Data Sufficiency Question 59

Directions : Each of the questions given below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements sufficient to answer the question. Read both the statements and

(Bank of Maharashtra PO Exam, 26.10.2016)

Give answer (1) if the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.

Give answer (2) if the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.

Give answer (3) if the data in statement I alone or in statement II alone are sufficient to answer the question.

Give answer (4) if the data in both the statements I and II are not sufficient to answer the question.

Give answer (5) if the data in both the statements I and II together are necessary to answer the question.

  1. How much time will train $M$ take to cross train $N$ (from the moment they meet) running in opposite directions (towards each other)?

I. Train M can cross a signal pole in 21 seconds. It can cross 440 metre long station in 43 seconds. II. The respective ratio of speeds of train $M$ and train $N$ is $2: 3$. The sum of the lengths of the train $M$ and train $N$ is 700 metre.

Show Answer

Correct Answer: 59. (5)

Solution: 59. (5) From statement I

If the length of train M be $x$ metre, then

Speed of train $\mathrm{M}=\frac{x}{21}$

$=\frac{x+440}{43}$

$\Rightarrow 43 x=21 x+21 \times 440$

$\Rightarrow 43 x-21 x=9240$

$\Rightarrow 22 x=9240$

$\Rightarrow x=\frac{9240}{22}=420$ metre

$\therefore$ Speed of train $\mathrm{M}$

$=\frac{420}{21}=20 \mathrm{~m} . / \mathrm{sec}$.

From both statements,

Speed of train $\mathrm{N}=\frac{3}{2} \times 20$

$=30 \mathrm{~m} . / \mathrm{sec}$.

$\therefore$ Relative speed

$=(30+20) \mathrm{m} . / \mathrm{sec}$.

$=50 \mathrm{~m} . / \mathrm{sec}$.

$\therefore$ Required time

$=\frac{700}{50}=14$ seconds