All questions of Time & Work for Interview Preparation Exam

Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,

Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 

∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t² + 15t
⇒ 3t² − 5t − 50 = 0
⇒ 3t² + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 

Thus, Total time taken by slower horse = 5 + 5 = 10 hr

So Option B is correct

Explanation : The total no. of days in which Anoop can dig the well is 10 days.

Anoop's one day efficiency is 10%.

On day one Anoop performs 10% of his work efficiency, then the next day he won't be able to perform because as per the question the efficiency of a worker falls by 10%.

thus, 10%-10% = 0.

Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins

Thus, 25 km/h is the correct answer. 

So Option A is correct

  ► 40% work in 40h → 60% work in 60h.
  ► Hence, working hours = 60 / 4 = 15 h

let usual efficiency of A = 2a and that of B = b

So total work = 12*(2a+b) = 24a+12b——1)

In 2nd case when efficiency of A is halved and efficiency of B get tripled , new efficiency of A = 2a/2 = a

New efficiency of B = b*3 =3b

Total work = 9(a+3b) = 9a+27b ——2)

From eq 1) and 2)

24a+12b = 9a + 27b

15a = 15b

Or a=b

So total work = 24a+12b = 24a+12a = 36a

Time taken by A alone to complete the work = 36a/2a = 18 days

Total time required to fill the tank: 1/10 + 1/20 + 1/25 - 1/100 = 18/100
Time required to fill the tank in 1hr = capacity/total time
= 50000*18/100
= 9000litre

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

►Again if we close 4 taps of set Z, the rate of emptying by set Z would be 2.5% per minute.

►A half filled tank would contain 50% of the capacity and hence would take 50 / 2.5 = 20 minutes to empty.

Let A's 1 day's work = x and B's 1 day's work = y.

Then, x + y =1/30 and 16x + 44y = 1.

Solving these two equations, we get: x = 1/60 and y = 1/60

 B's 1 day's work = 1/60.

Hence, B alone shall finish the whole work in 60 days.

We have,

B = 3/2*A

→ 

One day's work, (A+B) = 1/18

(2/3*B+B) = 1/18

5/3*B = 1/18

One day's work of B = 3/90

B alone can complete the work in,

= 90/3

= 30 days.

1 woman's 1 day's work = 1/70

1 child's 1 day's work = 1/140

(5 women + 10 children)'s day's work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7

5 women and 10 children will complete the work in 7 days.

Work done by A in 1 day = 1/90

Work done by B in 1 day = 1/40

Work done by C in 1 day = 1/12

Work done in 3 days = 1/90 + 1/40 + 1/12 = 43/360

in 8 * 3 = 24 days , work completed = 8 * 43/360 = 344/360

Remaining work = 1 - 344/360 = 16/360

in 25th day, A works and completes 1/90 work .

Remaining work = 16/360 - 1/90 = 12/360

in 26th day, B works and completes 1/40 work .

Remaining work = 12/360 - 1/40 = 1/120

in 27th day, C works and completes this entire 1/120 work

A worked 9 days by doing 1/90 work each day. Total work done by A = 9 * 1/90 = 1/10

B worked 9 days by doing 1/40 work each day. Total work done by B = 9 * 1/40 = 9/40

C worked 9 days by doing 1/12 work in the initial 8 days and 1/120 work in the 9th day.

Total work done by C = 8 * 1/12 + 1/120 = 81/120

Work done by A : Work done by B : Work done by C

= 1/10 : 9/40 : 81/120 

= 12 : 27 : 81

Total amount that they get = 240

Amount that A get = 240 * 12/(12+27+81) = Rs.24

Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.

So Option D is correct

Number of pages typed by Ravi in 1 hour = 32/6 = 16/3.

Number of pages typed by Kumar in 1 hour = 40/5 = 8.

Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3.

Therefore Time taken by both to type 110 pages = (110*3/40) hours

= 8 hours 15 minutes.

Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x     x = (4 * 20/5)

 x = 16 days.

Hence, Tanya takes 16 days to complete the work.

Let total work be 30 units (LCM of 10 and 15).

In one day, A can do 2 units of work and B can do 3 units of work.

In one day, both A and B can do 5 units of work.

In two days, A and B will complete 10 units of work. Remaining 20 units can be completed by A in 10 days (at rate of 2 units per day).

Hence, whole work will be completed in 12 days.

Since the amount of water flowing through each pipe is proportional to square of its diameter so if efficiency of longest pipe (3 cm) = 1/49

Then efficiency of pipe (2 cm) = 4/(49 x 9)

and efficiency of pipe (1 cm) = 1/ (49 x 9) 

Now let cistern is filled by all three pipes in x minutes.

