source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38597 | 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.
A sum of money is put out at compound interest for 2 years at 20%. It would fetch Rs.482 more if the interest were payable half-yearly, then it were pay able yearly. Find the sum? | [
"198",
"279",
"267",
"200"
] | D | P(11/10)4 - P(6/5)2
= 482
P = 200
Answer: D |
AQUA-RAT | AQUA-RAT-38598 | => 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.
Rose and Anna are partners in a business. Rose runs the business and receives 12% of the profit for managing the business, the rest is divided in proportion of their capitals. Rose invested Rs.12000 and Anna invested Rs.20000. the rest is divided in proportion of their capitals. Out of a total profit of Rs.9000, how much did Rose receive? | [
"5872",
"5839",
"5832",
"5838"
] | C | ratio of their investments Rs.12000:20000=3:5
12% of the profit for managing the business =12/100*9000=1080
remaining profit=9000-1080=7920
share profit in ratio of investment 7920/5 =1584
Moses's ratio share=3*1584=4752
total money Moses collects=4752+1080=5832
Answer: C |
AQUA-RAT | AQUA-RAT-38599 | +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 the value of x from this equation? 7(x - 1) = 21 | [
"2",
"4",
"6",
"8"
] | B | 1. Divide both sides by 7:
2. Simplify both sides:
x - 1 = 3
3. Add 1 to both sides:
x - 1 + 1 = 3 + 1
4. Simplify both sides:
x = 4
B |
AQUA-RAT | AQUA-RAT-38600 | 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....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A batsman in his 17th innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17thinnings? | [
"11",
"7",
"76",
"37"
] | D | 16x + 85 = 17(x + 3)
x = 34 + 3 = 37
Answer: D |
AQUA-RAT | AQUA-RAT-38601 | # Three Times Larger: Idiom or Error?
Having just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does “three times larger” mean? How about “300% more”? We’ll focus on one discussion that involved several of us, and referred back to other answers we’ve given.
## Percent increase vs. factor
The question, from 2006, started with the idea of a percent increase:
Percent Increase and "Increase by a Factor of ..."
A math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase.
But can't I also say there was an increase by a factor of 2? Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent? This is what is confusing me!
I'd never been confused about saying "increased by a factor of" and "increased by percent of" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know! Thank you for any help.
Unfortunately, Joseph didn’t directly quote from the page he had in mind, and we have never said exactly what he said; so we couldn’t be sure which page it was. Everything he said, however, was correct.
Doctor Rick was the first to answer:
Hi, Joseph.
Yes, this can be very confusing, because some statements about increases are ambiguous.
When we say "increased by a factor of 2," the word "factor" makes it clear that we mean "multiplied by 2."
When we say "increased by 10%," there is only one reasonable interpretation: the amount of the increase is 10% of the original amount. If we meant multiplication by 10%, that would be a decrease -- not an increase! Even when we say "increased by 100%," there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase.
The following is multiple choice question (with options) to answer.
If the numerator of a fraction E is tripled, and the denominator of a fraction is doubled, the resulting fraction will reflect an increase of what percent? | [
"16 1⁄6%",
"25%",
"33 1⁄3%",
"50%"
] | D | This question can be dealt with conceptually or by TESTing VALUES.
We're told that the numerator of a fraction E is tripled and the denominator of that same fraction is doubled. We're asked for the resulting increase, in percentage terms, of that fraction.
Let's TEST 2/1
If we triple the numerator and double the denominator, we end up with....
(2x3)/(1x2) = 6/2 = 3
Since we're increasing 2 to 3, we're increasing that fraction by (3-2)/2 = 1/2 of itself = 50%
D |
AQUA-RAT | AQUA-RAT-38602 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 13 km/hr and 23 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions? | [
"28 sec",
"30 sec",
"17 sec",
"23 sec"
] | B | Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 300 / (13 + 23)5/18 = 300* 18 / 36 * 5 = 30 seconds.
Answer : B |
AQUA-RAT | AQUA-RAT-38603 | 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.
You are given three coins: one has heads on both faces, the second has tails on both faces,
and the third has a head on one face and a tail on the other. You choose a coin at random and
toss it, and it comes up heads. The probability that the other face is tails is | [
"1/4",
"1/3",
"1/2",
"2/3"
] | C | the three coins are HH LL HT
total possiblity are 6
1st head probablity are 2/6,
2 nd head probablity are 0,
third head probablity are 1/6,
The total head probablity is 2/6+0+1/6=1/2,
Then tail probablity is 1-1/2=1/2
ANSWER:C |
AQUA-RAT | AQUA-RAT-38604 | # Math Help - Distance (Word Problem)
1. ## Distance (Word Problem)
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
2. Originally Posted by Beastkun
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
1. Let d denote the covered distance and t the elapsed time. Then you know:
$\dfrac dt = 50$
and
$\dfrac{d}{t-3}=60$
2. Solve for d.
Spoiler:
You should come out with d = 900
3. I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
4. Originally Posted by Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
Substitute d = 50t from the first equation into the second equation and solve for t and hence d.
5. What prove it is doing is simultaneous equations.
The following is multiple choice question (with options) to answer.
One hour after Yolanda started walking from X to Y, a distance of 31 miles, Bob started walking along the same road from Y to X. If Yolandaâs walking rate was 1 miles per hour and Bobâs was 2 miles per hour, how many miles had Bob walked when they met? | [
"19",
"20",
"22",
"21"
] | B | Let t be the number of hours that Bob had walked when he met Yolanda. Then, when they met, Bob had walked 4t miles and Yolanda had walked (t + 1) miles. These distances must sum to 31 miles, so 2t + (t + 1) = 31, which may be solved for t as follows
2t + (t + 1) = 31
2t + t + 1 = 31
3t = 30
T = 10 (hours)
Therefore, Bob had walked 2t = 2(10) = 20 miles when they met. The best answer is B. |
AQUA-RAT | AQUA-RAT-38605 | combinatorics, sets, matching
Put your students in a row: $a!$ possibilities. The first $b$ students form the first group, the next $b$ students the second group, etc. The order in the line-up between each group of students does not matter, so we have to divide by th1s number of possibilities which is $(c!)^b$. Again we divide by $(a-bc)!$ the number of permutations of the not chosen ones. Also, the order of the groups themselves does not matter, we additionally divide by $b!$.
Answer: $a!\;/\;{(c!)^b (a-bc)! \;b!}$
Another way of obtaining the same number. Choose a group of $c$ students from total $a$, then a group of $c$ students from the remaining $c-b$, etc. This can be done in ${a \choose c}{a-c\choose c}\dots{a-c(b-1)\choose c}$. Again I divide by the possible orderings of the groups which is $b!$.
Thus $\frac{a!}{c!(a-c)!}\frac{(a-c)!}{c!(a-2c)!}\dots \frac{(a-(b-1)c)!}{c!(a-bc)!} / b!$
Your first example $a=12$, $b=4$, $c=3$. You have 880, this number is 15400.
Second example $a=12$, $b=3$, $c=2$. 13680, we agree.
The following is multiple choice question (with options) to answer.
In a graduating class of 236 students, 144 took geometry and 119 took biology. What is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology? | [
"144",
"119",
"113",
"117"
] | D | Greatest Possible Number taken both should be 144 (As it is maximum for One)
Smallest Possible Number taken both should be given by
Total - Neither = A + B - Both
Both = A+B+Neither - Total (Neither must be 0 to minimize the both)
So 144 + 119 + 0 - 236 = 27
Greatest - Smallest is 144-27 = 117
So answer must be D. 117 |
AQUA-RAT | AQUA-RAT-38606 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
if a number cube is rolled once and a coin is tossed once. what is the probability that coin will show tails and composite number on on the cube | [
"1/6",
"2/6",
"3/6",
"4/6"
] | A | 1/2 * 2/6 = 1/6
ANSWER:A |
AQUA-RAT | AQUA-RAT-38607 | Author Message
TAGS:
### Hide Tags
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Joined: 26 Apr 2010
Posts: 122
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)
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Kudos [?]: 129 [0], given: 54
$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
A corporation four time of its annual bonus to 35 of its employees. What percent of the employees’ new bonus is the increase? | [
"5%",
"12%",
"25%",
"75%"
] | D | Let the annual bonus be X.
A corporation triples its annual bonus.
So new bonus = 4X.
Increase = 4X-X =3X
The increase is what percent of the employees’ new bonus
= (3X/4X)*100
= 75%
Hence D. |
AQUA-RAT | AQUA-RAT-38608 | 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 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. In how many hours B can complete the work ? | [
"10 hours",
"12 hours",
"16 hours",
"18 hours"
] | B | Explanation:
Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = (1/4)+(1/3) = 7/12
Work done by B in 1 hour = (7/12)–(1/2) = 1/12
=> B alone can complete the work in 12 hour
Option B |
AQUA-RAT | AQUA-RAT-38609 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
A trailer carries 3, 4 and 6 crates on a trip. Each crate weighs no less than 100 kg. What is the maximum weight of the crates on a single trip? | [
"100",
"625",
"600",
"7500"
] | A | Max No. of crates=6.
Max Weight=100Kg
Max. weight carried = 6*100=100Kg=A. |
AQUA-RAT | AQUA-RAT-38610 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Sushil got thrice as many marks in English as in Science. His total marks in English, Science and Maths are 162. If the ratio of his marks in English and Maths is 3:5, find his marks in Science? | [
"20",
"24",
"18",
"15"
] | C | S:E = 1:3
E:M = 3:5
------------
S:E:M = 3:9:15
3/27 * 162 = 18
ANSWER C |
AQUA-RAT | AQUA-RAT-38611 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
Which of the following numbers is the biggest => [-6/7 , -7/8 , -9/10 , -12/13 , -34/35] | [
"-6/7",
"-7/8",
"-9/10",
"-12/13"
] | A | Another way to solve this question :
[-6/7 , -7/8 , -9/10 , -12/13 , -34/35] can be written as
-1+1/7, -1+1/8, -1 + 1/10, -1+1/13, -1+1/35
Now, in order to make the number larger, we need to add something larger to -1.
