DIRECT AND INVERSE VARIATIONS - Q&A

Exercise 10 (A)

1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) The values of a and b from the table are:

(a) a = 30 and b = 270
(b) a = 30 and b = 15
(c) a = 45 and b = 15
(d) a = 30 and b = 540
Solution:
This is a case of direct variation because as x increases, y increases. We check the ratio $x/y$.
$3/18 = 1/6$
$8/48 = 1/6$
Since it is direct variation, the ratio $x/y$ must be constant ($1/6$).
For a: $5/a = 1/6 \Rightarrow a = 5 \times 6 = 30$
For b: $b/90 = 1/6 \Rightarrow b = 90 / 6 = 15$
So, $a = 30$ and $b = 15$.
Answer: (b)

(ii) In 5 hours, a machine makes 45 screws. In 12 hours, the same machine will make:
(a) 36 screws
(b) 90 screws
(c) 72 screws
(d) 108 screws
Solution:
This is direct variation (more time, more screws).
Let the number of screws be $x$.
Ratio: $5/45 = 12/x$
$1/9 = 12/x$
$x = 12 \times 9 = 108$
Answer: (d)

(iii) If the cost of 9 pens is ₹ 369, the cost of one dozen pens of the same make will be:
(a) ₹ 42
(b) ₹ 429
(c) ₹ 942
(d) ₹ 492
Solution:
1 dozen = 12 pens.
Cost of 9 pens = ₹ 369.
Cost of 1 pen = $369 / 9 = \text{₹ } 41$.
Cost of 12 pens = $12 \times 41 = \text{₹ } 492$.
Answer: (d)

(iv) If y varies directly to x and $y = 80$ when $x = 400$ the value of y when $x = 25$ is:
(a) 5
(b) 25
(c) 75
(d) 125
Solution:
Since y varies directly as x, $y/x = k$.
$80/400 = k \Rightarrow k = 1/5$.
Now, find y when $x = 25$:
$y/25 = 1/5$
$y = 25/5 = 5$.
Answer: (a)

(v) A man working 48 hours per week earns ₹ 2,400. If he works for 36 hours per week, he will earn:
(a) ₹ 3,200
(b) ₹ 900
(c) ₹ 1,800
(d) ₹ 1,080
Solution:
Earnings vary directly with hours worked.
Earnings per hour = $2400 / 48 = \text{₹ } 50$.
Earnings for 36 hours = $36 \times 50 = \text{₹ } 1,800$.
Answer: (c)

2. In which of the following tables, x and y vary directly:
(i)

3. If x and y vary directly, find the values of x, y and z:

4. A truck consumes 28 litres of diesel for moving through a distance of 448 km. How much distance will it cover in 64 litres of diesel ?
Solution:
Let the distance covered be $x$ km.
This is direct variation (more diesel, more distance).
Ratio: $\frac{\text{Diesel}_1}{\text{Distance}_1} = \frac{\text{Diesel}_2}{\text{Distance}_2}$
$28 / 448 = 64 / x$
$1 / 16 = 64 / x$ (Since $28 \times 16 = 448$)
$x = 16 \times 64$
$x = 1024$
Answer: 1024 km

5. For 100 km, a taxi charges ₹ 1,800. How much will it charge for a journey of 120 km?
Solution:
Let the charge be ₹ x.
Direct variation: $\frac{\text{Distance}_1}{\text{Charge}_1} = \frac{\text{Distance}_2}{\text{Charge}_2}$
$100 / 1800 = 120 / x$
$1 / 18 = 120 / x$
$x = 120 \times 18$
$x = 2160$
Answer: ₹ 2,160

6. If 27 identical articles cost ₹ 1,890, how many articles can be bought for ₹ 1,750?
Solution:
Let number of articles be x.
Direct variation: $\frac{\text{Articles}}{\text{Cost}} = \text{Constant}$
$27 / 1890 = x / 1750$
$27 / 1890 = 1 / 70$
$x = 1750 / 70$
$x = 25$
Answer: 25 articles

7. 7 kg of rice costs ₹ 1,120. How much rice can be bought for ₹ 3,680?
Solution:
Let quantity of rice be x kg.
Cost of 1 kg = $1120 / 7 = \text{₹ } 160$.
Rice for ₹ 3680 = $3680 / 160$
$x = 368 / 16 = 23$
Answer: 23 kg

