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S.chand Class 8 Maths Solution Chapter 3 Squares and Square roots Exercise 3B

Exercise 3B

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Question 1

Find by prime factorization the square root of the following numbers :

(i) 2916

(ii) 2704

(iii) 7056

(iv) 9025

(v) 9216

Sol :

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Question 2

Find the square root of the following fractions :

(i) $\frac{25}{49}$

(ii) $\frac{196}{484}$

(iii) $\frac{1225}{2025}$

(iv) $0.0009$

(v) $0.00000049$

Sol :

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Question 3

Simplify :

(i) $\sqrt{\left(5^{2}-4^{2}\right)}$

(ii) $\sqrt{\left(5^{2}+12^{2}\right)}$

(iii) $\left(\frac{1}{2}\right)^{2}+\sqrt{0.25}$

(iv) $\left(-\sqrt{\frac{4}{9}}\right)\left(-\sqrt{\frac{81}{100}}\right)$

Sol :

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Question 4

Find the square root of each of the following numbers by long division :

(i) 6724

(ii) 7921

(iii) 8649

(iv) 9801

(v) 15129

Sol :

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Question 5

Find the square root of each of the following decimal numbers by long division :

(i) $4556.25$

(ii) $6099.61$

(iii) $7903.21$

(iv) $9273.69$

Sol :

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Question 6

Evaluate :

(i) $\sqrt{\frac{2809}{4096}}$

(ii) $\sqrt{\frac{1849}{5776}}$

(iii) $\sqrt{1 \frac{869}{1156}}$

Sol :

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Question 7

Find the value of $\sqrt{9216}$ and from this value calculate $\sqrt{92.16}+9.216$.

Sol :

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Question 8

Find the square roots of 2116 and 1764 and hence find the value of $\frac{\sqrt{0.2116}+\sqrt{0.1764}}{\sqrt{0.2116}-\sqrt{0.1764}}$.

Sol :

Multiple Choice Questions (MCQs)

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Question 9

The value of $\frac{3}{\sqrt{0.09}}$ is :

(a) $\frac{1}{10}$

(b) $\frac{3}{10}$

(d) 10

Sol :

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Question 10

Of the numbers $0.16, \sqrt{0.16},(0.16)^{2}$ and $0.016$ the least number is

(a) $(0.16)^{2}$

(b) $\sqrt{0.16}$

(d) $0.16$

Sol :

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Question 11

$\sqrt{0.0025} \times \sqrt{2.25} \times \sqrt{0.0001}$ equals

(a) $0.000075$

(b) $0.0075$

(d) $0.00075$

Sol :

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Question 12

The square root of $\sqrt{2 \frac{1337}{3844}}$ is

(a) $2 \frac{35}{64}$

(b) $1 \frac{33}{62}$

(c) $1 \frac{17}{62}$

(d) $2 \frac{23}{62}$

Sol :

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Question 13

$\frac{\sqrt{59.29}-\sqrt{5.29}}{\sqrt{59.29}+\sqrt{5.29}}$ equals

(a) 1

(b) 0.65

(c) 0.45

(d) 0.54

Sol :

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Question 14

3600 soldiers are asked to stand in different rows. Every row has as many soldiers as there are rows. Find the number of rows.

Sol :

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Question 15

Find the perimeter of a square whose area is $6889 \mathrm{~m}^{2}$.

Sol :

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Question 16

A society collected ₹ 8836, each member contributing as many rupees as there were members. Find the number of members of the society.

Sol :

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Question 17

In a basket there are 1250 flowers. A man goes for worship and puts as many flowers as there are temples in the city. Thus he needs 8 baskets of flowers. Find the number of temples in the city.

Sol :

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Question 18

What should be subtracted from 6249 to get a perfect square number? What is this perfect square number? Also, find its square root.

Sol :

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Question 19

What least number must be added to 594 to make the sum a perfect square ?

Sol :

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Question 20

What would be added to 7912 to make the sum a perfect square ?

Sol :

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Question 21

Find the least number of six digits which is a perfect square.

Sol :

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Question 22

A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8341 , find the number of soldiers in each row.

Sol :

High Order Thinking Skills (HOTS)

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Question 23

Find $: \sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}$

Sol :

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Question 24

Find the value of $\sqrt{1191.16+70 \sqrt{129.96}}$.