In first option, we can see 1/7 is the largest of all the other given
Hence, A |
AQUA-RAT | AQUA-RAT-38612 | • In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52
The following is multiple choice question (with options) to answer.
The sum of the heights of two high-rises is x feet. If the second high rise is 37 feet taller than the first, how tall will the first high rise be ? | [
"(x+z)/2 + 37",
"2x−(37+z)",
"(x−37)/2 + z",
"(x - 37 )/2"
] | D | I will note h1 the height of high-rise 1 and h2 the height of high-rise 2. SO:
h1 + h2 = x
and h12= h21+ 37 =>
h1 + h1+ 37 = x => 2h1 = x-37 =? h1 = (x-37)/2
, CORRECT ANSWER D |
AQUA-RAT | AQUA-RAT-38613 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article offering a discount of 5% and earned a profit of 42.5%. What would have been the percentage of profit earned if no discount had been offered? | [
"28.5",
"27.675",
"30",
"50"
] | D | Giving no discount to customer implies selling the product on printed price. Suppose the cost price of the article is 100.
Then printed price = 100×(100+42.5)/(100−5)=150
Hence, required % Profit = 150 – 100 = 50%
Answer D |
AQUA-RAT | AQUA-RAT-38614 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
There are 10 books on a shelf: 5 English books, 4 Spanish books and 2 Portuguese books. What is the probability of choosing 2 books in different languages? | [
"31/90",
"3/10",
"1/3",
"38/45"
] | D | Probability = Favourable outcome / Total Outcome = 1- (Unfavourable outcome / Total Outcome)
Favourable Outcomes
Case-1: One book is English (out of 5 English books) and one book is Spanish (out of 4 Spanish books) i.e. 5C1*4C1
Case-2: One book is Spanish (out of 4 Spanish books) and one book is Portuguese (out of 2 Portuguese books) i.e. 4C1*2C1
Case-3: One book is English (out of 8 English books) and one book is Portuguese (out of 2 Portuguese books) i.e. 8C1*2C1
Total favourable Outcomes = 5C1*4C1+4C1*2C1+5C1*2C1 = 20+8+10 = 38
Total Possible Outcomes = 10C2 = 45
i.e. Probability = 38/45
D |
AQUA-RAT | AQUA-RAT-38615 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
3 people gave you 5 dollars. How many dollars do you have in all | [
"15",
"17",
"0",
"67"
] | A | If 3 people give you 5 dollars you will have 15
3x5=15 A |
AQUA-RAT | AQUA-RAT-38616 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
In a maths test, students were asked to find 5/16 of a certain number. One of the students by mistake found 5/6th of that number and his answer was 200 more than the correct answer. Find the number. | [
"125",
"280",
"384",
"400"
] | C | Explanation:
Let the number be x.
5*x/6 = 5*x/16 + 200
25*x/48 = 200
x = 384
ANSWER: C |
AQUA-RAT | AQUA-RAT-38617 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The Red Hotel has booked 1/4 of its rooms. The Blue Hotel has Has three times the capacity of the red hotel but has only booked 1/8 of its rooms. If all the bookings from the Red Hotel switched to the Blue Hotel, What fraction of the Blue Hotel's rooms would be booked? | [
"2/8",
"3/4",
"7/16",
"5/24"
] | D | We will pretend that the Red Hotel has a total of 8 rooms. This means that the Blue Hotel would have 24 rooms. From this, we can deduce that there is 2 bookings at the Red Hotel and 3 bookings at the Blue Hotel. If the two bookings from the Red Hotel moved to the blue hotel that would mean a total of 5 bookings at the Blue Hotel. This would make 5 of the 24 rooms would be filled. (Answer: D) |
AQUA-RAT | AQUA-RAT-38618 | 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.
A coin is weighted so that the probability of heads on any flip is 0.4, while the probability of tails is 0.6. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice? | [
"(0.4)^5 + 5(0.4)^4(0.6) + 10(0.4)^3(0.6)^2",
"(0.6)^5 + 4(0.6)^4(0.4) + 6(0.6)^3(0.4)^2",
"(0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2",
"(0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2"
] | A | Probability of Head, P(H) = 0.4
Probability of Tail, P(T) = 0.6
Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.4)^3*(0.6)^2
1 Tail and 4 Heads, Probability = 5C1*(0.4)^4*(0.6)^2
0 Tail and 5 Heads, Probability = (0.4)^5
Required Probability = Sum of all Favourable cases = (0.4)^5 + 5(0.4)^4(0.6) + 10(0.4)^3(0.6)^2
Answer: option A |
AQUA-RAT | AQUA-RAT-38619 | # 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.
train start from mumbai to delhi at 12.00am with speed of 60km/hr another train starts from delhi to mumbai at 1.30 with speed of 80km/hr find time when they meet. | [
"5am",
"6am",
"7am",
"8am"
] | B | 1 train 12.00 = 60km ; 2 train 1.30 = 80km;
1 train travel till 1.30 = 90 km.
1 2
150 80 -2.30
210 160 3.30
270 240 4.30
330 320 5.30
360 360 6.00
so they meat at 6am
ANSWER:B |
AQUA-RAT | AQUA-RAT-38620 | # Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube.
Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube. What I have so far: we express the prime factorizations of $35$ and $7$ as $5\cdot7$ and $7$, respectively. Then $n$ must be of the form $n=5^{\alpha}7^{\beta}$. Thus we see that $$35n=5^{\alpha+1}7^{\beta+1}$$ and $$\frac{n}{7}=5^{\alpha}7^{\beta-1}.$$ From here we proceed to see that if $35n$ is a perfect square, then $\alpha+1$ and $\beta+1$ are even, thus $\alpha$ and $\beta$ themselves must be odd, and so, $\alpha\equiv b\equiv1\bmod2$. Clearly, $n$ must be divisible by $7$ for $n/7$ to be an integral quantity. (This is where I'm stuck..)
I know $\alpha$ and $\beta-1$ are odd and that $n$ must be divisible by 7, however, I'm not sure what to fill in for the question marks below based on these conditions.
$$\alpha\equiv?\bmod3\quad\text{and}\quad\beta\equiv?\bmod3$$
Once I know why and how to fill in those question marks, I know it's a matter of solving a system of congruences via the chinese remainder theorem to finish the problem.
The following is multiple choice question (with options) to answer.
What is the smallest positive integer x such that 720-x is the cube of a positive integer | [
"5",
"7",
"47",
"9"
] | D | Given 738-x is a perfect cube
so we will take 729 = 9*9*9
738-x = 729
x =738-729 = 9
correct option is D |
AQUA-RAT | AQUA-RAT-38621 | $$3c-3\ge 2c+4$$
which comes out as $$c\ge 7$$. What follows is that then $$b\ge 8$$ and $$a\ge 9$$, so $$a+b\ge 17$$ and therefore $$c+d+e\ge 18$$, and so finally $$a+b+c+d+e\ge 17+18=35$$. So we have at least $$35$$ sweets altogether.
To prove that $$35$$ is the actual minimum, verify that all the conditions are satisfied with $$a=9$$, $$b=8$$, $$c=7$$, $$d=6$$ and $$e=5$$.
Lets go case by case:
Best case is one candy in the first jar and two candy in the second and keep on going. This does not work howver as $$1+2+3$$ is less than $$4+5$$.
So put two candy in the first and add one more to each jar from the previous ($$e=2$$ , $$d=3$$ , $$c=4$$ , $$b=5$$ , $$a=6$$). Does not work either. Then put three candy in the first and keep going. does not work ($$e=3 , d=4, c=5 , b=6, a=7$$).
Now put four in the first, five in the second... this works ($$e=4, d=5, c=6, b=7, e=8$$. Since $$4+5+6$$ is greater than $$7+8$$.
This is the lowest you can go since the above process implies that $$1$$ and $$2$$ $$3$$ can not be used and we used $$4$$ and all the other lowest numbers possible.
The following is multiple choice question (with options) to answer.
In a candy dish the ratio of red to yellow candies is 2:5, the ratio of red to green candies is 3:8, and the ratio of yellow ot blue candies is 9:2.what is the maximum total number T of yellow and green candies combined if the maximum combined number of red and blue candies is fewer than 85? | [
"144",
"189",
"234",
"279"
] | D | Given that Red is to yellow is 2:5, Red is to green is 3:8 and Yellow is to blue is 9:2.
Therefore, the total number of red, yellow, blue and green balls will be 18x, 45x, 10x and 48x respectively, where x is a constant.
If the combined number of red and blue balls is fewer than 85, i.e. max 84 balls, then the maximum number of yellow and green balls will be 279.
(10x+18x) < 85.
28x < 85, i.e. 28x <= 84 (Since number of balls cannot be in fraction). Thus, x<=3.
(45x+48x) = 93x.
Max (93x) T= 279.
Answer is D. |
AQUA-RAT | AQUA-RAT-38622 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A is half good a work man as B and together they finish a job in 32 days. In how many days working alone B finish the job? | [
"98 days",
"21 days",
"48 days",
"18 days"
] | C | WC = 1:2
2x + x = 1/32 => x = 1/96
2x = 1/96 => 48 days
Answer:C |
AQUA-RAT | AQUA-RAT-38623 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph? | [
"22",
"28",
"25",
"77"
] | C | D = 100 + 150 = 250
S = 36 * 5/18 = 10 mps
T = 250/10 = 25 sec
Answer: C |
AQUA-RAT | AQUA-RAT-38624 | 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.
