Time And Work — RRB NTPC Quantitative Aptitude

30 Questions • 60 Minutes • Quantitative Aptitude Mock Test in Hindi and English

Sample Questions from this Test

Question 1:

A can do a piece of work in 12 days and B can do it in 24 days. In how many days will they complete the work if they work together?
A.6
B.8
C.10
D.12

• Given: Time taken by A =12= 12 days, B =24= 24 days.

• Basic Method (Solution):
Step 1: Total work =LCM(12,24)=24= LCM(12, 24) = 24 units.
Step 2: Efficiency of A =2412=2= \frac{24}{12} = 2 units/day. Efficiency of B =2424=1= \frac{24}{24} = 1 unit/day.
Step 3: Total efficiency =2+1=3= 2 + 1 = 3 units/day.
Step 4: Time taken =243=8= \frac{24}{3} = 8 days.

• Shortcut Method: When A and B work together, time taken =xyx+y= \frac{xy}{x+y}. Here, 12×2412+24=28836=8\frac{12 \times 24}{12 + 24} = \frac{288}{36} = 8 days.

• Final Answer: 8

Question 2:

A, B, and C can complete a piece of work in 10, 12, and 15 days respectively. If they all work together, in how many days will the work be completed?
A.4
B.5
C.6
D.8

• Given: A=10A=10, B=12B=12, C=15C=15 days.

• Basic Method (Solution):
Step 1: Total work =LCM(10,12,15)=60= LCM(10, 12, 15) = 60 units.
Step 2: Efficiency: A=6010=6A = \frac{60}{10} = 6, B=6012=5B = \frac{60}{12} = 5, C=6015=4C = \frac{60}{15} = 4.
Step 3: Total efficiency per day =6+5+4=15= 6 + 5 + 4 = 15 units.
Step 4: Total time =6015=4= \frac{60}{15} = 4 days.

• Shortcut Method: LCM total-work method is the fastest shortcut here. Mental math: LCM=60LCM = 60. Sum of efficiencies =6+5+4=15= 6+5+4 = 15. Answer =60/15=4= 60/15 = 4.

• Final Answer: 4

Question 3:

A and B together can do a piece of work in 8 days. If A alone can do it in 12 days, in how many days can B alone do it?
A.16
B.20
C.24
D.28

• Given: (A+B)=8(A+B) = 8 days, A=12A = 12 days.

• Basic Method (Solution):
Step 1: Total work =LCM(8,12)=24= LCM(8, 12) = 24 units.
Step 2: (A+B)(A+B) efficiency =248=3= \frac{24}{8} = 3 units/day.
Step 3: AA's efficiency =2412=2= \frac{24}{12} = 2 units/day.
Step 4: BB's efficiency =(A+B)−A=3−2=1= (A+B) - A = 3 - 2 = 1 unit/day.
Step 5: Time for B =241=24= \frac{24}{1} = 24 days.

• Shortcut Method: Formula for remaining person: Time =xyy−x=12×812−8=964=24= \frac{xy}{y-x} = \frac{12 \times 8}{12 - 8} = \frac{96}{4} = 24 days.

• Final Answer: 24

Question 4:

If 15 men can complete a work in 20 days, how many days will 25 men take to complete the same work?
A.10
B.12
C.15
D.18

• Given: M1=15M_1 = 15, D1=20D_1 = 20, M2=25M_2 = 25, D2=?D_2 = ?

• Basic Method (Solution):
Step 1: Use MDH formula M1×D1=M2×D2M_1 \times D_1 = M_2 \times D_2.
Step 2: 15×20=25×D215 \times 20 = 25 \times D_2.
Step 3: 300=25×D2300 = 25 \times D_2.
Step 4: D2=30025=12D_2 = \frac{300}{25} = 12 days.

• Shortcut Method: Work is constant. Ratio of Men =15:25=3:5= 15:25 = 3:5. Time is inversely proportional to Men, so Time ratio =5:3= 5:3. If 5 units =20= 20 days, 1 unit =4= 4 days, so 3 units =12= 12 days.

• Final Answer: 12

Question 5:

40 men can build a 200m long wall in 15 days working 8 hours a day. How many days will 50 men take to build a 300m long wall working 6 hours a day?
A.16
B.20
C.24
D.18

• Given: M1=40,D1=15,H1=8,W1=200M_1=40, D_1=15, H_1=8, W_1=200. M2=50,H2=6,W2=300,D2=?M_2=50, H_2=6, W_2=300, D_2=?

• Basic Method (Solution):
Step 1: Use the full MDH formula: M1×D1×H1W1=M2×D2×H2W2\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}
Step 2: 40×15×8200=50×D2×6300\frac{40 \times 15 \times 8}{200} = \frac{50 \times D_2 \times 6}{300}
Step 3: Simplify left side: 4800200=24\frac{4800}{200} = 24.
Step 4: Simplify right side: 300×D2300=D2\frac{300 \times D_2}{300} = D_2.
Step 5: 24=D224 = D_2.

• Shortcut Method: Directly write D2=D1×M1M2×H1H2×W2W1=15×4050×86×300200D_2 = D_1 \times \frac{M_1}{M_2} \times \frac{H_1}{H_2} \times \frac{W_2}{W_1} = 15 \times \frac{40}{50} \times \frac{8}{6} \times \frac{300}{200}. Cancel out zeros and simplify: 15×45×43×32=2415 \times \frac{4}{5} \times \frac{4}{3} \times \frac{3}{2} = 24.