8. 6 notebooks cost ₹ 156, find the cost of 54 such notebooks.
Solution:
Cost of 1 notebook = $156 / 6 = \text{₹ } 26$.
Cost of 54 notebooks = $54 \times 26$
$54 \times 26 = 1404$
Answer: ₹ 1,404

9. 22 men can dig a 27 m long trench in one day. How many men should be employed for digging 135 m long trench of the same type in one day ?
Solution:
Let number of men be x.
More length requires more men (Direct Variation).
$\frac{\text{Men}_1}{\text{Length}_1} = \frac{\text{Men}_2}{\text{Length}_2}$
$22 / 27 = x / 135$
$x = (22 \times 135) / 27$
$x = 22 \times 5$ (Since $135 / 27 = 5$)
$x = 110$
Answer: 110 men

10. If the total weight of 11 identical articles is 77 kg, how many articles of the same type would weigh 224 kg?
Solution:
Let number of articles be x.
Weight of 1 article = $77 / 11 = 7$ kg.
Number of articles for 224 kg = $224 / 7$
$x = 32$
Answer: 32 articles

11. A train is moving with uniform speed of 120 km per hour.
(i) How far will it travel in 36 minutes ?
(ii) In how much time will it cover 210 km ?
Solution:
Speed = 120 km/hr.
(i) Time = 36 minutes = $36/60$ hours = $0.6$ hours.
Distance = Speed $\times$ Time = $120 \times 0.6 = 72$ km.
(ii) Distance = 210 km.
Time = Distance / Speed = $210 / 120$ hours.
Time = $7 / 4$ hours = $1 \frac{3}{4}$ hours.
$1 \frac{3}{4}$ hours = 1 hour 45 minutes.
Answer: (i) 72 km (ii) 1 hour 45 minutes

Exercise 10 (B)

1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) If a varies inversely to b and $b = 12$ when a is 8; the value of b when $a = 6$ is:
(a) 72
(b) 32
(c) 12
(d) 16
Solution:
Inverse variation means $a \times b = k$ (constant).
$8 \times 12 = k \Rightarrow k = 96$.
When $a = 6$, $6 \times b = 96$.
$b = 96 / 6 = 16$.
Answer: (d)

(ii) 12 men can make a certain number of screws in 30 days. The number of screws, of the same type will be made by 24 men in:
(a) 15 days
(b) 30 days
(c) 60 days
(d) none of these
Solution:
Number of men varies inversely with days required (for same work).
$M_1 \times D_1 = M_2 \times D_2$
$12 \times 30 = 24 \times D_2$
$360 = 24 \times D_2$
$D_2 = 360 / 24 = 15$.
Answer: (a)

(iii) A school has 8 periods in a days, each of 45 minutes duration. Assuming that the number of school hours to be the same and the school has 9 periods in a day, the duration of each period is:
(a) 72 minute
(b) 180/5 minute
(c) 40 minute
(d) 60 minute
Solution:
Total time is constant (Inverse variation between number of periods and duration).
$8 \times 45 = 360$ minutes total.
If 9 periods, let duration be x.
$9 \times x = 360$
$x = 360 / 9 = 40$ minutes.
Answer: (c)

(iv) From the given table find the values of a and b:

(v) A hostel has provisions for 200 students for 30 days. If 100 new students join the hostel, the same provisions will last for:
(a) 45 days
(b) 30 days
(c) 60 days
(d) 20 days
Solution:
Total students = $200 + 100 = 300$.
Inverse variation (More students, fewer days).
$M_1 \times D_1 = M_2 \times D_2$
$200 \times 30 = 300 \times D_2$
$6000 = 300 \times D_2$
$D_2 = 6000 / 300 = 20$.
Answer: (d)

2. Check whether x and y vary inversely or not
(i)

3. If x and y vary inversely, find the values of l, m and n:
(i)

4. 36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Solution:
Inverse variation.
$M_1 \times D_1 = M_2 \times D_2$
$36 \times 7 = x \times 42$
$252 = 42x$
$x = 252 / 42 = 6$.
Answer: 6 men

5. 12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Solution:
Inverse variation (More pipes, less time).
$P_1 \times T_1 = P_2 \times T_2$
$12 \times 42 = 21 \times T_2$
$T_2 = (12 \times 42) / 21$
$T_2 = 12 \times 2 = 24$.
Answer: 24 minutes