At Veridux Corporation, there are 250 employees. Of these, 90 are female, and the rest are males. There are a total of 40 managers, and the rest of the employees are associates. If there are a total of 160 male associates, how many female managers are there? | [
"40",
"20",
"25",
"30"
] | A | 250 Employees:
90 Male, 160 Female
40 Managers, 210 Associates
160 Male Associates implies 50 Female Associates which means the remaining 40 females must be Managers
A. 40 |
AQUA-RAT | AQUA-RAT-38625 | # 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 tradesman sold an article at a loss of 10%. If the selling price had been increased by $50, there would have been a gain of 10%. What was the cost price of the article ? | [
"$100",
"$250",
"$200",
"$150"
] | B | Let C.P. be $x
then, (110% of x)-(90% of x) = 50 or 20% of x=100
x/5 = 50
x=250
C.P. = $250
correct option is B |
AQUA-RAT | AQUA-RAT-38626 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are: | [
"39, 30",
"41, 32",
"42, 33",
"43, 34"
] | C | Let their marks be (x + 9) and x.
Then, x + 9 = 56 /100(x + 9 + x)
=25(x + 9) = 14(2x + 9)
= 3x = 99
=x = 33
So, their marks are 42 and 33.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-38627 | # 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.
Larry, Michael, and Peter have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed? | [
" 21",
" 42",
" 120",
" 504"
] | A | Larry, Michael, and Peter have five donuts to share.to get the answer, see how can we get sum5with 3 numbers.
1) 0,0,5 = 3 combinations or 3! /2!
2) 0,1,4 = 6 combinations or 3!
similarly
3) 0,2,3 = 6 combinations
4) 1,1,3 = 3 combination
5) 2,2,1 = 3 combination
total =21=A |
AQUA-RAT | AQUA-RAT-38628 | Since we are not interested on the numerical value of this equation, but that the result of the equation to be an integer, not a fraction, the n outside of the parenthesis is irrelevant.
So the equation could be written as, (3n ± 1)/2, and the result of this equation is always an integer for any odd or even value of n.
-
We know that every prime $> 3$ is of the form $6n+1$ or $6n-1$. Now find two numbers $a$ and $b$ such that $a+b = p$ and $a-b =1$, where $p$ is either $6n+1$ or $6n-1$. Suppose $p = 6n+1$ Solving $a+b=6n+1$ and $a-b = 1$, we get $a= 3n+1$ and $b=3n$ Thus $ab$ is a multiple of $6$. Therefore $4ab$ is a multiple of $24$. But $4ab = (a+b)^2 – (a-b)^2$. Thus we see that $p^2-1$ is a multiple of $24$.
-
Note that any prime $p$ which is $>3$ we have $p-1$ or $p+1$ should be divisible by $3.$ $$(2m+1)^2=4m(m+1)+1\equiv1\mod(8)$$ Hence $p^2-1=(p-1)(p+1)$ is divisible by $3\times 8=24.$
-
The great clue, I believe, is the perennially useful fact that p^2-1 is (p+1)(p-1). Since p>3, p must be odd and cannot be divisible by three; since there is one number divisible by three in any set of three consecutive numbers either p+1 or p-1 must be divisible by three; and since every other even number is divisible by four, one of these factors must be divisible by two and the other by four. Simple arithmetic follows: 4x3x2=24.
The following is multiple choice question (with options) to answer.
If n is a natural number, then (6n2 + 6n) is always divisible by: | [
"6 only",
"12 only",
"6 and 12 both",
"by 18 only"
] | C | (6n^2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.
Answer C) 6 and 12 both. |
AQUA-RAT | AQUA-RAT-38629 | Show Tags
14 Aug 2010, 09:44
As already discussed, the method is [(Last-First)/Increment ]+1
If you are refering to mgmat by any chance, this is discussed in Chap4 of Number properties
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Re: How many multiples of 10 are there between 1000 and 2000, inclusive? [#permalink]
Show Tags
06 Sep 2017, 22:57
Bunuel wrote:
Baten80 wrote:
How many multiples of 10 are there between 1000 and 2000, inclusive?
Let me to know the calculation process.
Ans.101
Hi, and welcome to Gmat Club! Below is a solution for your problem:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.
For our original question we would have: $$\frac{2,000-1,000}{10}+1=101$$.
Check this: http://gmatclub.com/forum/totally-basic ... 20multiple
Hope it helps.
Hi Bunuel, Thanks for the explanation. However, if the question would have been that how many multiples of 5 are there between -7 and 37, inclusive? Will we consider the last number to be 35 and the first number to be -5?
Math Expert
Joined: 02 Sep 2009
Posts: 55236
Re: How many multiples of 10 are there between 1000 and 2000, inclusive? [#permalink]
Show Tags
06 Sep 2017, 23:07
SinhaS wrote:
Bunuel wrote:
Baten80 wrote:
How many multiples of 10 are there between 1000 and 2000, inclusive?
Let me to know the calculation process.
Ans.101
Hi, and welcome to Gmat Club! Below is a solution for your problem:
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.
The following is multiple choice question (with options) to answer.
What is the average of all the multiples of ten from 10 to 200 inclusive? | [
"90",
"95",
"100",
"105"
] | D | The multiples of ten from 10 to 190 inclusive would be an evenly spaced set with 19 terms - 10, 20, 30,.......190
So average = (First term + Last Term)/2 = (10 + 200)/2 = 210/2 = 105
Hence, the correct answer is D. |
AQUA-RAT | AQUA-RAT-38630 | We can argue that the are 21/7= 3 times as many men now so they will require three times as much water, 3(13)= 39 quarts just for the same two weeks. And they need the water for 4/2= 2 times as many weeks so need 2(39)= 78 quarts of water.
The following is multiple choice question (with options) to answer.
Jill has 28 gallons of water stored in quart, half-gallon, and one gallon jars. She has equal numbers of each size jar holding the liquid. What is the total number of water filled jars? | [
"3",
"6",
"48",
"12"
] | C | Let the number of each size of jar = wthen 1/4w + 1/2w + w = 28 1 3/4w = 28w=16The total number of jars = 3w =48Answer: C |
AQUA-RAT | AQUA-RAT-38631 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 1136, 7636 and 15628 votes respectively. What percentage of the total votes did the winning candidate got? | [
"55%",
"64%",
"57%",
"58%"
] | B | Total number of votes polled = (1136 + 7636 + 15628) = 24400
So, Required percentage = 15628/24400 * 100 = 64%
ANSWER : B |
AQUA-RAT | AQUA-RAT-38632 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 500 meter long train moving with a speed of 63 km/hr take to cross a man walking with a speed of 3 km/hr in the direction of the train ? | [
"33",
"87",
"30",
"88"
] | C | Explanation:
Distance = 500 m
Speed of the train relative to man= ( 63 - 3 ) km/hr = 60 km/hr
= m/s = m/s
Time taken to pass the man = distance/speed = sec = 30 sec
Answer: C) 30 |
AQUA-RAT | AQUA-RAT-38633 | $$\frac{\left(\begin{array}{c}12\\7\end{array}\right)}{7^{5}}$$
If I got it right it will be $$\approx$$ 4.7123 %. Simulating $$10^7$$ weeks in Octave give me 4.7031% of the time.
• Just saw this because I'm doing the same problem, but this answer is actually incorrect. The correct probability is $\approx$ 22.85 %. – Ryker Feb 8 at 21:35
• Finally someone slightly awake. – mathreadler Feb 8 at 21:44
The following is multiple choice question (with options) to answer.
What will be the fraction of 10% | [
"1/10",
"1/50",
"1/75",
"1/25"
] | A | Explanation:
10*1/100 = 1/10
Option A |
AQUA-RAT | AQUA-RAT-38634 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Abhishek....
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Joined: 02 Sep 2009
Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A computer factory produces 3024 computers per month at a constant rate, how many computers are built every 30 minutes assuming that there are 28 days in one month? | [
"2.25.",
"3.125.",
"4.5.",
"5.225."
] | A | Number of hours in 28 days = 28*24
Number of 30 mins in 28 days = 28*24*2
Number of computers built every 30 mins = 3024/( 28*24*2) = 2.25
Answer A |
AQUA-RAT | AQUA-RAT-38635 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
A train travels 290 km in 4.5 hours and 400 km in 5.5 hours.find the average speed of train. | [
"80 kmph",
"69 kmph",
"70 kmph",
"90 kmph"
] | B | As we know that Speed = Distance / Time
for average speed = Total Distance / Total Time Taken
Thus, Total Distance = 290 + 400 = 690 km
Thus, Total Speed = 10 hrs
or,Average Speed = 690/10
or, 69 kmph.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38636 | 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.
6 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work? | [
"12/9 days",
"6 days",
"82/8 days",
"22/76 days"
] | B | 6M = 6W ---- 20 days
12M + 8W -----?
12W + 8 W = 20W ---?
6W ---- 20 20 -----?
6 * 20 = 20 * x => x = 6 days
Answer:B |
AQUA-RAT | AQUA-RAT-38637 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election, candidate Douglas won 54 percent of the total vote in Counties X and Y. He won 62 percent of the vote in County X. If the ratio of people who voted in County X to County Y is 2: 1, what percent of the vote did candidate Douglas win in County Y? | [
"25%",
"28%",
"32%",
"38%"
] | D | given voters in ratio 2:1
let X has 200 votersY has 100 voters
for X 62% voted means 62*200=124 votes
combined for XY has 300 voters and voted 54% so total votes =162
balance votes=162-124=38
As Y has 100 voters so 38 votes means 38% of votes required
Ans D |
AQUA-RAT | AQUA-RAT-38638 | 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.
