# AAO Exam-CT 9: Quant (Time and Work)

## AAO Exam-CT 9- Quant (Time and Work)

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1. The ratio of work done by 20 men and 30 women at the same time is 10 : 9. If 10 men and 20 women can do the work in 2 days then how many men are required to finish the work in 4 days.
• 12
• 13
• 16
• 10
• 11

Given:

The ratio of work done by 20 men and 30 women = 10 : 9

Time taken by 20 men = Time taken by 30 women

Formula Used:

Efficiency = (Total work)/(Time taken)

Calculation:

Let the efficiency of one man and one woman be M and W.

And let the time taken by 20 men be x days

Total work done by 20 men = (20 × M × x)

Total work done by 30 women = (30 × W × x)

According to the question,

(20 × M × x)/(30 × W × x) = 10/9

⇒ (2M)/(3W) = 10/9

⇒ M/W = 5/3

The efficiency of one man = 5 units/day

The efficiency of one woman = 3 units/day

Total work done by 10 men and 20 women = {(10 × 5) + (20 × 3)} × 2

⇒ 220 units

Number of men required to complete the work in 4 days = 220/(4 × 5)

⇒ 11

∴ 11 men are required to finish the work in 4 days.

2. Sunder, Ramu, and Chhotu can dig a well in 9, 18 and 36 days respectively. They all started digging together. Sunder worked continuously till well dug completely, Ramu leaves the work 3 days before its completion and Chhotu leaves the work 2 days before its completion. What is the time taken to dig the well?
• 45/7 days
• 7 days
• 55/7 days
• 44/7 days
• 6 days

Given

Sunder, Ramu, and Chhotu can dig a well respectively in = 9, 18 and 36 days

Formula used

Rate of Work = 1/Time Taken

Calculation

Let the well dug completely in days = x days

Work done by Sunder in 1 day = 1/9

Work done by Ramu in 1 day = 1/18

Work done by Chhotu in 1 day = 1/36

So According to question,

x/9 + (x-3)/18 + (x-2)/36 = 1

⇒ (4x + 2x - 6 + x - 2)/36 = 1

⇒ (7x - 8)36 = 1

⇒ 7x - 8 = 36

⇒ 7x = 36 + 8

⇒ x = 44/7

∴ Time taken to dig the well completely is 44/7 days.

3. Pipe A, B and C releases three different solution H2SO4, HCL and HNO2. These three pipes can fill the empty tank in 15 minutes, 25 minutes and 45 minutes respectively. After 5 minutes, if all the pipes were closed. what is the ratio of HNO2 in that mixed solution?
• 2/11
• 5/29
• 4/21
• 13/25
• None of these

Given:

Pipe A can fill the tank with H2SO4 in 15 minutes

Pipe B can fill the tank with HCL in 25 minutes

Pipe C can fill the tank with HNO2 in 45 minutes

Concept:

Pipe A can fill a tank in x hrs

⇒ Pipe A can fill a tank in 1 hr = 1/x

Calculation:

Pipe A can fill the tank with H2SO4 in 1 minute = 1/15

Pipe B can fill the tank with HCL in 1 minute = 1/25

Pipe C can fill the tank with HNO2 in 1 minute = 1/45

⇒ Pipe A, B, and C can fill the tank in 1 minute = {(1/15) + (1/25) + (1/45)} = 29/225

⇒ Pipe A, B, and C can fill the tank in 5 minute = 5 × {(1/15) + (1/25) + (1/45)} = 29/45

⇒ Part filled by C in 5 minutes = 5/45 = 1/9

∴ Ratio of HNO2 in the mixture after 5 minutes = (1/9) × (45/29) = 5/29

4. Total contract amount for completing a piece of work is Rs. 2100. A and B can complete the work in 40 days and 24 days respectively. They completed the work in 8 days with the help of C, then find the share of C.

• Rs. 960
• Rs. 940
• Rs. 980
• Rs. 1000
• None of these

Given:

Total contract amount = Rs. 2100

A and B can complete the work in 40 days and 24 days respectively

Calculations:

Consider, C’s 1 day’s work = 1/C

⇒ (A + B + C)’s 8 days work = 1

⇒ 8 [(1/40) + (1/24) + (1/C)] = 1

C ’s 1 day’s work 1/C = (1/8) – (1/40) – (1/24)

⇒ (15 – 3 – 5)/120 = 7/120

Ratio of working capacities = 1/40 : 1/24 : 7/120 = 3 : 5 : 7

Total (A + B + C)’s share = Rs. 2100

⇒ 15 parts = 2100

⇒ 1 part = 140

C’s share = 7 parts = 7 × 140 = Rs. 980.

∴ C's share is 980 Rs.

5. At Shradhha construction site, worker A can do the work in 12 days while worker B alone can do the work in 16 days. If worker A and B construct it together and A is getting a wage of Rs. 3600 for the whole work, then what is B's wage?
• Rs. 2200
• Rs. 2500
• Rs. 2700
• Rs. 3000
• Rs. 3200

Given:

Time taken by A to do the work = 12 days

Time taken by B to do the work = 16 days

Wage offered to A = Rs. 3600

Formula Used:

Rate of Work = 1/(Time Taken)

The ratio of A to B shares = (A's rate of work) : (B's rate of work)

Calculation:

A's 1 day work = (1/12) unit

B's 1 day work = (1/16) unit

The ratio of A to B shares = (1/12) : (1/16)

⇒ 4 : 3

A's share = 4 units

B's share = 3 units

According to the question,

4 units = Rs. 3600

⇒ 1 unit = Rs. 900

B's share = 3 × 900

⇒ Rs. 2700 ∴ B's share will be Rs. 2700.

