# Selina Concise Solutions for Chapter 7 Unitary Method (including Time and Work) Class 7 ICSE Mathematics

**Exercise 7 (A)**

**1. Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles ?**

**Answer**

Weight of 8 articles = 4.8 kg

Weight of 1 article = 4.8/8 kg

and weight of 11 articles = 4.8/8 × 11 kg

= 0.6 × 11

= 6.6 kg

**2. 6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?**

**Answer**

1 kg 260 g or 1.260 kg, no. of books = 6

and in 1 kg, no. of books = 6/1.260

**3. 8 men complete a work in 6 hours. In how many hours will 12 men complete the**

**same work ?**

**Answer**

8 men can complete a work in = 6 hours

1 man can complete the work in = 6×8 hours

12 men can complete the work in = (6×8)/12 = 4 hours

**4. If a 25 cm long candle burns for 45 minutes, how long will another candle of the**

**same material and same thickness but 5 cm longer than the previous one, burn ?**

**Answer**

25 cm long candle burn in = 45 minutes

1 cm long candle will burn in = 45/25 minutes

25 + 5 = 30 cm long candle will burn in

= (45×30)/25 minutes = 54 minutes

**5. A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages ?**

**Answer**

For typing 24 pages, time is required = 80 minutes

For typing 1 page, time is required = minutes

and for typing 87 pages, time is required

= (80×87)/24 minutes = 290 minutes

**6. Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support**

**the same family?**

**Answer**

Rs. 750, can support a family for = 15 days

Re. 1 will support for = 15/750 days

and Rs. 2,500 will support for = (15/750) × 2500 days = 50 days

**7. 400 men have provisions for 23 weeks. They are joined by 60 men. How long will**

**the provisions last ?**

**Answer**

400 men have provisions for = 23 weeks

1 man will have provisions for = 23 × 400 weeks

and 400 + 60 = 460 men will have provisions for = (23 × 400)/460 weeks = 20 weeks

**8. 200 men have provisions for 30 days. If 50 men left, the same provisions would**

**last for the remaining men, in how many days?**

**Answer**

200 men have provisions for = 30 days

1 man will have provisions for = 30 x 200 days

200 – 50 = 150 men will have provisions for = (30×200)/150 days = 40 days

**9. 8 men can finish a certain amount of provisions in 40 days. If 2 more men join**

**with them, find for how many days the same amount of provisions be sufficient ?**

**Answer**

8 men can finish a provision in = 40 days

1 man will finish in = 40 × 8 days

8+2 =10 men will finish in = (40×8)/10

= 32 days

**10. If interest on Rs. 200 be Rs. 25 in a certain time, what will be the interest on Rs**

**750 for the same time ?**

**Answer**

Interest on Rs. 200 is = Rs. 25

Interest on Rs. I will be = Rs. 25/200

and interest on Rs. 750 will be

= Rs. (25×750)/200 = Rs. 750/8 = Rs. 93.75

Interest on Rs. I will be = Rs. 25/200

and interest on Rs. 750 will be

= Rs. (25×750)/200 = Rs. 750/8 = Rs. 93.75

**11. If 3 dozen eggs cost Rs. 90, find the cost of 3 scores of eggs. (1 score = 20)**

**Answer**

3 dozen = 3 × 12 = 36 eggs,

3 scores = 3 × 20 = 60

The cost of 36 eggs is = Rs. 90

The cost of 1 egg will be = Rs. 90/36

∴ Cost of 60 eggs will be = Rs. (90×60)/36

= Rs. 150

**12. If the fare for 48 km is Rs. 288, what will be the fare for 36 km ?**

**Answer**

Fare for 48 km = Rs. 288

fare for 1 km = Rs. (288×36)/48 = Rs. 216

**13. What will be the cost of 3.20 kg of an item, if 3 kg of it costs Rs. 360 ?**

**Answer**

Cost of 3 kg of an item = Rs. 360

Cost of 1 kg of the item = Rs. 360/3

∴ Cost of 3.20 kg of the item = Rs. (360×3.20)/3

Cost of 1 kg of the item = Rs. 360/3

∴ Cost of 3.20 kg of the item = Rs. (360×3.20)/3

**14. If 9 lines of a print, in a column of a book contains 36 words. How many words**

**will a column of 51 lines contain ?**

**Answer**

In 9 lines of print, words are = 36

In 1 line of print, words will be = 36/9

∴ in 51 lines of print, words will be

= 36/9 ×51 = 204

In 1 line of print, words will be = 36/9

∴ in 51 lines of print, words will be

= 36/9 ×51 = 204

**15. 125 pupil have food sufficient for 18 days. If 25 more pupil join them, how long**

**will the food last now ? What assumption have you made to come to your answer ?**

**Answer**

Pupils in the beginning = 125

More pupils joined = 25

Total pupils = 125 + 25 = 150

Food is sufficient for 125 pupils for = 18 days

Food will be sufficient for 1 pupil for = 18×125 days (less pupil more days)

