B or C, which is B \cup C. So the first mistake is in the second sentence. You could show that A - (B \cup C) \subseteq A - (B \cap C). Because the former one is what is in A, but not in B nor C. The latter is what is in A, but not ... 2 That's a very creative idea. I don't know of an established set theory symbol for this (e.g. http://www.rapidtables.com/math/symbols/Set_Symbols.htm doesn't even define such an "operation") and it doesn't seem to be a standard symbol in \LaTeX. One problem I can see is that the symbol \ominus is quite widely used to denote the symmetric difference of ... 2 It's not awful, at least as an idea, but I would certainly say it's unnecessary, and I strongly dislike your particular notation choice. Basically it's a symbol that's too different and unintuitive to have to remember for too unimportant and uncommon a situation to be worthwhile. In my experience, once one gets to mathematics classes, textbooks, etc. that ... 2 In your post you only proved that B = \overline A \cup X \implies A\cap B \subseteq X \subseteq B, although
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 1701 what is the least possible value of k? | [
"8",
"10",
"12",
"14"
] | C | 1701 = 3*3*3*3*3*7
Thus k must include numbers at least up to the number 12 so that there are at least five appearances of 3 (that is: 3, 6, 9=3*3, and 12).
The answer is C. |
AQUA-RAT | AQUA-RAT-34288 | # 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, B, K start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr, 100 km/hr respectively. B starts two hours after A. If B and K overtake A at the same instant, how many hours after A did K start? | [
"4.2",
"4.8",
"5.6",
"6.4"
] | C | In 2 hours, A travels 60 km.
B can catch A at a rate of 10 km/hr, so B catches A 6 hours after B starts.
So A and B both travel a distance of 240 km.
C needs 2.4 hours to travel 240 km, so C leaves 5.6 hours after A.
The answer is C. |
AQUA-RAT | AQUA-RAT-34289 | $y = x$ if $x \geq -1$.
It also means
$y = -(x + 1) - 1$ if $x + 1 < 0$, i.e. $x < -1$
$y = -x - 2$ if $x < -1$.
So all the points which satisfy these two linear equations will be your solution.
The following is multiple choice question (with options) to answer.
What is the solution of the equations x - y = 10 and (x + y)-1 = 2 ? | [
"x = 3.2, y = 2.3",
"x = 5.25, y = 4.75",
"x = 2, y = 1.1",
"x = 1.2, y = 0.3"
] | B | Answer
x - y = 10 ...(i)
and 1/(x + y)-1=2
⇒ 1/ (x + y) = 2
⇒ 2(x + y) =1
⇒ x + y = 1/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 5.25
and y = 4.75
Correct Option: B |
AQUA-RAT | AQUA-RAT-34290 | Within a class of$M=30$children, knowing that the year counts$A=365$days... The probability that at least$n=2$children have their birthday the same day is: $$P(365,30,2)\simeq 70,6\%$$ The probability that at least$n=3$children have their birthday the same day is: $$P(365,30,3)\simeq 2,85\%$$ The probability that at least$n=4$children have their birthday the same day is: $$P(365,30,4)\simeq 0,0532\%$$ Nicolas Just like to point out that Trazom's answer is incorrect for the general case - the sets being counted in the outer sum overlap. I don't have enough reputation to comment. I wrote a blog post about the general case here : https://swarbrickjones.wordpress.com/2016/05/08/the-birthday-problem-ii-three-people-or-more/ Anyone looking for generalized birthday problem i.e. How many people are required such that M of them share same birthday with certain probability. This link explain various method for calculating probability of generalized birthday problem. http://mathworld.wolfram.com/BirthdayProblem.html Also this paper talk more about various kind of coincidences we face ib life, interesting read. https://www.stat.berkeley.edu/~aldous/157/Papers/diaconis_mosteller.pdf • "Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." – Shraddheya Shendre Sep 10 '17 at 5:35 I am looking at this question and the complicated answers and it's confusing me. Supposing I want to solve in a group of 100 people. what is the probability that at least 3 people share a birthday. So I start from very basic - if there are 3 people, the probability of them sharing a birthday is $$\frac{1}{365}
The following is multiple choice question (with options) to answer.
The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child? | [
"4 years",
"8 years",
"10 years",
"None of these"
] | A | Explanation:
Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20 => x = 4.
Age of youngest child = x = 4 years.
ANSWER IS A |
AQUA-RAT | AQUA-RAT-34291 | Also: explore what must happen if one of the variables, say x, is zero.
4. Aug 9, 2013
Thanks for the hints , but I used a different method to solve this equation. From eq. 1 I got y=(6x)/(5x-6) .