First, let's find the rate at which each burner discharges gas when opened at the given percentages.

For burner A, opened at 25%:
Rate of discharge = 1 cc/min * 25% = 0.25 cc/min

For burner B, opened at 50%:
Rate of discharge = 2 cc/min * 50% = 1 cc/min

Now, let's find the wastage of heat for each burner based on the given discharge rates.

For burner A, with a discharge rate of 0.25 cc/min:
Wastage = 10% (since it falls within the 0 - 0.5 cc/min range)

For burner B, with a discharge rate of 1 cc/min:
Wastage = 25% (since it falls within the 1 - 1.5 cc/min range)

Now, we can find the effective heat generated by each burner after accounting for wastage.

For burner A:
Effective heat = 1 kcal/cc * (1 - 10%) = 1 kcal/cc * 90% = 0.9 kcal/cc

For burner B:
Effective heat = 1 kcal/cc * (1 - 25%) = 1 kcal/cc * 75% = 0.75 kcal/cc

Next, we need to find the total amount of gas discharged by both burners until the gas cylinder is empty. Since the capacity of the cylinder is 1000 cc, we can find the time it takes for both burners to empty the cylinder.

Total rate of discharge = 0.25 cc/min (burner A) + 1 cc/min (burner B) = 1.25 cc/min

Time to empty the cylinder = 1000 cc / 1.25 cc/min = 800 minutes

Now, we can find the total heat generated by each burner during these 800 minutes.

Heat generated by burner A = 0.9 kcal/cc * 0.25 cc/min * 800 min = 180 kcal

Heat generated by burner B = 0.75 kcal/cc * 1 cc/min * 800 min = 600 kcal

Finally, we can find the total heat available for cooking by adding the heat generated by both burners.

Total heat available = 180 kcal (burner A) + 600 kcal (burner B) = 780 kcal

So, the amount of heat available for cooking would be 780 kcal.

Ratio of rates of working of A and B = 2 : 1
So, ratio of time taken = 1 : 2
B's 1 day's work = 1/12
∴ A's one day work = 1/3 (2 times of B's work)
(A + B)'s 1 day's work = (1/6) + (1/12) = 3/12 = 1/4
So, A and B together can finish work in 4 days.

A’s 1 day work = 1/24
B’s 1 day work = 1/6
C’s 1 day work = 1/12
(A+B+C)’s 1 day work = 1/24 + 1/6 + 1/12 = 7/24
The work will be completed by them is 24/7 days.

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b 

Work done by 6 men and 8 women in 1 day = 1/10 

=> 6m + 8b = 1/10

=> 60m + 80b = 1    (1)

Work done by 26 men and 48 women in 1 day = 1/2 

=> 26m + 48b =1/2

=> 52m + 96b = 1    (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day 

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

Shishu alone does the work in 30 hours 

So in 1 hour he does 1/30 of the work 

Mayank in 1 hour does 1/30 + 1/2*1/30= 1/30 +1/60 = 3/60 = 1/20 of the work 

Together in 1 hour they do 1/30 +1/20 = 5/60 = 1/12 of the work 

Together they can finish the work in 12 hours 

Shishu in 12 hours does 12/ 30 = 2/5 

Remaining work = 3/5 

3/5 X 12 = 36/5 = 7.2 hours

Let us assume that, first three pumps fills the tank in x hours .

so,

→ Efficiency of each pump = (1/x) m³ / hour .

then,

→ Efficiency of three pump = (3/x) m³ / hour .

 

now,

→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.

 

so,

→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .

 

now, given that,

→ Time taken by additional pump to fill the tank = 40 hours.

so,

→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .

 

and,

→ Additional pumps work for = 1 + 1 = 2 hours.

 

so,

→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .

 

therefore,

→ (13.5/x) + (1/10) = 1

→ (13.5/x) = 1 - (1/10)

→ (13.5/x) = (9/10)

→ x = 135/9 = 15 hours.

 

since given that, in last 1 hour they filled 15 m³ .