and food will be sufficient for 150 pupils = (18×125)/150 days (more pupil less days)

= (18×5)/6 = 15 days

**16. A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how**

**many hours per day must he work in order to make the same chair in 4 days ?**

**Answer**

A chair is completed in 3 days working per day = 8 hours

Then their will be completed in 1 day working for = 8×3 hours per day (less days more hours)

and it will be completed in 4 days working for = (8×3)/4 = 6 hours per day.

**17. A man earns ₹5,800 in 10 days. How much will he earn in the month of February**

**of a leap year?**

**Answer**

Note: Leap years has 29 days in the month of February.

Man earns in 10 days = ₹ 5800/10

∴ Man earns in February month of leap year

= (5800×29)/10 = ₹16820

Man earns in 10 days = ₹ 5800/10

∴ Man earns in February month of leap year

= (5800×29)/10 = ₹16820

**18. A machine is used for making rubber balls and makes 500 balls in 30 minutes.**

**How many balls will it make in 7/2 hours?**

**Answer**

30 minutes = 30/60 = 1/2 hours

In 1/2 hours ball makes by machine = 500

In 1 hours balls makes by machine

= 500 × 2/1 = 1000 balls

∴ In 7/2 hours, Balls make by machine

= 500 × 2×7/2 = 3500 balls

In 1/2 hours ball makes by machine = 500

In 1 hours balls makes by machine

= 500 × 2/1 = 1000 balls

∴ In 7/2 hours, Balls make by machine

= 500 × 2×7/2 = 3500 balls

**19. In a school’s hostel mess, 20 children consume a certain quantity of ration in 6**

**days. However, 5 children did not return to the hostel after holidays. How long**

**will the same amount of ration last now?**

**Answer**

Total number of children = 20

20 children consume a certain quantity of ration in = 6 days

1 children consume a certain quantity of ration in = 6 ×20 days

As 5 children did not return to the hostel after holidays.

Then number of children in hostel = 20-5 = 15

Hence, 15 children consume certain quantity 6×20

of ration in = (6×20)/15 days = 8 days

**Exercise 7 (B)**

**1. The cost of 3/5 kg of ghee is ₹ 96; find the cost of :**

(i) one kg ghee

(ii) 5/8 kg ghee.

(i) one kg ghee

(ii) 5/8 kg ghee.

**Answer**

Cost of 3/5 kg of ghee = ₹ 96

(i) ∴ Cost of 1 kg of ghee = ₹ 96 × 5/3 = ₹ 160

(ii) and cost of 5/8 of ghee = ₹ 160 × 5/8 = ₹ 100

(i) ∴ Cost of 1 kg of ghee = ₹ 96 × 5/3 = ₹ 160

(ii) and cost of 5/8 of ghee = ₹ 160 × 5/8 = ₹ 100

**Answer**

Cost of i.e. 7/2 m of cloth = Rs. 168

∴ Cost of 1 m of cloth = Rs. 168 × 2/7

= Rs. 48

and cost of = 13/3 m cloth = Rs. 48 × 13/3

= Rs. 208

∴ Cost of 1 m of cloth = Rs. 168 × 2/7

= Rs. 48

and cost of = 13/3 m cloth = Rs. 48 × 13/3

= Rs. 208

**3.A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?****Answer**

10 sec, are lost in = 8 hours

∴ 1 sec will be lost in = 8/10 hours

and 15 sec. will be lost in = 8/10 × 15 hours

= 12 hours

∴ 1 sec will be lost in = 8/10 hours

and 15 sec. will be lost in = 8/10 × 15 hours

= 12 hours

**4. In 2 days and 20 hours, a watch gains 20 sec ; find how much time will the watch**

**take to gain 35 sec. ?**

**Answer**

2 days 20 hours = 2×24 + 20 = 48 + 20 = 68 hours

Now 20 sec. are gained in = 68 hours

∴ 1 sec. will be gained in 68/20 hours

35 sec. will be gained in = 68/20 × 35

= 119 hours = 119 ÷ 24 days

= 4 days 23 hours

Now 20 sec. are gained in = 68 hours

∴ 1 sec. will be gained in 68/20 hours

35 sec. will be gained in = 68/20 × 35

= 119 hours = 119 ÷ 24 days

= 4 days 23 hours

**5. 50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?**

**Answer**

Land is same in both the cases.