An engineer undertakes a project to build a road 10 km long in 150 days and employs 30 men for the purpose. After 50 days, he finds only 2 km of the road has been completed. Find the (approximate) number of extra men he must employ to finish the work in time. | [
"22",
"30",
"15",
"18"
] | B | 30 workers working already
Let x be the total men required to finish the task in next 100 days
2 km done hence remaining is 8 km
Also, work has to be completed in next 100 days (150 - 50 = 100)
We know that, proportion of men to distance is direct proportion
and, proportion of men to days is inverse proportion
Hence, X = (30 * 8 * 50) / (2 *100)
thus, X = 60
Thus, more men needed to finish the task = 60-30=30
Answer : B |
AQUA-RAT | AQUA-RAT-38639 | 1. ## Last two digits...
I have to find the last two digits of 222227^2010.
So I know that this gets reduced to 27^2010 mod 100
But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong?
I have to find the last two digits of 222227^2010.
So I know that this gets reduced to 27^2010 mod 100
But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong?
$\phi(100)=40$ and $(27,100)=1\implies 27^{40}\equiv1\bmod{100}$
Now $2010=50\cdot40+10 \implies 27^{2010}=27^{50\cdot40+10}=\left(27^{40}\right)^{ 50}\cdot27^{10}\equiv1\cdot27^{10}=27^{10}\bmod{10 0}$.
Observe $10=8+2=2^3+2$
$27^2\equiv 29\bmod{100}$
$27^4=\left(27^2\right)^2\equiv 29^2\equiv 41\bmod{100}$
$27^8=\left(27^4\right)^2\equiv 41^2\equiv 81\bmod{100}$
Thus $27^{10}=27^{8+2}=27^8\cdot27^2\equiv 81\cdot29=2349\equiv49\bmod{100}$
Therefore the last two digits of $222227^{2010}$ are $49$.
The following is multiple choice question (with options) to answer.
Can you find the last number in the number sequence given below?
10 : 10 : 20 : 45 : 110 : 300 : ? | [
"920",
"930",
"940",
"950"
] | B | Solution:
930
Explanation:
New number = Last number * multiplication factor (increased by 0.5 every time) + sum factor (increased by 5 every time)
10
10 * 0.5 + 5 => 10
10 * 1.0 + 10 => 20
20 * 1.5 + 15 => 45
45 * 2.0 + 20 => 110
110 * 2.5 + 25 => 300
300 * 3.0 + 30 => 930 , desired answer
Answer B |
AQUA-RAT | AQUA-RAT-38640 | Leave the $99$ in the exponent of $2$, and calculate which of the twelve angles ($0,\frac\pi6,\dots,\frac{11\pi}6$) the $\frac{99\pi}6$ falls modulo $2\pi$.
The following is multiple choice question (with options) to answer.
When a number is divided by 6 &then multiply by 12 the answer is 9 what is the no.? | [
"4.5",
"5",
"4.8",
"5.6"
] | A | if $x$ is the number, x/6 * 12 = 9
=> 2x = 9
=> x = 4.5
A |
AQUA-RAT | AQUA-RAT-38641 | 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 completes a work in 4 days and B complete the same work in 12 days. If both of them work together, then the number of days required to complete the work will be? | [
"3 days",
"11 days",
"21 days",
"22 days"
] | A | If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 4 × 12/ 16 = 3 days.
Option 'A' |
AQUA-RAT | AQUA-RAT-38642 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
What is the dividend. divisor 17, the quotient is 9 and the remainder is 5 | [
"145",
"148",
"150",
"153"
] | D | D = d * Q + R
D = 17 * 9 + 5
D = 153 + 5
D = 158 |
AQUA-RAT | AQUA-RAT-38643 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a girl sitting at window seat in the slower train in 32 seconds. Find the length of the faster train ? | [
"170 m",
"100 m",
"270 m",
"320 m"
] | D | Explanation:
Relative speed = (72 - 36) x 5/18 = 2 x 5 = 10 mps.
Distance covered in 32 sec = 32 x 10 = 320 m.
The length of the faster train = 320 m.
ANSWER IS D |
AQUA-RAT | AQUA-RAT-38644 | Fact: there is an equal probability of the last passenger sitting in Seat 1 and Seat 100.
By the time the last passenger boards, the outcome of the lost boarding pass problem is already determined, as there is only one seat for them to choose. We must therefore look at the decisions faced by Passengers 1–99.
For Passenger 1, there is equal probability of choosing any of the 100 seats. By extension, the probability of him choosing his own assigned seat and the probability of him choosing the last passenger’s assigned seat are equal.
The only way Passengers 2–99 sit in Seat 1 or Seat 100 is if their assigned seat is occupied. In this case, there is also an equal probability of sitting in any available seat.
While both Seat 1 and Seat 100 are unoccupied, it is equally probable that either seat will be chosen. After one of these two seats is occupied, the other seat is guaranteed to remain unoccupied until the last passenger boards. The probability that Seat 100 is occupied by a previous passenger is 1/2.
Here's the statement of the question from Blitzstein, Introduction to Probability (2019 2 ed), Chapter 1, Exercise 61, p 42.
1. There are 100 passengers lined up to board an airplane with 100 seats (with each seat assigned to one of the passengers). The first passenger in line crazily decides to sit in a randomly chosen seat (with all seats equally likely). Each subsequent passenger takes their assigned seat if available, and otherwise sits in a random available seat. What is the probability that the last passenger in line gets to sit in their assigned seat? (This is a common interview problem, and a beautiful example of the power of symmetry.)
Why does this post require moderator attention?
Why should this post be closed?
The following is multiple choice question (with options) to answer.
On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship. If 50 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship's passengers held round-trip tickets? | [
"33.3%",
"40%",
"50%",
"60%"
] | B | 0.20P = RT + C
0.5(RT) = no C
=> 0.50(RT) had C
0.20P = 0.50(RT)
RT/P = 40%
Answer - B |
AQUA-RAT | AQUA-RAT-38645 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Two quarts containing 7⁄8 water and 1/8 formula are mixed with three quarts containing 7⁄10 water and 2⁄10 formula. Approximately what percent of the combined five-quart mixture is water? | [
"60%",
"55%",
"65%",
"77%"
] | D | 7/8 of 2 quarts = 14/8 quarts of water
7/10 of 3 quarts = 21/10 quarts of water
14/8 = 1.75
21/10 = 2.10
Total water = 1.75 + 2.10= 3.85
Total quarts = 5
3.85/5 = 77%
Answer : D |
AQUA-RAT | AQUA-RAT-38646 | # String probability (with conditional prob and combinations)
I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.
SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.
Q1. What is the probability that it contains exactly one letter that is a vowel?
A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)
Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?
A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))
SET2: How many distinct permutations of the letters in "letters"...
Q1. Begin with two vowels?
A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...
Q2. Begin with two e's or end with two t's?
A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.
Q3. Have the vowels together?
A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.
Thank you!
The following is multiple choice question (with options) to answer.
If two letters are chosen at random (without replacement) from the word COMPUTER, what is the probability that none of the letters are vowels? | [
"2/7",
"3/7",
"3/14",
"5/14"
] | D | The number of ways to choose two letters is 8C2 = 28
The number of ways to choose two consonants is 5C2 = 10.
P(no vowels) = 10/28 = 5/14
The answer is D. |
AQUA-RAT | AQUA-RAT-38647 | If your teacher insists on using Inclusion Exclusion, then let $$A_i$$ be the set of arrangements s.t. person $$i$$ gets his/her own gift. Then I-E gives:
$$answer = n^n - \sum_i |A_i| + \sum_{i
Now, $$|A_i| = n^{n-1}$$ since the other $$n-1$$ gifts are unconstrained and each can go to any of $$n$$ recipients. Similarly, $$|A_i \cap A_j| = n^{n-2}$$ because the other $$n-2$$ gifts are unconstrained and each can go to any of $$n$$ recipients. Similarly for higher terms. So:
$$answer = n^n - {n \choose 1} n^{n-1} + {n \choose 2} n^{n-2} - {n \choose 3} n^{n-3} + ... = \sum_{k=0}^n {n \choose k} (-1)^k n^{n-k}$$
But the last summation is exactly the binomial expansion of $$(n-1)^n$$.
The following is multiple choice question (with options) to answer.
The R students in a class agree to contribute equally to buy their teacher a birthday present that costs y dollars. If e of the students later fail to contribute their share, which of the following represents the additional number of dollars that each of the remaining students must contribute in order to pay for the present? | [
" y/R",
" y/(R-x)",
" xy/(R-x)",
" ey/(R(R-e))"
] | D | y/(R-e)- y/R
if we simplify this we get choice D, which is the correct answer. |
AQUA-RAT | AQUA-RAT-38648 | # 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 sale price of an article including the sales tax is Rs. 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 20%, then the cost price of the article is: | [
"500",
"277",
"466.7",
"297"
] | C | Explanation:
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (100 * 560)/120 = Rs. 466.7
Answer:C |
AQUA-RAT | AQUA-RAT-38649 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
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02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
One fourth of a solution that was 10 percent oats by weight was replaced by a second solution resulting in a solution that was 16 percent oats by weight. The second solution was what percent oats by weight ? | [
"34%",
"24%",
"22%",
"18%"
] | A | EQ can be written as ...
let sol be s... so (3/4)s*(.1)+(1/4)s*t=s*(.16).... we get t(% of oats in 2nd sol)=34%=A |
AQUA-RAT | AQUA-RAT-38650 | velocity, dimensional-analysis
Title: Why is meters/second the same as meters per second? In quantities such as speed where the derived (SI) unit is m/s, why do we pronounce it and interpret it as meters per second? My guess is that 1 m is associated with 1 second. Similarly, 5 m/s is pronounced and interpreted as 5 meters per second, because 5 meters are associated with 1 second. I am not sure whether this view is naive. It is an instantaneous value. The speed, in m/s (or any other unit) is:
$$\frac{ds}{dt} $$
where $s$ is distance/displacement. If speed is constant, it really doesn't matter: an object moving at 5 m/s will cover 5 m in every second.
