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TOUGH & TRICKY SET Of PROBLEMS

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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 ?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 21 Feb 2011, 08:33
I made a mistake ...

it is time of B - time of A = 6

changed the order ..

thnx
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 21 Feb 2011, 08:35
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 22 Feb 2011, 05:21
Very good set. Thanks!
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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).

# of even numbers between 1 and 96, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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...

IMO the answer should be A.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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
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Last edited by fivedaysleft on 02 Jul 2011, 11:02, edited 1 time in total.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 30 Jun 2011, 05:43
Traditional solution to 1:

We know,
Even numbers is an arithmetic progression (AP)
Even numbers are between 1 and K.
K is odd.
Sum of series is 79*80.

We infer,
First number - 2
Last number - K-1

AP formulae,
1. an = a1 + (n-1)*d, where an = last number, a1 = first number, n = number of elements, d = difference between to consecutive numbers

so, k-1=2 + (n-1)*2, solve for n
n = (k-1)/2

2. Sn = n/2 * (a1 + an)

So, 79*80 = (k-1)/4 * (2 + k-1)

Simplifying we get, 79*80 = (k-1)/2 * (k+1)/2

therefore k = 159

GMAT solution to 1:

I love Jax91's solution, which is great for GMAT

Sum of series of consecutive integers = n*(n+1)/2
Sum of series of odd/even integers = n*(n+1) / 4 (almost half the numbers)

if n = 159 and (n+1)/2 = 80 then n/2 is 79.5 (quick solve)
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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.

Please correct me if i missed anything.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 05 Sep 2011, 10:04
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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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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.


Thanks for the clarification.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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.

B's speed is 96/8 =12
Answer: A.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 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?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 27 Feb 2012, 08:01
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 27 Feb 2012, 08:11
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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.

Answer: B.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLMS   [#permalink] 27 Feb 2012, 08:11
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