• Final Answer: 24

Question 6:

A is twice as efficient as B. If they together complete a work in 14 days, in how many days can A alone complete the same work?
A.21
B.28
C.42
D.18

• Given: A's efficiency = 2×2 \times B's efficiency. Together time =14= 14 days.

• Basic Method (Solution):
Step 1: Let B's efficiency =1= 1 unit/day. Then A's efficiency =2= 2 units/day.
Step 2: Total efficiency (A+B)=2+1=3(A+B) = 2 + 1 = 3 units/day.
Step 3: Total work =Efficiency×Time=3×14=42= \text{Efficiency} \times \text{Time} = 3 \times 14 = 42 units.
Step 4: Time taken by A alone =Total WorkA’s efficiency=422=21= \frac{\text{Total Work}}{\text{A's efficiency}} = \frac{42}{2} = 21 days.

• Shortcut Method: Ratio of efficiency A:B=2:1A:B = 2:1. Total parts =3= 3. Work =3×14= 3 \times 14. A's time =3×142=21= \frac{3 \times 14}{2} = 21 days.

• Final Answer: 21

Question 7:

A can do a work in 10 days and B can do the same work in 15 days. They undertake to do it together for ₹5000. What is A's share of the wages?
A.₹2000
B.₹2500
C.₹3000
D.₹3500

• Given: A=10A=10 days, B=15B=15 days, Total wage =5000= 5000.

• Basic Method (Solution):
Step 1: Wages are distributed in the ratio of their efficiencies (work done).
Step 2: Total work =LCM(10,15)=30= LCM(10, 15) = 30.
Step 3: A's efficiency =30/10=3= 30/10 = 3. B's efficiency =30/15=2= 30/15 = 2.
Step 4: Ratio of wages =3:2= 3:2. Total parts =5= 5.
Step 5: A's share =35×5000=3000= \frac{3}{5} \times 5000 = 3000.

• Shortcut Method: Ratio of efficiencies is inverse of ratio of time taken. Time ratio A:B=10:15=2:3A:B = 10:15 = 2:3. Efficiency ratio =3:2= 3:2. A's share =35×5000=3000= \frac{3}{5} \times 5000 = 3000.

• Final Answer: ₹3000

Question 8:

A can do 12\frac{1}{2} of a work in 5 days and B can do 35\frac{3}{5} of the same work in 9 days. In how many days can both A and B together complete the work?
A.5 days
B.6 days
C.8 days
D.10 days

• Given: A does 1/21/2 work in 55 days. B does 3/53/5 work in 99 days.

• Basic Method (Solution):
Step 1: Time taken by A to complete full work =5×21=10= 5 \times \frac{2}{1} = 10 days.
Step 2: Time taken by B to complete full work =9×53=15= 9 \times \frac{5}{3} = 15 days.
Step 3: Total work =LCM(10,15)=30= LCM(10, 15) = 30 units.
Step 4: A's efficiency =3= 3, B's efficiency =2= 2. Total =5= 5.
Step 5: Time taken together =305=6= \frac{30}{5} = 6 days.

• Shortcut Method: Full days A=10,B=15A=10, B=15. Use product/sum formula: 10×1510+15=15025=6\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6 days.

• Final Answer: 6 days

Question 9:

Pipe A can fill a tank in 20 hours and Pipe B can empty the full tank in 30 hours. If both pipes are opened together, in how many hours will the empty tank be filled?
A.50
B.60
C.12
D.10

• Given: A fills in 20 hrs, B empties in 30 hrs.

• Basic Method (Solution):
Step 1: Total capacity of tank =LCM(20,30)=60= LCM(20, 30) = 60 units.
Step 2: Efficiency of A =+6020=+3= +\frac{60}{20} = +3 units/hr.
Step 3: Efficiency of B =−6030=−2= -\frac{60}{30} = -2 units/hr.
Step 4: Net efficiency =3−2=1= 3 - 2 = 1 unit/hr.
Step 5: Time taken =601=60= \frac{60}{1} = 60 hours.

• Shortcut Method: When one fills (xx) and one empties (yy), Net Time =xyy−x=20×3030−20=60010=60= \frac{xy}{y-x} = \frac{20 \times 30}{30 - 20} = \frac{600}{10} = 60 hours.

• Final Answer: 60

Question 10:

A and B can do a work in 20 days and 30 days respectively. They started the work together, but after 4 days A left. In how many days will B complete the remaining work?
A.15
B.16
C.20
D.24

• Given: A=20A=20, B=30B=30. They work together for 4 days. Then A leaves.

• Basic Method (Solution):
Step 1: Total work =LCM(20,30)=60= LCM(20, 30) = 60 units.
Step 2: Efficiencies: A=3A=3, B=2B=2. Total =5= 5.
Step 3: Work done in 4 days =5×4=20= 5 \times 4 = 20 units.
Step 4: Remaining work =60−20=40= 60 - 20 = 40 units.
Step 5: Time taken by B for remaining work =402=20= \frac{40}{2} = 20 days.

• Shortcut Method: Mentally calculate total work 60. In 4 days, they do 4×(3+2)=204 \times (3+2)=20. Left is 40. B's speed is 2. So 40/2=2040/2 = 20 days directly on paper.

• Final Answer: 20

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