Arithmetic - Time & Work - Previous Year CAT/MBA Questions
The best way to prepare for Arithmetic - Time & Work is by going through the previous year Arithmetic - Time & Work XAT questions. Here we bring you all previous year Arithmetic - Time & Work XAT questions along with detailed solutions.
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Some members of a social service organization in Kolkata decide to prepare 2400 laddoos to gift to children in various orphanages and slums in the city, during Durga puja. The plan is that each of them makes the same number of laddoos. However, on the laddoo-making day, ten members are absent, thus each remaining member makes 12 laddoos more than earlier decided.
How many members actually make the laddoos?
- (a)
100
- (b)
50
- (c)
90
- (d)
24
- (e)
40
Answer: Option E
Text Explanation :
Initially considering the number of members = a
The number of ladoos each member is required to make as per the original plan = b.
Given: a * b = 2400.
Given that 10 members were absent and each member had to make an additional 12 ladoos :
(a - 10) * (b + 12) = 2400.
ab - 10b + 12a - 120 = 2400.
Since a * b = 2400.
Hence 12a - 10b = 120.
Substituting b = 120/a
12a - 10 ∙ = 120
= 12a2 - 24000 - 120a = 0
The roots are a = 50, a = -40.
Hence a = 50, b = 48.
The number of people who took part in making ladoos = a - 10 = 40
Workspace:
X, Y, and Z are three software experts, who work on upgrading the software in a number of identical systems. X takes a day off after every 3 days of work, Y takes a day off after every 4 days of work and Z takes a day off after every 5 days of work.
Starting afresh after a common day off,
i) X and Y working together can complete one new upgrade job in 6 days
ii) Z and X working together can complete two new upgrade jobs in 8 days
iii) Y and Z working together can complete three new upgrade jobs in 12 days
If X, Y and Z together start afresh on a new upgrade job (after a common day off), exactly how many days will be required to complete this job?
- (a)
2 days
- (b)
3.5 days
- (c)
2.5 days
- (d)
4 days
- (e)
5 days
Answer: Option C
Text Explanation :
Let efficiency per day of X, Y and Z be x, y and z respectively.
Let the amount of work required for an upgrade = T units.
(i) X and Y working together can complete one new upgrade job in 6 days. In 6 days X will work for 5 days and Y will work for 5 days.
∴ 5x + 5y = T …(1)
(ii) Z and X working together can complete two new upgrade jobs in 8 days. X will work for 6 days and Z will work for 7 days.
∴ 6x + 7z = 2T …(2)
(iii) Y and Z working together can complete three new upgrade jobs in 12 days. Y will work for 10 days and Z will work for 10 days.
∴ 10y + 10z = 3T …(3)
Solving (1) , (2) and (3) we get
x = T/10, y = T/10 and z = T/5
If X, Y and Z work together their combined efficiency = T/10 + T/10 + T/5 = 2T/5
∴ Time required by three of them together to complete 1 upgrade = T/(2T/5) = 2.5 days.
Hence, option (c).
Workspace:
Abdul, Bimal, Charlie and Dilbar can finish a task in 10, 12, 15 and 18 days respectively. They can either choose to work or remain absent on a particular day. If 50 percent of the total work gets completed after 3 days, then, which of the following options is possible?
- (a)
Each of them worked for exactly 2 days.
- (b)
Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.
- (c)
Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.
- (d)
Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.
- (e)
Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.
Answer: Option E
Text Explanation :
Let the total units of work = LCM of (10, 12, 15 and 18) = 180
Therefore, efficiencies of Abdul, Bimal, Charlie and Dilbar individually in one day will be 18 units, 15 units, 12 units and 10 units per day respectively.
