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
2.
of cloth costs Rs. 168 ; find the cost of
of the same cloth.
of cloth costs Rs. 168 ; find the cost of
of the same cloth. 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
3.A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?
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
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.
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