Samacheer Kalvi 8th Maths Book Solutions Term 3 Chapter 2 | Life Mathematics Answers For Tamil Nadu Board

Samacheer Kalvi 8th Maths Book Solutions Term 3 Chapter 2 – Life Mathematics is available here. The Samacheer Kalvi 8th Maths book answers of Term 3 Chapter 2, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 8th Maths solutions. Solving the Samacheer Kalvi 8th Class Maths book Term 3 Chapter 2 questions for practise and referring to the solutions later to gauge self performance is the best way to get acquainted with answering a lot of questions of various difficulty level, thus preparing for the final exams.

Term 3 Chapter 2 of the Samacheer Kalvi 8th Maths guide will help the students to solve problems related to direct and inverse proportion, compound variation, time and work, sharing of the money for work.

Samacheer Kalvi 8th Maths Term 3 Chapter 2: Life Mathematics Book Exercise 2.1 Questions and Solutions

Question 1: 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 – Fewer men required

Hence, it is an inverse proportion.

More number of days – Fewer men required

Hence, it is an inverse proportion.

x = (210) * (12 / 14) * (18 / 20)

= 162 men

So, 162 men are required.

Question 2: 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 hours more a day?

Solution:

Let the required number of days be x.

To produce more soaps, more days are required.

Hence, it is a direct proportion.

If more hours are spent, then fewer days are required.

x = 6 * (14400 / 9600) * (15 / 18)

x = 15 / 2

So, 15 / 2 days will be needed.

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

Solution:

Let the total work done by LCM of (12,3,6) = 12 units.

Given A can do the work in 12 hours.

So, work per hour done by A = 12 / 12 = 1 unit.

Given, A and C together can do the work in 6 hours.

So, work per hour done by both A and C = 12 / 6 = 2 units.

So, C alone can complete (2 – 1) units = 1 unit of work per hour.

Given, B and C together can do the work in 3 hours.

So, work per hour done by both B and C = 12 / 3 = 4 units.

So, B alone can complete (4 – 1) units = 3 units of work per hour.

3 units of work B does in 1 hour.

From the unitary method, 1 unit of work is done by B in (1 / 3) hours.

Similarly, 12 units (required) work is done by B in (12 / 3) = 4 hours.

So, B takes 4 hours.

Question 4: 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:

Carpenter A takes 15 minutes to fit a chair

So, in 1 minute, the fraction of work done by A = 1 / 15

Carpenter B takes 3 minutes more than A for the same work to be done i.e., B will take 15 + 3 = 18 minutes

So, in 1 minute, the fraction of work done by B = 1 / 18

The fraction of work done by both A & B together in 1 minute will be,

= (1 / 15) + (1 / 18)

= (6 + 5) / 90

= 11 / 90

So, both A and B when working together will fit 1 chair in 90 / 11 minutes

∴ To fit the parts of 22 chairs, A and B will take = (90 / 11) * 22 = 180 minutes

Thus, it will take 180 minutes [3 hours] for them to fit the parts for 22 chairs.

Question 5: 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:

A can finish the total job in a thrice faster speed than B.

B can finish the total job in 24 days.

So, the number of days A will take to complete the total job is = 24 / 3 = 8 days

Let, the total amount of work = 1

In 8 days A does = 1 work

In 1 day A does = 1 / 8 work

In 24 days B does = 1 work

In 1 day B does = 1 / 24 work

So, in 1 day A + B will complete

= (1 / 8) + (1 / 24)

= (3 + 1) / (24)

= 4 / 24

= 1 / 6 work

Now, together

They complete 1 / 6 work in = 1 day

They complete 1 work in = 1 × 6 = 6 days

Together, they will finish the total job in 6 days.

Samacheer Kalvi 8th Maths Term 3 Chapter 2: Life Mathematics Book Exercise 2.2 Questions and Solutions

Question 1: 5 boys or 3 girls can do a science project in 40 days. How long will it take for 15 boys and 6 girls to do the same project?

Solution:

One day’s work of five boys = 1 / 40

One day’s work of single boy = 1 / (40 × 5) = 1 / 200

Now the total number of boys = 15 + 10 = 25

One day’s work of 25 boys = 25 × (1 / 200) = 1 / 8

The project can be completed in 8 days.

Question 2: Amutha can weave a saree in 18 days. Anjali is twice as good a weaver as Amutha. If both of them weave together, in how many days can they complete weaving the saree?

Solution:

Amutha takes time to weave a saree = 18 days

∵ Anjali is twice as good a weaver as Amutha, so she takes half time.

∴ Anjali takes time to weave a saree = 18 / 2 = 9 days

In one day Amutha can weave saree = 1 / 18

In one day Anjali can weave saree = 1 / 9

Together they can weave saree per day = (1 / 18) + (1 / 9)

= (1 + 2) / 18

= 3 / 18

= 1 / 6

They can together complete weaving the saree in 6 days.

Question 3: P and Q can do a piece of work in 12 days and 15 days, respectively. P started the work alone and then, after 3 days Q joined him till the work was completed. How long did the work last?

Solution:

P takes 12 days to complete work and Q takes 15 days to complete the same work.

So, in one day P can complete (1 / 12) of the work and Q can complete (1 / 15) of the work.

Now, P works for 3 days to complete (1 / 12 ) * 3 = (1 / 4) of the work.

So, the remaining (1 – [1 / 4]) = 3 / 4 of the work they will work together.

Now, P and Q together can do it in one day (1 / 12 ) + (1 / 15) = (9 / 60) = (3 / 20) of the work.

Therefore, P and Q together will complete 3 / 4 of the work in (20 / 3) * (3 / 4) = 5 days.

Hence, the work lasts for (5 + 3) = 8 days.

Question 4: A small–scale company undertakes an agreement to make 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?

Solution:

In the first phase of the work;

Produced pumps = 180

Days = 75

Worker = 40

Remaining pumps = 540 – 180 = 360

Remaining days = 150 – 75 = 75

In 75 days 180 pumps are made by = 40 workers

In 75 days 1 pump is made by = 40 / 180 workers

In 75 days 360 pumps are made by = (40 × 360) / 180 = 80 workers

So, more workers needed = 80 – 40 = 40

Question 5: X alone can do a piece of work in 6 days and Y alone in 8 days. X and Y undertook the work for 4800. With the help of Z, they completed the work in 3 days. How much is Z’s share?

Solution:

X alone can do a piece of work in 6 days

So, work done by X in 1 day = 1 / 6

Y alone can do a piece of work in 8 days

So, work done by Y in 1 day = 1 / 8

Let Z alone can do the piece of work in “a” days.

So, work done by Z in 1 day = 1 / a

It is also given that, with the help of Z, X and Y can do the piece of work in 3 days

So, work done by X + Y + Z in 1 day = 1 / 3

1 / 6 + 1 / 8 + 1 / a = 1 / 3

⇒ 1 / a = 1 / 3 – 1 / 8 – 1 / 6

⇒ 1 / a = [8 – 3 – 4] / 24

⇒ 1 / a = 1 / 24

⇒ a = 24 days

X and Y undertook the work for 4800

The wages are shared on the basis of work done and the work is done is reciprocal of the time taken to do that work,

The ratio of work done by X, Y & Z = 1 / 6 : 1 / 8 : 1 / 24 = 4:3:1

The share of Z from the total money will be,

= {1 / [4 + 3 +1]} × 4800

= {1 / 8} × 4800

= 600