Sol :

SChand Composite Mathematics Class 7 Chapter 3 Decimals Exercise 3B

Exercise 3B

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Question 1

(i) 8.37×10

=83.7

(ii) 4.6×10

=46

(iii) 49.87×100

=4987

(iv) 4.907×10

=49.07

(v) 0.0000078×1000

=0.0078

(vi) 0.00805×100

=0.805

(vii) 0.0893×10000

=893

(viii) 357.126×10000

=3571260

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Question 2

(i) 0.9×0.4

Sol : 0.36

(ii) 0.07×0.05

Sol : 0.0035

(iii) 0.008×3

Sol : 0.024

(iv) 0.006×0.009

Sol : 0.000054

(v) 0.8×0.09

Sol : 0.072

(vi) 2.63×0.0004

Sol : 0.001052

(vii) 150×0.007

Sol : 1.05

(viii) 52.8×0.007

Sol : 0.3696

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Question 3

(i) 5.9×2.3

Sol : 13.57

(ii) 0.009×0.049

Sol : 0.000441

(iii) 0.678×860

Sol : 583.08

(iv) 0.0095×81.7

Sol : 0.77615

(v) How many decimal places does $(1.4)^3$ have?

Sol :

=2.744

=3 decimal places

SChand New Learning Composite Mathematics Class 6 Chapter 3 Playing With Numbers Exercise 3B

Exercise 3B

Question 1

260, 7018, 20008, 148

Question 2

48, 594, 3096, 609513

Question 3

524, 71316, 8001008

Question 4

365, 7290, 21340, 845

Question 5

516, 7038, 800712

Question 6

464, 4176, 87544, 900392

Question 7

639, 29304, 38457

Question 8

670, 90020, 57390

Question 9

418, 7909, 17743, 10000001

Question 10

Divisible by 2 and 5 only.

Question 11

Divisible by 6, 8, 10 only.

Question 12

(a) Yes, it is also divisible by its factors because 6 & 9 are not co-prime.

(b) No, it is not divisible by their product because 6 & 9 are not co –prime.

Question 13

705 2 4 is divisible by 3.

Question 14

9 3 2760 is divisible by 9.

Question 15

(a) Yes

(b) Yes

(d) Yes

(e) Yes

(f) Yes

(g) No

(h) Yes

(i) No

Question 16

984 and 1736 are divisible by 8

Sum of 984 and 1736 = 2720, divisible by 8 differences of 1736 and 384= 752, divisible by 8.

SChand CLASS 10 Chapter 3 Shares and Dividends Exercise 3B

Exercise 3(B)

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Question 1

Find the number of shares that can be bought and the income obtained by investing:

(i) Rs. 50 in (Rs. 1) shares at Rs. 1.25 paying 8%. (ii) Rs. 240 in (Rs. 5) shares at Rs. 8 paying 9%.

Sol :

(i) Cost of bought (Re. 1) shares at Rs. 1.25 = Rs 50

No of bought Shares $=\frac{50}{1.25}$ = 40 Shares.

Nominal Value of Investment = 40 Shares×Re. 1 = Rs. 40

Income obtained by Investing $= 40 \times \frac{8}{100}$ = Rs. 3.20.

(ii) Cost of bought (Rs. 5) shares at Rs. 8 = Rs 240

No of bought Shares $= \frac{240}{8}$ = 30 Shares.

Nominal Value of Investment = 30 Shares×Rs. 5 = Rs. 150

Income obtained by Investing $= 150 \times \frac{9}{100}$ = Rs. 13.50.

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Question 2

A man bought 160 (Rs. 5) shares for Rs. 360. At what price did the shares stand? At what premium or discount were they quoted?

Sol :

A man bought 160 (Rs. 5) shares for Rs 360.

Price of the Shares Stand $= \frac{360}{160}$ = Rs. 2.25

Discount Rate per share = Rs. 5 – Rs. 2.25 = Rs. 2.25

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Question 3

A man sold 600 (Rs. 1) shares for Rs. 750. At what price did the shares stand? At what premium or discount were they quoted?

Sol :

A man bought 600 (Rs. 1) shares for Rs 750.

Price of the Shares Stand $= \frac{750}{600}$= Rs. 1.25

Premium Rate per Share = Rs. 1.25 – Rs. 1 = Rs. 0.25

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Question 4

A man buys 200 ten rupee shares at Rs. 12.50 each and receives a divided of 8 %. Find the amount invested by him and dividend receive by him in cash.

