Practice questions on Time and Work problem
1.If the work done by 8 men and 4 boys in 1 day is 7 times the work done by 1 man and 1 boy, then compare the work done by 1 man and 1 boy in 1 day? [CDS 2013]
Solution
Here a = 8, b = 1, x = 4, y = 1, n = 7
Therefore
Work done by 1 man / work done by 2 boy
= (ny - x) / ( a - nb)
= ( 7 × 1 - 4) / ( 8 - 7× 1)
= 3 / 1
2.If 14 men and 12 boys can finish work in 4 days, while 8 men and 16 boys can finish the same work in 5 days. Compare the 1 day work of 1 man and 1 boy.
Solution
Here a = 14, b = 12, x = 4, y = 5, c = 8, d = 16
One day work of 1 man / one day work of 1 boy
= ( yd - xb) / ( xa - yc)
= ( 5 × 16 - 4 × 12) / ( 4 × 14 - 5 × 8)
= 32 / 16
= 2 / 1
3.A stock of food is enough for 240 men for 48 days. How long will the same stock last for 160 men? [CDS 2012]
Solution
Given M1 = 240, D1 = 48,
M2 = 160, D2 =?
Therefore M1 D1= M2 D2
240 × 48 = 160 × D2
D2 = 240 × 48 / 160
= 72
4.45 people take 18 days to dig a pond. If the pond would have to be dug in 15 days, then the number of people to be employed will be [CDS 2012]
Solution
Given M1 = 45, D1 = 18
M2 =?, D2 = 15
Therefore M1 D1= M2 D2
45 × 18 = M2 × 15
M2 = 54
5.A certain number of men can do a piece of work in 80 days. If there were 10 men less, It could be finished in 20 days more. How many men are there in the starting?
Solution
Here a = 10, D =80, d = 20
Therefore number of men in starting
= a (D + d) / d
= 10 × ( 80 + 20)/ 20
= 50
6.20 workers working for 5 h per day complete a work in 10 days. If 25 workers are employed to work 10 h per day, what is the time required to complete the work? [CDS 2013]
Solution
Here M1 = 20, M2 = 25, T1 = 5, T2 = 10, D1 = 10, D2 =?
Therefore M1× D1 × T1= M2× D2× T2
20 × 10 × 5 = 25 × D2 × 10
D2 = 4
Therefore D2 = 4 days
7.A contractor undertook to do a certain piece of work in 18 days. He employed certain number of men but 12 of them being absent from the very 1st day, the rest could finish the work in 30 days. Find the number of men originally employed.
Solution
Let the number of men at the beginning = x
Given M1 = x, M2 = x - 12, D1 = 18, D2 =30
According to the formula
M1 D1 = M2 D2
x × 18 = (x - 12) × 30
2x = 60
x = 30
Shortcut tricks of Time and Work problem
Shortcut tricks of Unitary Method
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1.If the work done by 8 men and 4 boys in 1 day is 7 times the work done by 1 man and 1 boy, then compare the work done by 1 man and 1 boy in 1 day? [CDS 2013]
Solution
Here a = 8, b = 1, x = 4, y = 1, n = 7
Therefore
Work done by 1 man / work done by 2 boy
= (ny - x) / ( a - nb)
= ( 7 × 1 - 4) / ( 8 - 7× 1)
= 3 / 1
2.If 14 men and 12 boys can finish work in 4 days, while 8 men and 16 boys can finish the same work in 5 days. Compare the 1 day work of 1 man and 1 boy.
Solution
Here a = 14, b = 12, x = 4, y = 5, c = 8, d = 16
One day work of 1 man / one day work of 1 boy
= ( yd - xb) / ( xa - yc)
= ( 5 × 16 - 4 × 12) / ( 4 × 14 - 5 × 8)
= 32 / 16
= 2 / 1
3.A stock of food is enough for 240 men for 48 days. How long will the same stock last for 160 men? [CDS 2012]
Solution
Given M1 = 240, D1 = 48,
M2 = 160, D2 =?
Therefore M1 D1= M2 D2
240 × 48 = 160 × D2
D2 = 240 × 48 / 160
= 72
4.45 people take 18 days to dig a pond. If the pond would have to be dug in 15 days, then the number of people to be employed will be [CDS 2012]
Solution
Given M1 = 45, D1 = 18
M2 =?, D2 = 15
Therefore M1 D1= M2 D2
45 × 18 = M2 × 15
M2 = 54
5.A certain number of men can do a piece of work in 80 days. If there were 10 men less, It could be finished in 20 days more. How many men are there in the starting?
Solution
Here a = 10, D =80, d = 20
Therefore number of men in starting
= a (D + d) / d
= 10 × ( 80 + 20)/ 20
= 50
6.20 workers working for 5 h per day complete a work in 10 days. If 25 workers are employed to work 10 h per day, what is the time required to complete the work? [CDS 2013]
Solution
Here M1 = 20, M2 = 25, T1 = 5, T2 = 10, D1 = 10, D2 =?
Therefore M1× D1 × T1= M2× D2× T2
20 × 10 × 5 = 25 × D2 × 10
D2 = 4
Therefore D2 = 4 days
7.A contractor undertook to do a certain piece of work in 18 days. He employed certain number of men but 12 of them being absent from the very 1st day, the rest could finish the work in 30 days. Find the number of men originally employed.
Solution
Let the number of men at the beginning = x
Given M1 = x, M2 = x - 12, D1 = 18, D2 =30
According to the formula
M1 D1 = M2 D2
x × 18 = (x - 12) × 30
2x = 60
x = 30
Shortcut tricks of Time and Work problem
Shortcut tricks of Unitary Method
Computer Awareness
If you are a foodie, you can also view these links at ๐
Golden grillz
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