#1. If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?
A. 60
B. 45
C. 40
D. 30
Answer: Option A Solution: Step 1: Define Work Rates
– A and B together complete the work in 15 days. – B alone completes the work in 20 days. – Let A alone take \( x \) days to complete the work.
The work rates are:
– A and B together: \( \frac{1}{15} \) of the work per day. – B alone: \( \frac{1}{20} \) of the work per day.
Since A and B together complete \( \frac{1}{15} \) of the work each day, and B alone does \( \frac{1}{20} \) of the work per day, A’s contribution per day is:
\[ \frac{1}{15} – \frac{1}{20} \]
Step 2: Find A’s Work Rate
Find a common denominator for 15 and 20, which is 60:
Since A completes \( \frac{1}{60} \) of the work per day, the total time A needs to complete the entire work is:
\[ x = \frac{1}{\frac{1}{60}} = 60 \text{ days} \]
Final Answer: A alone can complete the work in 60 days.
#2. If A and B together can complete a work in 18 days, A and C together in 12 days, and B and C together in 9 days, then B alone can do the work in:
A. 18 days
B. 24 days
C. 30 days
D. 40 days
Answer: Option B Solution: We are given the time taken by different pairs of workers (A & B, A & C, B & C) to complete a piece of work. We need to find the time taken by B alone to complete the work.
Step 1: Assign Work Rates Let the total work be 1 unit and let the work rates of A, B, and C be:
– \( A + B \) completes the work in 18 days → Work done per day by \( A + B \) = \( \frac{1}{18} \) – \( A + C \) completes the work in 12 days → Work done per day by \( A + C \) = \( \frac{1}{12} \) – \( B + C \) completes the work in 9 days → Work done per day by \( B + C \) = \( \frac{1}{9} \)
Step 2: Add All Three Equations \[ (A + B) + (A + C) + (B + C) = \frac{1}{18} + \frac{1}{12} + \frac{1}{9} \] \[ 2A + 2B + 2C = \frac{1}{18} + \frac{1}{12} + \frac{1}{9} \]
Step 3: Solve for \( A + B + C \) Finding a common denominator: \[ \frac{1}{18} = \frac{1}{18}, \quad \frac{1}{12} = \frac{3}{36}, \quad \frac{1}{9} = \frac{4}{36} \] \[ \frac{1}{18} + \frac{1}{12} + \frac{1}{9} = \frac{2}{36} + \frac{3}{36} + \frac{4}{36} = \frac{9}{36} = \frac{1}{4} \] \[ 2(A + B + C) = \frac{1}{4} \] \[ A + B + C = \frac{1}{8} \]
Step 4: Solve for B’s Work Rate \[ B = (A + B + C) – (A + C) \] \[ B = \frac{1}{8} – \frac{1}{12} \]
Finding the common denominator: \[ \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{12} = \frac{2}{24} \]
\[ B = \frac{3}{24} – \frac{2}{24} = \frac{1}{24} \]
Step 5: Find B’s Time to Complete the Work Since B completes \(\frac{1}{24}\) of the work per day, the total time for B to complete the work alone is:
\[ \frac{1}{\frac{1}{24}} = 24 \text{ days} \] Final Answer: B alone can complete the work in 24 days.
#3. A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in
A. 5 days
B. 6 days
C. 9 days
D. 10 days
Answer: Option B Solution: We are given the following information:
– \( A + B \) can complete the work in 3 days → Work done per day by \( A + B \) = \( \frac{1}{3} \) – They work together for 2 days – After 2 days, B leaves, and A continues working alone – The remaining work is completed by A in 2 more days – We need to find how long **B alone** would take to complete the entire work.
Step 1: Work Done by A and B Together in 2 Days Since \( A + B \) can complete the entire work in 3 days, their work rate is:
\[ A + B = \frac{1}{3} \text{ work per day} \]
So, in 2 days, they complete:
\[ 2 \times \frac{1}{3} = \frac{2}{3} \text{ of the work} \]
Step 2: Work Remaining After 2 Days The total work is 1. Since \( \frac{2}{3} \) of the work is already done, the remaining work is:
\[ 1 – \frac{2}{3} = \frac{1}{3} \]
Step 3: Work Done by A in 2 More Day A alone completes the remaining \( \frac{1}{3} \)** of the work in 2 days. So, A’s work rate per day is:
Step 5: Find Time Taken by B Alone Since \( B \)’s work rate is \( \frac{1}{6} \), the time B alone would take to complete the work is:
\[ \frac{1}{\frac{1}{6}} = 6 \text{ days} \]
Final Answer: B alone can complete the work in 6 days.
#4. A can complete a piece of work in 18 days, B in 20 days and C in 30 days, B and C together start the work and forced to leave after 2 days. The time taken by A alone to complete the remaining work is:
A. 10 days
B. 12 days
C. 15 days
D. 16 days
Answer: Option C Solution: We are given:
– \( A \) can complete the work in 18 days → Work rate of \( A \) = \( \frac{1}{18} \) per day – \( B \) can complete the work in 20 days → Work rate of \( B \) = \( \frac{1}{20} \) per day – \( C \) can complete the work in 30 days → Work rate of \( C \) = \( \frac{1}{30} \) per day – \( B \) and \( C \) start the work and leave after 2 days – We need to find how long A alone takes to complete the remaining work.
Step 1: Work Done by B and C in 2 Days Total work = 1 unit Work rate of \( B + C \):
Step 2: Work Remaining Total work = 1, work completed by \( B + C \) = \( \frac{1}{6} \). Remaining work:
\[ 1 – \frac{1}{6} = \frac{5}{6} \]
Step 3: Time Taken by A to Complete the Remaining Work Work rate of \( A \) = \( \frac{1}{18} \). Time taken to complete \( \frac{5}{6} \) of the work:
Final Answer: A alone takes 15 days to complete the remaining work.
#5. Working 5 hours a day, A can Complete a work in 8 days and working 6 hours a day, B can complete the same work in 10 days. Working 8 hours a day, they can jointly complete the work in:
A. 3 days
B. 4 days
C. 4.5 days
D. 5.4 days
Answer: Option A Solution: We are given the working hours and days required by A and B to complete a piece of work. We need to find how long they will take to complete the work together if they work 8 hours per day.
Step 1: Find Work Done by A and B Per Hour – A’s work rate: A completes the work in 8 days, working 5 hours per day. Total hours A works = \( 8 \times 5 = 40 \) hours. So, A’s work rate per hour = \( \frac{1}{40} \).
– B’s work rate: B completes the work in 10 days, working 6 hours per day. Total hours B works = \( 10 \times 6 = 60 \) hours. So, B’s work rate per hour = \( \frac{1}{60} \).
Step 2: Find Combined Work Rate A and B together per hour:
This means they complete \( \frac{1}{3} \) of the work in 1 day.
Step 4: Find Total Days Required Since they complete \( \frac{1}{3} \) of the work per day, the total time required to complete the entire work:
\[ \frac{1}{\frac{1}{3}} = 3 \text{ days} \]
Final Answer: A and B together can complete the work in 3 days if they work 8 hours per day.
#6. Ganga and Saraswati, working separately can mow field in 8 and 12 hours respectively. If they work in stretches of one hour alternately. Ganga is beginning at 9 a.m., when will the moving be completed?
A. 6:20 PM
B. 6:30 PM
C. 6:36 PM
D. 6:42 PM
Answer: Option A Solution: Let’s break the problem down step by step:
Given data: – Ganga can mow the field in 8 hours, so her work rate is 1/8 of the field per hour. – Saraswati can mow the field in 12 hours, so her work rate is 1/12 of the field per hour. – They work alternately for one hour each, starting with Ganga at 9:00 AM.
Work done in two hours: – In the 1st hour, Ganga works and completes 1/8 of the field. – In the 2nd hour, Saraswati works and completes 1/12 of the field.
So, in 2 hours, the total work done is: \[ \frac{1}{8} + \frac{1}{12} \] Finding LCM of 8 and 12: \[ \text{LCM}(8,12) = 24 \] \[ \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{12} = \frac{2}{24} \] \[ \frac{3}{24} + \frac{2}{24} = \frac{5}{24} \]
So, in every 2-hour cycle, 5/24 of the field is completed.
Number of full cycles needed: The total work is 1 (whole field), so the number of full cycles required: \[ \frac{1}{5/24} = \frac{24}{5} = 4.8 \] This means 4 full cycles(which take \(4 \times 2 = 8\) hours) and an additional 0.8 of another cycle.
Work done in 4 full cycles: \[ 4 \times \frac{5}{24} = \frac{20}{24} \] This means after 8 hours, they have completed 20/24 of the field.
Remaining work: \[ 1 – \frac{20}{24} = \frac{4}{24} = \frac{1}{6} \] Now, Ganga starts the next hour (since she always starts each cycle).
– In the 9th hour (from 5:00 PM to 6:00 PM), Ganga works. – She completes 1/8 of the field, but we only need 1/6.
Since 1/8 < 1/6, she won’t finish in a full hour. The time required for her to complete 1/6 of the work is: \[ \left(\frac{1/6}{1/8}\right) \times 1 = \frac{1}{6} \times \frac{8}{1} = \frac{8}{6} = \frac{4}{3} \text{ hours} = 1 \text{ hour and } 20 \text{ minutes} \]
Final completion time: – Work starts at 9:00 AM. – After 8 hours, it is 5:00 PM. – Ganga works for 1 hour 20 minutes, so the work is completed at 6:20 PM.
Answer: The mowing will be completed at 6:20 PM.
#7. If 10 men can do a piece of work in 12 days, the time taken by 12 men to do the same piece of work will be:
A. 12 days
B. 10 days
C. 9 days
D. 8 days
Answer: Option B Solution: We can solve this using the work formula:
\[ \text{Work} = \text{Number of men} \times \text{Time} \]
Given: – 10 men can complete the work in 12 days. – So, the total amount of work is:
– We need to find the number of days (D) required for 12 men to complete the same work.
