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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Intext Questions

Exercise 1.1

Try This (Text book Page no. 1)

Question 1.

Find the indicated percentage value of the given numbers

Solution:

Think (Text book page no. 6)

Question 1.

An increase from 200 to 600 ¡s clearly a 200%increase. Isn’t it? (check!). With a lot of pride, the traffic police commiuêener of a city reported that the accidents had decreased by 200% ¡n one year. He cameupwith this number stating that the accidents had gone down from 600 last year to 2this year. Is the decrease from 600 to 200, the same 200% as above? Justify.

Solution:

Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600

% decrease = \(\frac{600-200}{600}\) x 100 = \(\frac{400}{600}\) x 100 = 66.67% decrease

Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try This (Text book page no. 7)

Question 1.

What percent of a day is 10 hours?

Solution:

In a day, there are 24 hours

10 hrs out of 24 hrs is \(\frac{10}{24}\)

As a percentage, we need to multiply by 100

∴ Percentage = \(\frac{10}{24}\) x 100 = 41.67%

Question 2.

Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.

Solution:

Let R get x, Q gets 50% of what R gets

∴ Q gets = \(\frac{50}{100}\) × x = \(\frac{x}{2}\)

P gets 50% of what Q gets

∴ P gets = \(\frac{50}{100}\) x \(\frac{x}{2}\) = \(\frac{x}{4}\)

Since 350 is divided among the three

∴ 350 = x + \(\frac{x}{2}\) + \(\frac{x}{4}\)

350 = \(\frac{4x+2x+x}{4}\) = \(\frac{7x}{4}\) = 350

x = \(\frac{350×4}{7}\) = 200

Q gets = \(\frac{x}{2}\) = \(\frac{200}{2}\) = 100,

P gets = \(\frac{x}{4}\) = \(\frac{200}{4}\) = 50

∴ P = 50

Q = 100

R = 200

Exercise 1.2

Think (Text book Page No. 13)

Question 1.

A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?

Solution:

Let cost price of marker board be 100

CP = 100 Marks it 15% above CP

∴ Marked price = MP = \(\frac{15}{100}\) x CP + CP = \(\frac{15}{100}\) 100 + 100 = 15 + 100 = 115

Discount % = 15%

∴ He sells it 97.75 which is less than his cost price. Therefore he loses

Loss = 97.75 – 100 = – 2.25

Try This (Text book Page No. 14)

Question 1.

By selling 5 articles, a man gains the cost price of 1 article. Find his gain percentage.

Solution:

Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.

Gain on 1 article = SP – CP; ⇒ Gain% = \(\frac{SP-CP}{CP}\) x 100

Gain on 2 articles = 2 x (SP – CP)

Gain on 5 articles = 5 x (SP – CP)

Given than gain on 5 articles is CP of 1 article

∴ 5(SP – CP) = CP

\(\frac{SP-CP}{CP}\) = \(\frac{1}{5}\)

Gain percentage \(\frac{SP-CP}{CP}\) x 100 = \(\frac{1}{5}\) x 100 % = 20%

Question 2.

By selling 8 articles, a shopkeeper gains the selling price of 3 articles. Find his gain percentage.

Solution:

Let cost price of 1 article be CP. Let selling price of 1 article be SP.

Gain on 1 article = SP – CP

Gain on 8 articles = 8 x (SP – CP)

Given that gain on 8 articles in selling price of 3 articles

8(SP – CP) = 3 x SP

∴ \(\frac{SP-CP}{CP}\) = \(\frac{3}{8}\)

∴ \(\frac{SP}{SP-CP}\) = \(\frac{8}{3}\)

Subtracting 1 on both sides

\(\frac{SP}{SP-CP}\) – 1 = \(\frac{8}{3}\) – 1 = \(\frac{SP-(SP-CP)}{SP-CP}\)= \(\frac{8-3}{3}\)

\(\frac{SP-SP-CP}{SP-CP}\) = \(\frac{5}{3}\) ⇒ \(\frac{CP}{SP-CP}\) = \(\frac{5}{3}\)

\(\frac{SP-CP}{CP}\) = \(\frac{3}{5}\) (taking reciprocals on both sides)

Gain% \(\frac{SP-CP}{CP}\) x 100 = \(\frac{3}{5}\) x 100 = 3 x 20 = 60%

Question 3.

If the C.P of 20 articles is equal to the S.P of 15 articles, find the profit or loss percentage.

Solution:

Given CP of 20 article = SP of 15 articles

∴ SP of 15 articles = CP of 20 articles

∴ SP of 1 article = \(\frac{1}{15}\) x CP of 20 articles

SP = \(\frac{1}{15}\) x 20 CP = \(\frac{20}{15}\) CP = \(\frac{4}{3}\) CP

∴ SP = \(\frac{4}{3}\) CP ⇒ SP is greater than CP

It is a profit.

Exercise 1.3

Try This (Text book Page No. 23)

Question 1.

Find the principal which gives ₹ 420 as C.I @ 20% p.a compounded half yearly for one year.

Solution:

CI = ₹420

Rate = ₹ 20% p.a

Principle = ₹ [required to find] Time period (n) = 1 year.

However, let us value of r to be 20% p.a so for half yearly, r is \(\frac{20}{2}\) = 10%

Formula for Amount (A) when compounded half yearly is

Question 2.

The price of a laptop depreciates @ 4% p.a. If its present price is ₹ 24,000, find its price after 3 years.

Solution:

Let original price of laptop be ‘P’, Rate of depreciation is 4% p.a,

Present price is ₹ 24,000 (D).

Formula for depreciation is

Price after 3 years from now is

P(1 – \(\frac{4}{100}\))^{n+3} = ? ⇒ (1 – \(\frac{4}{100}\))^{n} (1 – \(\frac{4}{100}\))^{3}

From (1)

24,000 x (1 – \(\frac{4}{100}\))^{3}

24,000 x \(\frac{96}{100}\) x \(\frac{96}{100}\) x \(\frac{96}{100}\) = 21233.66

Actvity I (text book Page No. 23)

Question 1.

Mukunthan invests 30,000/- for 3 months in a bank which gives C.I at the rate of 12% compounded monthly. A private company offers his S.l at the rate of 12% p.a What is the difference in the interests received by Mukunthan? Do by traditional method and verify your answer by calculator.

Solution:

Principal = 30,000

Time period = 3 months

In Bank rate of interest for CI = 12% compounded monthly

∴ A = (1 + \(\frac{r}{100}\))^{n} = 30,000(1 + \(\frac{12}{100}\))^{3}

30,000 x \(\frac{112}{100}\) x \(\frac{112}{100}\) x \(\frac{112}{100}\) = 42147.84

∴ CI = A – P = 42147.84 – 30,000

CI = 12147.84

In private company,

Rate of single Interest SI = 12% p.a

So, for 3 months, i.e \(\frac{3}{12}\) = \(\frac{1}{4}\) year,

∴ Difference in interest = CI – SI = 12,147.84 – 900 = 11247.84