Curriculum

Time and Work in arithmetic

If a person A can finish a piece of work in n days, then A’s 1 day work is:

$1/n$

For example, if A can finish a work in 10 days, then A’s 1 day work is:

$1/10$

If A is twice as good as B, then A takes half the time taken by B to finish the same work.

Men and Time Relationship

In work problems, the number of men and the time taken are inversely proportional.

If 10 men take 20 hours to finish a work, and 20 men are used, then:

$10 \times 20 = 20 \times x$

So,

$x = 10$

Therefore, if the number of men increases, the time taken decreases.

The relation is:

$Number\ of\ men \propto 1/Time\ taken$

Change in Number of Men

If the number of men doing a work changes in the ratio 2 : 3, then the time taken changes in the inverse ratio:

$3 : 2$

Formula:

$m_1 \times t_1 = m_2 \times t_2$

So,

$m_1/m_2 = t_2/t_1$

This means men and time are inversely related.

Problem 1: A and B Together

A can do a piece of work in 12 days and B alone can do it in 15 days. We need to find how many days both will take together.

A’s 1 day work:

$1/12$

B’s 1 day work:

$1/15$

Together, A and B’s 1 day work:

$1/12 + 1/15$

Taking LCM 60:

$5/60 + 4/60 = 9/60 = 3/20$

So, both together complete $3/20$ of the work in 1 day.

Therefore, total time taken:

$20/3\ days = 6⅔\ days$

So, A and B together can finish the work in 6⅔ days.

Problem 2: A+B, B+C, C+A

A and B together can do a piece of work in 12 days.

B and C together can do it in 15 days.

C and A together can do it in 20 days.

We need to find how many days A, B and C together can finish the work, and also separately.

Given:

$(A+B)'s\ 1\ day\ work = 1/12$

$(B+C)'s\ 1\ day\ work = 1/15$

$(C+A)'s\ 1\ day\ work = 1/20$

Adding all three:

$2(A+B+C) = 1/12 + 1/15 + 1/20$

LCM is 60:

$2(A+B+C) = 5/60 + 4/60 + 3/60$

$2(A+B+C) = 12/60 = 1/5$

So,

$A+B+C = 1/10$

Therefore, A, B and C together can finish the work in 10 days.

To find A alone:

$A = (A+B+C) - (B+C)$

$A = 1/10 - 1/15$

LCM is 30:

$A = 3/30 - 2/30 = 1/30$

So, A alone can finish the work in 30 days.

Similarly, B and C can also be found using the same method.

Problem 3: A Leaves After 5 Days

A can do a piece of work in 25 days and B can finish it in 20 days. They work together for 5 days, then A goes away. We need to find how many more days B will take to finish the work.

A’s 1 day work:

$1/25$

B’s 1 day work:

$1/20$

Together, their 1 day work:

$1/25 + 1/20$

LCM is 100:

$4/100 + 5/100 = 9/100$

In 5 days, A and B together complete:

$5 \times 9/100 = 45/100 = 9/20$

Remaining work:

$1 - 9/20 = 11/20$

B’s 1 day work is:

$1/20$

So, B will complete $11/20$ work in 11 days.

Therefore, B takes 11 more days.

Problem 4: Work and Wages

A can do a piece of work in 12 days, while B alone can do it in 15 days. They work together for 5 days, then C finishes the rest of the work in 2 days. They are paid ₹960 for the whole work. We need to divide the money.

A’s 1 day work:

$1/12$

B’s 1 day work:

$1/15$

A and B together:

$1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20$

The slide uses the combined work idea:

$(A+B+C)'s\ 1\ day\ work = 1/5$

Then,

$C's\ 1\ day\ work = 1/5 - 3/20$

$C's\ 1\ day\ work = 4/20 - 3/20 = 1/20$

The work ratio is:

$A : B : C = 1/12 : 1/15 : 1/20$

LCM is 60:

$5 : 4 : 3$

Total ratio parts:

$5 + 4 + 3 = 12$

A’s share:

$960 \times 5/12 = ₹400$

B’s share:

$960 \times 4/12 = ₹320$

C’s share:

$960 \times 3/12 = ₹240$

So, the money is divided as:

A = ₹400

B = ₹320

C = ₹240

Problem 5: Work, Wages and Daily Wages

A can do a piece of work in 10 days and B can do it in 15 days. A and B work together for 5 days. The rest of the work is finished by C in 2 days. They are paid ₹1920. We need to find how the money should be divided and their daily wages.