Sol :

Cost of bought 200 (₹ 10) shares at ₹ 12.50 each

Cost of bought 200 shares = 200×₹ 12.50 = ₹ 2,500.

Nominal value of investment = 200 shares×₹ 10 = ₹ 2,000.

The dividend received in cash $= ₹ 2,000 \times \frac{8}{100}$= ₹ 160.

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Question 5

A man bought 500 shares, each of face value Rs. 10 of a certain business concern, and during the first year after purchase received Rs. 400 as dividend on his shares. Find the rate of dividend on his shares.

Sol :

Face value of bought 500 shares = 500×₹ 10 = ₹ 5,000

Dividend received on bought 500 shares = ₹ 400

The rate of dividend received $= ₹ 400 \times \frac{ 100 }{5000 }$= 8%.

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Question 6

By purchasing Rs. 25 shares for Rs. 40 each a man gets 4 % profit on his investment. What rate percent is the company paying? What is his dividend if he buys 60 shares?

Sol :

Purchasing ₹ 25 shares for ₹ 40 each and gets 4% profit on his investment.

Profit (Dividend) on 1 shares $= ₹ 25 \times \frac{4}{100}$ = Rs. 1.60.

The paying rate $= ₹ 1.60\times \frac{100}{₹ 1.60} $= 6.4%.

Dividend on 60 shares = 60×₹ 1.60 = ₹ 96.

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Question 7

Mukul invests Rs. 9000 in a company paying a dividend of 6% per annum when a share of face value 100 stands at Rs. 150. What is his annual income? He sells 50 % of his shares when the price rises to Rs. 200. What is his gain on this transaction?

Sol :

Cost of bought (₹ 100) shares at ₹ 150 = ₹ 9,000

No. of bought shares $=\frac{₹9,000}{₹500}$= 60 Shares

Nominal value of investment = 60 shares×₹ 100 = ₹ 6,000

The income obtained by investing $= ₹ 6,000 \times \frac{6}{100}$= ₹ 360

Cost of sale of 50% shares = 30 shares×₹ 150 = ₹ 4,500

Selling price of 50% shares = 30 shares×₹ 200 = ₹ 6,000

Gain on sale of 50% shares = S.P. – C.P. = 6,000 – 4,500 = ₹ 1,500.

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Question 8

By investing Rs. 7500 in a company paying 10% dividend, an income of Rs. 500 is received. What price is paid for each Rs. 100 share?

Sol :

Nominal value of shares $=\text{Dividend on shares} \times \frac{100}{\text{Rate of dividend}}$ $= ₹ 500 \times \frac{100}{10}$= ₹ 5,000

No. of shares purchased $= \frac{\text{Nominal value of shares}}{\text{Face Value per share}}$ $=\frac{₹ 5,000 }{100}$ = 50 Shares

Cost of 50 shares = ₹ 7,500

Price paid per share to the company $= \frac{₹ 7,500 }{50}$= ₹ 150.

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Question 9

Arun owns 560 shares of a company. The face value of each share is Rs. 25. The company declares a dividend of 9%. Calculate: (i) the dividend Arun would receive, and (ii) the rate of interest, on his investment. Considering that Arun bought these shares @Rs. 30 per share in the market.

Sol :

(i) Nominal value of shares = No. of shares X Nominal value per share = 560×₹ 25 = ₹ 14,000

Dividend on shares $= \text{Nominal value of shares} \times \frac{\text{Rate of dividend}}{100}$ $= 14,000 \times \frac{ 9}{ 100}$ = ₹ 1,260

(ii) If Arun bought these 560 shares @ ₹ 30 per share in the market.

Cost of investment = 560 shares×₹ 30 = ₹ 16,800

Income on investment = ₹ 1,260

Interest on investment $= 1,260 \times \frac{100}{16,800} =7\frac{1}{2}$%.

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Question 10

What sum should Ashok invest in Rs 25 shares selling at Rs 36 to obtain an income of Rs. 720, if the dividend declared is 12%. Also find (i) the number of shares bought by Ashok (ii) the percentage return on his investment.