Using the formula:
\[ \text{Work} = \text{Number of men} \times \text{Time} \]
\[ 120 = 12 \times D \]
Solving for D:
\[ D = \frac{120}{12} = 10 \text{ days} \]
Answer: 12 men will complete the work in 10 days.
#8. To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?
A. 30 days
B. 35 days
C. 40 days
D. 45 days
Answer: Option A Solution: Step 1: Define Variables Let B take x days to complete the work alone. Since A takes 50% more time than B, A will take:
\[ x + \frac{50}{100} \times x = x + 0.5x = 1.5x \]
So, A alone completes the work in 1.5x days.
Step 2: Work Done in One Day – Work done by B in one day = \( \frac{1}{x} \) – Work done by A in one day = \( \frac{1}{1.5x} = \frac{2}{3x} \)
Since they together complete the work in 18 days, their combined efficiency is:
\[ \frac{1}{x} + \frac{2}{3x} = \frac{1}{18} \]
Step 3: Solve for x
\[ \frac{1}{x} + \frac{2}{3x} = \frac{1}{18} \]
Taking LCM of x and 3x, we get:
\[ \frac{3}{3x} + \frac{2}{3x} = \frac{1}{18} \]
\[ \frac{5}{3x} = \frac{1}{18} \]
Cross multiply:
\[ 5 \times 18 = 3x \]
\[ 90 = 3x \]
\[ x = \frac{90}{3} = 30 \]
Answer: B alone will complete the work in 30 days.
#9. If 10 men or 20 boys can make 260 mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20 days?
A. 260
B. 240
C. 280
D. 520
Answer: Option A Solution: We are given that:
– 10 men or 20 boys can make 260 mats in 20 days.
Step 1: Determine Work Rate Since 10 men can make 260 mats in 20 days, the rate of work of 10 men is:
\[ \frac{260}{20} = 13 \text{ mats per day} \]
So, the rate of work of 1 man is:
\[ \frac{13}{10} = 1.3 \text{ mats per day} \]
Similarly, since 20 boys can make 260 mats in 20 days, the rate of work of 20 boys is:
\[ \frac{260}{20} = 13 \text{ mats per day} \]
So, the rate of work of 1 boy is:
\[ \frac{13}{20} = 0.65 \text{ mats per day} \]
Step 2: Calculate Work Done by 8 Men and 4 Boys – The total work done by 8 men per day:
\[ 8 \times 1.3 = 10.4 \text{ mats per day} \]
– The total work done by 4 boys per day:
\[ 4 \times 0.65 = 2.6 \text{ mats per day} \]
– Combined work rate of 8 men and 4 boys:
\[ 10.4 + 2.6 = 13 \text{ mats per day} \]
Step 3: Calculate Total Mats in 20 Days \[ 13 \times 20 = 260 \text{ mats} \]
Thus, 8 men and 4 boys will make 260 mats in 20 days.
#10. A complete \( \frac{7}{10} \) of a work in 15 days, then he completed the remaining work with the help of B in 4 days. In how many day A and B can complete entire work together?
A. 10\( \frac{1}{2} \)days
B. 12\( \frac{2}{3} \)days
C. 13\( \frac{1}{3} \)days
D. 8\( \frac{1}{4} \)days
Answer: Option C Solution: We are given:
– A completes \( \frac{7}{10} \) of the work in 15 days. – The remaining \( \frac{3}{10} \) of the work is completed by A and B together in 4 days. – We need to find how many days A and B together take to complete the entire work.
tep 1: Find A’s Work Rate A completes \( \frac{7}{10} \) of the work in 15 days. So, A’s work rate per day:
Step 3: Find B’s Work Rate A and B together work at a rate of \( \frac{3}{40} \) per day. A alone works at a rate of \( \frac{7}{150} \) per day. Thus, B’s work rate per day is:
Final Answer A and B together can complete the entire work in \( 13 \frac{1}{3} \) days (or 13.33 days).
#11. A can complete a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work the work together but A left 8 days before the completion of the work and B 12 days before the completion of work. Only C worked up to the end. In how many days was the work completed?
A. 24 days
B. 25 days
C. 27 days
D. 30 days
Answer: Option A Solution: Let’s break the problem step by step:
Step 1: Determine the work rate of A, B, and C – A can complete the work in 36 days, so A’s work rate = \( \frac{1}{36} \) of the work per day. – B can complete the work in 54 days, so B’s work rate = \( \frac{1}{54} \) of the work per day. – C can complete the work in 72 days, so C’s work rate = \( \frac{1}{72} \) of the work per day.
Step 2: Assume the total work is 1 unit and let the total number of days required to complete the work be X days.
– A worked for \( X – 8 \) days. – B worked for \( X – 12 \) days. – C worked for the entire \( X \) days.
Step 3: Set up the equation Total work done by A, B, and C together should be equal to 1:
#12. While working 7 hour a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, 8 hour a day?
A. 3 days
B. 4 days
C. 2.1 days
D. 3.6 days
Answer: Option A Solution: Step 1: Determine A and B’s Work Rates
– A alone completes the work in 6 days, working 7 hours per day. – Total hours required by A = \( 6 \times 7 = 42 \) hours. – A’s work rate = \( \frac{1}{42} \) (fraction of work done per hour).
– B alone completes the work in 8 days, working 7 hours per day. – Total hours required by B = \( 8 \times 7 = 56 \) hours. – B’s work rate = \( \frac{1}{56} \) (fraction of work done per hour).
Step 2: Combined Work Rate of A and B
Total work done per hour when A and B work together:
\[ \frac{1}{42} + \frac{1}{56} \]
Find LCM of 42 and 56: \(\text{LCM}(42, 56) = 168\)
This means A and B together complete 1/3 of the work in one day.
Step 4: Total Days Required
Since they complete 1/3 of the work per day, they will finish the entire work in:
\[ 3 \text{ days} \]
Final Answer: A and B together can complete the work in 3 days, working 8 hours per day.
#13. Two pipes, P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes, P is turned off. In how many more minutes will Q fill the cistern?
A. 7 minutes
B. 7\( \frac{1}{2} \) minutes
C. 8 minutes
D. 8.25 minutes
Answer: Option D Solution: Step 1: Determine the Work Rate of Each Pipe
– Pipe P fills the cistern in 12 minutes. – Work rate of P = \( \frac{1}{12} \) (fraction of cistern filled per minute).
– Pipe Q fills the cistern in 15 minutes. – Work rate of Q = \( \frac{1}{15} \) (fraction of cistern filled per minute).
Step 2: Work Done in the First 3 Minutes
Both pipes P and Q are opened together for 3 minutes.
Total work done in 1 minute by both together:
\[ \frac{1}{12} + \frac{1}{15} \]
Find LCM of 12 and 15: \(\text{LCM}(12, 15) = 60\)
Final Answer: Pipe Q will take 8.25 minutes (8 minutes 15 seconds) to fill the remaining part of the cistern.
#14. Two pipes can fill an empty tank separately in 24 minutes and 40 minutes respectively and a third pipe can empty 30 gallons of water per minute. If all three pipes are open, empty tanks become full in one hour. The capacity of the tank (in gallons) is:
A. 800 gallons
B. 600 gallons
C. 500 gallons
D. 400 gallons
Answer: Option B Solution: Step 1: Define the Work Rate of Each Pipe
Let the capacity of the tank be X gallons.
– Pipe A fills the tank in 24 minutes – Work rate of A = \( \frac{X}{24} \) gallons per minute.
– Pipe B fills the tank in 40 minutes – Work rate of B = \( \frac{X}{40} \) gallons per minute.
– Pipe C empties the tank at a rate of 30 gallons per minute – Work rate of C = −30 gallons per minute.
Step 2: Set Up the Equation
Given that the tank is filled in 60 minutes (1 hour) when all three pipes are open, the combined rate of the three pipes must fill the tank in that time. This gives the equation:
Final Answer: The capacity of the tank is 600 gallons.
#15. A cistern can be filled by two pipes in 20 and 30 minutes respectively. Both pipes being opened, when the first pipe must be turned off so that the cistern may be filled in 10 minutes more.
A. After 10 minutes
B. After 12 minutes
C. After 20 minutes
D. After 8 minutes
Answer: Option D Solution: Let’s break this down step by step.
Step 1: Determine the Work Rate of Each Pipe – Pipe fills the cistern in 20 minutes – Work rate of A = \( \frac{1}{20} \) of the cistern per minute.
– Pipe B fills the cistern in 30 minutes – Work rate of B = \( \frac{1}{30} \) of the cistern per minute.
Step 2: Understand the Problem – Both pipes are opened together initially, and then Pipe A is turned off after some time. – The cistern must be filled in 10 minutes more than it would take with both pipes open the entire time.
Step 3: Set Up the Work Equation
Let T be the time that both pipes are open before Pipe A is turned off. After Pipe A is turned off, only Pipe B continues filling the cistern, and it takes 10 more minutes** to complete the filling. So, the total time is \( T + 10 \) minutes.
– Work done by both pipes in T minutes: \[ \text{Work done by A and B together} = T \times \left( \frac{1}{20} + \frac{1}{30} \right) \]
– After Pipe A is turned off, Pipe B continues filling for 10 more minutes: \[ \text{Work done by B alone} = 10 \times \frac{1}{30} \]
The total work done must equal 1 (the full cistern). So, we set up the equation:
Final Answer: The first pipe is open for 8 minutes before it is turned off.
#16. A tank can be filled with water by two pipes A and B together in 36 minutes. If the pipe B was stopped after 30 minutes, the tank is filled in 40 minutes. The pipe B can alone fill the tank in:
A. 45 minutes
B. 60 minutes
C. 75 minutes
D. 90 minutes
Answer: Option D Solution: Let’s break this problem down step by step.