A’s 1 day work:

$1/10$

B’s 1 day work:

$1/15$

A and B together:

$1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6$

In 5 days, A and B complete:

$5 \times 1/6 = 5/6$

Remaining work:

$1 - 5/6 = 1/6$

C completes $1/6$ work in 2 days.

So, C’s 1 day work:

$1/6 \div 2 = 1/12$

Work done by A in 5 days:

$5 \times 1/10 = 1/2$

Work done by B in 5 days:

$5 \times 1/15 = 1/3$

Work done by C in 2 days:

$2 \times 1/12 = 1/6$

So, the work ratio is:

$1/2 : 1/3 : 1/6$

LCM is 6:

$3 : 2 : 1$

Total ratio parts:

$3 + 2 + 1 = 6$

A’s share:

$1920 \times 3/6 = ₹960$

B’s share:

$1920 \times 2/6 = ₹640$

C’s share:

$1920 \times 1/6 = ₹320$

Daily wages:

A works 5 days, so A’s daily wage:

$960/5 = ₹192$

B works 5 days, so B’s daily wage:

$640/5 = ₹128$

C works 2 days, so C’s daily wage:

$320/2 = ₹160$

Therefore:

A = ₹960, daily wage ₹192

B = ₹640, daily wage ₹128

C = ₹320, daily wage ₹160

Problem 6: A is Thrice as Good as B

A is thrice as good a workman as B and is therefore able to finish a piece of work 60 days less than B. We need to find the time in which they can do the work together.

Suppose B finishes the work in $x$ days.

Since A is thrice as good as B, A takes one-third of B’s time.

So, time taken by A:

$x/3$

Given that A takes 60 days less than B:

$x - 60 = x/3$

Multiplying by 3:

$3x - 180 = x$

$2x = 180$

$x = 90$

So, B takes 90 days.

A takes:

$90/3 = 30\ days$

A’s 1 day work:

$1/30$

B’s 1 day work:

$1/90$

Together:

$1/30 + 1/90 = 3/90 + 1/90 = 4/90 = 2/45$

So, A and B together can finish the work in:

$45/2 = 22½\ days$

Problem 7: A, B and C with Leaving Days

A can do a piece of work in 10 days, B in 12 days, and C in 15 days. All begin together, but A leaves the work after 2 days, and B leaves 3 days before the work is finished. We need to find the total time taken.

A’s 1 day work:

$1/10$

B’s 1 day work:

$1/12$

C’s 1 day work:

$1/15$

A, B and C work together for the first 2 days.

Their 1 day work:

$1/10 + 1/12 + 1/15$

LCM is 60:

$6/60 + 5/60 + 4/60 = 15/60 = 1/4$

In 2 days, they complete:

$2 \times 1/4 = 1/2$

After that, A leaves.

B leaves 3 days before completion, so C alone works for the last 3 days.

C’s 3 days work:

$3 \times 1/15 = 1/5$

Remaining work after the first 2 days and C’s final 3 days:

$1 - (1/2 + 1/5)$

$= 1 - 7/10$

$= 3/10$

This $3/10$ work is done by B and C together.

B and C’s 1 day work:

$1/12 + 1/15$

LCM is 60:

$5/60 + 4/60 = 9/60 = 3/20$

Time taken by B and C to complete $3/10$ work:

$(3/10) \div (3/20)$

$= 3/10 \times 20/3 = 2\ days$

So, total time:

A+B+C together = 2 days

B+C together = 2 days

C alone = 3 days

Total time = 7 days

Problem 8: A Leaves, B Completes

A and B can do a piece of work in 45 days and 40 days respectively. They begin the work together, but A leaves after some days and B finishes the remaining work in 23 days. We need to find after how many days A left.

A’s 1 day work:

$1/45$

B’s 1 day work:

$1/40$

A and B together:

$1/45 + 1/40$

LCM is 360:

$8/360 + 9/360 = 17/360$

B’s 23 days work:

$23 \times 1/40 = 23/40$

Remaining work done by A and B together:

$1 - 23/40 = 17/40$

Now, A and B together do $17/360$ work in 1 day.

