# QUANT QUIZ ON TIME AND DISTANCE WITH SOLUTIONS ( SHORTCUTS)

**Formulae**

(i)

**Speed**= Distance/Time
(ii)

**x km/hr**= (x × 5/18 ) m/sec.**1.**A man covers a certain distance between his house and office on scooter.Having an average speed of 30 km/hr, he late by 10 min. However, with a speed of 40 km/hr. he reaches his office 5 min earlier. Find the distance between his house and office.

(1) 20 km (2) 40 km

(3) 30 km

**(4) 25 km****2.**A man car does a journey in 10 hrs, the first half at 21 km/hr and the second half at 24 km/hr. Find the distance.

(1) 220 km (2) 215 km

(3) 210 km (4) 224 km

**3.**Walking 3/4 of his usual speed, a person is 10 min late to his office. Find his usual time to cover the distance.

(1) 10 minutes (2) 50 minutes

(3) 30 minutes (4) None of these

**4.**Running 4/3 of his usual speed, a person improves his timing by 10 minutes. Find his usual time to cover the distance.

(1) 60 minutes (2) 40 minutes

(3) 25 minutes (4) 30 minutes

**5.**A train travelling 25 km an hour leaves Delhi at 9 a.m. and another train travelling 35 km an hour starts at 2 p.m. in the same direction. How many km from Delhi will they be together?

(1) 437 ½ km (1) 436 ½ km

(3) 435 ½ km (4) 434 ½ km

**6.**Two men A and B walk from P to Q, a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P to R.

(1) 18 km (2) 20 km

(3) 35 km (4) 15 km

**7.**A man sets out to cycle from Delhi to Rohtak, and at the same time another man starts from Rohtak to cycle to Delhi. After passing each other they complete their journeys in 3 1/3and 4 4/5hours respectively. At what rates does the second man cycle if the first cycles at 8 km per hour?

(1) 6 2/3 (2)
5 2/3

(3) 4 2/3 (4)
None of these

**8.**A monkey tries to ascend a greased pole 14 metres high. He ascends 2 metres in first minute and slipsdown 1 metre in the alternate minute. If he continues to ascend in this fashion, how long does he take to reach the top?

(1) 25 minutes (2) 20 minutes

(3) 15 minutes (4) None of these

**9.**A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. Another man leaves the point Q at 8 a.m. and reaches the point P at 12 noon. At what time do they meet?

(1) 10 a.m. (2) 9 a.m.

(3) 9: 30 a.m. (4) 8 a.m

**10.**Without any stoppage a person travels a certain distance at an average speed of 80 kmph, and with stoppages he covers the same distance at an average of 60 kmph. How many minutes per hour does he stop?

(1) 25minutes (2) 15 minutes

(3) 10 minutes (4) None of these

**Answers with Solution**:

**1.**(3) :

**Solution**: Let the distance be x km.

Time taken to cover x km at 30 km/hr =
x/30 hrs

Time taken to cover x km at 40 km/hr =
x/ 40 hrs.

Difference between the time taken = 15
min = ¼ hr.

x/30 – x/40 = ¼ or 4x – 3x = 30 or x = 30 km.

**Shortcut:**

**(Try this your self)**

Required distance

= Product of two speeds/ Difference of
two speeds x
Difference between arrival times.

**2.**(4)

**Solution**:

Let the distance be x km.Then x/2 km
is travelled at a speed of 21 km/hr and x/2 km at a speed of 24 km/hr.

Then time taken to travel the whole
journey

= x/(2x21) + x/(2 x 24) = 10 hrs.

so

**, x = 224 km**

**Short cut : (Try this your self)**

Distance =(2 x Time x S1 x S2 ) / (S1 +
S2 )

Where , S1 = Speed during first half and

S2 = Speed during second half of journey

**3.**(3) :

**Solution**:

Let the usual time be x min.

Time taken at ¾ of the usual speed =
4x/3 min

4x/3 – x = 10

x/3 = 10 Ã

**x = 30 min.****Shortcut**:

**(Try this your self)**

Usual time = Late time / ( 1 ÷ ¾ - 1) =
10/1/3

**= 30 minutes**

**4.**(2)

**Shortcut**:

Improved time / (1 - 1 + 4/3 ) = 40 minutes

**5.**(1) :

**Solution**:

The first train has a start of 25 / 5
km and the second train gains ( 35 – 25 ) or 10 km per hour.

The second train will gain 25 x 5 km
in (25 x5) / 10 or 12 ½ hours.

The required distance from Delhi = 12 ½ x
35 km =

**437 ½ km**

**Shortcut**: (try this )

Meeting point’s distance from starting
point

= (S1 x S2 x Difference in time) /
Difference in speed ( S1 and S2 are the speed )

**6.**(1)

**Solution**:

When B meets A at R, B has walked the
distance PQ + QR and A the distance PR. That is , both of them have together walked
twice the distance from P to Q, i.e 42 km.

Now the rates of A and B are 3 : 4 and
they have walked 42 km.

Hence the distance PR travelled by A =
3/7 of 42 km

**= 18 km**

**7.**(1)

**Solution**:

If two persons ( or train) A and B
start at the same time in opposite directions from two points and arrive at the pint a and b hrs
respectively after having met, then

A’s rate : B’s rate =

**√b : √a**
= (√ 4 4/5 ) / ( √3 1/3 ) = 6/5

2

^{nd}man’s rate = 5/6 x 8**= 6 2/3 km/hr.**

**8.**(1)

**Solution:**

In every 2 minutes he is able to ascend 2 – 1 =
1 metre. This way he ascends upto 12 metres because when he reaches at the top.
He does not slip down. thus, upto 12 metres he takes 12 x 2 = 24 minutes and for the last 2 metres he takes 1
minutes. Therefore he takes 24 + 1 = 25 minutes to reach the top.

**9.**(2)

**Solution**:

Let the distance PQ = A km.

And they meet x hrs after the first
man starts.

Average speed of first man = A/10 – 6 =
A /4 km/hr

Average speed of second man = A / 12 –
8 = A / 4 K/hr.

distance travelled by first man = Ax/4
km

They meet x hrs and the first man
starts. The second man, as he starts 2 hrs late, meets after ( x- 2 ) has from
has start. Therefore, the distance travelled by the second man = A(x-2) / 4 km

Now , Ax/4 + A (x-2) /4
km = A

x = 3 hrs.

They meet at 6 a.m. + 3 hrs

**= 9 a.m.**

**Shortcut**:

**(Try this your self)**

= Starting time + [ ( Time taken by first
) (2

^{nd}’s arrival time – 1^{st}starting time ) ] / (sum of time taken by both)**10.**(1)

**Solution**:

Let the total distance be x km.

Time taken at the sped of 80 km/hr =
x/80 hrs.

Time taken at the sped of 60 km/hr =
x/60 hrs.

He rested for ( x/60 – x/80) hrs = 20x / 60 x 80 = x/240 hrs.

his rest per hour = x/240 ÷ x/60 = ¼ hrs.

**= 15 minutes**.