Step 1: Define Work Rates – Let the total capacity of the tank be 1 unit of work (the entire tank). – Pipes A and B together can fill the tank in 36 minutes. Therefore, the combined rate of both pipes is: \[ \text{Rate of A and B together} = \frac{1}{36} \text{ (of the tank per minute)}. \] – Let the time it takes for Pipe B alone to fill the tank be B minutes. So, the rate of Pipe B is: \[ \text{Rate of B alone} = \frac{1}{B} \text{ (of the tank per minute)}. \] – Therefore, the rate of Pipe A alone is: \[ \text{Rate of A alone} = \frac{1}{36} – \frac{1}{B} \text{ (of the tank per minute)}. \]
Step 2: Work Done in the First 30 Minutes For the first 30 minutes , both pipes A and B are open, so they fill together: \[ \text{Work done in 30 minutes} = 30 \times \frac{1}{36} = \frac{30}{36} = \frac{5}{6} \text{ of the tank}. \]
Step 3: Work Done After Pipe B is Turned Off After 30 minutes, Pipe B is turned off, and Pipe A continues to fill the tank. Since the total time is 40 minutes, Pipe A works alone for the remaining 10 minutes: \[ \text{Work done by Pipe A in 10 minutes} = 10 \times \left( \frac{1}{36} – \frac{1}{B} \right). \]
The total work done by both pipes A and B together and by Pipe A alone must add up to 1 (the full tank). So: \[ \frac{5}{6} + 10 \times \left( \frac{1}{36} – \frac{1}{B} \right) = 1. \]
Step 4: Solve for \( B \) Now, let’s solve this equation step by step:
Final Answer: Pipe B can fill the tank alone in 90 minutes.
#17. A boy and a girl together fill a cistern with water. The boy pours 4 liters of water every 3 minutes and girl pours 3 litres every 4 minutes. How much time will it take to fill 100 litres of water in the cistern?
A. 36 minute
B. 42 minutes
C. 48 minutes
D. 44 minutes
Answer: Option C Solution: Let’s break down the problem step by step.
Step 1: Determine the Rate of Water Poured by the Boy and the Girl
– Boy’s rate: The boy pours 4 liters of water every 3 minutes. So, his rate of pouring is: \[ \text{Rate of the boy} = \frac{4}{3} \text{ liters per minute}. \]
– Girl’s rate: The girl pours 3 liters of water every 4 minutes. So, her rate of pouring is: \[ \text{Rate of the girl} = \frac{3}{4} \text{ liters per minute}. \]
Step 2: Determine Their Combined Rate The combined rate of the boy and the girl is the sum of their individual rates: \[ \text{Combined rate} = \frac{4}{3} + \frac{3}{4}. \]
To add these fractions, find the least common denominator (LCD). The LCD of 3 and 4 is 12, so:
Step 3: Determine the Time to Fill 100 Liters The combined rate is \( \frac{25}{12} \) liters per minute. To find the time required to fill 100 liters, divide the total volume by the combined rate:
Final Answer: It will take 48 minutes to fill the 100 liters of water in the cistern.
#18. A tank has a leak which would empty the completely filled tank in 10 hours. If the tank is full of water and a tap is opened which admits 4 litres of water per minutes in the tank, the leak takes 15 hours to empty the tank. how many litres does the tank hold?
A. 2400 litres
B. 4500 litres
C. 1200 litres
D. 1440 litres
Answer: Option D Solution: Let’s solve this step by step.
Step 1: Define Variables
Let the capacity of the tank be \( X \) liters.
– The leak alone would empty the tank in 10 hours, so the leak empties \( \frac{X}{10} \) liters per hour.
– The tap admits 4 liters per minute**, which means the tap admits \( 4 \times 60 = 240 \) liters per hour.
When the tank is full, the leak and the tap work together, and it takes 15 hours to empty the tank. So, the combined rate of the leak and the tap is \( \frac{X}{15} \) liters per hour.
Step 2: Set Up the Equation
The combined rate is the rate at which the tap is filling the tank minus the rate at which the leak is emptying the tank. Therefore, we have: \[ \text{Rate of tap} – \text{Rate of leak} = \text{Combined rate}. \]
Substitute the rates we know: \[ 240 – \frac{X}{10} = \frac{X}{15}. \]
Step 3: Solve the Equation
Now, let’s solve for \( X \). To eliminate the fractions, multiply through by 30 (the least common multiple of 10 and 15):
#19. A, B and C completed a work costing Rs. 1800. A work for 6 days, B for 4 days and C for 9 days. If their daily wages are in the ratio of 5 : 6 : 4, how much amount will be received by A?
A. Rs. 800
B. Rs. 600
C. Rs. 900
D. Rs. 750
Answer: Option B Solution: Let’s break down the problem step by step.
Step 1: Understand the Given Information
– The total cost of the work is Rs. 1800. – A worked for 6 days, B worked for 4 days, and C worked for 9 days. – The daily wages of A, B, and C are in the ratio of 5:6:4.
Step 2: Calculate the Total Wage Unit
Let the daily wages of A, B, and C be 5x, 6x, and 4x respectively, where x is the common factor.
Step 3: Calculate the Total Work Done (Total Wage)
The total work done is the sum of the work done by A, B, and C. Since their daily wages are in the ratio 5:6:4, we can express the total wages as:
– A’s total wages = 6 days × 5x = 30x. – B’s total wages = 4 days × 6x= 24x. – C’s total wages = 9 days × 4x= 36x.
The total wages for A, B, and C is: \[ 30x + 24x + 36x = 90x. \]
Step 4: Set up the Equation
The total cost of the work is Rs. 1800, so: \[ 90x = 1800. \]
Step 5: Solve for x
Now, solve for x: \[ x = \frac{1800}{90} = 20. \]
Step 6: Calculate A’s Share
A’s total wages = 30x = 30 × 20 = 600.
Final Answer: A will receive Rs. 600.
#20. A man and a boy received Rs. 800 as wages for 5 days for the work they did together. The man’s efficiency in the work was thrice times that of the boy. What are the daily wages of the boy?
A. Rs. 76
B. Rs. 56
C. Rs. 44
D. Rs. 40
Answer: Option D Solution: Let’s solve this problem step by step.
Step 1: Define Variables Let the daily wage of the boy be \( B \), and the daily wage of the man be \( M \).
Step 2: Relationship Between the Man’s and Boy’s Efficiencies We are told that the efficiency of the man is three times that of the boy. So, the man does three times as much work as the boy does in the same amount of time.
Step 3: Work Done by the Man and Boy Together If the man’s efficiency is three times that of the boy, the man’s daily wage will also be three times that of the boy because their wages are proportional to their efficiencies.
Thus, we can say: \[ M = 3B \quad \text{(since the man earns three times as much as the boy per day)}. \]
Step 4: Total Wages for 5 Days The total wages for the work done together in 5 days is Rs. 800. So, the combined daily wages of the man and the boy are: \[ M + B = \text{total daily wages of both} = \frac{800}{5} = 160 \text{ rupees per day}. \]
Step 5: Substitute the Relationship Between \( M \) and \( B \) Now substitute \( M = 3B \) into the equation for total daily wages: \[ 3B + B = 160. \] \[ 4B = 160. \]
Step 6: Solve for \( B \) \[ B = \frac{160}{4} = 40. \]
Final Answer: The daily wages of the boy are Rs. 40.
#21. Subhash can copy 50 pages in 10 hours; Subhash and Prakash together can copy 300 pages in 40 hours. In how much time can Prakash copy 30 pages?
A. 13 hour
B. 12 hour
C. 11 hour
D. 9 hour
E. 10 hour
Answer: Option B Solution: Let’s break the problem down step by step.
Step 1: Subhash’s Copying Rate – Subhash can copy 50 pages in 10 hours, so his rate of copying is: \[ \text{Subhash’s rate} = \frac{50 \text{ pages}}{10 \text{ hours}} = 5 \text{ pages per hour}. \]
Step 2: Subhash and Prakash’s Combined Copying Rate – Subhash and Prakash together can copy 300 pages in 40 hours. So, their combined rate is: \[ \text{Combined rate} = \frac{300 \text{ pages}}{40 \text{ hours}} = 7.5 \text{ pages per hour}. \]
Step 3: Prakash’s Copying Rate – We know Subhash’s rate is 5 pages per hour. So, the rate of Prakash is the difference between their combined rate and Subhash’s rate: \[ \text{Prakash’s rate} = 7.5 \text{ pages per hour} – 5 \text{ pages per hour} = 2.5 \text{ pages per hour}. \]
Step 4: Time for Prakash to Copy 30 Pages Now, we need to find how much time Prakash will take to copy 30 pages. The time can be calculated as: \[ \text{Time} = \frac{\text{Total pages}}{\text{Prakash’s rate}} = \frac{30 \text{ pages}}{2.5 \text{ pages per hour}} = 12 \text{ hours}. \]
Final Answer: Prakash can copy 30 pages in 12 hours.
#22. An engineer undertakes a project to build a road 15 km long in 300 days and employs 45 men for the purpose. After 100 days, he finds 2.5 km of the road has been completed. Find the (approx.) number of extra men he must employ to finish the work in time.
A. 43
B. 45
C. 55
D. 12
E. 60
Answer: Option D Solution: Let’s go through this step by step to solve the problem.
Step 1: Calculate Total Work Required The total length of the road is 15 km, and the project is to be completed in 300 days with 45 men.
We can calculate the total amount of work needed in terms of man-days, which is the product of the number of men and the number of days:
Step 2: Calculate the Work Done in the First 100 Days In the first 100 days, 2.5 km of the road has been completed.
Now, let’s calculate how many man-days were used to complete these 2.5 km. Since the total work of 15 km requires 13,500 man-days, the work done in terms of man-days for 2.5 km is:
Step 4: Calculate the Time Left to Finish the Work The engineer has 200 days left to complete the remaining work (300 – 100 = 200 days).
Step 5: Calculate the Number of Men Needed to Complete the Remaining Work To calculate how many men are needed to complete the remaining 12.5 km of road in 200 days, divide the remaining man-days by the remaining time:
Since we can’t have a fraction of a man, we round this up to 57 men.