So, time taken to do $17/40$ work:

$(17/40) \div (17/360)$

$= 17/40 \times 360/17$

$= 9\ days$

Therefore, A left after 9 days.

Problem 9: A, B and C Joining Later

A can do a piece of work in 120 days and B can do it in 150 days. They work together for 20 days. Then B leaves and A alone continues for 12 days. After that, A and C complete the remaining work in 48 days. We need to find how many days C alone will take.

A’s 1 day work:

$1/120$

B’s 1 day work:

$1/150$

A and B together:

$1/120 + 1/150$

LCM is 600:

$5/600 + 4/600 = 9/600 = 3/200$

In 20 days, A and B complete:

$20 \times 3/200 = 3/10$

A’s 12 days work:

$12 \times 1/120 = 1/10$

So far, work completed:

$3/10 + 1/10 = 4/10 = 2/5$

Remaining work:

$1 - 2/5 = 3/5$

This remaining $3/5$ work is completed by A and C together in 48 days.

So, A and C’s 1 day work:

$(3/5) \div 48 = 3/240 = 1/80$

C’s 1 day work:

$1/80 - 1/120$

LCM is 240:

$3/240 - 2/240 = 1/240$

Therefore, C alone can finish the work in 240 days.

Problem 10: A+B, B+C and A Twice as Good as C

A and B together can do a piece of work in 12 days.

B and C together can do it in 15 days.

A is twice as good as C.

We need to find in how many days B alone can finish the work.

Given:

$(A+B)'s\ 1\ day\ work = 1/12$

$(B+C)'s\ 1\ day\ work = 1/15$

Since A is twice as good as C:

$A's\ 1\ day\ work = 2 \times C's\ 1\ day\ work$

Let C’s 1 day work be:

$x$

Then A’s 1 day work is:

$2x$

Now:

$A+B = 1/12$

$B+C = 1/15$

Subtracting:

$(A+B) - (B+C) = 1/12 - 1/15$

$A - C = 5/60 - 4/60 = 1/60$

But:

$A = 2C$

So:

$2C - C = 1/60$

$C = 1/60$

Now B’s 1 day work:

$B = 1/15 - 1/60$

LCM is 60:

$B = 4/60 - 1/60 = 3/60 = 1/20$

Therefore, B alone can finish the work in 20 days.

Problem 11: Alternate Day Work

A and B working alone can finish a piece of work in 8 days and 7 days respectively. They work alternately for a day each, starting with A. We need to find how many days the work will be finished.

A’s 1 day work:

$1/8$

B’s 1 day work:

$1/7$

In 2 days, A and B together complete:

$1/8 + 1/7$

LCM is 56:

$7/56 + 8/56 = 15/56$

In the first 6 days, there are 3 such cycles.

Work done in 6 days:

$3 \times 15/56 = 45/56$

On the 7th day, it is A’s turn.

A does:

$1/8 = 7/56$

Total work done by the end of 7 days:

$45/56 + 7/56 = 52/56$

Remaining work:

$1 - 52/56 = 4/56 = 1/14$

Now it is B’s turn.

B’s 1 day work:

$1/7$

To complete $1/14$ work, B needs:

$(1/14) \div (1/7) = 1/2\ day$

Therefore, total time:

$7 + 1/2 = 7½\ days$

Problem 12: Men Required

A certain number of men complete a piece of work in 60 days. If there were 8 more men, the work would be finished in 10 days less, that is, in 50 days. We need to find the original number of men.

Let the original number of men be:

$x$

Original work:

$x \times 60$

With 8 more men:

$(x + 8) \times 50$

Since work is the same:

$x \times 60 = (x + 8) \times 50$

$60x = 50x + 400$

$10x = 400$

$x = 40$

Therefore, the original number of men is 40.

Problem 13: Men and Boys

4 men or 6 boys can finish a piece of work in 20 days. We need to find how many days 6 men and 11 boys can finish it.