Now 50 men can mow land in = 3 days

∴ 1 man will mow it in = 3 x 50 days

and 15 men will mow it in = (3×50)/15 = 10 days

**6. The wages of 10 workers for a six days week are Rs, 1,200. What are the one day**

**wages: (i) of one worker ? (ii) of 4 workers ?**

**Answer**

(i) 10 workers can earn in 6 days = Rs. 1,200

1 worker will earn in 6 days = Rs. 1200/10

1 worker will earn in 1 day = Rs. 1200/(10×6) = Rs. 20

(ii) ∴ 4 workers will earn in 1 day = Rs. 20 × 4 = Rs. 80

1 worker will earn in 6 days = Rs. 1200/10

1 worker will earn in 1 day = Rs. 1200/(10×6) = Rs. 20

(ii) ∴ 4 workers will earn in 1 day = Rs. 20 × 4 = Rs. 80

**7. If 32 apples weigh 2 kg 800 g. How many apples will there be in a box, containing**

**35 kg of apples ?**

**Answer**

Apples in a box = 35 kg

Now, If weight is 2kg 800 g = 2.800 kg.

then number of apples = 32

if weight is 1 kg, then number of apples = 32/2.800

and if weight is 35 kg, then number of apples = (32×35)/2.800

= (32 ×35×1000)/2800 = 400

Now, If weight is 2kg 800 g = 2.800 kg.

then number of apples = 32

if weight is 1 kg, then number of apples = 32/2.800

and if weight is 35 kg, then number of apples = (32×35)/2.800

= (32 ×35×1000)/2800 = 400

**8. A truck uses 20 litres of diesel for 240 km. How many litres will be needed for**

**1200 km?**

**Answer**

For 240 km, diesel is needed = 20 litres

∴ for 1 km, diesel will be needed 20

**9. A garrison of 1200 men has provisions for 15 days. How long will the provisions**

**last if the garrison be increased by 600 men ?**

**Answer**

1200 men has provision for = 15 days

1 man will have that provision for = 15×1200 days

∴1200 + 600 = 1800 men will has that provisions for = (15×1200)/1800 days

= 10 days

**10. A camp has provisions for 60 pupil for 18 days. In how many days, the same**

**provisions will finish off if the strength of the camp is increased to 72 pupil ?**

**Answer**

60 pupil have provision for = 18 days 1 pupil will have provision for = 18 x 60 days (less

pupils more days)

and 72 pupils will have provision for = (18×60)/72 days

= 15 days.

**Exercise 7 (C)**

**1. A can do a piece of work in 6 days and B can do it in 8 days. How long will they**

**take to complete it together ?**

**Answer**

A can do a work in = 6 days

∴ A's one day's work = 1/6

B can do the same work in = 8 days

∴ B is one day's work = 1/8

∴ A and B's both one day's work

∴ A's one day's work = 1/6

B can do the same work in = 8 days

∴ B is one day's work = 1/8

∴ A and B's both one day's work

**2. A and B working together can do a piece of work in 10 days B alone can do the**

**same work in 15 days. How long will A alone take to do the same work ?**

**Answer**

A and B together can do a work in 10 days

and B can do the same work in 15 days

A and B's one day's work = 1/10

and B's one day's work = 1/15

∴ A's one day's work = 1/10 - 1/15

= (3-2)/30 = 1/30

Hence A can do the same work in 30 days

and B can do the same work in 15 days

A and B's one day's work = 1/10

and B's one day's work = 1/15

∴ A's one day's work = 1/10 - 1/15

= (3-2)/30 = 1/30

Hence A can do the same work in 30 days

**3. A can do a piece of work in 4 days and B can do the same work in 5 days. Find,**

**how much work can be done by them working together in : (i) one day (ii) 2 days.**

**What part of work will be left, after they have worked together for 2 days ?**

**Answer**

A can do a piece of work in 4 days

and B can do the same work in 5 days.

