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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
21 Feb 2011, 07:49

Bunuel wrote:

SOLUTION: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Let x be the speed of B. Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2 x=12

Answer: A.

eqns : ta1-tb1 = 6 tb2-ta2 =2

hence, we get the Shouldn't it be Ta=432/Sb+ 6 = 336/Sb - 2 Sb = 24 m/s

Is it not like this ? _________________

Argument : If you love long trips, you love the GMAT. Conclusion : GMAT is long journey.

What does the author assume ? Assumption : A long journey is a long trip.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
21 Feb 2011, 07:56

Expert's post

maddy2u wrote:

Bunuel wrote:

SOLUTION: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Let x be the speed of B. Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2 x=12

Answer: A.

eqns : ta1-tb1 = 6 tb2-ta2 =2

hence, we get the Shouldn't it be Ta=432/Sb+ 6 = 336/Sb - 2 Sb = 24 m/s

Is it not like this ?

As you can see 24 is not even among answer choices. OA for this question is A (12). Also discussed here: race-100472.html _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
05 Apr 2011, 06:45

Bunuel wrote:

anilnandyala wrote:

9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25% B 50% C 62.5% D. 72.5% E. 75%

N=n*(n+1)*(n+2)

N is divisible by 8 in two cases: When n is even: No of even numbers (between 1 and 96)=48 AND When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

my doubt is total even numbers is 48 + 12 , but the 12 numbers divisible by 8 already included in 48 . so why de we need to add 12

thanks in advance

See the bold part in the end of the solution:

\(n(n+1)(n+2)\) is divisible by 8 in two cases:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Hope it's clear.

Dear Bunuel Though i should not imagine myself developing easiest method than you but still I did it by simple method as below

first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12 so for each value there will be 3 pairs I.e. 12 x 3 =36 similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)

i added up 36+24=60 and hence the required probability = 60/96=.625

please correct me if i am wrong _________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
06 Apr 2011, 22:10

Bunuel wrote:

SOLUTION: 6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

since it has been mentioned that we need to draw a red and a white ball in 2 successive draws... doesn't it imply that they want the first one to be red and the second one to be white? if that is the case multiplication by 2 isn't necessary...

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
30 Jun 2011, 04:20

ALTERNATE METHOD :

this is a graphical method...seems easier... D(min)=d(perpendicular dist from origin)-1(ie radius of circle)

draw a perpendicular to the line from the origin (imagine it)....the equation of new line will be y=(-4/3)*x (no constant since it passes thru 0,0)

now find point of intersection of two lines...y=(3/4)x-3 and y=(-4/3)x gives x=36/25 and y=-48/25 distance from origin = (x^2 + y^2)^.5 = 12/5 = 2.4

thus minimum dist = 2.4-1 = 1.4

ANS : A _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

Last edited by fivedaysleft on 02 Jul 2011, 11:02, edited 1 time in total.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Jul 2011, 01:17

Hi i the 9th problem is jus dodging my brain how manyy ever times i try! i am actually struck up at the step where it is mentioned n+1/8 =p and then p=12.3 => p=12 ! i am really confused at this point...how is p=12?? could some 1 clear this doubt of mine?

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Jul 2011, 04:51

saikrishna2589 wrote:

Hi i the 9th problem is jus dodging my brain how manyy ever times i try! i am actually struck up at the step where it is mentioned n+1/8 =p and then p=12.3 => p=12 ! i am really confused at this point...how is p=12?? could some 1 clear this doubt of mine?

you have to find the number of (n+1)s for which (n+1) is divisible by 8

so (n+1) can take the values 8,16,24....96

so, total number of values (n+1) can take is 96/8 which is 12.

i think you are trying to find the absolute value of p instead of finding the number of value p can take... theres the error.

hope my explanation helped _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
15 Jul 2011, 03:01

Isn't the answer to the first question 79? The sum of n terms in an a.p is n/2{2a + (n-1)d} But this question asks the sum of even terms, which means: n/2{2(2) + (n-1)2} n/2(2n + 2) n(n+1) which is 79*80. Therefore, n equals 79.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
05 Sep 2011, 05:26

fluke wrote:

The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.

Bunuel wrote:

5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Let A's speed: A m/s Let B's speed: B m/s

1/10th of a minute = 6sec 1/30th of a minute = 2sec

Total distance of the race track = 480m

First heat; A's distance = 480m A's speed = A m/s A's time = t secs

B's distance = 480-48=432m B's speed = B m/s B's time = t+6 secs

A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds) \(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))

Second heat; A's distance = 480m A's speed = A m/s A's time = t secs

B's distance = 480-144=336m B's speed = B m/s B's time = t-2 secs

A's time = B's time + 2 \(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd

Using 1st and 2nd, \(\frac{432}{B}-6=\frac{336}{B}+2\)

\(\frac{432}{B}-\frac{336}{B}=8\)

\(\frac{96}{B}=8\)

\(B=12\)

B's speed : 12m/s

Ans: "A"

I dont have any Question on calculations made above. But i do have a clarification to ask.