Given:

$4\ men = 6\ boys$

So,

$1\ man = 6/4\ boys = 3/2\ boys$

Therefore:

$6\ men = 6 \times 3/2 = 9\ boys$

Now, 6 men and 11 boys together are equivalent to:

$9\ boys + 11\ boys = 20\ boys$

If 6 boys can finish the work in 20 days, then total work:

$6 \times 20 = 120\ boy-days$

With 20 boys, time taken:

$120/20 = 6\ days$

Therefore, 6 men and 11 boys can finish the work in 6 days.

Problem 14: Men, Women, Hours and Days

2 men and 3 women working 7 hours a day finish a work in 5 days.

4 men and 4 women working 3 hours a day finish the same work in 7 days.

We need to find the number of days required for 7 men only, working 4 hours a day, to finish the work.

Let 1 man’s 1 hour work be:

$M$

Let 1 woman’s 1 hour work be:

$W$

First condition:

$(2M + 3W) \times 7 \times 5$

$= (2M + 3W) \times 35$

Second condition:

$(4M + 4W) \times 3 \times 7$

$= (4M + 4W) \times 21$

Both are equal because the work is same:

$35(2M + 3W) = 21(4M + 4W)$

$70M + 105W = 84M + 84W$

$21W = 14M$

So,

$3W = 2M$

The total work can be converted in terms of men’s work.

Using the first condition:

$70M + 105W$

Since:

$3W = 2M$

Then:

$W = 2M/3$

So:

$105W = 105 \times 2M/3 = 70M$

Total work:

$70M + 70M = 140M-hours$

Now, 7 men working 4 hours per day do:

$7 \times 4 = 28M-hours\ per\ day$

Number of days:

$140/28 = 5\ days$

Therefore, 7 men working 4 hours per day can finish the work in 5 days.

Problem 15: Men, Boys and Earnings

2 men with 3 boys can earn ₹252 in 7 days.

4 men with 7 boys can earn ₹624 in 8 days.

We need to find in what time 6 men with 8 boys can earn ₹1020.

Let 1 man’s 1 day earning be:

$x$

Let 1 boy’s 1 day earning be:

$y$

From the first condition:

$(2x + 3y) \times 7 = 252$

$2x + 3y = 36$

Equation 1:

$2x + 3y = 36$

From the second condition:

$(4x + 7y) \times 8 = 624$

$4x + 7y = 78$

Equation 2:

$4x + 7y = 78$

Multiplying Equation 1 by 2:

$4x + 6y = 72$

Equation 2:

$4x + 7y = 78$

Subtracting:

$y = 6$

Substitute in Equation 1:

$2x + 3(6) = 36$

$2x + 18 = 36$

$2x = 18$

$x = 9$

So, 1 man earns ₹9 per day and 1 boy earns ₹6 per day.

Earning of 6 men and 8 boys per day:

$6x + 8y$

$= 6(9) + 8(6)$

$= 54 + 48 = 102$

Time required to earn ₹1020:

$1020/102 = 10\ days$

Therefore, 6 men and 8 boys earn ₹1020 in 10 days.

Problem 16: Men Dropping Off

25 men are employed to do a piece of work which they could finish in 20 days, but the men drop off by 5 at the end of every 10 days. We need to find in what time the work will be completed.

Total work:

$25 \times 20 = 500\ man-days$

First 10 days:

$25 \times 10 = 250\ man-days$

So, half the work is completed.

Remaining work:

$500 - 250 = 250\ man-days$

After 10 days, 5 men leave, so remaining men:

$25 - 5 = 20\ men$

Next 10 days:

$20 \times 10 = 200\ man-days$

Remaining work:

$250 - 200 = 50\ man-days$

After another 10 days, 5 more men leave, so remaining men:

$20 - 5 = 15\ men$

Time taken by 15 men to complete 50 man-days:

$50/15 = 10/3 = 3⅓\ days$

Total time:

$10 + 10 + 3⅓ = 23⅓\ days$

Therefore, the work will be completed in 23⅓ days.

Problem 17: A Can Do Half as Much Again as B

A can do half as much again as B can do in the same time. B alone can do a piece of work in 18 days. We need to find in how many days both together can finish the work.

“Half as much again as B” means A’s work is:

$B + half\ of\ B = 3/2\ of\ B$