6. In a fort, 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate ?
Solution:
Initial: 150 men for 45 days.
After 10 days, remaining provisions are for 150 men for ($45 - 10$) = 35 days.
Remaining men = $150 - 25 = 125$ men.
Let provisions last for x days for 125 men.
$M_1 \times D_1 = M_2 \times D_2$
$150 \times 35 = 125 \times x$
$x = (150 \times 35) / 125$
$x = (6 \times 35) / 5 = 6 \times 7 = 42$.
Answer: 42 days

7. 72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Solution:
Inverse variation.
$72 \times 25 = 30 \times x$
$x = (72 \times 25) / 30$
$x = (72 \times 5) / 6$
$x = 12 \times 5 = 60$.
Answer: 60 days

8. If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours ?
Solution:
Inverse variation.
$56 \times 180 = x \times 70$
$x = (56 \times 180) / 70$
$x = 0.8 \times 180$ (Wait, $56/70 = 8/10 = 0.8$)
$x = (8 \times 180) / 10 = 8 \times 18 = 144$.
Answer: 144 workers

9. A car takes 6 hours to reach a destination by travelling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour ?
Solution:
Distance is constant. Speed and Time are inversely proportional.
$S_1 \times T_1 = S_2 \times T_2$
$50 \times 6 = 75 \times T_2$
$300 = 75 \times T_2$
$T_2 = 300 / 75 = 4$.
Answer: 4 hours

Exercise 10 (C)

1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) A can do a piece of work in 2 days and B can do the same work in 3 days. If A and B work together, the amount of work done by them in 1 day is:
(a) 30
(b) 1.5
(c) 2/3
(d) 5/6
Solution:
A's 1 day work = $1/2$.
B's 1 day work = $1/3$.
(A+B)'s 1 day work = $1/2 + 1/3 = (3+2)/6 = 5/6$.
Answer: (d)

(ii) If 1/20 of a work can be done in 5 days, the amount of work done in one day will be :
(a) 1/100
(b) 100
(c) 4
(d) 5
Solution:
Work done in 5 days = $1/20$.
Work done in 1 day = $(1/20) / 5 = 1/100$.
Answer: (a)

(iii) Ritu can knit a sweater in 4 days and Manish can knit the same sweater in 6 days. If they together knit the same sweater, the number of days taken by them will be:
(a) 10 days
(b) 5 days
(c) 12/5 days
(d) 5/12 days
Solution:
Ritu's rate = $1/4$ sweater/day.
Manish's rate = $1/6$ sweater/day.
Combined rate = $1/4 + 1/6 = (3+2)/12 = 5/12$.
Time taken = $1 / (5/12) = 12/5$ days.
Answer: (c)

(iv) A and B working together can complete a work in 4 days. If A alone can do the same work in 6 days, then B alone can do the same work in:
(a) 12 days
(b) 10 days
(c) 2 days
(d) 24 days
Solution:
(A+B)'s 1 day work = $1/4$.
A's 1 day work = $1/6$.
B's 1 day work = (A+B)'s work - A's work
$= 1/4 - 1/6 = (3-2)/12 = 1/12$.
So B takes 12 days.
Answer: (a)

2. A can do a piece of work in 10 days and B in 15 days. How long will they take to finish it working together?
Solution:
A's 1 day work = $1/10$.
B's 1 day work = $1/15$.
(A+B)'s 1 day work = $1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6$.
Time taken = $6$ days.
Answer: 6 days

3. A and B together can do a piece of work in $6\frac{2}{3}$ days, but B alone can do it in 10 days. How long will A take to do it alone?
Solution:
Time for A+B = $6 \frac{2}{3} = 20/3$ days.
(A+B)'s 1 day work = $1 / (20/3) = 3/20$.
B's 1 day work = $1/10$.
A's 1 day work = $3/20 - 1/10 = 3/20 - 2/20 = 1/20$.
A alone will take 20 days.
Answer: 20 days

4. A can do a work in 15 days and B in 20 days. If they work together on it for 4 days, what fraction of the work will be left?
Solution:
A's work in 1 day = $1/15$.
B's work in 1 day = $1/20$.
(A+B)'s 1 day work = $1/15 + 1/20 = (4+3)/60 = 7/60$.
Work done in 4 days = $4 \times (7/60) = 28/60 = 7/15$.
Work left = $1 - 7/15 = 8/15$.
Answer: 8/15