 

hence,

→ 3 * (1/15) + (1/20) = 15 m³

→ (1/5) + (1/20) = 15

→ (4 + 1)/20 = 15

→ (5/20) = 15

→ (1/4) = 15

→ 1 = 60 m³ (Ans.) (Option A)

Let the rate of each filling pipe be 'f ' litres/hr and the rate of each draining pipes be 'd' litres/hr
As per the first condition given in the question, Capacity of tank will be = (6f - 5d ) x 6 …(i)
And as per the second condition given in the question, Capacity of tank will be = (5f - 6d ) x 60 …(ii)
Since the capacity of tank is same in both the cases, on equating (i) and (ii), we have
    ► (6f - 5d) x 6 = (5f - 6d) x 60
    ► or, 6f - 5d = 50f - 60d
    ► or, 44f = 55d
    ► or, 4f = 5d
Therefore, f = 1.25d , putting this value in eq. (i),
Capacity of the tank = (6f - 5d ) x 6 = (7.5d - 5d) x 6 = 15d
When 2 filling and 1 draining pipes work simultaneously, the effective rate of filling the tank = (2f - d ),
Now putting f =1.25d in this eq. we have
Effective rate = (2.5d - d ) = 1.5d
Therefore time required to fill the empty tank = 15d / 1.5d = 10 hours.

Suppose A, B and C take x,x/2 and x/3 days respectively to finish the work.

Then, (1/x + 2/x + 3/x) = 1/2

= 6/x = 1/2

= x = 12

So, B takes (12/2) = 6 days to finish the work.

Given, 15/H+5/R=1/30  ——-1)

And 5/H+15/R=1/60  ——–2)

By solving eq 1) & eq 2) we get

1/(H )=1/(40�12)

Thus H = 40�12

So time taken by 15 humans = 40�12/15=32

Since A finishes 6 / 7^th of the work in 2z hours.
B would finish 12 / 7 of the work in 2z hours.
Thus, to do 1/7^th of the work (which represents the remaining work), B would require
     ► 2z /12 = z / 6 hours.

Let your efficiency of doing some work be t days, then the efficiency of your roommate is 2t days.
So, we have, 

t = 45 days thus, 2t = 90 days

Given, 

it rained 7 times 

there were 5 clear afternoons

there were 6 clear mornings

Therefore a total of 9 days of vacation.

To solve this problem, let's first find out the total time Anoop takes to read all four passages and answer all the questions.

Let the time he takes to read one passage be T. Since there are four passages, he takes 4T time to read all the passages. It is given that he can answer 12 questions in the time he takes to read one passage. So, the time he takes to answer one question is T/12.

There are a total of 5+8+8+6 = 27 questions. The time he takes to answer all the questions is 27 * (T/12) = 27T/12 = 9T/4.

Now, the total time spent on the section is the sum of the time spent on reading all the passages and answering all the questions: 4T + 9T/4 = 25T/4.

To cut down on his total time spent on the section by 20%, the new total time should be 80% of the original time, which is 0.8 * (25T/4) = 5T.

Since the time spent on answering the questions remains constant, the time spent on reading should reduce to 5T - 9T/4 = 11T/4. The new time he takes to read one passage is (11T/4) / 4 = 11T/16.

Now, let's find out the percentage increase in reading speed. The original time to read one passage is T, and the new time is 11T/16. Since speed is inversely proportional to time, the new speed will be 16/11 times the original speed.

The percentage increase in speed is [(16/11 - 1) * 100] = [(5/11) * 100] = 45.45%.

So, Anoop should increase his reading speed by 45.45% to cut down on his total time spent on the section by 20%.

►Let us assume that time to read one passage = x

►So, according to the question, 3x + 21( x / 12) - 13 ⇒ x = 4

►Therefore, total time taken for answering all questions = 4 x 4 + (4 x 27 / 12) = 25 min

►Time spent on first passage = 4 + (4 x 5 / 12) = 17 / 3 min

►Required percentage = 22 .6%

LCM(10,50)=50

 

Suppose total work is 50 units.

Work done by A and B together in 1 day 

= 50/10 = 5 unit.

Work done by C in 1 day

= 50/50 = 1 unit.

Work done by A,B,C together in 1 day 

= 5 + 1 = 6 unit.

Since work done by A is equal to work done by B and C together,

work done by B and C together in 1 day 

= 6/2 = 3 unit.

Work done by B in 1 day 

= 3 − 1 = 2 unit.

Therefore, B alone can do the work in 50/2 = 25 days.

Let 1 man's 1 day work = x and 1 woman's 1 day work = y.

Then, 4x + 6y = 1/8 and 3x + 7y = 1/10

Solving these two equations, we get:

x = 11/400 and y = 1/400

10 woman's 1 day work = (1/400 x 10) = 1/40.

Hence, 10 women will complete the work in 40 days.

Let us make a table of the units of work everyday :

    ► Total length =113 m

B's 10 day's work = (1/15 * 10) = 2/3

Remaining work = (1 - 2/3) = 1/3.

Now, 1/18 work is done by A in 1 day.

Therefore

 1/3 work is done by A in(18 * 1/3)= 6 days.