Sol :

Nominal value of shares $= Dividend on shares \times \frac{100}{ Rate of dividend} $

$= 720 \times \frac{100}{ 12 }$ = ₹ 6,000

No. of shares purchased $= \frac{\text{Nominal value of shares}}{\text{ Face value per share} $ $=\frac{6,000}{ 25}$ = 240 shares

Sum paid for investment = 240 shares×₹ 36 = ₹ 8,640

Return on investment $= \text{Annual income from shares } \times \frac{100}{\text{Cost of Investment}$ $= 720\times \frac{100}{8,640}$= 8.33%

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Question 11

Mr. Sharma has 60 shares of nominal value of Rs. 100 and he decides to sell them when they are at a premium of 60%. He invests the proceeds in shares of nominal value Rs. 50 quoted at 4% discount, paying 18% dividend annually. Calculate: (i) the sale proceeds (ii) the number of shares he buys (iii) the annual dividend from these shares

Sol :

(i) Sales proceeds of 60 shares = Nominal value of 60 shares + 60% Premium = ( 60 x 100 ) + 60% of ₹ 6,000 = 6,000 + 3,600 = ₹ 9,600.

(ii) Rate per new share = ₹ 50 – 4% Discount 0f ₹ 50 = 50 – 2 = ₹ 48

No. of purchased new shares $= \frac{₹ 9,600 }{ 48} $= 200 shares

(iii) Nominal value of shares = No. of shares×Nominal value per share = 200×₹ 50 = ₹ 10,000

Annual dividend on shares $=\text{ Nominal value of shares }\times \frac{\text{Rate of dividend} }{ 100 }$ $= 10,000 \times \frac{18 }{100}$ = ₹ 1,800.

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Question 12

A man invests a sum of money in Rs. 100 shares, paying 15% dividend, quoted at 20% premium. If his annual dividend is Rs. 540, calculate (i) his total investment, (ii) the rate of return of his investment.

Sol :

Nominal value of shares $= Dividend on shares \times \frac{100}{ Rate of dividend} $

$= ₹ 540 \times \frac{100}{ 15}$ = ₹ 3,600

(i) Cost of total investment = Nominal value of shares + 20% Premium = ₹ 3,600 + 20% of ₹ 3,600 = 3,600 + 720 = ₹ 4,320.

(ii) Rate of return on investment $= \text{Annual dividend on shares } \time \frac{100}{ \text{Cost of Investment }}$ $= 540 \times \frac{ 100 }{4,320} $= 12.5%

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Question 13

A lady holds 1800 hundred rupee shares of a company that pays 15% dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what percentage return would she have got on her investment? Give your answer to the nearest integer.

Sol :

Nominal value of shares = No. of shares×Nominal value per share = 1,800×₹ 100 = ₹ 1,80,000

Annual dividend on shares $= \text{Nominal value of shares} \times \frac{\text{Rate of dividend}}{ 100}$ $= 1,80,000 \times \frac{15}{100}$= ₹ 27,000.

Cost of total investment = Nominal value of shares + 40% Premium = ₹ 1,80,000 + 40% of ₹ 1,80,000 = 1,80,000 + 72,000 = ₹ 2,52,000

Rate of return on investment $= \text{Annual dividend on shares}\times \frac{100 }{Cost of Investment}$ $= 27,000\times \frac{100}{ 2,52,000}$ = 10.71% = 11% ( to the nearest integer ).

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Question 14

A man invests Rs. 11200 in a company paying 6% dividend when its Rs. 100 share can be bought for Rs. 140. Find (i) his annual income. (ii) the percentage income on his investment.

Sol :

(i) Cost of bought (₹ 100) shares at ₹ 140 = ₹ 11,200

No. of bought shares $= \frac{₹ 11,200 }{₹ 140 }$= 80 Shares

Nominal value of investment = 80 shares×₹ 100 = ₹ 8,000

∴ The income obtained by investing $= ₹ 8,000 \times \frac{6}{100}$ = ₹ 480

(ii) Rate of return on investment $= \text{Annual dividend on shares} \times \frac{100}{\text{Cost of Investment}}$ $= 480\times \frac{100}{11,200}$= 4 2/7%.

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Question 15

A company with 10000 shares of Rs. 100 each, declares an annual dividend of 5%. (i) What is the total amount of dividend paid by the company? (ii) What would be the annual income of a man, who has 72 shares in the company? (iii) If he received only 4% of his investment, find the price he had paid for each share.