While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat). ( As if the start time taken into consideration is the point from when both A and B are in action ). Why is that we are going by this. I mean why are we not considering the head start time provided to B.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
05 Sep 2011, 09:55

Expert's post

vinodmallapu wrote:

fluke wrote:

The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.

Bunuel wrote:

5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Let A's speed: A m/s Let B's speed: B m/s

1/10th of a minute = 6sec 1/30th of a minute = 2sec

Total distance of the race track = 480m

First heat; A's distance = 480m A's speed = A m/s A's time = t secs

B's distance = 480-48=432m B's speed = B m/s B's time = t+6 secs

A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds) \(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))

Second heat; A's distance = 480m A's speed = A m/s A's time = t secs

B's distance = 480-144=336m B's speed = B m/s B's time = t-2 secs

A's time = B's time + 2 \(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd

Using 1st and 2nd, \(\frac{432}{B}-6=\frac{336}{B}+2\)

\(\frac{432}{B}-\frac{336}{B}=8\)

\(\frac{96}{B}=8\)

\(B=12\)

B's speed : 12m/s

Ans: "A"

I dont have any Question on calculations made above. But i do have a clarification to ask.

While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat). ( As if the start time taken into consideration is the point from when both A and B are in action ). Why is that we are going by this. I mean why are we not considering the head start time provided to B.

Please correct me if i missed anything.

When A gives B a head start of 48 m, it means B starts from 48 m ahead of the start line while A starts from the start line. They both start the race at the same time and run till A reaches the finish line. Thereafter, only B runs for 1/10th of a minute. There is no head start time provided to B, only head start distance. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
05 Sep 2011, 10:04

2

This post received KUDOS

Expert's post

Bunuel wrote:

5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Alternative Approach here:

A starts from the start line and runs till the finish line in both the races. He would take the same time in both the cases then. On the other hand, in the first race, B takes 6 secs more than A while in the second race, B takes 2 secs less than A. So there is a time difference of 8 secs in the time taken by B in the two cases. B travels (144 - 48 =) 96 m less in the second race and taken 8 secs less in the second race. This means that in the first race, B runs a distance of 96m in 8 secs. So speed of B = 96/8 = 12 m/sec _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
05 Sep 2011, 20:32

VeritasPrepKarishma wrote:

When A gives B a head start of 48 m, it means B starts from 48 m ahead of the start line while A starts from the start line. They both start the race at the same time and run till A reaches the finish line. Thereafter, only B runs for 1/10th of a minute. There is no head start time provided to B, only head start distance.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
30 Oct 2011, 16:47

SOLUTION: 5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Ans: 1/10th min= 6 sec & 1/30th min = 2 sec Now, B takes 6 sec more(+) when given 48 m of start. B takes 2 sec less(-) when given 144 m of start. 144-48=96 & 6+2 =8 So difference of 96 m start is used by B to not only reduce the extra 6 sec he took ealier instead he completes it 2 sec before i.e. savings of 8 sec.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
27 Feb 2012, 07:33

Bunuel wrote:

SOLUTION OF 8-10 8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20 B. 25 C. 40 D. 45 E. 75

Average=50, Sum of temperatures=50*5=250 As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25

Answer: B.

Isnt the correct answer A. it says 45 is the lowest temperature. Doest that statement imply that all other numbers have to be greater than 45. Hence at least 45,00001?

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
27 Feb 2012, 08:11

Expert's post

flokki wrote:

Bunuel wrote:

SOLUTION OF 8-10 8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20 B. 25 C. 40 D. 45 E. 75

Average=50, Sum of temperatures=50*5=250 As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25

Answer: B.

Isnt the correct answer A. it says 45 is the lowest temperature. Doest that statement imply that all other numbers have to be greater than 45. Hence at least 45,00001?

No, it doesn't. There can be more than 1 temperatures equal to 45.

COMPLETE SOLUTION:

Given: T(min)=45 and average=50 --> sum of temperatures=50*5=250.

We want to maximize the range --> in order to maximize the range we need to maximize the highest temperature --> in order to maximize the highest temperature we should make all other temperatures as low as possible, so equal to 45 (lower limit) --> T(max)=250-4*45=70 --> Range=T(max)-T(min)=70-45=25.

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