Practice Basic problems formulas - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is

(a)

(b)

(c)

(d)

Explanation:

If the required distance be = x km, then

$x/4 - x/5 = {10 + 5}/60$

${5x - 4x}/20 = 1/4$

$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 5 km.

Using Rule 10,
If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

Here, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10

Distance =${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

=${(4 × 5)(5 + 10)}/{5 - 4}$

=20 × $15/60$ = 5 kms


Q-2)   An athlete runs 200 metres race in 24 seconds. His speed (in km/ hr) is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = $\text"Distance"/\text"Time"$

= $200/24$ m/s

$200/24$ m/s = $200/24 × 18/5$

= 30 km/h [Since, x m/s = $18/5$ x km/h]


Q-3)   A man crosses a road 250 metres wide in 75 seconds. His speed in km/hr is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = $\text"Distance"/\text"Time" = 250/75$

= $10/3$ m/sec = $10/3 × 18/5$ km/hr

[Since, 1 m/s = $18/5$ km/hr]

= 2 × 6 km/hr. = 12 km/hr.


Q-4)   The speed 3$1/3$ m/sec when expressed in km/hour becomes

(a)

(b)

(c)

(d)

Explanation:

Since, 1 m/sec = $18/5$ kmph

$10/3$ m/sec

= $18/5 × 10/3$ = 12 kmph


Q-5)   A train is moving with the speed of 180 km/hr. Its speed (in metres per second) is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = 180 kmph

= ${180 × 5}/18$ m/sec = 50 m/sec

[Since, 1 km/hr = $5/18$ m/s]


Q-6)   A car goes 10 metres in a second. Find its speed in km/hour.

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed of car = 10 m/sec.

Required speed in kmph

=${10 × 18}/5 = 36$ km/hr


Q-7)   A man riding his bicycle covers 150 metres in 25 seconds. What is his speed in km per hour ?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = $150/25$ = 6 m/sec

= $6 × 18/5 = 108/5$ = 21.6 kmph


Q-8)   A man covers $2/15$ of the total journey by train, $9/20$ by bus and the remaining 10 km on foot. His total journey (in km) is

(a)

(b)

(c)

(d)

Explanation:

Let the total journey be of x km, then

${2x}/15 + {9x}/20 + 10 = x$

$x - {2x}/15 - {9x}/20$ = 10

${60x - 8x - 27x}/60$ = 10

${25x}/60$ = 10

$x = {60 × 10}/25$ = 24 km


Q-9)   A man walking at the rate of 5 km/hr. crosses a bridge in 15 minutes. The length of the bridge (in metres) is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed of the man = 5km/hr

= $5 × 1000/60$ m/min = $250/3$ m/min

Time taken to cross the bridge

= 15 minutes

Length of the bridge

= speed × time

= $250/3 × 15$m = 1250m


Q-10)   Two men start together to walk a certain distance, one at 4 km/h and another at 3 km/h. The former arrives half an hour before the latter. Find the distance.

(a)

(b)

(c)

(d)

Explanation:

If the required distance be x km, then

$x/3 - x/4 = 1/2$

${4x - 3x}/12 = 1/2$

$x/12 = 1/2$ ⇒ x = 6 km

Using Rule 9,

Here $S_1 = 4, t_1 = x, S_2 = 3, t_2 = x + 1/2$

$S_1t_1 = S_2t_2$

$4 × x = 3(x + 1/2)$

$4x - 3x = 3/2 x = 3/2$

Distance= $4 × 3/2$ = 6 kms