5. A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they altogether do it?
Solution:
A's rate = $1/6$. B's rate = $1/12$. C's rate = $1/24$.
(A+B+C)'s 1 day work = $1/6 + 1/12 + 1/24$.
LCM = 24.
$= (4 + 2 + 1) / 24 = 7/24$.
Time taken = $24/7$ days = $3 \frac{3}{7}$ days.
Answer: $3 \frac{3}{7}$ days

6. A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take to mow the field by himself?
Solution:
(A+B)'s 1 day work = $1/56$.
(A+B+C)'s 1 day work = $1/42$.
C's 1 day work = (A+B+C)'s work - (A+B)'s work
$= 1/42 - 1/56$.
LCM of 42 and 56 = 168.
$= (4 - 3) / 168 = 1/168$.
C alone takes 168 days.
Answer: 168 days

7. A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in $10\frac{2}{3}$ days. In how many days can A and C do it working together?
Solution:
A's rate = $1/24$.
(A+B)'s rate = $1/16$.
(A+B+C)'s rate = $1 / (32/3) = 3/32$.

First, find C's rate:
C's rate = (A+B+C)'s rate - (A+B)'s rate
$= 3/32 - 1/16 = 3/32 - 2/32 = 1/32$.

Now find (A+C)'s rate:
(A+C)'s rate = A's rate + C's rate
$= 1/24 + 1/32$.
LCM of 24 and 32 is 96.
$= (4 + 3) / 96 = 7/96$.
Time for A and C = $96/7$ days = $13 \frac{5}{7}$ days.
Answer: $13 \frac{5}{7}$ days

8. A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work?
Solution:
A's rate = $1/20$, B's rate = $1/15$.
Combined rate = $1/20 + 1/15 = (3+4)/60 = 7/60$.
Work done in 6 days = $6 \times (7/60) = 42/60 = 7/10$.
Remaining work = $1 - 7/10 = 3/10$.
B does $1/15$ work in 1 day.
Time for B to do $3/10$ work = $\text{Work} / \text{Rate} = (3/10) / (1/15) = (3/10) \times 15 = 45/10 = 4.5$ days.
Answer: 4.5 days (or $4 \frac{1}{2}$ days)

9. A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Solution:
(A+B)'s rate = $1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6$.
Work done in 2 days = $2 \times 1/6 = 1/3$.
Remaining work = $1 - 1/3 = 2/3$.
A's rate = $1/15$.
Time for A = $(2/3) / (1/15) = (2/3) \times 15 = 30/3 = 10$ days.
Answer: 10 days

10. A can do a piece of work in 10 days, B in 18 days, and A, B and C together in 4 days. In what time would C do it alone?
Solution:
(A+B+C)'s rate = $1/4$.
A's rate = $1/10$, B's rate = $1/18$.
C's rate = $1/4 - (1/10 + 1/18)$.
$1/10 + 1/18 = (9 + 5)/90 = 14/90 = 7/45$.
C's rate = $1/4 - 7/45$.
LCM of 4 and 45 is 180.
$= (45 - 28) / 180 = 17/180$.
Time for C = $180/17$ days = $10 \frac{10}{17}$ days.
Answer: $10 \frac{10}{17}$ days

11. A can do 1/3 of a work in 5 days and B can do 1/4 of the same work in 10 days. Find the number of days in which both working together will complete the work.
(Note: The fraction for A was inferred as 1/3 from typical problem structures and context, as the OCR text was partial. B is 1/4).
Solution:
If A does $1/3$ work in 5 days, A does full work in $5 \times 3 = 15$ days.
If B does $1/4$ work in 10 days, B does full work in $10 \times 4 = 40$ days.
A's rate = $1/15$. B's rate = $1/40$.
Combined rate = $1/15 + 1/40$.
LCM = 120.
$= (8 + 3) / 120 = 11/120$.
Total time = $120/11$ days = $10 \frac{10}{11}$ days.
Answer: $10 \frac{10}{11}$ days

12. One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?
Solution:
Filling rate = $1/3$.
Emptying rate = $1/5$.
Net rate = $1/3 - 1/5 = (5-3)/15 = 2/15$.
Time to fill = $15/2$ hours = $7.5$ hours.
Answer: 7.5 hours (or 7 hours 30 minutes)