Sol :

(i) Nominal value of shares in the company = No. of shares×Nominal value per share = 10,000×₹ 100 = ₹ 10,00,000

∴ Annual dividend paid by the company $=\text{Nominal value of shares}\times \frac{\text{Rate of dividend}}{100}$ $= 10,00,000 \times \frac{5}{100}$ = ₹ 50,000

(ii) Nominal value of 72 shares = No. of shares×Nominal value per share = 72×₹ 100 = ₹ 7,200

∴ Annual income of a man on 72 shares $= \text{Nominal value of shares} \times \frac{\text{Rate of dividend}}{100 }$= 7,200×5100 = ₹ 360

(iii) Income on investment at 4% $ = 7,200 \times \frac{4}{100}$= 288

Price paid for each share $= Annual dividend on shares \times \frac{100}{\text{Income of Investment}}$ $= 360 \times \frac{100}{288}$= ₹ 125.

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Question 16

A man invests Rs. 1680 in buying shares of nominal value of Rs. 24 and selling at 12% premium. The dividend on the shares is 15% per annum. (i) Calculate the number of shares he buys. (ii) Calculate the dividend he receives annually.

Sol :

(i) Cost of investment per share = Nominal value of shares + 12% Premium = ₹ 24 + 12% of ₹ 24 = 24 + 2.88 = ₹ 26.88

Cost of total investment = ₹ 1,680

No. of purchased shares $=\frac{ ₹ 1,680 }{26.88}$ = 62.5 Shares

(ii) Nominal value of 62.5 shares = No. of shares×Nominal value per share = 62.5×₹ 24 = ₹ 1,500

∴ Annual dividend received on shares $= \text{Nominal value of shares} \times \frac{\text{Rate of dividend }}{100} $ $=\frac{ 1,500 \times 15 }{100}$ = ₹ 225.

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Question 17

A man invests Rs. 7425 on buying shares of face value of Rs. 90 each at a premium of 10% in a company. If he earns Rs. 1350 as dividend at the end of the year, find (i) the number of shares he has in the company. (ii) the dividend percentage per share that he received.

Sol :

(i) Cost of investment per share = Face value per shares + 10% Premium = ₹ 90 + 10% of ₹ 90 = 90 + 9 = ₹ 99

Cost of total investment = ₹ 7,425

No. of purchased shares $= \frac{₹ 7,425}{99}$= 75 shares

(ii) Nominal value of 75 shares = No. of shares×Nominal value per share = 75×₹ 90 = ₹ 6,750

Dividend percentage per share $= \text{Annual dividend on shares }\times \frac{ 100}{\text{Nominal Value}} $= 1,350 X 100 / 6,750 = 20%.

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Question 18

Abhishek sold a certain number of shares of Rs. 20 paying 8% dividend at Rs. 18 and invested the proceeds in Rs. 10 shares, paying 12% dividend at 50 % premium. If the change in his annual income is Rs. 120, find the number of shares sold by him?

Sol :

Let, Abhishek sell X shares.

Selling price of X shares = X×₹ 18 = 18X

Face value of X shares = X×₹ 20 = 20X

Dividend on X shares = 8% of 20X $= 20X \times \frac{8}{100} =\frac{ 8X}{5}$

Now, Rate per new share = ₹ 10 + 50% Premium = 10 + 5 = 15

No. of purchased new shares = Sale Proceeds / Rate per new share $= \fac{18X}{15}= \frac{6X}{5}$

Nominal value of purchased new shares $= \frac{6X}{5} \times 10$ = 12X

Dividend on purchased new shares = 12% of 12X $= 12X \times \frac{12}{100} =\frac{ 36X}{25}$

Change in annual income = ₹ 120

$\frac{8X}{5}− \frac{36X}{25} = 120 $

⇒ X = 750

No. of shares sold by Abhishek = X = 750 shares

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Question 19

A person invested Rs. 8000 and Rs. 10000 in buying shares of two companies which later on declared dividends of 12% and 8% respectively. He collects the dividends and sells out his shares at a loss of 2% and 3% respectively. Find his total earning from the above transaction.

Sol :

Dividend received = 12% of ₹ 8,000 + 8% of ₹ 10,000 = 960 + 800 = ₹ 1,760

Loss on sale of shares=2% of ₹8,000 + 3% of ₹ 10,000 = 160 + 300 = ₹ 460

∴Total earning = Dividend received – Loss on sale of shares = 1,760 – 460 = ₹ 1,300

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Question 20

A person invested 20%, 30% and 25% of his saving in buying shares of three different companies A, B and C, which declared dividends of 10%, 12% and 15% respectively. If his total income on account of dividends be Rs. 2337.50, find his saving and the amount which he invested in buying shares of each company.

Sol :

Let, Total saving be X.

