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A, B and C compete in x metre race. C completes the race 14 metres ahe
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20 Apr 2020, 09:47
20 Apr 2020, 09:47
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A, B and C compete in x metre race. C completes the race 14 metres ahead of A. B finishes 20 metres ahead of C and 32 metres ahead of A. What is A’s speed?
A. 9/10th of C's speed
B. 5/8th of B's speed
C. 14/15th of A's speed
D. 10/7th of B's speed
E. 10/9th of B's speed
Are You Up For the Challenge: 700 Level Questions
A. 9/10th of C's speed
B. 5/8th of B's speed
C. 14/15th of A's speed
D. 10/7th of B's speed
E. 10/9th of B's speed
Are You Up For the Challenge: 700 Level Questions
Re: A, B and C compete in x metre race. C completes the race 14 metres ahe
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20 Apr 2020, 10:31
20 Apr 2020, 10:31
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When C runs x metres A has done x-14 metres
When B has run x metres A has done x-32 metres and c has done x-14 metres
\frac{A }{B} = \frac{(x-14)}{(x)} = \frac{(x-32)}{(x-20)}
\(x^2 - 14x - 20x +280 = x^2 - 32x\)
\(2x=280\)
\(x=140\)
Now we have the ratios for A B and C
A's speed is 0.9C's speed or Option A.
When B has run x metres A has done x-32 metres and c has done x-14 metres
\frac{A }{B} = \frac{(x-14)}{(x)} = \frac{(x-32)}{(x-20)}
\(x^2 - 14x - 20x +280 = x^2 - 32x\)
\(2x=280\)
\(x=140\)
Now we have the ratios for A B and C
A's speed is 0.9C's speed or Option A.
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Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
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Re: A, B and C compete in x metre race. C completes the race 14 metres ahe
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20 Apr 2020, 10:43
20 Apr 2020, 10:43
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Bunuel
A, B and C compete in x metre race. C completes the race 14 metres ahead of A. B finishes 20 metres ahead of C and 32 metres ahead of A. What is Amar’s speed?
A. 9/10th of C's speed
B. 5/8th of B's speed
C. 14/15th of A's speed
D. 10/7th of B's speed
E. 10/9th of B's speed
Are You Up For the Challenge: 700 Level Questions
A. 9/10th of C's speed
B. 5/8th of B's speed
C. 14/15th of A's speed
D. 10/7th of B's speed
E. 10/9th of B's speed
Are You Up For the Challenge: 700 Level Questions
What is basically being said is:
When C completes the race, A is 14 m behind C => When C covers x meters, A covers (x-14) meters
=> Speed ratio of C and A = x : (x-14) ... (i)
When B completes the race, C is 20 m behind B => When B covers x meters, C covers (x-20) meters
=> Speed ratio of B and C = x : (x-20) ... (ii)
When B completes the race, A is 32 m behind B => When B covers x meters, A covers (x-32) meters
=> Speed ratio of B and A = x : (x-32) ... (iii)
From (ii) and (iii): Speed ratio of C and A = (x-20) : (x-32) ... (iv)
From (i) and (iv):
x / (x-14) = (x-20) / (x-32)
From Addendo: x / (x-14) = (x-20) / (x-32) = [x - (x-20)] / [(x-14) - (x-32)] = 20/18 = 10/9
=> x / (x-14) = 10/9
=> 9x = 10x - 140
=> x = 140
Speed ratio of C and A = (x-20) : (x-32) = 120 : 108 = 10 : 9
=> Speed of A = 9/10 of Speed of C
Answer A