Test yourself

1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) The value of a when

(ii) The value of b when

(iii) 15 note books can be bought for ₹ 240. The number of note books that can be bought for ₹ 160 is:
(a) 16
(b) 10
(c) 18
(d) 15
Solution:
Direct variation.
$15 / 240 = x / 160$
$1/16 = x / 160$
$x = 10$.
Answer: (b)

(iv) 6 men can do a certain piece of work in 15 days. The number of men required to complete the same work in 10 days:
(a) 30
(b) 25
(c) 60
(d) 9
Solution:
Inverse variation.
$6 \times 15 = x \times 10$
$90 = 10x \Rightarrow x = 9$.
Answer: (d)

(v) If x is in inverse variation with y and $x = 4$ when $y = 6,$ the value of x when $y = 12$ is:
(a) 2
(b) 18
(c) 8
(d) 12
Solution:
$x \times y = k \Rightarrow 4 \times 6 = 24$.
$x \times 12 = 24 \Rightarrow x = 2$.
Answer: (a)

(vi) Statement 1: Work done varies inversely to the number of persons at work.
Statement 2: Time taken to finish a work varies directly to the number of persons at work.
Which of the following options is correct?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.
Solution:
Statement 1 is False. More persons do *more* work (Direct), or for fixed work, time varies inversely. "Work done" implies amount of work. Usually, Work $\propto$ Men. So Direct. False.
Statement 2 is False. More persons take *less* time. Inverse.
Answer: (b)

The following questions are Assertion-Reason based questions. Choose your answer based on the codes given below.
(1) Both A and R are correct, and R is the correct explanation for A.
(2) Both A and R are correct, and R is not the correct explanation for A.
(3) A is true, but R is false.
(4) A is false, but R is true.

(vii) Assertion (A): In the following table, p and q are in direct variation.

(viii) Assertion (A): If 3 pipes can fill a tank in 6 hours, then 6 pipes will fill the same tank in half the time.
Reason (R): Time required to complete a certain work = $\frac{\text{Work to be completed}}{\text{One day's work}}$.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
Assertion A: 3 pipes -> 6 hours. Double the pipes (6) -> Half the time (3 hours). True.
Reason R: This is the standard formula for Time. True.
Does R explain A? Yes, doubling pipes doubles "one day's work" (rate), which halves the time (since Time = Work/Rate).
Answer: (a)

(ix) Assertion (A): In the following table, p and q are in inverse variation.

(x) Assertion (A): The cost of 16 bulbs is ₹ 144. The number of bulbs that can be bought for ₹ 270 is 30.
Reason (R): In inverse variation, the ratio of one kind of like terms is equal to the inverse ratio of the second kind of like terms.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
Check A: Cost of 1 = $144/16 = 9$. Bulbs for $270 = 270/9 = 30$. A is True.
Check R: This statement describes Inverse Variation. But cost and quantity are Direct Variation. The reasoning logic is sound for *inverse* variation, but the problem is about *direct* variation. R is a true statement about inverse variation theory, but irrelevant/wrong context for A? No, the options ask if R is true/false. R is a standard definition of inverse variation property ($x_1/x_2 = y_2/y_1$). So R is True.
Is R the explanation? No, this is a Direct Variation problem.
Answer: (b) (Both true, but R is not explanation).

2. If x varies directly as y and $x = 150$ when $y = 50.$ Find:
(i) x, when $y = 12.5$ (ii) y, when $x = 75$
Solution:
$x/y = k \Rightarrow 150/50 = 3$. So $x = 3y$.
(i) If $y = 12.5$, $x = 3 \times 12.5 = 37.5$.
(ii) If $x = 75$, $75 = 3y \Rightarrow y = 25$.
Answer: (i) 37.5 (ii) 25

3. If x varies inversely as y and $y = 300$ when $x = 60,$ Find:
(i) x, when $y = 90$ (ii) y, when $x = 300$
Solution:
$xy = k \Rightarrow 60 \times 300 = 18000$.
(i) $x \times 90 = 18000 \Rightarrow x = 200$.
(ii) $300 \times y = 18000 \Rightarrow y = 60$.
Answer: (i) 200 (ii) 60

4. Total length of 153 iron bars is 680 m. What will be the total length of 135 similar bars.
Solution:
Direct variation.
Length of 1 bar = $680 / 153$.
Length of 135 bars = $(680 / 153) \times 135$.
$153$ is divisible by 9 ($1+5+3=9$). $153/9 = 17$.
$135$ is divisible by 9 ($1+3+5=9$). $135/9 = 15$.
So, $(680 / 17) \times 15$.
$680 / 17 = 40$.
$40 \times 15 = 600$.
Answer: 600 m