Dividend received = 10% of (20% of X) + 12% of (30% of X) + 15% of (25% of X)

⇒$ 2,337.50 = \frac{X}{50} +\frac{ 9X}{250} + \frac{3X}{80} $

⇒ X = 25000

Total saving = X = ₹ 25,000

Investment in company ‘A’ = 20% of ₹ 25,000 = ₹ 5,000

Investment in company ‘B’ = 30% of ₹ 25,000 = ₹ 7,500

Investment in company ‘C’ = 25% of ₹ 25,000 = ₹ 6,250

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 3 Squares and Square roots,Cube and Cube roots Exercise 3B

Exercise 3B

Q1 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

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Question 1

Find the two square roots of the following numbers.

(a) 4

Sol : 2,-2

(b) 81

Sol : 9,-9

Sol : 14,-14

(d) 400

Sol : 20,-20

(e) 441

Sol : 21,-21

(f) 0.0049

Sol : 0.07, -0.07

(g) 0.0001

Sol : 0.01, -0.01

(h) $3\frac{1}{16}$

Sol : $\frac{7}{4},-\frac{7}{4}$

(i) $2\frac{1}{4}$

Sol : $\frac{3}{2},-\frac{3}{2}$

(j) 0.0289

Sol : 0.17 , -0.17

Q2 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

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Question 2

Evaluate the following.

(a) √36

Sol :

=2×2×3×3

=2×3

(b) -√9

Sol : =-(3×3)

=-3

Sol : =11×11×11

(d) -√225

Sol :

=-(5×5×3×3)

=-(5×3)

=-15

(e) √361

Sol :

=19×19

=19

(f) √900

Sol :

=30×30

=30

(g) √0.09

Sol :

=0.3×0.3=0.3

(h) √0.0256

Sol :

=0.16×0.16

=0.16

(i) √2.25

Sol :

=1.5×1.5

=1.5

(j) $\sqrt{4\frac{25}{36}}$

Sol :

$=\frac{13}{6}$

Q3 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

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Question 3

Solve

(a) x² = 1

Sol : x=±1

(b) 12x² = 108

Sol :

$x^2=\frac{108}{2}=9$

x=±3

Sol :

x²=-(16)

x=±4

(d) x²=$\frac{16}{25}$

Sol :

x²=$\pm \frac{4}{5}$

Q4 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

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Question 4

Find the square root of each of the following numbers by the prime factorisation method.

(a) 484

Sol :

=2×2×11×11

=2×11=22

(b) 2500

Sol :

=2×2×5×5×5×5

=2×5×5

=50

Sol :

=5×5×9×9

=5×9

=45

(d) 2916

Sol :

=2×2×3×3×9×9

=2×3×9=54

(e) 2401

Sol :

=7×7×7×7

=7×7=49

(f) 6084

Sol :

=2×2×3×3×13×13

=2×3×13=78

(g) $1\frac{184}{441}$

Sol :

$=\frac{625}{441}=\frac{25\times 25}{21\times 21}$

$=\frac{25}{21}$

(h) 0.1936

Sol :

$=\frac{1936}{10000}=\frac{44\times 44}{10\times 10\times 10 \times 10}$

$=\frac{44}{100}$=0.44

(i) 0.0576

Sol :

$=\frac{576}{10000}=\frac{24\times 24}{10\times 10\times 10\times 10}$

$=\frac{24}{100}$=0.24

(j) 40.96

Sol :

$=\frac{4096}{100}=\frac{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times }{10\times 10}$

$=\frac{64}{100}$=0.64

Q5 |Ex-3B |Class 8 |Squares and Square roots,Cube ,Cube roots |S.Chand New Learning Composite

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Question 5

Find the smallest whole number for each of the following numbers by which it should be multiplied so as to get a perfect square number.

(a) 768

Sol :

=2×2×2×2×2×2×2×2×3

∴Multiplied by 3

(b) 200

Sol :

=5×5×2×2×2

∴Multiplied by 2

Sol :

=2×2×2×2×2×2×3×3×5

∴Multiplied by 5

(d) 16807

Sol :

=7×7×7×7×7

∴Multiplied by 7

(e) 1331

Sol :

=11×11×11

∴Multiplied by 11

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Question 6

Find the smallest whole number for each of the following numbers by which it should be divided so as to get a perfect square number.