Step 6: Calculate the Extra Men Needed Currently, there are 45 men working on the project. Therefore, the number of extra men needed is:
\[ \text{Extra men} = 57 – 45 = 12. \]
Final Answer: The engineer must employ 12 extra men to finish the work on time.
#23. There is provision of food in fort for 1200 soldiers for 60 days. After 15 days, 200 soldiers leave the fort. Remaining food will last for how many days?
A. 56 days
B. 50 days
C. 54 days
D. 48 days
Answer: Option C Solution: Let’s solve this step by step.
Step 1: Total Food Provision in Terms of Soldier-Days Initially, there is enough food for 1200 soldiers for 60 days. The total food provision in terms of soldier-days is: \[ \text{Total food provision} = 1200 \times 60 = 72,000 \, \text{soldier-days}. \]
Step 2: Food Consumption in the First 15 Days For the first 15 days, 1200 soldiers are consuming food. The total amount of food consumed in the first 15 days is: \[ \text{Food consumed in 15 days} = 1200 \times 15 = 18,000 \, \text{soldier-days}. \]
Step 3: Remaining Food After 15 days, the remaining food is: \[ \text{Remaining food} = 72,000 – 18,000 = 54,000 \, \text{soldier-days}. \]
Step 4: Number of Soldiers Remaining After 15 days, 200 soldiers leave the fort, so the remaining number of soldiers is: \[ \text{Remaining soldiers} = 1200 – 200 = 1000 \, \text{soldiers}. \]
Step 5: How Many Days Will the Remaining Food Last? Now, we can calculate how many days the remaining food will last for 1000 soldiers. The number of days the remaining food will last is: \[ \text{Days remaining} = \frac{54,000 \, \text{soldier-days}}{1000 \, \text{soldiers}} = 54 \, \text{days}. \]
Final Answer: The remaining food will last for 54 days.
#24. A and B working together completed a job in 5 days. If A works twice as efficiently as he actually did and B works \( \frac{1}{3} \) of actual efficiency, the work would have completed in 3 days. Find the for A to complete the job alone.
A. 6\( \frac{1}{2} \)
B. 6.25
C.6\( \frac{3}{4} \)
D.12\( \frac{1}{2} \)
E. None of these
Answer: Option B Solution:
Let’s solve the problem step by step.
Step 1: Define Variables Let: – \( A_{\text{eff}} \) be A’s actual efficiency (amount of work A does in 1 day), – \( B_{\text{eff}} \) be B’s actual efficiency (amount of work B does in 1 day).
Step 2: Set Up the First Equation A and B together complete the job in 5 days. Hence, the total work done per day by both A and B is: \[ A_{\text{eff}} + B_{\text{eff}} = \frac{1}{5}. \] This means that A and B together complete \( \frac{1}{5} \) of the job per day.
Step 3: Set Up the Second Equation The problem states that if A works twice as efficiently and B works at one-third of his actual efficiency, the job would be completed in 3 days.
In this case: – A’s efficiency becomes \( 2A_{\text{eff}} \), – B’s efficiency becomes \( \frac{1}{3} B_{\text{eff}} \).
Thus, the total work done per day in this case is: \[ 2A_{\text{eff}} + \frac{1}{3} B_{\text{eff}} = \frac{1}{3}. \] This means that A and B together complete \( \frac{1}{3} \) of the job per day in this scenario.
Step 4: Solve the System of Equations Now, we have the following two equations: 1. \( A_{\text{eff}} + B_{\text{eff}} = \frac{1}{5} \), 2. \( 2A_{\text{eff}} + \frac{1}{3} B_{\text{eff}} = \frac{1}{3} \).
Substituting from the First Equation From the first equation, solve for \( B_{\text{eff}} \): \[ B_{\text{eff}} = \frac{1}{5} – A_{\text{eff}}. \]
Substitute this expression for \( B_{\text{eff}} \) into the second equation: \[ 2A_{\text{eff}} + \frac{1}{3} \left( \frac{1}{5} – A_{\text{eff}} \right) = \frac{1}{3}. \] Simplify the equation: \[ 2A_{\text{eff}} + \frac{1}{15} – \frac{1}{3} A_{\text{eff}} = \frac{1}{3}. \]
Multiply the entire equation by 15 to eliminate the fractions: \[ 30A_{\text{eff}} + 1 – 5A_{\text{eff}} = 5. \] Simplify further: \[ 25A_{\text{eff}} + 1 = 5. \] \[ 25A_{\text{eff}} = 4. \] \[ A_{\text{eff}} = \frac{4}{25}. \]
Step 5: Calculate the Time for A to Complete the Job Alone Now that we know A’s efficiency is \( A_{\text{eff}} = \frac{4}{25} \), we can calculate the time it would take for A to complete the job alone. The time for A to complete the entire job is the reciprocal of A’s efficiency: \[ \text{Time for A to complete the job alone} = \frac{1}{A_{\text{eff}}} = \frac{1}{\frac{4}{25}} = \frac{25}{4} = 6.25 \, \text{days}. \]
Final Answer: A alone would take 6.25 days to complete the job.
#25. Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to \( \frac{7}{8} \)of full flow in pipe A and \( \frac{5}{6} \) of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.
A. 8 min
B. 3 min
C. 5.6 min
D. 4.5 min
E. None of these
Answer: Option D Solution: Let’s break down the problem step by step. Step 1: Define the Variables
Let: – A’s full flow rate = \( \frac{1}{12} \) (because pipe A can fill the cistern in 12 minutes, so in 1 minute, it fills \( \frac{1}{12} \) of the cistern). – B’s full flow rate = \( \frac{1}{16} \) (because pipe B can fill the cistern in 16 minutes, so in 1 minute, it fills \( \frac{1}{16} \) of the cistern).
When there is obstruction: – The flow of water in pipe A is restricted to \( \frac{7}{8} \) of its full flow, so the rate of water flow from pipe A becomes \( \frac{7}{8} \times \frac{1}{12} = \frac{7}{96} \). – The flow of water in pipe B is restricted to \( \frac{5}{6} \) of its full flow, so the rate of water flow from pipe B becomes \( \frac{5}{6} \times \frac{1}{16} = \frac{5}{96} \).
Step 2: Find the Total Flow Rate During the Obstruction When both pipes A and B are open but the flow is restricted, the combined flow rate is the sum of the individual restricted flow rates: \[ \text{Combined flow rate during obstruction} = \frac{7}{96} + \frac{5}{96} = \frac{12}{96} = \frac{1}{8}. \] So, during the obstruction, both pipes together fill \( \frac{1}{8} \) of the cistern per minute.
Step 3: Flow After the Obstruction is Removed After the obstruction is removed, both pipes A and B are operating at full flow, so their combined flow rate becomes: \[ \text{Combined flow rate after obstruction is removed} = \frac{1}{12} + \frac{1}{16}. \] To find the sum, first find the least common denominator: \[ \frac{1}{12} + \frac{1}{16} = \frac{4}{48} + \frac{3}{48} = \frac{7}{48}. \] So, after the obstruction is removed, both pipes together fill \( \frac{7}{48} \) of the cistern per minute.
Step 4: Time Taken to Fill the Remaining Cistern After the Obstruction We are told that the cistern is filled in 3 minutes after the obstruction is removed. The amount of work done in 3 minutes is: \[ \text{Work done in 3 minutes} = 3 \times \frac{7}{48} = \frac{21}{48} = \frac{7}{16}. \] Thus, \( \frac{7}{16} \) of the cistern is filled after the obstruction is removed.
Step 5: Calculate the Remaining Work Before the Obstruction Was Removed Since the total capacity of the cistern is 1 (the full cistern), the amount of work done during the obstruction time (before the flow was fully restored) is: \[ \text{Work done before obstruction was removed} = 1 – \frac{7}{16} = \frac{9}{16}. \] This \( \frac{9}{16} \) of the cistern was filled during the obstruction period, when the combined flow rate was \( \frac{1}{8} \).
Step 6: Calculate the Time During the Obstruction Let the time during the obstruction be \( t \) minutes. During this time, the work done is: \[ \text{Work done during obstruction} = t \times \frac{1}{8} = \frac{9}{16}. \] Solving for \( t \): \[ t = \frac{9}{16} \div \frac{1}{8} = \frac{9}{16} \times 8 = 4.5 \, \text{minutes}. \]
Final Answer: The obstruction lasted for 4.5 minutes before the full flow was restored.
#26. Three pipes A,B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?
A. 12 min
B. 16 min
C. 30 min
D. 15 min
E. None of these
Answer: Option C Solution: Let’s solve this step by step:
Step 1: Understand the Problem
-Pipe A can fill the cistern in 10 minutes, so the rate of A is \( \frac{1}{10} \) of the cistern per minute. -Pipe B can fill the cistern in 15 minutes, so the rate of B is \( \frac{1}{15} \) of the cistern per minute. – Pipe C is a waste pipe that empties the cistern, and we need to determine how long it takes for C to empty the cistern when opened alone.
Let the time taken by pipe C to empty the full cistern be \( x \) minutes. So, the rate of pipe C is \( \frac{1}{x} \) of the cistern per minute (since it empties the cistern).
Step 2: Combined Rate of A and B Together
When pipes A and B are open together, they fill the cistern at the combined rate: \[ \text{Rate of A and B together} = \frac{1}{10} + \frac{1}{15}. \] To add these fractions, we need to find the least common denominator (LCD): \[ \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}. \] So, pipes A and B together fill \( \frac{1}{6} \) of the cistern per minute.
Step 3: Work Done When C is Left Open
Now, pipes A and B are opened, but pipe C is also left open by mistake. The man leaves the cistern and returns when the cistern should have been full, but when he returns, he finds that it was not full because C was still open.
