7
Q:

The distance of the college and home of Rajeev is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached  to college  at the normal time. What is the changed (or increased) speed of Rajeev?

 A) 28 km/h B) 30 km/h C) 40 km/h D) 20 km/h

Explanation:

Let the normal speed be x km/h, then

$\frac{80}{x}-\frac{80}{\left(x+4\right)}=1$

$⇒$${x}^{2}+4x-320=0$

$⇒$x (x + 20) - 16 (x + 20) = 0

(x + 20 ) (x - 16) =0

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore increased speed = 20 km/h

Q:

If a girl cycles at 10 kmph, then she arrives at a certain place at 1 p.m. If she cycles at 15 kmph, she will arrive at the same place at 11 a.m. At what speed must she cycle to get there at noon?

 A) 14 kmph B) 13 kmph C) 12 kmph D) 11 kmph

Explanation:

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be 't'.

Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = speed x time = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = 12 kmph.

10 383
Q:

In a daily morning walk three persons step off together. their steps measure 75 cm, 80 cm and 85 cm respectively. What is the minimum distance each should walk so that thay can cover the distance in complete steps  ?

 A) 222 m 44 cm B) 204 m C) 201 m 21 cm D) 208 m

Explanation:

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.

7 255
Q:

Three friends Rudra, Siva and Anvesh start to run around a circular stadium. They complete a revolution in 24, 36 and 30 seconds respectively. After how many minutes will they meet at the starting point ?

 A) 60 B) 120 C) 360 D) 6

Explanation:

For this we have to find the LCM of 24, 36 and 30

LCM of 24, 36 and 30 = 360 sec

360/60 min = 6 minutes.

10 462
Q:

Kamal consistently runs 240 meters a day and on Saturday he runs for 400 meters. How many kilometers will he have to run in four weeks ?

 A) 5.75 kms B) 7.36 kms C) 8.2 kms D) 6.98 kms

Explanation:

Total running distance in four weeks = (24 x 240) + (4 x 400)

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms

4 490
Q:

K and L starts walking towards each other at 4 pm at speed of 3 km/hr and 4 km/hr respectively. They were initially 17.5 km apart. At what time do they meet ?

 A) 6:00 am B) 6:30 pm C) 5:45 am D) 5:52 pm

Explanation:

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

So they meet at => 4 + 2.5 = 6:30 pm

9 683
Q:

P, Q and R start simultaneously from A to B. P reaches B, turns back and meet Q at a distance of 11 km from B. Q reached B, turns back and meet R at a distance of 9 km from B. If the ratio of the speeds of P and R is 3:2, what is the distance between A and B ?

 A) 99 B) 100 C) 89 D) 1

Explanation:

Let, Distance between A and B = d

Distance travelled by P while it meets Q = d + 11

Distance travelled by Q while it meets P = d – 11

Distance travelled by Q while it meets R = d + 9

Distance travelled by R while it meets Q = d – 9

Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11

The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9

But given Ratio of speeds of P & R => P : R = 3 : 2

=> $\frac{\left(d+11\right)\left(d+9\right)}{\left(d-11\right)\left(d-9\right)}$ = 3/2

=>  d = 1, 99

=> d = 99 satisfies.

Therefore, Distance between A and B = 99

7 454
Q:

In what time a 360 m. long train moving at the speed of 44 km/hr will cross a 140 m. long bridge ?

 A) 36 sec B) 39 sec C) 41 sec D) 43 sec

Explanation:

Speed = 44 kmph x 5/18 = 110/9 m/s

We know that, Time = distance/speed

Time = (360 + 140) / (110/9)

= 500 x 9/110 = 41 sec.

6 541
Q:

Karthik could cover a distance of 200 km in 22 days while resting for 2 hrs. per day. In how many days (approx) he will cover a distance of 250 km while resting for 2 hrs. per day and moving with 2/3rd of the previous speed ?

 A) 41.25 days B) 37.5 days C) 39.75 days D) 40 days

Explanation:

Given Karthik can cover the distance of 200 kms resting 2 hrs per day in 22 days.

Let the initial speed be '1'

Hence, Time = 22(24hrs - 2)

Noe new speed = 2/3

Let the number of days he take be 'D'

Therefore, $\frac{\mathbf{22}\left(\mathbf{24}\mathbf{-}\mathbf{2}\right)\mathbf{x}\mathbf{3}}{\mathbf{200}}\mathbf{=}\frac{\mathbf{D}\left(\mathbf{24}\mathbf{-}\mathbf{2}\right)\mathbf{x}\mathbf{2}}{\mathbf{250}}$

D = 11 x 3 x 5/4 = 41.25 days. = 41 1/4 days.