(a) 3125

Sol :

=5×5×5×5×5

∴Divided by 5

(b) 1800

Sol :

=2×2×2×5×5×3×3

∴Divided by 2

Sol :

=2×2×2×2×3×3×7

∴Divided by 7

(d) 6912

Sol :

=2×2×2×2×2×2×2×2×3×3×3

∴Divided by 3

(e) 2925

Sol :

=13×15×15

∴Divided by 13

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Question 7

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Sol :

⇒8,15 & 20

⇒L.C.M of 8,15,20 is 120

∴Prime factor of 120=2×2×[2×3×5]

=30

∴The factor 30 remains unpaired so to make 120 a perfect square it should be multiplied by 30

∴The smallest square number=120×30

=3600

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Question 8

Find the least square number, which is exactly divisible by 3, 4, 5, 6 and 8.

Sol :

L.C.M of 3,4,5,6,8=120

∴Prime factor of 120=2×2×[2×3×5]

=30

∴Least square number=120×30

=3600

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Question 9

The area of a square plot is $101 \frac{1}{400}$ meter square, Find the length of one side of the plot.

Sol :

ATQ

$a^2=101\frac{1}{400}$ m²

∴$a=\sqrt{101\frac{1}{400}}=\sqrt{\frac{40401}{400}}$

∴$a=\pm \frac{201}{20}=10\frac{1}{20}$

∴One side of plot$=10\frac{1}{20}$ m

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Question 10

Find the value of $=\sqrt{162+\sqrt{38 + \sqrt{121}}}$

Sol :

$=\sqrt{162+\sqrt{38+11}}$

$=\sqrt{162+\sqrt{49}}=\sqrt{162+7}$

=√169=13

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Question 11

Given that √3136 = 56, find the value of √31.36 + √0.3136

Sol :

=√31.36 + √0.3136

$=\sqrt{\frac{3136}{100}}+\sqrt{\frac{3136}{10000}}$

$=\frac{56}{10}+\frac{56}{10}$

$=\frac{560+560}{1000}$

$=\frac{616}{100}$=6.16

S.chand Composite Mathematics class 8 Chapter 3 Square and Square Roots Exercise 3B

Exercise 3 (B)

Q1 | Ex-3B | Square and Square Roots | Class 8 | Schand Composite Mathematics | myhelper

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Question 1

Find the square root of each of the following numbers by division method.

(i) 3249

Sol :

$\begin{array}{r|l} &57 \\\hline5&3249\\5&25\\\hline 107 & ~~749 \\ 7&~~749\\\hline &~~~~0 \end{array}$

(ii) 6889

Sol :

$\begin{array}{r|l} &83 \\\hline 8&6889\\8&64\\\hline 163 & ~~489 \\ 3&~~489\\\hline &~~~~0 \end{array}$

(iii) 15129

Sol :

$\begin{array}{r|l} &123 \\\hline 1&15129\\1&1\\\hline 22 & ~~51 \\ 2&~~44\\\hline 243&~~~~~~729\\&~~~~~~729\\ \hline &~~~~~~~~0 \end{array}$

(iv) 75625

Sol :

$\begin{array}{r|l} &275 \\\hline 2&75625\\2&4\\\hline 47 & 356 \\7&329\\\hline 545&~~2725\\5&~~2725\\ \hline &~~~~~0 \end{array}$

(v) 166464

Sol :

$\begin{array}{r|l} &408 \\\hline 4&166464\\4&16\\\hline 80 & ~~~~64 \\0&~~~~00\\\hline 808&~~6464\\8&~~6464\\ \hline &~~~~~0 \end{array}$

(vi) 9548100

Sol :

$\begin{array}{r|l} &3090 \\\hline 3&9548100\\3&9\\\hline 60 & ~~54 \\0&~~00\\\hline 609&~~5481\\9&~~5481\\ \hline &~~~~~0 \end{array}$

(vii) $ \dfrac{1089}{3481} $

Sol :

$=\frac{\sqrt{1089}}{\sqrt{3481}}$

$\begin{array}{r|l} &33 \\\hline 3&1089\\3&9\\\hline 63 & 189 \\3&189\\\hline &~~0 \end{array}$

$\begin{array}{r|l} &59 \\\hline 5&3481\\5&25\\\hline 109 & ~~981 \\9&~~981\\\hline &~~0 \end{array}$

$=\frac{33}{59}$

(viii) $ \dfrac{7569}{14884} $

Sol :

$=\frac{\sqrt{7569}}{\sqrt{14884}}$

$\begin{array}{r|l} &87 \\\hline 8&7569\\8&64\\\hline 167& 1169 \\17&1169\\\hline &~~0 \end{array}$