The combined rate of A, B, and C together is: \[ \text{Rate of A, B, and C together} = \frac{1}{6} – \frac{1}{x}. \]
Step 4: Work Done After C is Closed
When the man returns, he finds that pipe C was still open, so he closes it. From that moment, the cistern is filled in 2 minutes with only pipes A and B open.
We know that the rate of A and B together is \( \frac{1}{6} \), and in 2 minutes, the cistern is filled by: \[ 2 \times \frac{1}{6} = \frac{1}{3}. \] Thus, after closing C, the cistern is filled by \( \frac{1}{3} \) of its capacity in 2 minutes.
Step 5: Calculate the Total Work Done
When the man was away, the work done during the time \( t \) when C was still open should be equal to the remaining work (i.e., \( 1 – \frac{1}{3} = \frac{2}{3} \)).
The work done during this time is: \[ t \times \left( \frac{1}{6} – \frac{1}{x} \right) = \frac{2}{3}. \]
Step 6: Combine the Work Done and Solve for \( x \)
We now know that the total work done in the time \( t \) (when C was open) plus the work done in the final 2 minutes (after closing C) should equal the full work of filling the cistern, which is 1.
So: \[ t \times \left( \frac{1}{6} – \frac{1}{x} \right) + \frac{1}{3} = 1. \] Now, subtract \( \frac{1}{3} \) from both sides: \[ t \times \left( \frac{1}{6} – \frac{1}{x} \right) = 1 – \frac{1}{3} = \frac{2}{3}. \] We already know that the work done by A and B together in 2 minutes is \( \frac{1}{3} \), and the remaining work, \( \frac{2}{3} \), was done while C was still open.
Step 7: Solve for \( x \)
The next step is to solve for \( x \) by using this equation, which relates the rate of pipe C (emptying the tank) to the total work done. Let me calculate that.
Let’s continue solving for \( x \) (the time taken by pipe C to empty the cistern):
Step 7: Solve for \( x \)
We have the equation: \[ t \times \left( \frac{1}{6} – \frac{1}{x} \right) = \frac{2}{3}. \] We also know that the work done by pipes A and B in 2 minutes is \( \frac{1}{3} \). So the total work done during the time \( t \), when C was still open, is \( \frac{2}{3} \).
The total time \( t \) can be related to the total work done during the time that C was open. The rate at which A and B fill the cistern is \( \frac{1}{6} \), and the rate of C is \( \frac{1}{x} \).
Let’s isolate \( t \) from the equation: \[ t \times \left( \frac{1}{6} – \frac{1}{x} \right) = \frac{2}{3}. \]
Dividing both sides by \( \left( \frac{1}{6} – \frac{1}{x} \right) \) to solve for \( t \), we get: \[ t = \frac{\frac{2}{3}}{\frac{1}{6} – \frac{1}{x}}. \]
Let me simplify this equation and solve for \( x \).
Let’s simplify the equation for \( t \):
We have the equation: \[ t = \frac{\frac{2}{3}}{\frac{1}{6} – \frac{1}{x}}. \] Now, simplify the denominator: \[ \frac{1}{6} – \frac{1}{x} = \frac{x – 6}{6x}. \] Thus, the equation for \( t \) becomes: \[ t = \frac{\frac{2}{3}}{\frac{x – 6}{6x}}. \] Simplifying further: \[ t = \frac{2}{3} \times \frac{6x}{x – 6}. \] Now multiply: \[ t = \frac{12x}{3(x – 6)} = \frac{4x}{x – 6}. \]
Now substitute this expression for \( t \) into the work equation. We know that the work done during time \( t \) is \( \frac{2}{3} \). So, substitute into: \[ \frac{4x}{x – 6} \times \left( \frac{1}{6} – \frac{1}{x} \right) = \frac{2}{3}. \] Simplify the left-hand side: \[ \frac{4x}{x – 6} \times \left( \frac{x – 6}{6x} \right) = \frac{2}{3}. \] The terms \( (x – 6) \) and \( x \) cancel out, leaving: \[ \frac{4}{6} = \frac{2}{3}. \] This simplifies to: \[ \frac{2}{3} = \frac{2}{3}, \] which is true.
Thus, the equation holds, and the time taken by pipe C to empty the cistern is \( \boxed{30 \text{ minutes}} \).
#27. There is a group of 5 boys and 2 girls. The two groups working together can do four times as much work as a boy and a girl. Ratio of working capacities of a boy and a girl is:
A. 2 : 1
B. 2 : 3
C. 1 : 3
D. 1 : 2
Answer: Option A Solution: Let’s assume the working capacity of a boy is \( B \) and the working capacity of a girl is \( G \). According to the problem:
– The total work done by 5 boys and 2 girls is four times the work done by 1 boy and 1 girl. – Mathematically, we can write this as: \[ 5B + 2G = 4(B + G). \]
Step 1: Expand the Equation
Expanding both sides of the equation: \[ 5B + 2G = 4B + 4G. \]
Step 2: Rearrange the Equation
Rearranging terms to bring like terms on one side: \[ 5B – 4B = 4G – 2G, \] \[ B = 2G. \]
Step 3: Interpret the Result
The ratio of the working capacities of a boy and a girl is: \[ \frac{B}{G} = \frac{2}{1}. \]
Thus, the ratio of the working capacities of a boy to a girl is 2:1.
#28. A group of 12 men can do a piece of work in 14 days and other group of 12 women can do the same work in 21 days. They begin together but 3 days before the completion of work, man’s group leaves off. The total number of days to complete the work is:
A. \( \frac{65}{4} \)
B. \( \frac{93}{3} \)
C. 247.5
D. 60
Answer: Option C Solution: Let’s solve the problem step by step:
Step 1: Calculate the rate of work for men and women
– Rate of work for 12 men: The group of 12 men can complete the work in 14 days, so their rate of work is: \[ \text{Rate of 12 men} = \frac{1}{14} \text{ work per day}. \] Thus, the rate of work for 1 man is: \[ \text{Rate of 1 man} = \frac{1}{14 \times 12} = \frac{1}{168}. \]
– Rate of work for 12 women: The group of 12 women can complete the work in 21 days, so their rate of work is: \[ \text{Rate of 12 women} = \frac{1}{21} \text{ work per day}. \] Thus, the rate of work for 1 woman is: \[ \text{Rate of 1 woman} = \frac{1}{21 \times 12} = \frac{1}{252}. \]
Step 2: Determine the work done in the first 3 days by both men and women
Together, the rate of work of the men and the women is: \[ \text{Rate of men and women together} = \frac{1}{168} + \frac{1}{252}. \] Let’s find a common denominator for \( 168 \) and \( 252 \), which is \( 504 \): \[ \frac{1}{168} = \frac{3}{504}, \quad \frac{1}{252} = \frac{2}{504}. \] Thus, the combined rate of work is: \[ \frac{3}{504} + \frac{2}{504} = \frac{5}{504}. \]
In the first 3 days, the total work done is: \[ 3 \times \frac{5}{504} = \frac{15}{504} = \frac{5}{168}. \]
Step 3: Work remaining after 3 days
After 3 days, the remaining work is: \[ 1 – \frac{5}{168} = \frac{163}{168}. \]
Step 4: Work done by women alone after men leave
After 3 days, the men leave, and only the women continue working. The rate at which the women work is \( \frac{1}{252} \) per day. Let \( d \) be the number of days the women work after the men leave. The work done by the women is: \[ d \times \frac{1}{252}. \] The total work done by the women is equal to the remaining work, so: \[ d \times \frac{1}{252} = \frac{163}{168}. \] Solving for \( d \): \[ d = \frac{163}{168} \times 252 = \frac{163 \times 252}{168}. \] Simplifying: \[ d = \frac{163 \times 3}{2} = 244.5 \text{ days}. \]
Step 5: Total time to complete the work
The total time to complete the work is the 3 days when both the men and women worked together, plus the 244.5 days when only the women worked: \[ \text{Total time} = 3 + 244.5 = 247.5 \text{ days}. \]
Thus, the total number of days to complete the work is 247.5 days.
#29. Vimal can do a piece of work in 20 days, Vimal and Kamal together can do in 12 days. If Kamal does the work only for half a day daily then in how many days the work will be completed ?
A. 14
B. 17
C. 12
D. 15
Answer: Option D Solution: Let’s break down the problem step by step.
Step 1: Calculate the rate of work for Vimal and Kamal
– Vimal can complete the work in 20 days, so Vimal’s rate of work is: \[ \text{Rate of Vimal} = \frac{1}{20} \text{ work per day}. \]
– Vimal and Kamal together can complete the work in 12 days, so their combined rate of work is: \[ \text{Rate of Vimal and Kamal together} = \frac{1}{12} \text{ work per day}. \]
Step 2: Find Kamal’s rate of work
To find Kamal’s rate of work, subtract Vimal’s rate from the combined rate of Vimal and Kamal: \[ \text{Rate of Kamal} = \frac{1}{12} – \frac{1}{20}. \]
Let’s find the least common denominator (LCD) of 12 and 20, which is 60. Rewriting the fractions: \[ \frac{1}{12} = \frac{5}{60}, \quad \frac{1}{20} = \frac{3}{60}. \] Now, subtract the two fractions: \[ \text{Rate of Kamal} = \frac{5}{60} – \frac{3}{60} = \frac{2}{60} = \frac{1}{30}. \] So, Kamal can complete the work in 30 days.
Step 3: Work done by Kamal for half a day daily
If Kamal works only for half a day daily, then Kamal’s effective work rate per day is half of \( \frac{1}{30} \): \[ \text{Effective rate of Kamal} = \frac{1}{2} \times \frac{1}{30} = \frac{1}{60}. \]
Step 4: Total work rate when both Vimal and Kamal work together
Vimal works every day at the rate of \( \frac{1}{20} \) per day, and Kamal works half a day at the rate of \( \frac{1}{60} \). The combined rate of work per day is: \[ \text{Combined rate} = \frac{1}{20} + \frac{1}{60}. \] Let’s find the least common denominator (LCD) of 20 and 60, which is 60: \[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{60} = \frac{1}{60}. \] Now, add the rates: \[ \text{Combined rate} = \frac{3}{60} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}. \]
Step 5: Calculate the total time to complete the work
The combined rate of \( \frac{1}{15} \) means that both Vimal and Kamal together will complete the work in 15 days.
