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A boy walks at a speed of 10 kmph from his home and reaches his school
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21 Mar 2021, 22:21
21 Mar 2021, 22:21
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Question Stats:
36% (02:18) correct
64% (02:37) wrong based on 25 sessions
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A boy walks at a speed of \(10\) kmph from his home and reaches his school late by \(15\) minutes. Next time he increases his speed by \(2\) kmph, but still is late by \(5\) minutes.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
distance of the school from his home |
\(10\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: A boy walks at a speed of 10 kmph from his home and reaches his school
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22 Mar 2021, 03:23
22 Mar 2021, 03:23
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Distance to school = x
Assume fixed/correct time to reach school
case1
x/10 - 15/60 = correct time
case2
x/12 - 5/60 = correct time
x/10 - 15/60 = x/12 - 5/60
x = 10
Assume fixed/correct time to reach school
case1
x/10 - 15/60 = correct time
case2
x/12 - 5/60 = correct time
x/10 - 15/60 = x/12 - 5/60
x = 10
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: A boy walks at a speed of 10 kmph from his home and reaches his school
[#permalink]
22 Mar 2021, 22:50
22 Mar 2021, 22:50
Explanation:
The boy goes faster or at same speed, the distance from his home to his school is a constant
i.e. \((S_1)(T_1) = (S_2)(T_2) = D\)
Let the actual time be \(T\)
\((10)(T + \frac{15}{60}) = (12)(T + \frac{5}{60})\)
\(10T + \frac{10}{4} = 12T + 1\)
\(2T = \frac{3}{2}\)
\(T = \frac{3}{4}\) hours
Now, \(D = (10)(\frac{3}{4} + \frac{15}{60}) = (12)(\frac{3}{4} + \frac{5}{60}) = 10\)
Col. A: \(10\)
Col. B: \(10\)
Hence, option C
The boy goes faster or at same speed, the distance from his home to his school is a constant
i.e. \((S_1)(T_1) = (S_2)(T_2) = D\)
Let the actual time be \(T\)
\((10)(T + \frac{15}{60}) = (12)(T + \frac{5}{60})\)
\(10T + \frac{10}{4} = 12T + 1\)
\(2T = \frac{3}{2}\)
\(T = \frac{3}{4}\) hours
Now, \(D = (10)(\frac{3}{4} + \frac{15}{60}) = (12)(\frac{3}{4} + \frac{5}{60}) = 10\)
Col. A: \(10\)
Col. B: \(10\)
Hence, option C
