either some data is missing or i m solving it the wrong way.. can u post the options??
the options are (a)45 or 23 minutes (b)63 or 12 minutes (c)60 minutes (d)19 minutes (e)25 minutes
and the solution as suggested by someone on another forum is Sat Apr 24, 2010 9:10 pm Let the distance the crew rows be d. x - speed of crew in still water. y - speed of the stream.
Solution According to question d/(x - y) = 84. And d/x - d/(x + y) = 9 or dy/{x*(x + y)} = 9. Dividing second by first we get that {y*(x - y)}/{x*(x + y)} = 9/84 = 3/28. Divide both numerator and denominator by x^2. Let y/x = k So we get that {k*(1 - k)}/(1 + k) = 3/28. From above we get a quadratic equation 28k^2 - 25k + 3 = 0. Or k = 1/7 or k = . What the question wants is d/(x + y) or (d/x)/(1 + y/x) = (d/x)/(1 + k) Now (d/x)/(1 - k) = 84. So (d/x)/(1-1/7) = 84, if we take the first value of k = 1/7. Or d/x = 84 * 6/7 = 72. So d/(x+y) = 72 - 9 = 63 If k = , then (d/x)/(1 - ) = 84 or d/x = 21. So d/(x+y) = 21 - 9 = 12. So d/(x+y) is either 63 or 12 minutes. The correct answer is (b). CAN SOMEBODY SUGGEST A SHORTER WAY?
brain teaser: Albert and Bertram plan to use,at any time,four out of the 5 tyres that they decided to take along with them in a 4800 km car journey.They plan to interchange the tyres in such a way that each tyre is used for the same number of kilometers.What is the number of kilometers for which each tyre will be used??
brain teaser: Albert and Bertram plan to use,at any time,four out of the 5 tyres that they decided to take along with them in a 4800 km car journey.They plan to interchange the tyres in such a way that each tyre is used for the same number of kilometers.What is the number of kilometers for which each tyre will be used??
Selecting 4 tyres from 5 will give 5 combinations as follows.Marking the tyres as 1-5,we get :
1234 1235 1245 1345 2345
So each combination will be used for 4800/5 = 960 kms.So every tyre will be used for 960*4 = 3840 kms.
the options are (a)45 or 23 minutes (b)63 or 12 minutes (c)60 minutes (d)19 minutes (e)25 minutes
and the solution as suggested by someone on another forum is Sat Apr 24, 2010 9:10 pm Let the distance the crew rows be d. x - speed of crew in still water. y - speed of the stream.
Solution According to question d/(x - y) = 84. And d/x - d/(x + y) = 9 or dy/{x*(x + y)} = 9. Dividing second by first we get that {y*(x - y)}/{x*(x + y)} = 9/84 = 3/28. Divide both numerator and denominator by x^2. Let y/x = k So we get that {k*(1 - k)}/(1 + k) = 3/28. From above we get a quadratic equation 28k^2 - 25k + 3 = 0. Or k = 1/7 or k = . What the question wants is d/(x + y) or (d/x)/(1 + y/x) = (d/x)/(1 + k) Now (d/x)/(1 - k) = 84. So (d/x)/(1-1/7) = 84, if we take the first value of k = 1/7. Or d/x = 84 * 6/7 = 72. So d/(x+y) = 72 - 9 = 63 If k = , then (d/x)/(1 - ) = 84 or d/x = 21. So d/(x+y) = 21 - 9 = 12. So d/(x+y) is either 63 or 12 minutes. The correct answer is (b). CAN SOMEBODY SUGGEST A SHORTER WAY?
backing solving will b a shorter method.. but i hope this type of question wont appear in gmat with this kind of weird options..
5. If P = (n)(n - 1)(n - 2) . . . (1) and n > 2, what is the largest value of integer n where P has zero as its last 6 digits and a non-zero digit for its millions place? (A) 29 (B) 30 (C) 34 (D) 35 (E) 39
OA is C, which i think is wrong. it should be A.
6. John is choosing a number n randomly from all integers from 56 to 150 inclusive. What is the probability that the number he chooses will be one where n(n + 1) is divisible by 5? (A)1/ 5 (B)19/95 (C)2/5 (D)19/94 (E)3/5
5. If P = (n)(n - 1)(n - 2) . . . (1) and n > 2, what is the largest value of integer n where P has zero as its last 6 digits and a non-zero digit for its millions place? (A) 29 (B) 30 (C) 34 (D) 35 (E) 39
OA is C, which i think is wrong. it should be A.
C is the correct answer. IMO, A cant be the answer My thinking is as follows: for having 6 zeros, u will need to have at least three, "5*2" number pairs in the units place because one "5*2" pair results in 10 i.e. a "0" in the units place. The reamining three zeros will be given by numbers ending in "0" themselves viz. 30, 20 and 10. So u will only have five zeros if u use 29 instead of 30. Now 30 is the least value of n and 34 is the greatest because if u move to 35, u will have made another "5*2" pair, i.e. added another zero...which means that u now have seven zeros instead of the required 6. Hope this makes sense!
