Quantitative Techniques: Practice Paper on Time and Work
Quantitative Techniques: Practice Paper on Time and Work
Akira can do paint a room in 5 days but with the help of Heer, she can do the work in 3 days. How long would Heer take to paint the room alone?
P takes twice the time than Q to complete the work and if together work they finish the task in 3 days. How much time would P take to complete the task?
A and B can do the work in 20 days, B and C can do the work in 12 days while C and A can do the work in 15 days. If they work together then how many days they can do the work.
Ram can do 2/5 work in 10 days then he calls in Shyam to complete the rest of the work and they complete the job in 9 days. How long Shyam would take to complete the job alone?
P and Q can do a job in 45 and 40 days work in respectively. They started working together and P left the work after some days and Q completed the remaining work in 25 days. How many days did P work with Q?
A can-do 1/6th of the work in 4 days while B can do 1/4th of the same work in 4 days. How much A would get paid out of Rs. 100 if they worked together?
A alone can do a work in 9 days while B can do it in 12 days. If starting from A they worked on alternate days then when will be the work completed?
Eight men can do a work in 15 days. After 7 days 3 of them left, In how many days the remaining work will be completed?
If 16 men and 12 boys can do a work in 5 days; 24 men and 13 boys can do a work in 4 days then Find the ratio of their daily work done?-
P can do a wok in 15 days, Q can do the same work in 10 days. They work for 5 days and then c completes the work in 3 days. If a total of Rs. 1500 was paid for the job than how much were P and R paid in one day?
NOTE: There can be different approaches to a problem, In this article only ratio based approach has been used to avoid confusion and I would suggest to use the similar method all the time and work problem because with change one might end up marking the wrong answer. But if with your logic your answer is right throughout then please stick to it. Also, there is no specific way to solve a problem in Time and work the only trick is to solve using logic.
Answer and Solutions:
Ans: 7.5 days
Solution: Let x be no. of days Heer would take to complete the work.
Amount of work Akira can do in one day = 1/5
Akira and Heer together can do the work in 3 days
In one day Akira and Heer can do 1/3 of work. Hence, 1/5 + 1/ x = 1/ 3
1/x = 1/3 – 1/5 = 2/15
x = 15/2 = 7.5 days.
Ans: 9 days
Solution: Let x be the no. of days Q takes to complete the work, then the time taken by P would be 2x.
In one day P can do 1/2x work while Q can do 1/x work.
Together they can do 1/3 of the work.
1/2x + 1/x = 1/3
Solving above equation would give x = 4.5 days
Hene P can do the work in 2x days = 2*4.5 = 9 days.
Ans: 10 days
Solution: Let days taken by A, B and C be a, b and c respectively.
A and B together can do a work in 20 days hence, eq1: 1/a+ 1/b = 1/20
B and C together can do a work in 12 days hence, eq2: 1/b+ 1/c = 1/12
A and C together can do a work in 15 days hence, eq3: 1/a+ 1/c = 1/15
As we need to find their cumulative time we will add the above equations 1 2 and 3,
1/a + 1/b + 1/b + 1/c + 1/a + 1/c = 1/20 + 1/12 + 1/15
2(1/a + 1/b + 1/c ) = (3+5+4) / 60
1/a + 1/b + 1/c = 1/10
Hence A B and C can together work 10 days to complete the job.
Ans: 100 days
Solution: Ram did 2/5th of the work in 10 days, therefore he could complete the job in 10*5/2 = 25 days. One day work done by Ram = 1/25
For remaining work i.e., 3/5th 0of the work was done by both Ram and Shyam in 9 days. Thus If they worked whole work together from the beginning they would have completed the work in 9*5/3 days = 15 days. Their together one day work = 1/15
Work done by Shyam in one day = 1/15 – 1/25 = 1/100
Hence Shyam would take 100 days to complete the work.
Ans: 135/17 days
Solution: Let the no. of days P and Q worked together be x days.
P’s one day work = 1/45; Q’s one day work = 1/40
As the rest of the work is done by Q and initial was done by P and q together hence they ultimately complete the work.
(1/45 + 1/40 )1/x + 25/40 = 1
Solving the above equation gives x=135/17 days.
Ans: Rs. 40
Solution: A can-do 1/6th work in 4 days hence complete work in 24 days; Similarly B would take 16 days to complete the work.
Together they would do 1/24 + 1/16 = 5/48 work in one day. Hence 48/5 days to complete the work i.e., 48/5 * 1/24 would be A’s part of work = 2/5th work.
Therefore payment would be 2/5th of Rs. 100 = Rs. 40
Ans: 11th Day
Solution: One day work of A = 1/9
One day work by B = 1/12; In pair, if we count the days they can complete 1/9+1/12= 7/36 work in 2 days.
Hence in 4 pairs i.e., 8 days they will complete 7*4/36 = 7/9 part of work.
Remaining work = 1-7/9 = 2/9
On 9th day a’s turn will be there hence he will do 1/9th work,
hence remaining work = 2/9-1/9 = 1/9
On 10th day B will work and he will do 1/12th work, as 1/12 is less than 1/9
Hence remaining work = 1/9 – 1/12 = 1/36
As 1/36 is less than 1/9 when a will work on 11th day he will complete the work.
Ans: 64/5 days
Solution: 8 men take 15 days to complete the work,
Work done by 1 man in 1 day = 1/8*15 = 1/120
For 7 days all 8 men worked hence work done in 7 days = 1*7*5/120 = 7/15th work
Remaining work =1-7/15 = 8/15
3 men left only 5 workings hence 1*5/120 = 1/24
8/15th work is done by the remaining 5 workers in 8/15 *24 = 64/5 days.
Solution: Let one day’s one man work done be x and one day one boy’s work be y,
Then, 16x +12y = 5 & 24x +13y = 4
Solving above equations, x = 1/200 and y = 1/100
Hence the ratio of work done in one day by man to a boy = 1:2
Ans: Rs. 188.33
Solution: One day P’s work = 1/15
One day Q’s work = 1/10
Part of work done by P = 1/15 * 5 = 1/3
Part of Work done by Q = 1/10 * 5 = 1/2
Part of work done by R = 1 - 1/3 - 1/2 = 1/6
So ratio of their share’s in wages = 1/3 : 1/2 : 1/6 = 2:3:1
P’s Share = 2/6 * 1500 = 500; P worked for 5 day hence daily wage = 500/5 = Rs.100
R’s Share = 1/6 *1500 = 250; R works for 3 days hence daily wage = 250/3 = Rs. 83.33
P and R’s daily wage = Rs.100+83.33 = Rs. 188.33