Let us check options now:
Option (a): Total work done = 2 × (18 + 15 + 12 + 10)
= 110 units
Option (b): Total work done = 1 × (15 + 10) + 2 × 12 + 3 × 18
= 25 + 24 + 54 = 103 units
Option (c): Total work done = 2 × (18 + 2) + 1 × 10 + 3 × 15
= 40 + 10 + 45 = 95 units
Option (d): Total work done = 2 × (18 + 10) + 1 × 12 + 3 × 15
= 56 + 12 + 45 = 113 units
Option (e): Total work done = 1 × (18 + 12) + 2 × 15 + 3 × 10
= 30 + 30 + 30 = 90 units
Hence, option (e).
Workspace:
Four two-way pipes A, B, C and D can either fill an empty tank or drain the full tank in 4, 10, 12 and 20 minutes respectively. All four pipes were opened simultaneously when the tank is empty. Under which of the following conditions the tank would be half filled after 30 minutes?
- (a)
Pipe A filled and pipes B, C and D drained
- (b)
Pipe A drained and pipes B, C and D filled
- (c)
Pipes A and D drained and pipes B and C filled
- (d)
Pipes A and D filled and pipes B and C drained
- (e)
None of the above
Answer: Option A
Text Explanation :
Let the capacity of the tank = LCM(4, 10, 12, 20) = 60 units
∴ Efficiency of A, B, C and D = 15, 6, 5, 3 units/minute respectively.
We need to check options now,
Option 1: Work Done in a minute = 15 – 6 – 5 – 3 = 1 unit
∴ After 30 minutes, the tank will be half filled.
We don’t have to check other options.
Hence, option (a).
Workspace:
A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 8 hours while an outlet pipe can empty the full tank in 12 hours. If all pipes are left open simultaneously, it takes 6 hours to fill the empty tank. What is the relationship between M and N?
- (a)
M : N = 1 : 1
- (b)
M : N = 2 : 1
- (c)
M : N = 2 : 3
- (d)
M : N = 3 : 2
- (e)
None of the above
Answer: Option E
Text Explanation :
Let the total capacity of the tank be 24 units.
∴ The work done by an inlet pipe = 24/8 = 3 units/hour and
Work done by an outlet pipe = 24/12 = 2 units/hour
Now, work done by M inlet and N outlet pipes in an hour = 3M – 2N
It taken 6 hours for these pipes to fill the tank completely.
∴ 6(3M – 2N) = 24
∴ 3M – 2N = 4
Now when N = 1, M = 2.
Also when N = 4, M = 4
Similarly we get infinite values of N for which we get different values of M and the ratio M : N cannot be determined uniquely.
Hence, option (e).
Workspace:
Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?
- (a)
Pipe A was open for 19 hours.
- (b)
Pipe A was open for 19 hours 30 minutes.
- (c)
Pipe B was open for 19 hours 30 minutes.
- (d)
Pipe C was open for 19 hours 50 minutes.
- (e)
The situation is not possible.
Answer: Option C
Text Explanation :
Let the capacity of the tank be 24x litres.
Pipes A and B fill 3x and 2x litres per hour while pipe C empties 6x litres in an hour.
Let radius of the cone be r and height be h.
∴ πr2h = 72x
For first 19 hours, water inside the cone = 24x + 57x + 38x – 114x = 5x litres
∆ABE ∼ ∆ACD
If AC = 2AB, CD = 2BE
∴ BE = r/2 and AB = h/2
After 50% reduction in the height of the water, volume
Option 1: Pipe A was open for 19 hours.
i.e., B and C were open for 1 more hour.
∴ 2x – 6x = –4x
The cone will have 5x – 4x = x litres of water.
∴ Option 1 is eliminated.
Option 2: Pipe A was open for 19 hours 30 minutes.
i.e., B and C were open for 1 more hour and A for 30 more minutes.
∴ 2x – 6x + 1.5x = –2.5x
The cone will have 5x – 2.5x = 2.5x litres of water
∴ Option 2 is eliminated.
Option 3: Pipe B was open for 19 hours 30 minutes.
i.e., A and C were open for 1 more hour and B for 30 more minutes.
∴ 3x – 6x + 1x = –2x
The cone will have 5x – 2x = 3x litres of water.
∴ Option 3 would be the possible option.
Hence, option (c).
Workspace:
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