∴ A's one day's work = 1/4

and B's one day's work = 1/5

(i) A and B's both one day's work = 1/4 + 1/5

= (5+4)/20 = 9/20

(ii) A and B's 2 day's work = (9/20) × 2 = 9/10

∴ work left after 2 day's = 1 - (9/10

= (10-9)/10 = 1/10

and B can do the same work in 5 days.

∴ A's one day's work = 1/4

and B's one day's work = 1/5

(i) A and B's both one day's work = 1/4 + 1/5

= (5+4)/20 = 9/20

(ii) A and B's 2 day's work = (9/20) × 2 = 9/10

∴ work left after 2 day's = 1 - (9/10

= (10-9)/10 = 1/10

**4. A and B take 6 hours and 9 hours respectively to complete a work. A works for 1**

**hour and then B works for two hours.**

**(i) How much work is done in these 3 hours ?**

**(ii) How much work is still left ?**

**Answer**

A take 6 hours to finish work

and B take 9 hours to finish the same work

∴ A's 1 hour's work = 1/6

and B's 1 hour's work = 1/9

and B's 2 hours work = 1/9 × 2 = 2/9

(i) Now A's 1 hours work + B's 2 hours work

= 1/6 +2/9 = (3+4)/18 = 7/18

(ii) Work left = 1 - 7/18 = (18-7)/18 = 11/18

and B take 9 hours to finish the same work

∴ A's 1 hour's work = 1/6

and B's 1 hour's work = 1/9

and B's 2 hours work = 1/9 × 2 = 2/9

(i) Now A's 1 hours work + B's 2 hours work

= 1/6 +2/9 = (3+4)/18 = 7/18

(ii) Work left = 1 - 7/18 = (18-7)/18 = 11/18

**5. A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long**

**will they take to do it working together ?**

**Answer**

A can do a piece or work in 12 days

B can do the same work in 15 days

C can do the same work in 20 days

∴ A's 1 day's work = 1/12

B's 1 day's work = 1/15

C's 1 day's work = 1/20

∴ (A + B + C)'s together 1 day's work

= 1/12 + 1/15 + 1/20

= (5 + 4+3)/60 = 12/60 = 1/5

∴ Thy can do the work in 5 days.

B can do the same work in 15 days

C can do the same work in 20 days

∴ A's 1 day's work = 1/12

B's 1 day's work = 1/15

C's 1 day's work = 1/20

∴ (A + B + C)'s together 1 day's work

= 1/12 + 1/15 + 1/20

= (5 + 4+3)/60 = 12/60 = 1/5

∴ Thy can do the work in 5 days.

**6. Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can**

**empty it in 15 hours. How long will it take to fill the empty cistern, if all of them**

**are opened together ?**

**Answer**

First tap can fill a cistern in 10 hours

Second tap can fill the cistern in 8 hours

Third tap can empty the cistern in 15 hours

∴ First tap's 1 hour's work = 1/10

Second tap's 1 hour's work = 1/8

and third tap's 1 hour's work = 1/15

If all of them are opened together, then their one hour's work = 1/10 + 1/8 - 1/15

= (12 + 15 -8)/120 = (27 - 8)/120 = 19/120

∴ They can fill the cistern in = 120/19 hours

Second tap can fill the cistern in 8 hours

Third tap can empty the cistern in 15 hours

∴ First tap's 1 hour's work = 1/10

Second tap's 1 hour's work = 1/8

and third tap's 1 hour's work = 1/15

If all of them are opened together, then their one hour's work = 1/10 + 1/8 - 1/15

= (12 + 15 -8)/120 = (27 - 8)/120 = 19/120

∴ They can fill the cistern in = 120/19 hours

**7 . Mohit can complete a work in 50 days, whereas Anuj can complete the same work**

**in 40 days.**

**Find:**

**(i) work done by Mohit in 20 days.**

**(ii) work left after Mohit has worked on it for 20 days.**

**(iii) time taken by Anuj to complete the remaining work.**

**Answer**

Mohit can complete a work in 50 days and Anuj can complete the same work in 40 days

