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# Exercise 4.4 (Compound Variation, Time and Work)

8th Maths : Chapter 4 : Life Mathematics : Compound Variation, Time and Work : Exercise 4.4 :

Exercise 4.4

1. Fill in the blanks:

(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job working together is __________days. [Answer: 2 days]

(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in ________ days. [Answer: 5]

(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in ________ days. [Answer: 8]

(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in ___________days. [Answer: 25]

(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for 200000. The amount that A will get is ________.  [Answer: ₹ 1,20,000]

2. 210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?

Solution:

Let the required number of men be x. More working hours less men required.

It is inverse proportion

Multiflying factor is 12/14

Also more number of days  ⇒  less men

It is an inverse proportion

Multiplying factor is 18/20

x = 210 × [12/14] × [18/20] = 162 men

162 men are required.

3. A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?

Solution:

Let the required number of cement bags be x. Number of days more  More cement bags.

It is direct variation.

The Multiflying factor = 18/12

Number of machines more  More cement bags.

It is direct variation.

The Multiflying factor = 24/36

x = 7000 × [18/12] × [24/36] x = 7000 cement bags

7000 cement bags can be made.

4. A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 more hours a day?

Solution:

Let the required number of days be x. To produce more soaps more days required.

It is direct proportion.

Multiflying factor = 14400/9600

If more hours spend, less days required.

It is indirect proportion.

Multiflying factor = 15/18

x = 6 × (14400/9600) × (15/18) x = 15/2 or 7 1/2 days

15/2 days will be needed.

5. If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?

Solution:

Let the number of lorries required more = x. As the goods are more More lorries are needed to transport.

It is direct proportion.

Multiflying factor = 180/135

Again if more days less number of lorries enough.

It is direct proportion.

Multiplying factor = 5/4

6 + x = 6 × [180 / 135] × [5 / 4] 6 + x = 10

x = 10 − 6

x = 4

4 more lorries are required.

6. A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?

Solution: Solution 2:

Time taken by A to complete the work = 12 hrs.

A's 1 hr work = 1/12 ……… (1)

(B + C) complete the work in 3 hrs.

(B + C)'s 1 hour work = 1/3  ……… (2)

(l) + (2) =>

(A+ B + C)'s 1 hour work = (1/12) + (1/3) = [1 + 4] / 12 = 5/12

Now (A + C) complete the work in 6 hrs.

(A+ C)'s 1 hour work = 1/6

B's 1 hour work = (A+ B + C)'s 1 hour work − (A + C)'s 1 hr work

= (5/12) – (1/6) = [5 – 2] / 12 = 3/12 = 1/4

B alone take 4 days to complete the work.

7. A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?

Solution:  Solution 2:

(A + B) complete the work in 12 days.

(A + B)'s 1 day work = 1/12 ……(1)

(B + C) complete the work in 15 days

(B + C)'s 1 day work = 1/15 ……….(2)

(A + C) complete the work in 20 days

(A + C)'s 1 day work = 1/20 ………(3)

Now (l) + (2) + (3) =>

[(A + B) + (B + C) + (A + C)]'s 1 day work = 1/12 + 1/15 + 1/20

(2A + 2B + 2C)’s 1 day work = 5/60 + 4/60 + 3/60

2(A + B + C)’s 1 day work = [5 + 4 + 3] / 60

(A + B + C)’s 1 day work = 12 / [60 × 2]

(A + B + C)’s 1 day work = 1/10

Now A’s 1 day work = (A + B + C)’s 1 day work − (B + C)'s 1 day work

= 1/10 – 1/15 = 3/30 – 2/30 = 1/30.

A takes 30 days to complete the work.

B's 1 day work = (A + B + C)'s 1 day's work − (A + C)'s 1 day's work

= 1/10 – 1/20 = 6/60 – 3/60

= [6 – 3] / 60 = 3/60 =  1/20

B takes 20 days to complete the work.

C's 1 day work = (A + B + C)'s 1 day work − (A + B)'s 1 day work

= 1/10 – 1/12 = 6/60 – 5/60 = [6 – 5] / 60 = 1/60

C takes 60 days to complete the work.

8. Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 minutes more than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?

Solution: Solution 2:

Time taken by A to fit a chair = 15 minutes

Time taken by B = 3 minutes more than A

= 15 + 3 = 18 minutes

A's 1 minute work = 1/15

B's 1 minute work = 1/18 (A + B)'s 1 minutes work = 1/15 + 1/18

12/180 + 10/180 = 22/180 = 11/90

Time taken by (A + B) to fit a chair

= 1 / (11/90) = 90/11 minutes

Time taken by (A + B) to fit 22 chairs

= (90/11) × 22 = 180 minutes

9. A can do a work in 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work, if they work together.

Solution: Solution 2:

A completes the work in 45 days.

A's 1 day work = 1 / 45

A's 15 days work = 15/45 = 1/3

Remaining work = 1 – (1/3) = [3 – 1] / 3 = 2/3

B finishes 2/3 rd work in 24 days

B's 1 day work = (2/3) / 24

= 2 / [3 × 24] = 1/36

(A+B)'s 1 day work = [1 / 45] + [1 / 36]

= [4 + 5] / 180

= 9 / 180 = 1/20

Let x days required

x / 20 = 80 / 100  x = [80 / 100] × 20 = 16 days.

10. A is thrice as fast as B. If B can do a piece of work in 24 days, then find the number of days they will take to complete the work together.

Solution:

If B does the work in 3 days, A will do it in 1 day.

B complete the work in 24 days. They together complete the work in 6 days.

Solution 2:

If B does the work in 3 days, A will do it in 1 day.

B complete the work in 24 days.

A complete the same work in 24/3 = 8 days.

(A + B) complete the work in ab / [ a + b] days

= ( [24 × 8] / [24 + 8] ) days = ( [24 × 8] / 32 ) days = 6 days.

They together complete the work in 6 days.

Exercise 4.4

1. (i) 2 (ii) 5 (iii) 8 (iv) 25 (v) ₹1,20,000

2. 162 men

3. 7000 cement bags

4. 7 1/2 days

5. 4 more lorries

6. 4 hours

7. A- 30 days, B -20 days, C-60 days

8. 180 minutes

9. 16 days

10. 6 days

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8th Maths : Chapter 4 : Life Mathematics : Exercise 4.4 (Compound Variation, Time and Work) | Questions with Answers, Solution | Life Mathematics | Chapter 4 | 8th Maths