It will be very different if the speed varies. Imagine a dropping stone, subject to a 10m/s/s acceleration. At 1 s, the instantaneous speed is 10 m/s, but that speed occurs for exactly zero time.
The following is multiple choice question (with options) to answer.
Express a speed of 18 kmph in meters per second? | [
"5 mps",
"10 mps",
"6 mps",
"7 mps"
] | A | 18 * 5/18
= 5 mps
Answer:A |
AQUA-RAT | AQUA-RAT-38651 | 5. Originally Posted by ANDS!
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
Thanks.
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
Why aren't these included in the answer given in the post above?
6. Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
7. Soroban only considered these 2 cases:
Originally Posted by Soroban
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
The following is multiple choice question (with options) to answer.
In a garden, there are five blue flowers, five red flowers,five green flowers, and five pink flowers. What is the probability that a florist will choose three flowers of the same color when randomly picking three flowers? | [
"11/10",
"3/95",
"31/10",
"3/55"
] | B | P(all the same color) = P(1st flower is ANY colorAND2nd flower is same as firstAND3rd flower is also the same color)
= P(1st flower is ANY color)xP(2nd flower is same as 1st)ANDP(3rd flower is the same color)
= 1x4/20x3/19
= 3/95
Answer: B |
AQUA-RAT | AQUA-RAT-38652 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
An electrical appliances store sold this month 400% more than the average of all the other months in the year. The sales total for this month was approximately what percent of total sales for this year? | [
"14%",
"21%.",
"31%.",
"37%."
] | B | The average sales for all the other months (11 months) of the year is x.
So, the total sales of those 11 months will be= 11x
Now, this month's sales, which is 400% more than the average of other months' sales, is 5x, (x+400% of x).
Thus, this month's' sales are (5x/16x)*100=31.25% of the total yearly sales. [Total sales=11x+5x=16x]
Answer is B |
AQUA-RAT | AQUA-RAT-38653 | • +1 though irritated that you just scooped me! Apr 20, 2016 at 17:16
• Thanks for the answer. I have made an appropriate system of equations and solved it but my results differ from those from the answers in my textbook. I got $3x^3-2x^2+3x+1$ (confirmed by WolframAlpha: m.wolframalpha.com/input/?i=%28x^2-x-3%29%28ax%2Bb%29%2B13x-2%3D%28x^2-2x%2B5%29%28cx%2Bd%29-1-7x&x=0&y=0 ) but according to the textbook the answer is $3x^3-5x^2+6x+4$ could you please verify if I have done this correctly? Apr 20, 2016 at 18:03
• It's easy enough to check. The remainders on dividing $3x^3 - 2x^2 + 3 x + 1$ by $x^2-x-3$ and $x^2 -2x+5$ are $13x+4$ and $-4x-19$ respectively. So you must have done something wrong. The textbook's answers are correct. Apr 20, 2016 at 19:46
The following is multiple choice question (with options) to answer.
If m=9^(x−1), then in terms of m, 3^(4x−3) must be which of the following? | [
"m/3",
"9m",
"3m^2",
"m^2/3"
] | C | m = 9 ^ (X-1)
m = 3 ^ (2x-2)
m^2 = 3 ^ (4x-4)
3m^2 = 3 ^ (4x-3)
Answer C |
AQUA-RAT | AQUA-RAT-38654 | 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 can do a piece of work in 12 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in? | [
"15 5/8 days",
"10 5/8 days",
"12 5/8 days",
"19 5/8 days"
] | B | B
(x – 5)/12 + x/20 = 1
x = 10 5/8 days |
AQUA-RAT | AQUA-RAT-38655 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
the egg vendor calls on his 1st customer and sells his half his eggs and half an egg.to the 2nd customer he sells half of what he had left and half an egg,and to the 3rd customer he sells half of what he had then left and half an egg.by the way he did not break any eggs.in the end 3 eggs were remainig.how many did he start out with? | [
"21",
"31",
"41",
"51"
] | B | He start out with 31eggs.
In the end, he have 3eggs left.
So,
3*2 +1= 7
7*2 +1= 15
15*2 +1= 31
31eggs he had initially.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38656 | homework-and-exercises
I understand that F is directly proportional to m and F is also directly proportional to a as well. Therefore as m and a increases F also increases, but the formula could also be F=m+a, in this case also F increases as m and a increases.
Like if we redefine every formulas in terms of addition instead of product then what difference would it make? and if i see a phenomena that increases with respect to a as well as b then why should I multiply them together? The deep answer is the group structure of the physical theory. A mathematical group is a set of objects that can be pairwise combined ("multiplied") so the result is another object in the group, there is an invariant object that leaves other objects unchanged (often called "1"), and inverses, so that $a \times a^{-1}=1$. Normal multiplication (the objects can be real numbers or rational numbers) is an obvious example, but many other things can obey group structures. Adding numbers is another group structure where the invariant object is zero and the inverses negative numbers, rotations make up other groups, and so on... The point is that this is much more fundamental than multiplication seen as repeated addition (which only works for positive integers).
We know from observation how objects move under the influence of different forces and masses, and can make a mathematical model of it. Double the force, and the acceleration doubles. The proportionality of acceleration to force by definition means $a = k F$ for some constant $k$. Similarly we see that there should be a mass term, and by setting the units of mass and force right we can make $k=1$ and get a simple formula. This formula obeys the group structure of real number multiplication: I can get the same acceleration by doubling the force and mass simultaneously, there is some mass (1 kg) that makes any force accelerate exactly by the same number as we denote the force, and so on.
The following is multiple choice question (with options) to answer.
Consider the formula M = 4c^2/d. If c is doubled and d is increased by a factor of 2, by what factor is M increased? | [
"1",
"2",
"3",
"4"
] | B | We have,
M = 4c^2/d
If c is doubled and d is increased by a factor of 2, the formula changes to:
M = 4*{(2c)^2}/(2d)
M = 4*{4c^2}/(2d)
M = 4*[2c^2/d]
At first, M = 4c^2/d
After changing c and d, M = 2*4c^2/d
Hence, M changes by a factor of 2.
Answer = B = 2 |
AQUA-RAT | AQUA-RAT-38657 | Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have
\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}
And so we have our case where
\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}
The following is multiple choice question (with options) to answer.
A team of seven entered for a shooting competition. The best marks man scored 85 points. If he had scored 92 points, the average scores for. The team would have been 84. How many points altogether did the team score? | [
"288",
"581",
"168",
"127"
] | B | Explanation:
7 * 84 = 588 - 7 = 581
Answer:B |
AQUA-RAT | AQUA-RAT-38658 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
P beats Q by 125 meter in a kilometer race. Find Q's speed if P's speed is 24 meter/sec. | [
"21 meter/sec",
"7 meter/sec",
"14 meter/sec",
"18 meter/sec"
] | A | P's speed = 24 m/s
P's distance = 1000m (as it is a km race)
t = 1000/24 = 41.7 sec
Q's distance = 1000-125= 875m
Q's time to complete this distance is also 62.5 sec.
Q's speed = dist/time = 875/41.7 = 21 m/s
Hence A is the answer. |
AQUA-RAT | AQUA-RAT-38659 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible? | [
"3,024",
"4,536",
"5,040",
"9,000"
] | B | No. of ways select first digit (other than 0) * No of wasy select second digit (exclude first digit selected) * no of ways select 3rd digit (exclude first 2) * no of ways to select 4 th digit (excllude first 3 digits)
= 9*9*8*6= 4536
ANSWER:B |
AQUA-RAT | AQUA-RAT-38660 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
90, 180, 12, 50, 100, 100, ?, 3, 25, 4, 25, 2, 6, 30, 3 | [
"100",
"85",
"75",
"66"
] | C | 30*3=90
6*30=180
6*2=12
25*2=50
4*25=100
25*4=100
3*25=75 ans is 75
ANSWER:C |
AQUA-RAT | AQUA-RAT-38661 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
Of the science books in a certain supply room, 40 are on botany, 45 are on zoology, 90 are on physics, 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same area science? | [
"186",
"225",
"294",
"302"
] | C | The worst case scenario is to remove all the books on botany, zoology, and geology.
Also we can remove 79 books each from physics and chemistry.
The total is 40 + 45 + 50 + 79 + 79 = 293.
The next book removed must either be physics or chemistry, ensuring 80 books in that subject.
The answer is C. |
AQUA-RAT | AQUA-RAT-38662 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A person purchased a TV set for Rs. 16000 and a DVD player for Rs. 8500. He sold both the items together for Rs. 31150. What percentage of profit did he make? | [
"30.05%",
"13.15%",
"40%",
"27.15%"
] | D | The total CP = Rs. 16000 + Rs. 8500 = Rs. 24500 and SP = Rs. 31150
Profit(%) = (31150 - 24500 )/24500* 100 = 27.15%
ANSWER:D |
AQUA-RAT | AQUA-RAT-38663 | Thus,
200 x 0.3 = 60 received student loans
200 x 0.4 = 80 received scholarships
We are trying to determine what percent of those surveyed said that they had received neither student loans nor scholarships.