5. 12 men can repair a road is 25 days; how long will 30 men will take to do so?
Solution:
Inverse variation.
$12 \times 25 = 30 \times x$
$300 = 30x \Rightarrow x = 10$.
Answer: 10 days

6. The price of oranges is ₹ 90 per dozen. Manoj can buy 12 dozen oranges with the money he has if the price of orange is increased by ₹ 30, how many oranges can Manoj buy?
Solution:
Initial price = ₹ 90/dozen.
Total Money = $90 \times 12 = \text{₹ } 1080$.
New price = $90 + 30 = \text{₹ } 120$/dozen.
New quantity = Total Money / New Price = $1080 / 120 = 9$ dozens.
The question asks "how many oranges", which could mean number of oranges or dozens. Usually dozens in this context, but $9 \times 12 = 108$ oranges.
Answer: 9 dozens (or 108 oranges)

7. A and B can do a work in 8 days, B and C in 12 days, and A and C in 16 days. In what time can they do it, all working together?
Solution:
(A+B)'s 1 day work = $1/8$.
(B+C)'s 1 day work = $1/12$.
(A+C)'s 1 day work = $1/16$.
Adding all: $2(A+B+C) = 1/8 + 1/12 + 1/16$.
LCM of 8, 12, 16 is 48.
$2(A+B+C) = (6 + 4 + 3) / 48 = 13/48$.
(A+B+C) = $13 / 96$.
Time taken = $96 / 13$ days = $7 \frac{5}{13}$ days.
Answer: $7 \frac{5}{13}$ days

8. A and B complete a piece of work in 24 days. B and C do the same work in 36 days, and A, B and C together finish it in 18 days. In how many days will:
(i) A alone,
(ii) C alone,
(iii) A and C together, complete the work?
Solution:
Given:
(1) $A+B = 1/24$
(2) $B+C = 1/36$
(3) $A+B+C = 1/18$

(i) A alone = (A+B+C) - (B+C) = $1/18 - 1/36 = (2-1)/36 = 1/36$.
A takes 36 days.
(ii) C alone = (A+B+C) - (A+B) = $1/18 - 1/24 = (4-3)/72 = 1/72$.
C takes 72 days.
(iii) A and C together = $1/36 + 1/72 = (2+1)/72 = 3/72 = 1/24$.
They take 24 days.
Answer: (i) 36 days (ii) 72 days (iii) 24 days

9. A and B can do a piece of work in 40 days, B and C in 30 days, and C and A in 24 days.
(i) How long will it take them to do the work, working together?
(ii) In what time can each finish it working alone?
Solution:
(A+B) = $1/40$. (B+C) = $1/30$. (C+A) = $1/24$.
2(A+B+C) = $1/40 + 1/30 + 1/24$.
LCM of 40, 30, 24 is 120.
2(A+B+C) = $(3 + 4 + 5) / 120 = 12/120 = 1/10$.
(A+B+C) = $1/20$.
(i) Together they take 20 days.
(ii)
A alone = $1/20 - 1/30 = (3-2)/60 = 1/60$ (60 days).
B alone = $1/20 - 1/24 = (6-5)/120 = 1/120$ (120 days).
C alone = $1/20 - 1/40 = (2-1)/40 = 1/40$ (40 days).
Answer: (i) 20 days (ii) A: 60 days, B: 120 days, C: 40 days

10. 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. How long did the work last?
Solution:
Let total time be $x$ days.
A worked for 2 days. Work = $2/10 = 1/5$.
B worked for $(x-3)$ days. Work = $(x-3)/12$.
C worked for $x$ days. Work = $x/15$.
Total work = 1.
$1/5 + (x-3)/12 + x/15 = 1$
Multiply by 60 (LCM):
$12 + 5(x-3) + 4x = 60$
$12 + 5x - 15 + 4x = 60$
$9x - 3 = 60$
$9x = 63$
$x = 7$.
Answer: 7 days

11. Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
Solution:
Let first pipe (P) be turned off after $t$ minutes.
Pipe Q runs for the full 16 minutes.
Work done by P + Work done by Q = 1.
$(t / 24) + (16 / 32) = 1$
$t / 24 + 1/2 = 1$
$t / 24 = 1/2$
$t = 12$.
Answer: 12 minutes