I substituted that in eq. 2 to get x in terms of z. ( x=(126z)/(126+45z) ) then I substituted that into eq. 3 to get a quadratic equation in z with solutions z=0,7. From z=7 , I got x= 2 . Then I got y=3 from the relationship between x and y obtained earlier from eq. 1 . But I felt that the arithmetic involved was quite daunting and there must be a better way to solve it. I tried using the method that pasmith recommended but am feeling a bit lost. Can anyone please explain a simpler method to solve this system ?
5. Aug 9, 2013
### Ray Vickson
Let $a = xy, \;b = yz, \; c = zx \:\text{ and }\: p = xyz.$ Multiply the first equation by z, the second by x and the third by y to get
$$6b+6c=5p\\ 21a+21c=10p \\ 14a+14b=9p$$
This is a simple 3x3 linear system which can be solved using grade-school methods, to get
$a = p/7,\: b = p/2, \: c = p/3.$ Assuming $x,y,z,\neq 0$ we have
$$xy = xyz/7 \Longrightarrow z = 7 \\ yz = xyz/2 \Longrightarrow x = 2\\ zx = xyz/3 \Longrightarrow y=3$$
6. Aug 10, 2013
The following is multiple choice question (with options) to answer.
If X+Y = 2X-2Z, X-2Y = 4Z and X+Y+Z = 21, what is the value of Y/Z? | [
"-4.5.",
"-2.",
"-1.7.",
"3."
] | C | Given : X+Y = 2X-2Z, --> eq 1
X-2Y = 4Z --> eq 2
X+Y+Z = 21 -->eq 3.
Now we are asked to find the value of Y/Z.
Now let choose some option randomly,
Take C , which is 3, I randomly chose this value as it is positive :)
Y/Z = 3 , this is the value we need to get.
Then Y = 3Z -- eq 4
Sub this in eq 1.
X + 3Z = 2X - 2Z
=> X = 5Z -->eq 5.
Sub X and Y value in eq 3.
5Z + 3Z + Z = 21
=> Z = 21/9. Now sub this value to get Y , we get Y = 21/3.
Then Y/Z = 21/3 / 21/9 = we get 3 again. So this option is correct.
Answer: option C is correct. |
AQUA-RAT | AQUA-RAT-34292 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
The probability that a convenience store has no iced tea is 50%. If Karl is stopping by 2 convenience stores on his way to work, what is the probability thatat least oneof the stores will not have a can of iced tea? | [
"1/8",
"1/4",
"1/2",
"3/4"
] | D | p=1-1/2^2(all stores have iced tea)=3/4
D |
AQUA-RAT | AQUA-RAT-34293 | 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.
The length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"267 m",
"256 m",
"245 m",
"627 m"
] | C | Speed
= (45 * 5/18) m/sec = (25/2) m/sec. Time
= 30 sec. Let the length of bridge be x meters. Then, (130 + X)/30 = 25/2 ==> 2(130 + X)
= 750 ==> X
= 245 m.
Answer: C |
AQUA-RAT | AQUA-RAT-34294 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
A and B working saperately can do a piece of work in 12 and 18 hours respectively. If they work for 6 hour alternately with B beginning, in how many hours will the work be completed? | [
"9 hours",
"15 hours",
"13 ½ hours",
"12 ½ hours"
] | B | Explanation:
A and B 2 hours work = 1/12 + 1/18 = 5/36
work done by A and B during 6 pairs
i.e 12 hours = 6 * 5/36 = 5/6
Remaining work = 1 – 5/6 = 1/6
1/6 of work and this work done by B only = 18* 1/6 = 3 hours
Hence the total time take is 15 hours
Answer: Option B |
AQUA-RAT | AQUA-RAT-34295 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
At the end of a business conference, ten people present shake hands with each other once. How many handshakes will be there all together? | [
"20",
"45",
"55",
"90"
] | B | Explanation : . Total no. of handshakes = (9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 45
Answer B |
AQUA-RAT | AQUA-RAT-34296 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
Stay hungry, Stay foolish
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1059
Location: India
GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3400
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
What is the smallest of six consecutive odd integers whose average (arithmetic mean) is c + 2? | [
"c - 5",
"c - 3",
"c- 1",
"c"
] | B | Since the numbers are consecutive odd integers, mean = median = 3rd integer + 4th integer /2
And 1st integer= 3rd integer- 4
let's say 3rd integer = n and 4th integer = n+2
2n+2/2= c+2
n= c+1
1st integer= c+1-4= c-3
B is the answer |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.