Let’s fill all this information into a table. Note that each row sums to create a row total, and each column sums to create a column total. These totals also sum to give us the grand total, designated by 200 at the bottom right of the table.
Statement One Alone:
25 percent of those surveyed said that they had received scholarships but no loans.
Using statement one we can determine the number of students who received scholarships but no loans.
200 x 0.25 = 50 students who received scholarships but no loans.
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
We are given that 50 percent of those surveyed who said they had received loans also said that they had received scholarships. From the given information we know that 60 students received loans; thus, we can determine the number of these 60 students who also received scholarships.
60 x 0.5 = 30 students who received loans who also received scholarships
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement two is sufficient to answer the question.
_________________
# Jeffrey Miller
Jeff@TargetTestPrep.com
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Re: In a survey of 200 college graduates, 30 percent said they [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
At a certain college, 60 percent of the total number of students are freshmen. If 40 percent of the fresh-men are enrolled in the school of liberal arts and, of these, 20 percent are psychology majors, what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts? | [
"4%",
"8%",
"4.2%",
"4.8%"
] | D | Let's say there is a TOTAL of100students at this college.
60 percent of the total number of students are freshmen.
# of freshmen = 60% of 100 =60
40 percent of the fresh-men are enrolled in the school of liberal arts...
Number of liberal arts freshmen = 40% of 60 =24
...and, of these, 20 percent are psychology majors...
Number of liberal arts freshmen who are psychology majors = 20% of 24=4.8
What percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts?
4.8/100= 4.8%
Answer:D |
AQUA-RAT | AQUA-RAT-38664 | # Two points of a square $K$ determine a diagonal of another square that is contained in $K$
Let $$K:=[0,1]^2$$ be a square on $$\mathbb{R}^{2}$$. We select 2 random points $$A$$, $$B$$ $$\in [0,1]^{2}$$ in this square. What is the probability that the square whose diagonal is the line segment $$AB$$, is contained in $$K$$?
I found that if we fix coordinates of $$A=(x,y)\in [0,1]^{2}$$, then the probability equals $$\int\limits_{0}^{1}\int\limits_{0}^{1}[1-(x-y)^{2}-(1-x-y)^{2}\times\textbf{1}(x+y<1)]dxdy \, ,$$ where: $$\textbf{1}(x+y<1):=1$$ if $$x+y<1$$ and $$\textbf{1}(x+y<1):=0$$ otherwise.
Some attempts(or some elements of stream of consciousness)
I tackled other problems from geometric probability, but this problem cannot be solved by standard methods(i.e. by finding dependency between given information in question, then making, at least, rough plot on cartesian coordinate system and integrate the area under the graph of detected dependencies within specific constraints). I have completely, even intuitively, idea how to come towards such solution.
The following is multiple choice question (with options) to answer.
In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's y-coordinate will be greater than its x-coordinate? | [
"7/12",
"5/12",
"3/8",
"1/3"
] | C | Now, rectangle R has an area of 3*4=12. All point that has y-coordinate greater than x-coordinate lie above the line y=xy=x, so in yellow triangle, which has an area of 1/2*3*3=4.5. So, the probability equals to favorable outcomes/total=yellow triangle/rectangle R=4.5/12=3/8.
Answer: C. |
AQUA-RAT | AQUA-RAT-38665 | kinematics, geometry
Title: Path of wheels of a bicycle Why are the wheels of a bicycle moving in concentric circles with the center O?
I know that the velocity of the back wheel is parallel to the frame of the bicycle and the velocity of the front wheel is parallel to the direction in which the wheel points.
FIXED DIAGRAM Because that is where the lines normal to the wheels and through their centers intersect. If the center of rotation was at any other point, then there would be a component of velocity perpendicular to each wheel (wheel slip) and ideal wheels roll without slip.
The following is multiple choice question (with options) to answer.
A shop sells bicycles and tricycles. In total there are 9 cycles (cycles include both bicycles and tricycles) and 21 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels. | [
"b=2,t=3",
"b=5,t=3",
"b=6,t=3",
"b=2,t=5"
] | C | Let b be the number of bicycles and let t be the number of tricycles.
Set up the equations
b+t=9............(1)
2b+3t=21............(2)
Rearrange equation (1) and substitute into equation (2)
2b+27-3b=21
-b=-6
b=6
Calculate the number of tricycles t
t=9−b
=9−6
=3
There are 3 tricycles and 6 bicycles.
Answer is C. |
AQUA-RAT | AQUA-RAT-38666 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
By selling 15 pencils for a rupee a man loses 60%. How many for a rupee should he sell in order to gain 60%? | [
"2.15",
"4.2",
"5.18",
"3.75"
] | D | 40% --- 15
160% --- ?
40/160 * 15 = 3.75
Answer: D |
AQUA-RAT | AQUA-RAT-38667 | (Now... what can we say about $A$ and the set $\{\{1, 2, 3, 4, 5\}, 1, 2, 3, 4, 5\}?)$
The following is multiple choice question (with options) to answer.
A set of numbers has the property that for any number t in the set,t+3 is in the set. If -1 is in the set, which of the following must also be in the set?
I. 2
II. 5
III. 7 | [
"II only",
"I only",
"I,II only",
"II,III only"
] | C | question is asking for must be there elements.
According to the question if t is there t+3 must be there.
if -1 is the starting element the sequence is as follows.
S ={-1,2,5,8,11....}
if -1 is not the starting element the sequence is as follows
S = {...-4,-1,2,5,8,11...}
By observing the above two sequences we can say that 2,5 must be there in set S.
Answer : C |
AQUA-RAT | AQUA-RAT-38668 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
Nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them? | [
"Rs.117",
"Rs.100",
"Rs.110",
"Rs.107"
] | A | Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.
Answer is A. |
AQUA-RAT | AQUA-RAT-38669 | 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$$
Hope it helps .
The following is multiple choice question (with options) to answer.
In what time will a train 200 meters long completely pass an electric pole, if it travels at a speed of 80kmph. | [
"7",
"9",
"5",
"10"
] | B | Sol.
Speed = [ 80x5/18] m/sec =22.2 m / sec.
Time taken = (200/22.2)sec = 9 sec.
Answer B |
AQUA-RAT | AQUA-RAT-38670 | Finally $L_3$ is: $$L_3 = c_0 + 2^{-1}c_{-1} + 2^{-2}c_{-2} + 2^{-3}c_{-3} = 4 + 0.125 = 4.125$$
This approach guarantees that $L_3 < L < L_3 + 2^{-3}$ giving an absolute error of $2^{-4} = 0.0625$ and a percent error of $1.47\,\%$ (in the worst case).
Our best guess $L^\star$ is $$L_3 + 2^{-4} = 4.15625$$ while the actual value $L$ was $\simeq 4.16992$.
The actual percent error is: $$\frac{L - L^\star}{L} \simeq 0.33\,\%$$
• omg, you are a machine – RollRoll Aug 1 '18 at 18:27
Here is one way for your specific problem. As you noted, $5^2=25$ and now keep squaring, instead of just multiplying by $5$. There is also a known trick to quickly square numbers ending in $5$, i.e. if $x = a5$ (where $a$ is any positive integer) then $x^2 = (a*(a+1))25$, for example if $5^2=25$ then $a=2$ and $a(a+1)=2\cdot 3=6$ so $$5^4=25^2 = 625$$ and $62\cdot 63 = 3906$ so $$5^8 = 625^2 = 390625,$$ so the answer will be between $8$ and $9$.
Your mistake was that $5^3=125 \ne 75$...
The following is multiple choice question (with options) to answer.
A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation? | [
"34%",
"44%",
"54%",
"64%"
] | D | Let the number be x.
Then, error =(5/3)x - (3/5) x =(16/15)x.
Error% =((16/15)x) x (3/5x)x 100% = 64%.
Answer:D |
AQUA-RAT | AQUA-RAT-38671 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
Rose went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items? | [
"19",
"19.7",
"19.9",
"20.1"
] | B | Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (25−30/100−x)6/100 = 30/100
⇒ 6(25 −0.3 −x) = 30
⇒ (25 − 0.3 − x) = 5
⇒ x = 25 − 0.3 − 5 = 19.7
B) |
AQUA-RAT | AQUA-RAT-38672 | Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 52344
### Show Tags
31 Mar 2014, 00:26
1
karimtajdin wrote:
walker wrote:
jimjohn wrote:
1. z is prime and is 3. So, x=5 SUFF.
Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
Notice that we are told that x, y, and z are integers greater than 1, hence x=25 and y=1 is not possible.
Check here for a complete solution: if-x-y-and-z-are-integers-greater-than-1-and-57122.html#p1346892
Similar question to practice: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Hope it helps.
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Joined: 14 Mar 2014
Posts: 2
### Show Tags
31 Mar 2014, 05:02
Bunuel wrote:
we are told that x, y, and z are integers greater than 1
Oh! Can't believe I missed that! It makes sense now . Thanks!
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Posts: 195
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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01 Sep 2014, 16:11
Can anyone explain whether my approach is valid?
5^2*z = 3*x^y
(x^y)/z = (5^2)/3 = (5^2a)/(3a)
x^y = 5^2a
z = 3a
(1) z is prime, so a = 1 and x^y = 25 => x = 5
S
The following is multiple choice question (with options) to answer.
If y ≠ 3 and 2h/y is a prime integer greater than 2, which of the following must be true?
I. h = y
II. y = 1
III. h and y are prime integers. | [
" None",
" I only",
" II only",
" III only"
] | A | Note that we are asked which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
So we should try to make the listed statements not true for some h and y (which satisfy y≠3 and 2h/y=prime>2).
