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rsansiya111
asked
in Quantitative Aptitude
Feb 24

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A car 'X' in Pune starts towards Mumbai at 40 Km/hr. Fifteen minutes later, a car 'Y' in Mumbai starts towards Pune which is 150 Km away, at 55 Km/hr. Which car is nearer to Pune when they meet?

(A) X is nearer

(B) Y is nearer

(C) Both are at equal distance

(D) Not sufficient information to compute answer

(A) X is nearer

(B) Y is nearer

(C) Both are at equal distance

(D) Not sufficient information to compute answer

1 vote

Answer will be C.

first convert Km/hrs into Km/min.then X speed will be 4/6 and speed of Y will be 11/12.(just divide by 60)

After 15 min-

X will cover 10km.

Now at this time instant distance between X and Y will be 140.

Lets suppose at time instant X they both meet.

so equation will be

$\tfrac{4}{6}x=M$

$\tfrac{11}{12}x=140-M$

When we solve this question x(time) will be= $\tfrac{1680}{9}$

When we calculate distance covered by X in time we will get=58.947.

Total distance covered by x=58.947+10=68.947KM

Total distance covered by Y=81.052.

Distance of Y from Pune=150-81.052=68.94KM

**So Answer will be C.**