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11 Jun 2020, 21:07
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Difficulty:
95%
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Question Stats:
38% (02:22) correct
62%
(02:27)
wrong
based on 155
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A man goes from home to office at an average speed of a km/hr. and on reaching office he realizes that he forgot his phone charger on his table at home. So, he immediately travels back to home at speed of b km/hr and reaches home. He maintained an average speed of c km/hr during his complete travel from home to office and then from office to home. is a≤10?
1) a ≤ b
2) c = 20 km/hr.
1) a ≤ b
2) c = 20 km/hr.
Source: https://www.GMATinsight.com
11 Jun 2020, 23:08
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Distance is same, say d, when the speed is a and b. The average speed is c while traveling twice the distance or 2d.
The time taken has to be same => \(\frac{d}{a}+\frac{d}{b}=\frac{2d}{c}\)
\(\frac{1}{a}+\frac{1}{b}=\frac{2}{c}\)
1) a ≤ b
we do not know the value of any of the variable.
Insuff
2) c = 20 km/hr.
\(\frac{1}{a}+\frac{1}{b}=\frac{2}{20}=\frac{1}{10}\)
As none of a and b are 0 or negative, they have to be greater than 10 to get the right side as 1/10.
Suff. a>10
B
12 Jun 2020, 01:17
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First statement is not sufficient in itself.
From second statement
C= (2ab)÷ (a+b)
10= ab÷ (a+b)
b =10a÷(a-10)
As a and b are positive so a has to be greater then 10.
Posted from my mobile device
From second statement
C= (2ab)÷ (a+b)
10= ab÷ (a+b)
b =10a÷(a-10)
As a and b are positive so a has to be greater then 10.
Posted from my mobile device