I. h = y --> not necessarily true: h=3 and y=2 (2h/y=3=prime>2);
II. y=1 --> not necessarily true: h=3 and y=2 (2h/y=3=prime>2);
III. h and y are prime integers --> not necessarily true: h=10 and y=4 (2h/y=5=prime>2).
Answer: A. |
AQUA-RAT | AQUA-RAT-38673 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg? | [
"5/6",
"5/9",
"5/1",
"5/3"
] | A | Let us say the ratio of the quantities of cheaper and dearer varieties
= x : y
By the rule of allegation, x/y
= (87.5 - 7.50) / (7.50 - 6) =5/6
Answer: A |
AQUA-RAT | AQUA-RAT-38674 | $4m^2n^2 + (m^4-2m^2n^2+n^4) = (m^2+n^2)^2$
$m^4+2m^2n^2+n^4=(m^2+n^2)^2$
$(m^2+n^2)^2 = (m^2 + n^2)^2$
-
HINT $\$ By difference of squares $\rm \ (m^2+n^2)^2 - (m^2-n^2)^2 =\ (2\:m^2)\ (2\:n^2)\: =\: (2\:m\:n)^2$
The following is multiple choice question (with options) to answer.
If m > 1 and n = 2^(m−2), then 4^m = | [
"16n^2",
"4n^2",
"n^2",
"n^2/4"
] | A | n = 2^(m-2) = 2^m/4
2^m = 4n
4^m = (2^m)^2 = (4n)^2 = 16n^2
The answer is A. |
AQUA-RAT | AQUA-RAT-38675 | special-relativity, time-dilation
#2 already had a head start of 1.4434 s, at t = 5 in frame S, clock #2 will read 1.4434 + 4.3301 = 5.7735 s, and that's the moment at which the firecracker goes off next to it.
The following is multiple choice question (with options) to answer.
If a light flashes every 6 seconds, how many times will it flash in 1/2 of an hour? | [
"301 times",
"351 times",
"401 times",
"451 times"
] | A | In 1/2 of an hour there are 30*60 = 1800 seconds
The number of 6-second intervals = 1800/6 = 300
After the first flash, there will be 300 more flashes for a total of 301.
The answer is A. |
AQUA-RAT | AQUA-RAT-38676 | Statement 2: $$a>20$$. Nothing mentioned about $$b$$. Insufficient
Combining 1 & 2: $$a>20 =>2a>40$$. Add $$35$$ to both sides
$$2a+35>40+35 => a+b>75$$. Sufficient
Option C
Is the average (arithmetic mean) of a and b greater than 30? [#permalink] 04 Jan 2018, 11:07
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The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of a and b is 50 and the average of b and c is 70, what is the value of c − a? | [
"40",
"50",
"90",
"140"
] | A | -(a + b = 100)
b + c=140
c-a=40
A.40 |
AQUA-RAT | AQUA-RAT-38677 | Let $y_2$ be any common divisor of $(a \cdot b)$ and $(a+b)$. Let $p_y$ be any odd prime factor of $y_2$. Then $p_y$, being prime, divides at least one of $a,b$. WLOG let us assume $p_y|a$. Then $a + b \equiv 0 + b \equiv b$ $($mod $p_y)$. Recall that $p_y|(a+b)$, therefore $0 \equiv a + b \equiv b$ $($mod $p_y)$. But this means $p_y$ is a common nontrivial divisor of $a$ and $b$, contradicting that $GCD(a,b)=1$.
Thus, our assumption that $p_y$ exists must have been wrong. Then this implies that $GCD(a \cdot b, a + b) = 1$.
Now, let $p_3$ be any common PRIME divisor of $(2 \cdot a \cdot b)$ and $(a+b)$. We already showed $p_3$ does not divide $a \cdot b$, therefore $p_3$ divides the only remaining factor of $2 \cdot a \cdot b$, which is $2$. Thus, $p_3 = 2$ (since $2$ is prime). This implies that any common divisor is a power of $2$. Notice that this also implies that both $a$ and $b$ are odd (whenever the GCD we are looking for is greater than 1). Thus, $4$ does not divide $(2 \cdot a \cdot b)$ so the only possible nontrivial common divisor is $2$. (IOW, other than $2$, the only possible divisor is $1$).
The following is multiple choice question (with options) to answer.
If y is an integer and x = 3y + 2, which of the following CANNOT be a divisor of x? | [
"4",
"6",
"8",
"9"
] | B | Just to add some more to Bunuel's explanation, if a number is a multiple of 3, it will be in a form:
n = 3*k, where k is an integer
And if a number is a multiple of 6, it will be in a form :
m = 6*l, where l is an integer
so m = 2 * 3 * l = (2*l) * 3 = 3 * p, where p is an integer.
In other words, the prime factorization of m must have 3 as a factor, if it's a multiple of 6, but the expression can't have that (as explained by Bunuel), so it can't be a multiple of 6 either.
Answer is B |
AQUA-RAT | AQUA-RAT-38678 | The statement
"(the integer referred to by the decimal symbol 1) + (the integer referred to by the decimal symbol 1) = (the integer referred to by the decimal symbol 2)"
is indeed always true. Whether or not this is expressed in symbols as
"1 + 1 = 2"
depends on how you choose to represent integers.
The following is multiple choice question (with options) to answer.
If 1/a + 1/a + 1/a is an integer, which of the following must also be an integer? | [
"12/a",
"3/a",
"6/a",
"9/a"
] | B | 1/a + 1/a + 1/a
=3/a
Answer B |
AQUA-RAT | AQUA-RAT-38679 | # Formulation and computation of “the” unique median of an even-sized list
Consider an even-sized set of numbers $X = \{x_k\}$, such as $X = \{1, 2, 7, 10\}$.
The median $m$ is defined as:
$$m = \mathrm{arg \min_x} \sum_k \lvert x_k - x\rvert^1$$
Any $m \in [2, 7]$ is a minimizer of this function, and is therefore "a" median of this list.
Now, it is common practice to take the average of 2 and 7 and call it "the" median.
But that's lame, and I think I have invented (?) a more logical way to find a unique median $m^*$:
$$m^* = \lim_{\epsilon \to 0^+} \mathrm{arg \min_x} \sum_k \left\lvert x_k - x\right\rvert^{1+\epsilon}$$
Diferentiation to find the minimum only gets us so far:
$$\sum_k \mathrm{sgn}{\left(x_k - m^*\right)}\left\lvert x_k - m^*\right\rvert^{\epsilon} = 0$$ This expression can be solved numerically for smaller and smaller $\epsilon$ to give $m^* \approx 4.85$ in this example, and I suspect the "correct" median is in fact $m^* = 34/7$, but I don't know how to prove it.
I have 3 questions:
1. First of all, is this a well-known and/or useful approach? Does it have a name?
I came up with the new formulation myself, but I've never seen it used anywhere.
2. Is there some way to directly find the exact value of $m^*$, without numerical optimization?
If not, is there a better/faster approach than brute-force numerical optimization techniques?
The following is multiple choice question (with options) to answer.
List I: { y, 2, 4, 7, 10, 11}
List II: {3, 3, 4, 6, 7, 10}
If the median W of List I is equal to the sum of the median of list II and the mode of list II, then y equals | [
"5",
"7",
"8",
"W=9"
] | D | mode of list ii =3
median W of list ii =4+6/2 =5
sum of mode + mean = 3+ 5=8
now to make 8 as the median we need to find a value of y such that if the no. of terms in list 1 are odd then y=8 else if even then 7+y /2=8
here its even so
7+y/2=8 from this y= 9 (D) |
AQUA-RAT | AQUA-RAT-38680 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If y exceeds x by 26%, then x is less than y by? | [
"16 2/8%",
"20 8/3%",
"20 40/63%",
"76 2/3%"
] | C | X=100 y=126
126------26
100-------? => 20 40/63%
Answer:C |
AQUA-RAT | AQUA-RAT-38681 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
5
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3
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.5 the sixth number is? | [
"11.4",
"11.3",
"11.8",
"12.1"
] | D | 1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.5 = 69
63 + 69 = 132 – 119.9 = 12.1
6th number = 12.1
Answer: D |
AQUA-RAT | AQUA-RAT-38682 | (A) 9
(B) 12
(C) 18
(D) 24
(E) 27
Weight of 2nd piece = 4 pound
Since the weight is directly proportional to the square of its length., we may write
$$\frac{16}{36^2}$$ = $$\frac{4}{x^2}$$
Solving above, we get x = 18
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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the [#permalink]
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30 Jun 2017, 04:02
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?
(A) 9
(B) 12
(C) 18
(D) 24
(E) 27
$$20$$ pounds wire is cut into two pieces = $$16$$ pounds $$+$$ $$4$$ pounds
The piece weighs $$16$$ pounds is $$36$$ feet long.
Ratio of weight and length of $$16$$ pounds piece $$= \frac{16}{36^2} = \frac{4 * 4}{36 * 36} = \frac{1}{81}$$
Therefore required ratio of other part would also be $$\frac{1}{81}$$
Ratio $$= \frac{4}{x^2} =$$ $$\frac{1}{81}$$
$$x^2 = 81*4$$ $$=> x = \sqrt{81*4}$$
$$x = 9 * 2 = 18$$
Hence length of $$4$$ pounds wire $$= 18$$
The following is multiple choice question (with options) to answer.
A 70 cm long wire is to be cut into two pieces so that one piece will be 2/3th of the other, how many centimeters will the shorter piece be? | [
"35",
"20",
"28",
"36"
] | C | 1: 2/3 = 3: 2
2/5 * 70 =28
Answer:C |
AQUA-RAT | AQUA-RAT-38683 | Note: To show that the area of the polygon converges to the area of the circle, note that the area between the polygon and the circle is bounded by $n r \sin \frac{\theta_n}{2} r ( 1 - \cos \frac{\theta_n}{2})$. A calculation along the above lines shows that this converges to $0$.