Thus, the work will be completed in 15 days.
#30. There are three boats A, B and C, working together they carry 60 people in each trip. One day an early morning A carried 50 people in few trips alone. When it stopped carrying the passengers B and C started carrying the people together. It took a total of 10 trips to carry 300 people by A, B and C. It is known that each day on an average 300 people cross the river using only one of the 3 boats A, B and C. How many trips it would take to A to carry 150 passengers alone?
A. 15
B. 30
C. 25
D. 10
E. None of these
Answer: Option A Solution: Combined efficiency of all the three boats = 60 passengers /trip Now, consider option (A) 15 trips and 150 passengers means efficiency of A = 10 passengers per trip A’s efficiency = 10 passengers per trip Then, (B + C) combined efficiency = 50 passengers per trip Since, combined efficiency is 60 so option (A) is correct
#31. Raj can do a piece of work in 20 days. He started the work and left after some days, when 25% work was done. After it Abhijit joined and completed it working for 10 days. In how many days Raj and Abhijit can do the complete work, working together?
A. 6
B. 8
C. 10
D. 12
Answer: Option B Solution: Let’s break the problem down step by step.
Step 1: Raj’s Work Rate – Raj can complete the work in 20 days. So, Raj’s rate of work is:
\[ \text{Raj’s rate of work} = \frac{1}{20} \text{ of the work per day}. \]
Step 2: Work Done by Raj – Raj worked for some days, and during that time, 25% of the work was completed. Since 25% of the work equals \(\frac{1}{4}\) of the total work, we can calculate how many days Raj worked:
\[ \text{Work done by Raj} = \frac{1}{4}. \]
Since Raj works at a rate of \(\frac{1}{20}\) of the work per day, the number of days Raj worked is:
Step 4: Abhijit’s Work – After Raj left, Abhijit joined and worked for 10 days to finish the remaining \(\frac{3}{4}\) of the work. – This means Abhijit’s rate of work can be calculated as:
\[ \text{Abhijit’s rate of work} = \frac{\frac{3}{4}}{10} = \frac{3}{40} \text{ of the work per day}. \]
Step 5: Combined Work Rate of Raj and Abhijit – Now, to find out how many days Raj and Abhijit together can complete the work, we add their rates:
\[ \text{Combined rate of Raj and Abhijit} = \frac{1}{20} + \frac{3}{40}. \]
So, Raj and Abhijit together can complete \(\frac{1}{8}\) of the work per day.
Step 6: Time Taken for Raj and Abhijit to Complete the Work Together – Since Raj and Abhijit together can complete \(\frac{1}{8}\) of the work per day, the time taken to complete the entire work is:
\[ \text{Time taken to complete the work} = \frac{1}{\frac{1}{8}} = 8 \text{ days}. \]
Final Answer: Raj and Abhijit can complete the entire work together in 8 days.
#32. If 10 persons can do a job in 20 days, then 20 person with twice the efficiency can do the same job in:
A. 5 days
B. 40 days
C. 10 days
D. 20 days
Answer: Option A Solution: Let’s break this problem down step by step:
Step 1: Work Done by 10 Persons – 10 persons can complete the job in 20 days. – The total amount of work is the product of the number of persons and the number of days taken, which is:
This means the job requires 200 person-days to complete.
Step 2: Work Done by 20 Persons with Twice the Efficiency – Now, if we have 20 persons, each with **twice the efficiency**, the total efficiency of the group increases. – If each person is twice as efficient, then the work done by one person in a day is doubled. – So, the rate of work for each person is doubled, which means the total rate for 20 persons would be:
Step 3: Time Taken to Complete the Job – Since the total work is 200 person-days, and the combined rate of 20 persons with twice the efficiency is 40 person-days per day, the time taken to complete the job is:
Final Answer: The job will be completed in 5 days by 20 persons with twice the efficiency.
#33. There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, if the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted, if there was no leakage ant it was completely used for eating purpose?
A. 80 days
B. 72 days
C. 100 days
D. 120 days
Answer: Option B Solution: Let x kg of oil be used for the eating purpose, daily, then (x + 11) × 50 = (x +15) × 45 → x = 25 Thus, Total quantity of oil, = (25 +11) × 50 = 1800 Hence, required number of days, = \( \frac{1800}{25} \)= 72 days
#34. If 2 men or 3 women or 4 boys can do a piece of work in 52 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in:
A. 48 days
B. 36 days
C. 45 days
D. 50 days
Answer: Option A Solution: Let’s solve this problem step by step:
Step 1: Define the rates of work for each person – We are given that: – 2 men can do the work in 52 days. – 3 women can do the work in 52 days. – 4 boys can do the work in 52 days.
We will use the concept of “work per day” for each group.
Step 2: Work rate for men, women, and boys The total amount of work can be represented as \( W \).
– For 2 men, the total work \( W \) is completed in 52 days, so the rate of work for 2 men is:
\[ \text{Rate of 2 men} = \frac{W}{52} \quad \text{(work per day)}. \]
Step 3: Rate of work for 1 man, 1 woman, and 1 boy working together Now, let’s find the combined rate of work for 1 man, 1 woman, and 1 boy working together. We add their individual rates:
To add these fractions, we first find a common denominator. The least common denominator of 104, 156, and 208 is 624. Now, express each fraction with this denominator:
Step 4: Time taken to complete the work The time taken to complete the work is the inverse of the combined rate. Therefore, the time taken for 1 man, 1 woman, and 1 boy to complete the work is:
Final Answer: The work will be completed by 1 man, 1 woman, and 1 boy in 48 days.
#35. If one pipe A can fill a tank in 20 minutes, then 5 pipes, each of 20% efficiency of A, can fill the tank in:
A. 80 min
B. 100 min
C. 20 min
D. 25 min
Answer: Option C Solution: Let’s solve this step by step:
Step 1: Understand the efficiency of the pipes – Pipe A can fill the tank in 20 minutes. So, the rate at which Pipe A fills the tank is:
\[ \text{Rate of A} = \frac{1}{20} \text{ of the tank per minute}. \]
– The efficiency of each of the 5 pipes is 20% of Pipe A’s efficiency. Therefore, the rate of each of these pipes is:
\[ \text{Rate of each pipe} = 0.20 \times \frac{1}{20} = \frac{1}{100} \text{ of the tank per minute}. \]
Step 2: Calculate the combined rate of the 5 pipes – Since there are 5 pipes, each with a rate of \( \frac{1}{100} \) of the tank per minute, the combined rate of the 5 pipes is:
\[ \text{Combined rate of 5 pipes} = 5 \times \frac{1}{100} = \frac{5}{100} = \frac{1}{20} \text{ of the tank per minute}. \]
Step 3: Time taken to fill the tank The time taken to fill the tank is the inverse of the combined rate. So, the time taken for 5 pipes to fill the tank is:
Final Answer: The 5 pipes, each with 20% efficiency of Pipe A, can fill the tank in 20 minutes.
#36. If m men can do a work in r days, then the number of days taken by (m + n) men to do it is :
A.\( \frac{m + n}{mn} \)
B.\( \frac{m + n}{mr} \)
C.\( \frac{mr}{m + n} \)
D.\( \frac{r(m + n)}{mn} \)
Answer: Option C Solution:
To solve this, let’s break it down: Step 1: Understand the given information – \( m \) men can complete the work in \( r \) days. This means the total amount of work can be expressed as:
\[ \text{Total work} = m \times r \, \text{(since work = number of workers × number of days)}. \]
Step 2: Find the rate of work for \( m \) men The rate of work for \( m \) men is:
\[ \text{Rate of work for } m \text{ men} = \frac{\text{Total work}}{r} = \frac{m \times r}{r} = m \, \text{(work per day)}. \]
Step 3: Find the time taken by \( m + n \) men Now, if there are \( m + n \) men, the total rate of work becomes:
\[ \text{Rate of work for } (m + n) \text{ men} = m + n. \]
To find the time taken by \( (m + n) \) men to complete the work, we use the formula:
\[ \text{Time taken} = \frac{\text{Total work}}{\text{Rate of work for } (m + n) \text{ men}} = \frac{m \times r}{m + n}. \]
Final Answer: The number of days taken by \( (m + n) \) men to do the work is: \[ \boxed{\frac{m \times r}{m + n}}.\]
#37. X takes 4 days to complete one-third of a job. Y takes 3 days to complete one-sixth of the job and Z takes 5 days to complete half the job. If all of them work together for 3 days and X and Z quit, how long will it take for Y to complete the remaining work done.
A. 6 days
B. 8.1 days
C. 5.1 days
D. 7 days
E. None of these
Answer: Option C Solution: Let’s break this problem down step by step.
Step 1: Work rates of X, Y, and Z – X’s rate: X takes 4 days to complete one-third of the job. So, X’s work rate is:
\[ \text{X’s rate} = \frac{1}{3} \div 4 = \frac{1}{12} \text{ of the job per day}. \]
– Y’s rate: Y takes 3 days to complete one-sixth of the job. So, Y’s work rate is:
\[ \text{Y’s rate} = \frac{1}{6} \div 3 = \frac{1}{18} \text{ of the job per day}. \]
– Z’s rate**: Z takes 5 days to complete half of the job. So, Z’s work rate is:
\[ \text{Z’s rate} = \frac{1}{2} \div 5 = \frac{1}{10} \text{ of the job per day}. \]
Step 2: Work done by X, Y, and Z together in 3 days Now, we can calculate how much work X, Y, and Z together will complete in 3 days. The combined rate of work for all three is:
To add these, we need a common denominator. The least common denominator (LCD) of 12, 18, and 10 is 180. Let’s express each rate with the denominator 180:
Step 4: Time taken by Y to complete the remaining work Now, only Y is left to complete the remaining work. Y’s rate of work is \(\frac{1}{18}\) of the job per day. To find the time taken for Y to complete the remaining \(\frac{17}{60}\) of the job, we use the formula:
Final Answer: It will take 5.1 days for Y to complete the remaining work.