6. Pipe X, Y, and Z can fill a reservoir in 8 hrs. After working at it together for 2 hrs, Z is closed but X and Y can fill the remaining part in 12 hrs. How much time will taken by Z to fill the reservoir?

• 21 hrs
• 10 hrs
• 16 hrs
• 25 hrs
• None of these

Given:

Pipe X, Y and Z can fill the reservoir in 8 hrs

Three pipes together work for 2 hrs

The remaining part X and Y can fill in 12 hrs

Concept used:

Let, Pipe A can fill a tank in x hrs

⇒ Pipe A can fill a tank in 1 hr = 1/x

Calculation:

Pipe X, Y and Z can fill a reservoir in 8 hrs

⇒ Pipe X, Y and Z fill the reservoir in 1 hr = 1/8

⇒ Pipe X, Y and Z fill the reservoir in 2 hr = 2/8 = 1/4

⇒ Remaining part to be filled = (1 - 1/4) = 3/4

(X + Y) can fill the reservoir in 12 hrs = 3/4

⇒ (X + Y) can fill the reservoir in 1 hr = 1/16

⇒ Z alone can fill the cistern in (1/8) - (1/16) =  (1/16) = 16 hrs

∴ The time in which Z will fill the reservoir is 16 hrs.

7. A, B and C take 12, 15 and 20 hours to fill the tank. They start together to fill the tank. A is removed 7 hours before the tank is filled and B is removed 4 hours before the tank is filled. The tank is filled by C. The time required for tank to fill (in hours) is?
• 9.25
• 8.25
• 9.5
• 9
• 8.5

Given:-

Time for A to fill the tank = 12 hours

Time for B to fill the tank = 15 hours

Time for C to fill the tank = 20 hours

Concept used:-

Considering capacity of tank equal to LCM of filling and emptying pipes / leaks.

Calculation:-

Capacity of tank = LCM of (12, 15 and 20) = 60 units

Work done by pipe A in 1 hour = 60 / 12 = 5 units

Work done by pipe B in 1 hour = 60 / 15 = 4 units

Work done by pipe C in 1 hour = 60 / 20 = 3 units

Let the total time be x hours

According to question,

5 (x – 7) + 4 (x – 4) + 3x = 60

12x – 51 = 60

12x = 111 X = 9.25 hours

8. A and B can do a work in 40 days and 25 days respectively. They start the work together, but A left after some days and B finished the remaining work in 12 days. After how many days did A leave?

• 8
• 10
• 12
• 14
• None of these

Given:

A takes time to do a work = 40 days.

B takes time to do the same work = 25 days.

They start the work together.

A left after some days and B finished the remaining work in 12 days.

Formula:

A will do work in one day = (1/40) of the work.

B will do work in one day = (1/25) of the work.

Calculation:

If they work together then they will do work in one day = (1/40 + 1/25) of the work

⇒13/200 of the work

B has completed the work in 12 days after A leaving the work in the middle.

B can do in one day = 1/25 of the work.

So in 12 days, he can do = 12/25 of work.

So work completed by A and B together is = 1 - 12/25 = 13/25

They together can do in one day = 13/200 of work

They have completed together 13/25 work

So the number of days they work together = (13/25) / (13/200)

⇒ 13/25 × 200/13 = 8 days

A left the work after 8 days.

LCM of (40, 25) = 200 units

Let the total work be 200 units

Efficiency of A = 200/40 = 5 units per day

Efficiency of B = 200/25 = 8 units per day

Work done  by B in 12 days = 12 × 8 = 96 units

Remaining work by A and B in days = (200 - 96)/13 = 8 days

So, A left the work after 8 days

9. A is twice efficient as B. A and B together do the same work in as much time as C and D can do together. The ratio of the number of alone working days of C  to D is 2 : 3 and if B worked 16 days more than C then number of days which A worked alone?
• 18 days
• 17 days
• 16 days
• 14 days
• None of these

Given

Number of work C and D alone = 2 : 3

Formula Used

Work = Time × Efficiency

Calculation

Let the working days for A, B, C and D are

A = x, B = 2x, C = 2y, D = 3y

Now, 1/x + 1/2x = 1/2y + 1/3y

⇒ 3/2x = 5/6y

⇒ 18y = 10x

⇒ x : y = 9 : 5

And 2x – 2y = 16

Now, solving both the equations, we get,

⇒ x = 18 days.

∴Number of days which A worked alone is 18 days.

10. A works twice as fast as B. B works twice as fast as C. A works 16 times as his efficiency and completes the whole in 12 days, how many days would be required for B and C together if B works 32 times of his efficiency and C with 64 times of his efficiency to complete the same job?

• 5 days
• 6 days
• 8 days
• 10 days
• 9 days

Given:

A works twice as fast as B

B works twice as fast as C

Concept Used:

Work Done = Time Taken × Rate of Work.

Calculation:

Let x be a time taken.

The ratio of efficiency of A : B : C is 4 : 2 : 1 .

the ratio of time taken for A : B : C is 1 : 2 : 4 .

According to the question

A works 16 times as his efficiency and completes the whole in 12 days

Total Work from A = 16 × 4 × 12

⇒ 768

Required for B if he works 32 efficiencies and C with 64 times of his efficiency

Total Work = {(32 × 2) + 64} × x

⇒ 768 = 128x

⇒ x = 6

∴  Together to complete the same job in 6 days.

Alternate Method

Let efficiency of A, B and C be 4x, 2x and x respectively

Total work if A works 16 times of his efficiency = 4x × 16 × 12

Efficiency of 32 times of B = 2x × 32

Efficiency of 64 times of C = x × 64

Required time taken = (4x × 16 × 12) ÷ (64x + 64x)

⇒ (64x × 12) ÷ 128x

⇒ 6 days

∴ Required time taken is 6 days