5. If P = (n)(n - 1)(n - 2) . . . (1) and n > 2, what is the largest value of integer n where P has zero as its last 6 digits and a non-zero digit for its millions place? (A) 29 (B) 30 (C) 34 (D) 35 (E) 39
OA is C, which i think is wrong. it should be A.
6. John is choosing a number n randomly from all integers from 56 to 150 inclusive. What is the probability that the number he chooses will be one where n(n + 1) is divisible by 5? (A)1/ 5 (B)19/95 (C)2/5 (D)19/94 (E)3/5
for the 1st question.. me too getting 29 (option A) as the answer.. number of '5's will be deciding factor.. number of 2s will be redundant.. 5 -- 1 five 10 -- 1 five 15 -- 1 five 20 -- 1 five 25-- 2 fives sufficient.. till 29! we'll have 6 zeroes in da end.. 30! will have 7 zeroes.. answer shd b option A.
for second question.. for n(n+1) to bedivisibe by 5, n has to b 59 or 60, 64 or 65, 79 or 80, ...... so on.. there are 19 pairs = 38 total numbers that john can pick = 95
A certain city with a poplulation of 132000is to be devided into 11 distrcits and no disctrict is to have a population more than 10% greater than any other district. what is the minimum possible population that the least populated district could have? A) 10700 B) 10800 C) 10900 D)11000 E)11100
the main condition none of the district shud have population 10% greater than the other district.
Lets assume the least populated district = x and all the other districts to be 1.1x.
So the total population will become 1.1x*10 + x = 11x
So, 11x = 132000
So x = 11000
-Deepak.
Deepak I didnt get why all other districts to be 1.1x ?? Also, 1.1x*10 + x = 11x ?? it should be 11x+x = 12x and then x = 11000. Could you plz explain it?
Deepak I didnt get why all other districts to be 1.1x ?? Also, 1.1x*10 + x = 11x ?? it should be 11x+x = 12x and then x = 11000. Could you plz explain it?
Marking all the 10 districts as 1.1x is to ensure that the least populated district is getting the minimum possible value.
I have marked the calculation wrongly. It must be 12x So, 12x = 132000 and x = 11000
Arvind,Babulal,Chunilal and Dahyalal contributed some money and bought a treadmill at a cost of Rs.2640.Arvind contributed one-half of what the other three contributed in all.Babulal contributed one-third of what the other three together contributed.Chunilal contributed one-fourth of the sum that the other three gave together.How many rupees did Dahyalal contribute??
Arvind,Babulal,Chunilal and Dahyalal contributed some money and bought a treadmill at a cost of Rs.2640.Arvind contributed one-half of what the other three contributed in all.Babulal contributed one-third of what the other three together contributed.Chunilal contributed one-fourth of the sum that the other three gave together.How many rupees did Dahyalal contribute??
A+B+C+D = 2640
Given that A = 1/2(B+C+D).In otherwords, A = 1/2(2640-A) or 3/2 A = 1320
Or A = 880
Similarly, B = 1/3(A+C+D) or 4/3 B = 880 or B = 660 Similarly C = 528 & D = 572
a) tanya prepared 4 different letters to be sent to 4 different addresses. for each letter, she prepared an envelope with its correct address. If 4 leters are to be put in 4 envelopes at random, what is the probability that only 1 letter will be put into envelope with its correct address?
a) tanya prepared 4 different letters to be sent to 4 different addresses. for each letter, she prepared an envelope with its correct address. If 4 leters are to be put in 4 envelopes at random, what is the probability that only 1 letter will be put into envelope with its correct address?
1) 1/24 2) 1/8 3) 1/4 4) 1/3 5) 3/8
tanya prepared 4 different letters to be sent to 4 different addresses. for each letter, she prepared an envelope with its correct address. If 4 leters are to be put in 4 envelopes at random, what is the probability that only 1 letter will be put into envelope with its correct address?
1) 1/24 2) 1/8 3) 1/4 4) 1/3 5) 3/8
Ans: 4
The 4 letters have to be kept in 4 different envelopes. this can be done in 24 ways I.e 4,3,2,1
Only 1 letter has to be kept in the correct envelope chosing 1 letter out of 4 is 4C1 and that as to be kept in the right one only one way. the remaining 3 have to be kept in the wrong envelopes. So, 2 ways for the first of remaining 3 letters and only one for the rest of the 2 letters left. So the final eqn becomes like this
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?
1:600 2:800 3:1000 4:1200 5:1500
I got this. Ans is 1: 600 pages.
first find time to complete same task by B alone: 1/B = (1/24)- (1/60) B completes the task in 40 min. lets say there are x pages in the task. A takes 60 min to complete it so its rate is x/60 pages per minute. rate of B is x/60 + 5 pages per minute. 1 minute------> x/60+5 pages 40 minutes------> 40 (x/60+5) = x (total task x) that gives x=600 pages.
certain junior class has 1000 students and certain senior class has 800 students. among these students there are 60 sibling pairs consisting 1 junior and 1 senior. If one student is to be selected at random from each class what is the probability that the 2 students selected will be a sibling pair?