AAO Exam-CT 9: Quant (Time and Work)
AAO Exam-CT 9- Quant (Time and Work)
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12
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13
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16
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10
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.
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45/7 days
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7 days
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55/7 days
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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.
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2/11
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4/21
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13/25
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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.
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Rs. 960
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Rs. 940
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Rs. 1000
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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.
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Rs. 2200
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Rs. 2500
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Rs. 3000
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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?
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21 hrs
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10 hrs
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25 hrs
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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.
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8.25
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9.5
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9
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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?
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10
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12
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14
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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
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17 days
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16 days
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14 days
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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?
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5 days
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8 days
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10 days
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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
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