-
This is very nice – Alyosha Jun 8 '13 at 12:31
You can actually convince yourself geometrically that
$$A'(r)=2 \pi r (*)$$
Intuitively, the rate of change of the area of the circle is the circumference.
Formally
$$A'(r) = \lim_{\Delta r \to 0} \frac{A(r+\Delta r) -A(r)}{\Delta r}$$
Now, geometrically it is pretty clear (but not really easy to prove mathematically) that the area of a corona between circles satisfies
$$2 \pi r_\text{in} (r_\text{out}-r_\text{in}) < \text{Area} < 2 \pi r_\text{out} (r_{out}-r_{in})$$
Using these inequalities it is easy to calculate the above limit which leads to $(*)$.
The formula you provided solves $(*)$ for $A(r)$.
-
Consider a circle of radius $r$ with $O$ as its center.
Now, consider an arc $XY$of the circle which subtends an angle of $\theta$ at the center.
let, $dA$ be the area of the segment $XOY$
if we take infinitesimally small angle $\delta\theta$ then we have, $dA=\frac{1}{2}(r^2)(\sin{\delta\theta})$
we know that $\lim_{\delta\theta ->0} \frac{\sin\delta\theta}{\delta\theta} =1$
The following is multiple choice question (with options) to answer.
If the area of a circle decreases by 30%, then the radius of a circle decreases by | [
"20%",
"16%",
"36%",
"64%"
] | B | If area of a circle decreased by x % then the radius of a circle decreases by
(100−10√100−x)%=(100−10√100−30)%
= (100−10√70)%
= 100 - 84 = 16%
Answer B |
AQUA-RAT | AQUA-RAT-38684 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio of two numbers is 2 : 5 and their H.C.F. is 5. Their L.C.M. is | [
"50",
"22",
"56",
"27"
] | A | Explanation:
Let the numbers be 2x and 5x. Then, their H.C.F. = x. So, x = 5.
So, the numbers 10 and 25.
L.C.M. of 10 and 25 = 50.
Option A |
AQUA-RAT | AQUA-RAT-38685 | +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.
60 is subtracted from 50% of a number, the result is 10. Find the number? | [
"150",
"120",
"266",
"288"
] | B | (50/100) * X – 50 = 10
5X = 600
X = 120
Answer:B |
AQUA-RAT | AQUA-RAT-38686 | 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.
Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.130. How much was lent at 5%? | [
"2000",
"2777",
"2688",
"1000"
] | A | (x*5*1)/100 + [(2500 - x)*6*1]/100 = 130
X = 2000
Answer: A |
AQUA-RAT | AQUA-RAT-38687 | Kudos [?]: 53125 [5] , given: 8043
Re: problem solving question on ratios [#permalink] 16 Dec 2010, 13:47
5
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spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134
Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$
$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.
$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).
The following is multiple choice question (with options) to answer.
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats? | [
"2 : 3 : 4",
"6 : 7 : 8",
"6 : 8 : 9",
"6 : 8 : 7"
] | A | let 100% seats are there in all then mathematics increased by 40% then
5*140/100=7
physics is increased by 50% then
7*150/100=21/2
biology increased by 75% then
8*175/100=14
then ratios of increasing seats are 7:21/2:14
14:21:28
2:3:4
ANSWER:A |
AQUA-RAT | AQUA-RAT-38688 | = 0 \\ a^7+ b^7 + c^7 = -7mk^2$$ Then your identity reads $$\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\ \left( -m \right) \left( -mk^2 \right) = (mk)^2$$
The following is multiple choice question (with options) to answer.
7M-20=2M,THEN M+7 IS EQUAL TO? | [
"9",
"10",
"11",
"12"
] | C | 7m-20=2m
so, 5m=20
so, m+7=11
ANSWER:C |
AQUA-RAT | AQUA-RAT-38689 | eq2= a1*b - a*a4 - a*a3 + a2*b + a7*b - a8*c;
eq3=a1*d - a3*c - a6*c - a2*c + a4*d + a5*d;
eq4=a^2*a3 + a^2*a4 + a1*b^2 + a2*b^2 + a7*b^2 + a8*c^2;
eq5=a*a3*c - a2*b*c - a*a4*d + a1*b*d-1;
eq6=a2*c^2 + a3*c^2 + a6*c^2 + a1*d^2 + a4*d^2 + a5*d^2;
eq7=a1*b^3 - a^3*a4 - a^3*a3 + a2*b^3 + a7*b^3 - a8*c^3;
eq8=a1*d^3 - a3*c^3 - a6*c^3 - a2*c^3 + a4*d^3 + a5*d^3;
eq9=a^2*a4*d - a2*b^2*c - a^2*a3*c + a1*b^2*d;
eq10=- a3*a^2*c - a4*a^2*d + a2*b*c^2 + a1*b*d^2;
eq11 =d^2*b^2*a1+c^2*b^2*a2+a^2*c^2*a3+d^2*a^2*a4;
The following is multiple choice question (with options) to answer.
Let A = {a, b, c, d, e, f} and B = {b, d, f, g}.Find B-A | [
"{A}",
"{G}",
"{L}",
"{B}"
] | B | A = {a, b, c, d, e, f}
B = {b, d, f, g}
correct answer :B - A = {g)
B |
AQUA-RAT | AQUA-RAT-38690 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Deepak deposited $9,000 to open a new savings account that earned 12 percent annual interest, compounded semi-annually. If there were no other transactions in the account, what the amount of money in Deepak account one year after the account was opened? | [
" $10,200",
" $10,092",
" $10,105",
" $9,405"
] | B | Approach #1:
12 percent annual interest compounded semi-annually --> 6% in 6 moths.
For the first 6 moths interest was 6% of $9,000, so $540;
For the next 6 moths interest was 6% of $9,000,plus6% earned on previous interest of $540, so $540+$12=$552;
Total interest for one year was $540+$552=$1092, hence balance after one year was $9,000+ $1,092=$10,092.
Answer: B. |
AQUA-RAT | AQUA-RAT-38691 | In a bag there are 10 Black balls, 8 White balls and 5 Red balls.
Three balls are chosen at random and one is found to be Black.
Find the probability that remaining two are white.
. $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$
I see it as a Conditional Probability problem . . .
Given that at least one ball is Black,
. . find the probability that we have one Black and two White balls.
Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$
There are ${23\choose3} = 1771$ possible ways to choose 3 balls.
To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways.
. . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$
The opposite of "at least 1 Black" is "NO Blacks".
There are: . ${13\choose3} = 286$ ways to choose no Blacks.
So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls.
. . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$
The following is multiple choice question (with options) to answer.
A certain bag contains 100 balls — 50 white, 20 green, 10 yellow, 17 red, and 3 purple. If a ball is to be chosen at random, what is the probability that the ball will be neither red nor purple? | [
"0.9",
"0.75",
"0.6",
"0.8"
] | D | According to the stem the ball can be white, green or yellow, so the probability is (white + green + yellow)/(total) = (50 + 20 + 10)/100 = 80/100 = 0.8.
Answer is D |
AQUA-RAT | AQUA-RAT-38692 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article offering a discount of 5% and earned a profit of 38.7%. What would have been the percentage of profit earned if no discount had been offered? | [
"28.5",
"46",
"30",
"Data inadequate"
] | B | Giving no discount to customer implies selling the product on printed price. Suppose the cost price of the article is 100.
Then printed price = 100×(100+38.7)/(100−5) =146
Hence, required % Profit = 146 – 100 = 46%
Answer B |
AQUA-RAT | AQUA-RAT-38693 | 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.
P alone can complete a piece of work in 12 days. Work done by Q alone in one day is equal to one-third of the work done by P alone in one day. In how many days can the work be completed if P and Q work together? | [
"6 (1/4) days",
"9 days",
"7 (3/4) days",
"8 days"
] | B | Work done by P alone in one day = 1/12th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/12 of the total) = 1/36 of the total.
Work done by P and Q, working together in one day = 1/12 + 1/36 = 4/36=1/9 of the total
They would take 9 days to complete the work working together.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38694 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
Legendary Member
Posts: 2898
Joined: 07 Sep 2017
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
### GMAT/MBA Expert
GMAT Instructor
Posts: 16162
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Location: Vancouver, BC
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
A restaurant meal cost $18.5 and there was no tax. If the tip was more than 30 percent but less than 10 percent of the cost of the meal, then the total amount paid must have been between | [
"$24 and $20",
"$25 and $23",
"$38 and $40",
"$37 and $39"
] | A | The total amount for the meal was between 18.5*1.3=24.05 and 18.5*1.1=~20.35. Only option which covers all possible values of the meal is A.
Answer: A. |
AQUA-RAT | AQUA-RAT-38695 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price? | [
"Rs.1000",
"Rs.1140",
"Rs.860",
"Rs.760"
] | A | Explanation:
90%
104%
--------
14% ---- 140
100% ---- ? => Rs.1000
ANSWER IS A |
AQUA-RAT | AQUA-RAT-38696 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
A man travels 250km by train at 50km/hr, 200km by ship at 20km/hr, 300km by aeroplane at 100km/hr and 100km by car at 50km/hr. What is the average speed for the entire distance? | [
"35.3km/hr",
"42.5km/hr",
"48.9km/hr",
"52.6km/hr"
] | B | Total distance traveled = 250+200+300+100 = 850km
Total time taken = 250/50 + 200/20 + 300/100 + 100/50 = 20 hrs
Average speed =850/20 = 42.5km/hr approximately ;
Answer is B |
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