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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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