#38. Two typist of varying skills can do a job in 6 minutes if they work together. If the first typist typed alone for 4 minutes and then the second typist typed alone for 6 minutes, they would be left with \( \frac{1}{5} \) of the whole work. How many minutes would it take the slower typist to complete the typing job working alone ?
A. 10 minutes
B. 15 minutes
C. 12 minutes
D. 20 minutes
E. 17 minutes
Answer: Option B Solution: Let’s solve this step by step:
Step 1: Define the work rates of the typists Let the rate of work for the first typist be \( r_1 \) (work per minute), and the rate of work for the second typist be \( r_2 \).
– When they work together, they can complete the entire job in 6 minutes. So, their combined work rate is:
\[ r_1 + r_2 = \frac{1}{6} \text{ (since together, they complete the job in 6 minutes)}. \]
Step 2: Work done when the typists work separately – The first typist works alone for 4 minutes, so the amount of work done by the first typist is:
\[ \text{Work done by first typist} = 4 \times r_1. \]
– The second typist works alone for 6 minutes, so the amount of work done by the second typist is:
\[ \text{Work done by second typist} = 6 \times r_2. \]
Step 3: Total work done and remaining work According to the problem, after both typists have worked as described, \( \frac{1}{5} \) of the work remains. Therefore, the work done by both typists together in the 10 minutes (4 minutes by the first typist and 6 minutes by the second typist) is:
Step 5: Time taken by the slower typist to complete the job The slower typist is the one with the smaller work rate. In this case, \( r_2 = \frac{1}{15} \), so the time taken by the slower typist to complete the entire job working alone is:
Final Answer: The slower typist will take 15 minutes to complete the typing job working alone.
#39. A and B completed a work together in 5 days. had A worked at twice the speed and B at half the speed, it would have taken them four days to complete the job. How much time would it take for A alone to do the
work?
A. 10 days
B. 20 days
C. 25 days
D. 15 days
E. 24 days
Answer: Option A Solution: Let’s solve this step by step.
Step 1: Define the rates of work for A and B Let the rate of work of A be \( r_A \) (fraction of the work A completes in one day), and the rate of work of B be \( r_B \) (fraction of the work B completes in one day).
– A and B together complete the work in 5 days, so their combined work rate is:
\[ r_A + r_B = \frac{1}{5} \text{ (since they finish the job in 5 days)}. \]
Step 2: Work rates when A works at twice the speed and B works at half the speed – If A works at twice the speed, then A’s rate becomes \( 2r_A \). – If B works at half the speed, then B’s rate becomes \( \frac{r_B}{2} \).
In this case, they would complete the work in 4 days, so their combined work rate is:
\[ 2r_A + \frac{r_B}{2} = \frac{1}{4} \text{ (since they finish the job in 4 days)}. \]
Step 3: Set up the system of equations We now have the following system of equations:
Step 4: Solve the system of equations We’ll first eliminate the fraction in the second equation. Multiply the second equation by 2 to clear the fraction:
Step 5: Time taken for A alone to complete the work The time taken by A to complete the entire work is the reciprocal of \( r_A \):
\[ \text{Time for A alone} = \frac{1}{r_A} = \frac{1}{\frac{1}{10}} = 10 \text{ days}. \]
Final Answer: It would take 10 days for A alone to complete the work.
#40. Two forest officials in their respective division were involved in the harvesting of tendu leave. One division had an average output of 21 tons from a hectares and other division, which had 12 hectares of land less, dedicated to tendu leaves, got 25 tons of tendu from a hectare. As a result, the second division harvested 300 tons of tendu leaves more than the first. How many tons of tendu leaves did the first division
A. 3150
B. 3450
C. 3500
D. 3600
E. None of these
Answer: Option A Solution: Let’s solve this step by step:
Step 1: Define variables for the areas and outputs Let: – \( x \) be the number of hectares of land in the first division. – The second division has 12 hectares less, so it has \( x – 12 \) hectares of land.
Step 2: Write the equations for the total output – The first division harvested 21 tons of tendu leaves per hectare, so the total output of the first division is:
\[ \text{Output of the first division} = 21 \times x \text{ tons}. \]
– The second division harvested 25 tons of tendu leaves per hectare, so the total output of the second division is:
\[ \text{Output of the second division} = 25 \times (x – 12) \text{ tons}. \]
Step 3: Set up the equation for the difference in output We are told that the second division harvested 300 tons more than the first division. So:
\[ 25 \times (x – 12) – 21 \times x = 300. \]
Step 4: Solve for \( x \) First, expand both sides of the equation:
\[ 25x – 300 – 21x = 300. \]
Simplify:
\[ 4x – 300 = 300. \]
Now, solve for \( x \):
\[ 4x = 600, \]
\[ x = 150. \]
Step 5: Calculate the total output of the first division Now that we know \( x = 150 \), the first division has 150 hectares of land. The total output of the first division is:
\[ \text{Output of the first division} = 21 \times 150 = 3150 \text{ tons}. \]
Final Answer: The first division harvested 3150 tons of tendu leaves.
#41. The charges per hour of internet surfing is increased by 25% then find the percentage decrease in the time period of surfing user (a net savy) who can afford only 10% increase in expenditure:
A. 22%
B. 12%
C. 15%
D. 9.09%
Answer: Option B Solution: Let’s break this problem down step by step:
Step 1: Define the variables Let: – \( p \) be the original charge per hour for internet surfing. – \( t \) be the original time (in hours) the user can afford to surf.
The total expenditure for the user is the product of the charge per hour and the time spent surfing:
\[ \text{Total expenditure} = p \times t. \]
Step 2: New charges and total expenditure after increase – The charges per hour increase by 25%, so the new charge per hour is:
\[ \text{New charge per hour} = p \times (1 + 0.25) = 1.25p. \]
– The user can afford only a 10% increase in expenditure. Therefore, the new expenditure is:
\[ \text{New expenditure} = \text{Original expenditure} \times (1 + 0.10) = p \times t \times 1.10. \]
Step 3: New time spent surfing Let the new time period of surfing be \( t’ \). The new expenditure can also be expressed as:
\[ \text{New expenditure} = 1.25p \times t’. \]
Since the new expenditure is also \( p \times t \times 1.10 \), we can set the two expressions equal:
\[ 1.25p \times t’ = p \times t \times 1.10. \]
Step 4: Solve for \( t’ \) Cancel out \( p \) from both sides:
\[ 1.25 \times t’ = t \times 1.10. \]
Now, solve for \( t’ \):
\[ t’ = \frac{1.10}{1.25} \times t = 0.88 \times t. \]
Step 5: Calculate the percentage decrease in time The new time \( t’ \) is 88% of the original time, meaning there is a 12% decrease in time. The percentage decrease in time is:
Final Answer: The percentage decrease in the time period of surfing is 12%.
#42. Ram Lal is a renowned packager of fruits in Varanasi. He packs 70 mangoes or 56 guavas every day working 7 hours a day. His wife also helps him. She packs 30 mangoes or 24 guavas working 6 hours per day. Ram Lal has to pack 3300 mangoes and 2400 guavas with help of his wife. They works alternately, each day 10 hours. His wife started packaging firs day and works every alternate days. Similarly, Ram Lal started his work second day and and worked alternatively till the completion of the work. In how many days the work will finished?
A. 85\( \frac{3}{2} \)
B. 85
C. 44
D. 80
E. None of thes
Answer: Option C Solution: Let’s work through this step by step:
Step 1: Calculate the work rate for Ram Lal and his wife
– Ram Lal’s work rate: – Ram Lal packs 70 mangoes or 56 guavas per day, working 7 hours per day. – His rate for packing mangoes is:
\[ \text{Ram Lal’s rate for mangoes} = \frac{70 \text{ mangoes}}{7 \text{ hours}} = 10 \text{ mangoes per hour}. \]
– His rate for packing guavas is:
\[ \text{Ram Lal’s rate for guavas} = \frac{56 \text{ guavas}}{7 \text{ hours}} = 8 \text{ guavas per hour}. \]
-Ram Lal’s wife’s work rate: – She packs 30 mangoes or 24 guavas per day, working 6 hours per day. – Her rate for packing mangoes is:
\[ \text{Wife’s rate for mangoes} = \frac{30 \text{ mangoes}}{6 \text{ hours}} = 5 \text{ mangoes per hour}. \]
– Her rate for packing guavas is:
\[ \text{Wife’s rate for guavas} = \frac{24 \text{ guavas}}{6 \text{ hours}} = 4 \text{ guavas per hour}. \]
Step 2: Calculate the work done in 10 hours by each person
– Ram Lal’s total work in 10 hours: – Mangoes: \( 10 \times 10 = 100 \) mangoes. – Guavas: \( 10 \times 8 = 80 \) guavas.
– Wife’s total work in 10 hours: – Mangoes: \( 10 \times 5 = 50 \) mangoes. – Guavas: \( 10 \times 4 = 40 \) guavas.
Step 3: Work done in a 2-day cycle
They alternate working each day, with his wife starting on the first day. So, the work done in 2 days will be:
– Total mangoes packed in 2 days: – Day 1 (Wife): 50 mangoes. – Day 2 (Ram Lal): 100 mangoes. – Total: \( 50 + 100 = 150 \) mangoes.
– Total guavas packed in 2 days: – Day 1 (Wife): 40 guavas. – Day 2 (Ram Lal): 80 guavas. – Total: \( 40 + 80 = 120 \) guavas.
