Lakshman Patel RJIT
asked
in Quantitative Aptitude
Mar 31, 2020
recategorized
Sep 12, 2020
by Lakshman Patel RJIT

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An hour after Aishwarya started from her college towards Ananya’s home, a distance of $53$ km, Ananya started from her home on the same road towards Aishwarya’s college. If Aishwarya’s speed was $4$ km per hour and Ananya’s was $3$ km per hour, how many km from Ananya’s home did the two meet?

- $24$
- $22$
- $21$
- $19.5$

1 vote

Aishwarya $\Longrightarrow \underbrace{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}_{53\;\text{km}} \Longleftarrow $ Ananya

In $1$ hour Aishwarya distance $D_{1} = 4\;\text{km/hour} \times 1\;\text{hour} = 4\;\text{km}$

Remaining distance $D = 53\;\text{km}-D_{1}\;\text{km} = 53\;\text{km}-4\;\text{km} = 49\;\text{km}$

Relative speed $S_{r} = 4\;\text{km/hour} + 3\;\text{km/hour} = 7\;\text{km/hour}$

Time when they meet $T = \dfrac{49\;\text{km}}{7\;\text{km/hour}} = 7\;\text{hours}$

Ananya distance in $7\;\text{hours} = 7\;\text{hour} \times 3\;\text{km/hour} = 21\;\text{km}.$

$\text{(OR)}$

Aishwarya distance in $7\;\text{hours} = 7\;\text{hour} \times 4\;\text{km/hour} = 28\;\text{km},$ and distance traveled by Ananya $= 49 - 28 = 21\;\text{km}.$

Thus they meet $21\;\text{km}$ from the Ananya home.

So, the correct answer is $(C).$

In $1$ hour Aishwarya distance $D_{1} = 4\;\text{km/hour} \times 1\;\text{hour} = 4\;\text{km}$

Remaining distance $D = 53\;\text{km}-D_{1}\;\text{km} = 53\;\text{km}-4\;\text{km} = 49\;\text{km}$

Relative speed $S_{r} = 4\;\text{km/hour} + 3\;\text{km/hour} = 7\;\text{km/hour}$

Time when they meet $T = \dfrac{49\;\text{km}}{7\;\text{km/hour}} = 7\;\text{hours}$

Ananya distance in $7\;\text{hours} = 7\;\text{hour} \times 3\;\text{km/hour} = 21\;\text{km}.$

$\text{(OR)}$

Aishwarya distance in $7\;\text{hours} = 7\;\text{hour} \times 4\;\text{km/hour} = 28\;\text{km},$ and distance traveled by Ananya $= 49 - 28 = 21\;\text{km}.$

Thus they meet $21\;\text{km}$ from the Ananya home.

So, the correct answer is $(C).$