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problem solving Q

by siddhartha123 » Mon Dec 12, 2011 1:57 am
Dear all,

I was struck up with the following q's can any one
Help me out

1) A and b ran a race of 48o .In the first heat,A gives
B a head start of 48 m and beat him by 1/10 th of a
Minute.In the second heat , A gives B a head start of
144 m and is beaten by 1/30 of a minute. What is
B's speed in m/s

A)12 B)14 C)16 D)18 E)20

Ans:a
[spoiler]

2) $252 is divided among a,b,c so that for every $2 that A
Receieved ,B receives $5 and for every $3 that B recieves ,c
Receives $ 7.find the share of A?

A)15 B)20 C)17 d)27 e)23

Ans: D[/spoiler]
M.siddhartha
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by neelgandham » Mon Dec 12, 2011 9:35 am
Question 1

Total distance = 480 metres
Let Sa, Sb be the speeds of A and B respectively.

First heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-48)/Sb
A wins the race by 6 secs, Implies
((480-48)/Sb) - (480/Sa) = 6
48*((9/Sb)-(10/Sa)) = 6 - (equation 1)

Second heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-144)/Sb
A wins the race by 2 secs, Implies
((480/Sa)-(480-144)/Sb) = 2
48*((10/Sa)-(7/Sb)) = 2 - (equation 2)

Add equation 1 and 2
Sb = 12m/sec
Option A
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by neelgandham » Mon Dec 12, 2011 9:42 am
Question 2

$252 is divided among a,b,c so that for every $2 that A
Receieved ,B receives $5 and for every $3 that B recieves ,c
Receives $ 7.find the share of A?

A)15 B)20 C)17 d)27 e)23

Ratio of the amount A received to the amount B received = 2:5 = 6:15
Ratio of the amount B received to the amount C received = 3:7 = 15:35

Total parts = 6+15+35 = 56
56 parts make 252 (total amount)
6 parts make ?

? = 252*6/56 = 27

Option D

Moving it to the appropriate forum!
Last edited by neelgandham on Mon Dec 12, 2011 3:28 pm, edited 1 time in total.
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by GMATGuruNY » Mon Dec 12, 2011 11:55 am
I posted a solution here:

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by cpschott » Mon Dec 12, 2011 3:22 pm
ratio for second problem

should say 3:7
not
Ratio of the amount B received to the amount C received = 5:7 = 15:35


5:7 not= 15:35
answer is correct since this is just a typo i think
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by neelgandham » Mon Dec 12, 2011 3:29 pm
Thanks for correcting! Yes, it is a typo
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by [email protected] » Tue Dec 13, 2011 10:04 pm
Question 1

Total distance = 480 metres
Let Sa, Sb be the speeds of A and B respectively.

First heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-48)/Sb
A wins the race by 6 secs, Implies
((480-48)/Sb) - (480/Sa) = 6
48*((9/Sb)-(10/Sa)) = 6 - (equation 1)

Second heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-144)/Sb
A wins the race by 2 secs, Implies
((480/Sa)-(480-144)/Sb) = 2
48*((10/Sa)-(7/Sb)) = 2 - (equation 2)

Add equation 1 and 2
Sb = 12m/sec
Option A

Can you show me how did you add the two equations to obtain 12m/s? Thank you!
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by neelgandham » Wed Dec 14, 2011 1:43 am
Total distance = 480 metres
Let Sa, Sb be the speeds of A and B respectively.

First heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-48)/Sb
A wins the race by 6 secs, Implies
((480-48)/Sb) - (480/Sa) = 6
48*((9/Sb)-(10/Sa)) = 6 - (equation 1)

Second heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-144)/Sb
A wins the race by 2 secs, Implies
((480/Sa)-(480-144)/Sb) = 2
48*((10/Sa)-(7/Sb)) = 2 - (equation 2)

Add equation 1 and 2
Sb = 12m/sec
Option A
Can you show me how did you add the two equations to obtain 12m/s? Thank you!
Sure I will.
Equation 1 : 48*((9/Sb)-(10/Sa)) = 6
Equation 2 : 48*((10/Sa)-(7/Sb)) = 2
Let (10/Sa) = A
then
Equation 1 : 48*((9/Sb)-A) = 6
Equation 2 : 48*(A-(7/Sb)) = 2
Implies
Equation 1 : (48*(9/Sb))-48*A) = 6
Equation 2 : 48*A-(48*(7/Sb)) = 2
Add the equations
48*(9/Sb)-48*A + 48*A-48*(7/Sb) = 6+2
48*(9/Sb)-48*(7/Sb)=8
48*((9/Sb)-(7/Sb))=8
((9/Sb)-(7/Sb)) = 8/48
(9-7)/Sb = 1/6
2/Sb = 1/6
Sb = 6*2
Sb = 12m/sec

Let me know if you still find it difficult!
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by siddhartha123 » Thu Dec 15, 2011 9:21 am
neelgandham wrote:Question 1

Total distance = 480 metres
Let Sa, Sb be the speeds of A and B respectively.

First heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-48)/Sb
A wins the race by 6 secs, Implies
((480-48)/Sb) - (480/Sa) = 6
48*((9/Sb)-(10/Sa)) = 6 - (equation 1)

Second heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-144)/Sb
A wins the race by 2 secs, Implies
((480/Sa)-(480-144)/Sb) = 2
48*((10/Sa)-(7/Sb)) = 2 - (equation 2)

Add equation 1 and 2
Sb = 12m/



Thanks neel ;
M.siddhartha
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by siddhartha123 » Thu Dec 15, 2011 9:23 am
neelgandham wrote:Question 1

Total distance = 480 metres
Let Sa, Sb be the speeds of A and B respectively.

First heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-48)/Sb
A wins the race by 6 secs, Implies
((480-48)/Sb) - (480/Sa) = 6
48*((9/Sb)-(10/Sa)) = 6 - (equation 1)

Second heat
Time taken by A to complete the race = 480/Sa
Time taken by B to complete the race = (480-144)/Sb
A wins the race by 2 secs, Implies
((480/Sa)-(480-144)/Sb) = 2
48*((10/Sa)-(7/Sb)) = 2 - (equation 2)

Add equation 1 and 2
Sb = 12m/



Thanks neel ;
M.siddhartha
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