CAT 1996 | Question: 141

Question

A man travels form A to B at a speed of x kmph. He then rests at B or x hours. He then travels from B to C at a speed of $2x$ kmph and rests at C for $2x$ hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hours. If distances A-B, B-C, C-D are all equal to $12$ km. the time for which he rested at B could be:

• A. $3$ hours
• B. $6$ hours
• C. $2$ hours
• D. $4$ hours

Answer 1

Total time taken by the man to travel from A to D = 16 hr and total distance traveled = 36 km.

The time that he would have taken had he not rested in between will be (16-x-2x) = (16-3x).

But, this time should be equal to the addition of the times that he takes to travel individual segments. This is given as:

12/x + 12/2x + 12/4x = 21/x

Therefore, 21/x = 16-3x

x = 3 or 7/3

Option A is the Correct Answer.

Answer 2

Total time is the sum of all travel times and rest times.

Time (in hr) from $A$ to $B$ $= \frac{\text{Distance}}{\text{Speed}}= \frac{12}{x}$  
Rest at $B$ $=x$  (This is asked)
Time from $B$ to $C$ $=\frac{12}{2x}$
Rest at $C$ $=2x$  
Time from $C$ to $D$ $=\frac{12}{4x}$ 

Total time equation:

$\frac{12}{x}+x+\frac{12}{2x}+2x+\frac{12}{4x}=16$

$\frac{12}{x}+x+\frac{6}{x}+2x+\frac{3}{x}=16$

$\frac{21}{x}+3x=16$

Multiply by $x$:
$3x^{2}-16x+21=0$

Discriminant formula:
$D=b^{2}-4ac$

Here $a=3$, $b=-16$, $c=21$

$D=(-16)^{2}-4(3)(21)=256-252=4$

$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{16\pm2}{6} = 3 \text{ or } \frac{7}{3}$ 

Answer: $\text{A) 3 hours}$

Answer 3

answer a 3hours