Question

Akash cycles to his school at an average speed of 12 km/h. With this speed he takes 20 minutes to reach school,but is late by 5 minutes. How should he adjust his speed so that he reaches in time?

Answer 1

Answer:-

$\sf{Akash \ should \ increase \ his \ speed \ by}$

$\sf{4 \ km \ hr^{-1}, \ i.e \ his \ speed \ from}$

$\sf{12 \ km \ hr^{-1} \ to \ 16 \ km \ hr^{-1}}$

$\sf{to \ reach \ school \ on \ time.}$

Given:

• Akash cycles to his school at an average speed of 12 km/hr

• With this speed he takes 20 minutes to reach school.

• He gets 5 minutes late.

To find:

• How should he adjust his speed so that he reaches in time?

Solution:

$\sf{Here,}$

$\sf{Speed_{1}=12 \ km \ hr^{-1},}$

$\sf{Time=20 \ minutes}$

$\sf{=\frac{20}{60}=\frac{1}{3} \ hr}$

$\boxed{\sf{Distance=Speed\times \ Time}}$

$\sf{\therefore{Distance=12\times\frac{1}{3}}}$

$\sf{\therefore{Distance=4 \ km}}$

$\sf{New, \ time \ required=20-5}$

$\sf{\therefore{Time_{2}=15 \ minutes}}$

$\sf{\therefore{Time_{2}=\frac{15}{60}=\frac{1}{4} \ hr}}$

$\boxed{\sf{Speed=\frac{Distance}{Time}}}$

$\sf{\therefore{Speed_{2}=\frac{4}{\frac{1}{4}}}}$

$\sf{\therefore{Speed_{2}=16 \ km \ hr^{-1}}}$

$\sf{Difference \ between \ Speed_{2} \ and \ Speed_{1}}$

$\sf{=16-12=4 \ km \ hr^{-1}}$

$\sf\purple{\tt{\therefore{Akash \ should \ increase \ his \ speed \ by}}}$

$\sf\purple{\tt{4 \ km \ hr^{-1}, \ i.e \ his \ speed \ from}}$

$\sf\purple{\tt{12 \ km \ hr^{-1} \ to \ 16 \ km \ hr^{-1}}}$

$\sf\purple{\tt{to \ reach \ school \ on \ time.}}$