12
Q:

# A and B runs around a circular track. A beats B by one round or 10 minutes. In this race, they had completed 4 rounds. If the race was only of one round, find the A's time over the course:

 A) 8 min B) 7.5 min C) 12.5 min D) 12 min

Explanation:

B runs around the track in 10 min.

i.e ,    Speed of B = 10 min per round

$\inline \fn_jvn \therefore$ A beats B by 1 round

Time taken  by A to complete 4 rounds

= Time taken by B to complete 3 rounds

= 30 min

$\inline \fn_jvn \therefore$ A's speed = 30/4 min per round

= 7.5 min per round

Hence, if the race is only of one round A's time over the course = 7 min 30 sec

Q:

A man rows his boat 60 km downstream and 30 km upstream taking 3 hrs each time. Find the speed of the stream ?

 A) 5 kmph B) 10 kmph C) 15 kmph D) 45 kmph

Explanation:

Speed of the boat downstream  $\inline \fn_jvn \small s=\frac{d}{t}$ = $\inline \fn_jvn \small \frac{60}{3} = 20 kmph$

Speed of the boat upstream $\inline \fn_jvn \small s= \frac{d}{t} = \frac{30}{3} = 10 kmph$

$\fn_jvn&space;\small&space;\therefore$ The speed of the stream = $\inline \fn_jvn \small \frac{(speed of downstream - speed of upstream)}{2} = 5kmph$.

6 144
Q:

Mr. Karthik drives to work at an average speed of 48 km/hr. Time taken to cover the first 60% of the distance is 20 minutes more than the time taken to cover the remaining distance. Then how far is his office ?

 A) 40 km B) 50 km C) 70 km D) 80 km

Explanation:

Let the total distance be 'x' km.
Time taken to cover remaining 40% of x distance is   $\inline \fn_jvn t1 = \frac{40 \times x}{100\times 48}$
But given time taken to cover first 60% of x distance is   $\inline \fn_jvn t2 = t1 + \frac{20}{60}hrs$
$\fn_jvn&space;\Rightarrow$ $\inline \fn_jvn t2 = \frac{60 \times x}{100 \times 48}$

$\inline \fn_jvn \small \therefore \frac{60\times x}{100 \times 48}=\frac{40\times x}{100\times 48}+\frac{20}{60}$ $\fn_jvn&space;\small&space;\Rightarrow$ x=80 km.

6 64
Q:

Two bike riders ride in opposite directions around a circular track, starting at the same time from the same point. Biker A rides at a speed of 16 kmph and biker B rides at a speed of 14 kmph. If the track has a diameter of 40 km, after how much time (in hours) will the two bikers meet?

 A) 6.52 B) 8.14 C) 4.18 D) 5.02

Explanation:

Distance to be covered = $\inline \fn_jvn \small \prod D$ = 40 $\inline \fn_jvn \small \prod$ km

Relative speed of bikers = 16 + 14 = 30 kmph.

Now, $\inline \fn_jvn \small time = \frac{distance}{speed}$ =  $\inline \fn_jvn \small \frac{40\prod }{30}$ = 4.18 hrs.

7 145
Q:

A man is riding a bike with front and back wheel circumference of 40 inches and 70 inches respectively. If the man rides the bike on a straight road without slippage, how many inches will the man have travelled when the front wheel has made 15 revolutions more than the back wheel?

 A) 1100 B) 1300 C) 1400 D) 1200

Explanation:

Given the ratio of the circumference of front wheel  is 40 inches and back wheel is 70 inches

Distance covered = Circumference of the wheel × No. of Revolutions made by the wheel

If n is the number of revolutions made by back wheel, the number of revolutions made by front wheel is n+15

Distance covered by both the wheels is the same

40*(n+3)=70n

n=20

Front wheel    :      Back Wheel

Circumference       40      :      70

Revolutions           35      :      20

Distance covered      $\inline&space;\small&space;\Rightarrow$       40×35 = 70×20 = 1400 inches.

7 83
Q:

Two trains 85 m and 75 m long are running in same direction with speeds of 62 km/hr and 44 km/hr respectively. In what time will the first train cross the second train ?

 A) 22 sec B) 32 sec C) 42 sec D) 52 sec

Explanation:

Given that,

1st train speed = 62 km/hr

2nd train speed  = 44 km/hr

Hence the relative speed of two trains is $\inline \fn_jvn \Rightarrow$62-44 = 18 km/hr $\inline \fn_jvn \Rightarrow$ $\inline \fn_jvn 18\times \frac{5}{18}$ m/s = 5 m/s.

There fore, the first train crosses the second train i.e (85+75)mts of distance with a speed of 5m/s in,

time$\inline&space;\dpi{100}&space;\fn_phv&space;\frac{distance}{speed}$ $\inline \fn_jvn \Rightarrow$ $\inline&space;\dpi{100}&space;\fn_phv&space;\frac{(85+75)}{5}$ = 32 sec.