13. A car completes a certain journey in 8 hours. It covers half the distance at 40 km/hr. and the rest at 60 km/hr. The length of the journey is :
a. 350 km
b. 420 km
c. 384 km
d. 400 km
Answer: C
Explanation:
Let the total journey be x km.
$\displaystyle\frac{x}{2}.\displaystyle\frac{1}{{40}} + \displaystyle\frac{x}{2}.\displaystyle\frac{1}{{60}} = 8$
$ \Rightarrow \displaystyle\frac{x}{8} + \displaystyle\frac{x}{{120}} = 8$
3x + 2x = 1920$ \Rightarrow $5x = 1920 $ \Rightarrow $ x = 384
14. Ravi travelled 1200 km by air which formed $\left( {\displaystyle\frac{2}{5}} \right)$ of his trip. One third of the whole trip, he travelled by car and the rest of the journey he performed by train. The distance travelled by train was :
a. 1600 km
b. 800 km
c. 1800 km
d. 480 km
Answer: B
Explanation:
Let the total distance be x km.
Then, $\displaystyle\frac{2}{5}x = 1200 \Rightarrow x = \displaystyle\frac{{1200 \times 5}}{2} = 3000$
Distance travelled by car = $\left( {\displaystyle\frac{1}{3} \times 3000} \right)$ = 1000 km
Distance travelled by train = [(3000-(1200+1000)] km = 800 km.
15. Walking at $\displaystyle\frac{3}{4}$ of his usual speed a man is $1\displaystyle\frac{1}{2}$ hr late. Find his usual travel time.
a. 4.5 hr
b. 3 hr
c. 6 hr
d. 2.5 hr
Answer: A
Explanation:
Let usual time be t hours. We know that S × T = D, But now his speed is $\dfrac{3}{4}(s)$. So time to cover the same distance becomes $\dfrac{4}{3}(t)$.
But given that he has taken $1\displaystyle\frac{1}{2}$ hrs or $\displaystyle\frac{3}{2}$ hrs extra.
$ \Rightarrow \dfrac{4}{3}t = t + \dfrac{3}{2}$
$ \Rightarrow \dfrac{4}{3}t - t = \dfrac{3}{2}$
$ \Rightarrow \dfrac{1}{3}t = \dfrac{3}{2}$
$\therefore$ t = 4.5 hr.
Alternative method:
We know that when his speed got reduced to $\dfrac{3}{4}(s)$, his time went up by $1\displaystyle\frac{1}{2}$ or $\dfrac{3}{2}$ hr.
$ \Rightarrow $ s × t = $\dfrac{{3s}}{4}$(t +$\dfrac{3}{2}$ ) ($\because$ distance D = st is constant)
$ \Rightarrow $ $t = \dfrac{3}{4} \times \left( {t + \dfrac{3}{2}} \right)$
$ \Rightarrow t = \dfrac{3}{4}t + \dfrac{9}{8}$
$ \Rightarrow t - \dfrac{3}{4}t = \dfrac{9}{8}$
$ \Rightarrow \dfrac{1}{4}t = \dfrac{9}{8}$
$ \Rightarrow t = \dfrac{9}{2} = 4.5$ hr.
16. Two cyclists cover the same distance in 15 km/hr and 16 km/hr, respectively. Find the distance travelled by each, if one takes 16 min longer than the other does.
a. 50 km
b. 60 km
c. 64 km
d. 70 km
Answer: C
Explanation:
Let the required distance be $x$ km.
Then, difference between the times = $\dfrac{x}{{15}} - \dfrac{x}{{16}} = \dfrac{{16}}{{60}}$
$\Rightarrow \dfrac{{16x - 15x}}{{15 \times 16}} = \dfrac{{16}}{{60}}$
$\Rightarrow \dfrac{x}{{15 \times 16}} = \dfrac{{16}}{{60}}$
$ \Rightarrow x = 64 $
Hence, the required distance = 64 km
Alternative Method:
If the speeds are in the ratio 15 : 16 then the ratio of the times to cover the same distance would be in the ratio 16 : 15 or 16x and 15x respectively.
But we know that difference in the times is 16 min or 16x - 15x = 16 min.
So x = 16 min
From the above derivation, second person takes 15 × 16 min = 4 hours to cover the distance at the speed of 16 km/hr.
So distance = 4 × 16 = 64 km
17. Train A took 35 minute to cover a distance of 50 km. If the speed of train B is 25% faster than train A, it will cover the same distance in:
a. 20 min
b. 28 min
c. 30 min
d. 25 min
Answer: B
Explanation:
Let, speed of train A = 100 km/h
Then, speed of train B = 125 km/h
Ratio of speeds of trains A and B = 100 : 125 = 4 : 5
Therefore, Ratio of time taken by them to cover equal distance = 5 : 4
Given, time taken by train A = 35 minutes.
Time take by train B = $\displaystyle\frac{4}{5}$ × 35 minute = 28 minutes.
Alternative Method:
Speed of train B is faster by 25% = $\displaystyle\frac{1}{4}$
Therefore, Decrease in time taken by train B = $\displaystyle\frac{1}{5}$ × 35 minute = 7 minutes.
Therefore, Time taken by train B = 35 - 7 = 28 minutes.
18. By walking at $\displaystyle\frac{3}{2}$ of his usual speed, a man reaches his office 20 minutes earlier than usual . His usual time is :
a. 30 min
b. 60 min
c. 75 min
d. 90 min
Answer: B
Explanation:
Let the usual time = $x$ minutes.
At a speed of $\displaystyle\frac{3}{2}$ of the usual speed, the time taken is (2/3)rd of the usual time.
(usual time) - $\left( {\dfrac{2}{3}{\rm\text{of usual time}}} \right)$ = 20 min.
$ \Rightarrow x - \dfrac{2}{3}{{x}} = 20$
$ \Rightarrow \dfrac{x}{3} = 20 \Rightarrow x = 60$ min.