Step 4: Calculate the number of cycles needed to finish the work
– For mangoes: \[ \frac{3300 \text{ mangoes}}{150 \text{ mangoes per 2-day cycle}} = 22 \text{ cycles}. \]
– For guavas: \[ \frac{2400 \text{ guavas}}{120 \text{ guavas per 2-day cycle}} = 20 \text{ cycles}. \]
The longer of the two cycle requirements will dictate the time taken to complete the work. Since 22 cycles are required for mangoes, we will finish the work after 22 cycles.
Step 5: Calculate the total number of days
Each cycle takes 2 days, so for 22 cycles, the total number of days is:
Final Answer: The work will be finished in 44 days.
#43. A group of men decided to do a job in 4 days. But since 20 men dropped out every day, the job completed at the end of the 7 day. How many men were there at the beginning?
A. 240
B. 140
C. 280
D. 150
E. None of these
Answer: Option B Solution: Let’s solve this step by step:
Step 1: Define the variables Let: – \( x \) be the number of men who originally decided to do the job.
Step 2: Total work required The total work required to complete the job is the same, no matter how many men are working on it. Since the job was initially planned to be completed in 4 days, the total work can be represented as the number of men multiplied by the number of days:
\[ \text{Total work} = x \times 4. \]
Step 3: Work done with decreasing number of men Since 20 men drop out every day, the number of men working each day changes. Let’s calculate the total amount of work done:
– Day 1: \( x \) men work. – Day 2: \( x – 20 \) men work. – Day 3: \( x – 2(20) = x – 40 \) men work. – Day 4: \( x – 3(20) = x – 60 \) men work. – Day 5: \( x – 4(20) = x – 80 \) men work. – Day 6: \( x – 5(20) = x – 100 \) men work. – Day 7: \( x – 6(20) = x – 120 \) men work.
Step 4: Total work done over 7 days The total work done over the 7 days is the sum of the work done each day:
Step 5: Set up the equation We know that the total work is also equal to \( x \times 4 \) (from the original plan):
\[ 7x – 420 = 4x. \]
Step 6: Solve for \( x \)
\[ 7x – 4x = 420, \]
\[ 3x = 420, \]
\[ x = \frac{420}{3} = 140. \]
Final Answer: The number of men at the beginning was 140.
#44. Two persons having different productivity of labour, working together can reap a field in 2 days. If onethird of the field was reaped by the first man and rest by the other one working alternatively took 4 days. How long did it take for the faster person to reap the whole field working alone?
A. 3
B. 6
C. 8
D. 12
E. None of these
Answer: Option A Solution: Let’s solve this problem step by step.
Step 1: Define the variables – Let \( A \) be the first person (slower worker), and \( B \) be the second person (faster worker). – Let the total amount of work required to reap the entire field be represented as 1 unit of work.
Step 2: Work done by both persons together – When both work together, they can complete the field in 2 days. So, the combined rate of work of \( A \) and \( B \) is:
\[ \text{Rate of } A + \text{Rate of } B = \frac{1 \text{ unit}}{2 \text{ days}} = \frac{1}{2}. \]
This means together they can reap \( \frac{1}{2} \) of the field per day.
Step 3: Work done when they alternate – The first person, \( A \), reaps \( \frac{1}{3} \) of the field by working alone. After this, \( B \) and \( A \) alternate working. – It takes 4 days for them to complete the field when alternating, so in 4 days, the total work done is 1 unit. – From the problem, we know that: – \( A \) works on the first day and on alternate days. – \( B \) works on the second day and on alternate days.
Step 4: Work breakdown over 4 days Let’s break the work done over the 4 days.
-Day 1: \( A \) works alone, and he completes \( \frac{1}{3} \) of the field. – Day 2: \( B \) works alone, completing a certain amount of the field. – Day 3: \( A \) works again, completing some more of the field. – Day 4: \( B \) works again, completing the remaining amount of the field.
Let’s denote the work rates of \( A \) and \( B \) by \( r_A \) and \( r_B \), respectively. We have the following equations:
– \( r_A + r_B = \frac{1}{2} \) (from the combined rate when both work together). – The total work done in 4 days is 1 unit: \[ \text{Work by } A \text{ in 4 days} = \frac{1}{3} + 2r_A \quad (\text{since A works on days 1 and 3}). \] \[ \text{Work by } B \text{ in 4 days} = 2r_B \quad (\text{since B works on days 2 and 4}). \]
So, the total work is:
\[ \frac{1}{3} + 2r_A + 2r_B = 1. \]
Step 5: Solve the system of equations We now have the following system of equations: 1. \( r_A + r_B = \frac{1}{2} \), 2. \( \frac{1}{3} + 2r_A + 2r_B = 1 \).
Substitute \( r_B = \frac{1}{2} – r_A \) from the first equation into the second equation:
Final Answer: It would take 3 days for the faster person (B) to reap the whole field working alone.
#45. A group of workers was put on a job. From second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?
A. 15
B. 14
C. 12
D. 10
E. 8
Answer: Option D Solution: Let initially X number of workers be there. Now, Using work equivalence method, X + (X – 1) + (X – 2) + . . . . . + 1 = X × 55% of X \( \frac{X × (X + 1)}{2} \)= \( \frac{5x}{100} \) [series is in AP. Sum of AP = {No. of terms (first term + last term)/2}] X = 10 workers.
#46. A single reservoir supplies the petrol to the whole city, while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume. When the reservoir is full and if 40, 000 litres of petrol is used daily, the supply fails in 90 days. If 32, 000 litres of petrol used daily, it fails in 60 days. How much petrol can be used daily without the supply ever failing?
A. 64000 litres
B. 56000 litres
C. 60000 litres
D. 78000 litres
E. None of these
Answer: Option B Solution: Let X litres be the per day filling and L litres be the capacity of the reservoir, then 90X + L = 40000 × 90 ———– (1) 60X + L = 32000 × 60 ———– (2) Solving the equation, X = 56000 litres Thus, 56000 litres per day can be used without the failure of supply.
#47. Working together B and C take 50% more number of days than A, B and C together take and A and B working together, take more number of days than A, B and C take together. If A, B and C all have worked together till the completion of the work and B has received Rs. 120 out of total earnings of Rs. 450, then in how many days did A, B and C together complete the whole work?
A. 2 days
B. 4 days
C. 6 days
D. 8 days
E. 10 days
Answer: Option B Solution: Ratio of efficiencies of A, B and C, = 5x : 4x : 6x Number of days required by A and B\( \frac{100}{9x} \)—1 Number of days required by A, B and C\( \frac{100}{15x} \)—2 \( \frac{100}{9x} \)- C\( \frac{100}{15x} \)=C\( \frac{8}{3} \) Number of days required by A, B and C =\( \frac{100}{15x} \) = \( \frac{100}{15×\( \frac{15}{3} \)}\ ) = 4 days
#48. Two men and women are entrusted with a task. The second man needs three hours more to cope up with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as second man and the woman working together. The first man working alone,would spend eight hours less than the double period of the time second man would spend working alone. How much time would the two men and the women need to complete the task if they all asked together?
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours
E. 5 hours
Answer: Option A Solution: Difference in times required by the first man (A) and second man (B) = 3 hours. Also, if t and t are the respective times, then t – t = 3 . . . . . . . . . ..(1) Also, B alone be take = (t + 3) h According to the question, 2t – t = 8 2 × (t + 3) – t = 8 [Using equation (1)] t = 2 hours. Now B and woman together take 2 hours and A also take 2 hours, so time required will be half when all 3 work together. So in 1 hour work would be completed.
#49. At a Tech Pvt Ltd. there are some engineering students employed as trainee engineers, belong to two eminent institutions of India. One group belongs to IIT and another to NIT. Each student of IIT works for 10 hours a day till 60 days and each student of NIT works for 8 hours a day till 80 days on the two same project. The ratio of students of IIT and that of NIT is 4:5 respectively. Students of which institution is slower in work and by how much?
A. IITian is 20% less efficient
B. NITian is 33.33% less efficien
C. NIT is 40% less efficient
D. IIT is 33.33% less efficient
E. Cannot be determined
Answer: Option C Solution: Let’s calculate the total work done by the students of IIT and NIT to determine which group is slower.
Step 1: Work done by IIT students – Number of hours worked by one IIT student per day: 10 hours – Number of days worked by IIT students: 60 days
So, total work done by one IIT student is: \[ \text{Total work of one IIT student} = 10 \, \text{hours/day} \times 60 \, \text{days} = 600 \, \text{hours} \]
Step 2: Work done by NIT students – Number of hours worked by one NIT student per day: 8 hours – Number of days worked by NIT students: 80 days
So, total work done by one NIT student is: \[ \text{Total work of one NIT student} = 8 \, \text{hours/day} \times 80 \, \text{days} = 640 \, \text{hours} \]
Step 3: Compare the work done The ratio of students of IIT to NIT is given as 4:5. Let’s assume the number of IIT students is 4x and the number of NIT students is 5x.
Thus, the total work done by all IIT students is: \[ \text{Total work of IIT students} = 4x \times 600 \, \text{hours} = 2400x \, \text{hours} \]
The total work done by all NIT students is: \[ \text{Total work of NIT students} = 5x \times 640 \, \text{hours} = 3200x \, \text{hours} \]
Step 4: Determine which group is slower The IIT students work 600 hours per student, while NIT students work 640 hours per student. Since 640 hours is greater than 600 hours, NIT students do more work individually, meaning the IIT students are slower.
Step 5: How much slower are the IIT students? The difference in work done per student is: \[ 640 \, \text{hours} – 600 \, \text{hours} = 40 \, \text{hours} \]
Thus, IIT students are 40 hours slower in terms of total work done per student.
#50. 42 women can do a piece of work in 18 days, How many women would be required do the same work in 21 days.
A. 35
B. 36
C. 37
D. 38
Answer: Option B Solution: Let X be the number of women required to finish the work in 21 days. Now, using Work Equivalence Method: 42 × 18 = X × 21 X = 36. Number of women required = 36