∴ Mohit's one day's work = 1/50

and Anuj's one day's work = 1/40

(i) Mohit's 20 day's work

= 1/50 × 20 = 2/5

(ii) Work left = 1 - 2/5

= (5-2)/5 = 3/5

Anuj can do 3/5 work in = 40× 3/5 = 24 days

∴ Mohit's one day's work = 1/50

and Anuj's one day's work = 1/40

(i) Mohit's 20 day's work

= 1/50 × 20 = 2/5

(ii) Work left = 1 - 2/5

= (5-2)/5 = 3/5

Anuj can do 3/5 work in = 40× 3/5 = 24 days

**8. Joseph and Peter can complete a work in 20 hours and 25 hours respectively.**

**Find :**

**(i) work done by both together in 4 hrs.**

**(ii) work left after both worked together for 4 hrs.**

**(iii) time taken by Peter to complete the remaining work.**

**Answer**

Joseph can do a work in = 20 hours

Peter can do the same work in = 25 hours

Now joseph's 1 hour's work = 1/20

and Peter's 1 hour's work = 1/25

Both's 1 hour's work = 1/20 + 1/25

= (5+4)/100 = 9/100

(i) Both's 4 hours work = 9/100 × 4 = 9/25

(ii) Work left over = 1 - 9/25

= (25 -9)/25 = 16/25

(iii) Peter can do 16/25 work in = 25 × 16/25 = 16 hours

Peter can do the same work in = 25 hours

Now joseph's 1 hour's work = 1/20

and Peter's 1 hour's work = 1/25

Both's 1 hour's work = 1/20 + 1/25

= (5+4)/100 = 9/100

(i) Both's 4 hours work = 9/100 × 4 = 9/25

(ii) Work left over = 1 - 9/25

= (25 -9)/25 = 16/25

(iii) Peter can do 16/25 work in = 25 × 16/25 = 16 hours

**9. A is able to complete of a certain work in 10 hrs and B is able to complete of the**

**same work in 12 hrs.**

**Find:**

**(i) how much work can A do in 1 hour ?**

**(ii) how much work can B do in 1 hour ?**

**(iii) in how much time will the work be completed, if both work together.**

**Answer**

A can do 1/3 of a work in = 10 hours

∴ A can do full work in (10 × 3)/1 = 30 hours

B can do 2/5 of the work in = 12 hours

∴ B can do the whole work in = (12× 5)/2 = 30 hours

(i) Now A's 1 hour's work = 1/30

(ii) B's 1 hour's work = 1/30

(iii) Both's 1 hour's work = 1/30 +1/30 = 2/30 = 1/15

∴ Both can finish the work in 15 hours

∴ A can do full work in (10 × 3)/1 = 30 hours

B can do 2/5 of the work in = 12 hours

∴ B can do the whole work in = (12× 5)/2 = 30 hours

(i) Now A's 1 hour's work = 1/30

(ii) B's 1 hour's work = 1/30

(iii) Both's 1 hour's work = 1/30 +1/30 = 2/30 = 1/15

∴ Both can finish the work in 15 hours

**10. Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same**

**chair in 4 days. If they work together, in how many days will they prepare :**

**(i) one chair ?**

**(ii)14 chairs of the same kind?**

**Answer**

Shaheed's 1 day's work = 1/3

and Shaif's 1 day's work = 1/4

Both one day's work = 1/3 + 1/4 = (4+3)/12 = 7/12

∴ Both can prepare the chair in = 12/7 days

One chair is prepared in = 12/7 days

∴ 14 chairs will be prepared in = 12/7 × 14 = 24 days

and Shaif's 1 day's work = 1/4

Both one day's work = 1/3 + 1/4 = (4+3)/12 = 7/12

∴ Both can prepare the chair in = 12/7 days

One chair is prepared in = 12/7 days

∴ 14 chairs will be prepared in = 12/7 × 14 = 24 days

**11. A, B and C together finish a work in 4 days. If A alone can finish the same work in**

**8 days and B in 12 days, find how long will C take to finish the work.**

**Answer**

A, B and C finish work together in = 4 days.

A, B and C finish work together in 1 day = 1/4

A' one day work = 1/8

B's one day work = 1/12

∴ C's one day work = 1/4 - (1/8 + 1/12)

= [6 - (3+2)]/24 = 1/24

∴ C can finish the work in = 24 days

A, B and C finish work together in 1 day = 1/4

A' one day work = 1/8

B's one day work = 1/12

∴ C's one day work = 1/4 - (1/8 + 1/12)

= [6 - (3+2)]/24 = 1/24

∴ C can finish the work in = 24 days