$\begin{array}{r|l} &122 \\\hline 1&14884\\1&1\\\hline 22& ~~48 \\2&~~44\\\hline &~~~~484\\&~~~~484\\ \hline &~~~~~~0 \end{array}$

$=\frac{87}{122}$

(ix) $ 2\dfrac{337}{9216} $

Sol :

$=\sqrt{\left(\frac{9216 \times 2+337}{9216}\right)}$

$=\sqrt{\frac{18769}{9216}}$

$=\frac{\sqrt{18769}}{{\sqrt{9216}}}$

$\begin{array}{r|l} &137 \\\hline 1&18769\\1&1\\\hline 23& ~~87 \\3&~~69\\\hline 267&~~1869\\7&~~1869\\\hline& ~~~~0 \end{array}$

$\begin{array}{r|l} &96 \\\hline 9&9216\\9&81\\\hline 186& ~~1116 \\6&~~1116\\\hline &~~~~0 \end{array}$

$=\frac{137}{96}=1\frac{41}{96}$

(x) $ 9\dfrac{4185}{5776} $

Sol :

$=\sqrt{\frac{5776 \times 9+4185}{5776}}$

$=\sqrt{\frac{56169}{5776}}$

$=\frac{\sqrt{56169}}{\sqrt{5176}}$

$\begin{array}{r|l} &237 \\\hline 2&56169\\1&4\\\hline 43& 161 \\3&129\\\hline 467&~~3269\\7&~~3269\\\hline& ~~~~0 \end{array}$

$\begin{array}{r|l} &76\\\hline 7&5776\\7&49\\\hline 146& 876\\6&876\\\hline &~~0 \end{array}$

$=\frac{237}{76}$

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Question 2

Find the perimeter of a square field whose area is 13689 m²

Sol :

Perimeter of square= 4×side

Area of square=side×side=side²

13689=side²

side=$\sqrt{13689}$

$\begin{array}{r|l} &117 \\\hline 1&13689\\1&1\\\hline 21& ~~36 \\1&~~21\\\hline 227&~~1589\\7&~~1589\\\hline& ~~~~0 \end{array}$

side=117

Perimeter of square=4×117=468

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Question 3

What should be subtracted from 18246 to get a perfect square number ? What is this perfect square number ? Also, find its square root.

Sol :

$\begin{array}{r|l} &135 \\\hline 1&18246\\1&1\\\hline 23& ~~82 \\3&~~69\\\hline 265&~~1346\\5&~~1325\\\hline& ~~~~~~21 \end{array}$

As it can be seen that square of 135 is less than by 21 (remainder)

If we subtract the remainder 21 from the number 18.246, we get a perfect square.

∴Required perfect square=18246-21=18.225

Square root$=\sqrt{18225}=135$

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Question 4

What should be added to 14841 to make the sum a perfect square ?

Sol :

$\begin{array}{r|l} &121 \\\hline 1&14841\\1&1\\\hline 22& ~~48 \\2&~~44\\\hline 241&~~441\\1&~~241\\\hline& ~~~~~~200 \end{array}$

We observe that the given number is greater than square of 121 and less than square of 122.

So, number to be added$=(122)^{2}-14841$

=14884-14841=43

Resulting number=14841+43=14884

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Question 5

Find the least number which must be subtracted from 63520 to make it a perfect square.

Sol :

$\begin{array}{r|l} &252 \\\hline 2&63520\\2&4\\\hline 45& 235 \\5&225\\\hline 502&~~1020\\&~~1004\\ \hline &~~~~~~16 \end{array}$

As it can be seen that square of 252 is less than 63520 by 16 (remainder)

To make it a perfect square, we have to subtract=63520-16=63504

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Question 6

Find:

(i) Find the least number of six digits which is a perfect square.

Sol :

The least six digit number is 100000

$\begin{array}{r|l} &316 \\\hline 3&100000\\3&9\\\hline 61& 100 \\1&~~61\\\hline 626&~~3900\\&~~3756\\ \hline &~~~~144 \end{array}$

We can see that square of 316 is less than 100000 by 144

Perfect square=100000-144

=99856=$(316)^2$

But it is not a six digit number.

So, we have to choose next one

$=(317)^2$

=100489

(ii) Find the greatest number of six digits which is a perfect square.

Sol :

The greatest six digit number is 999999

$\begin{array}{r|l} &999 \\\hline 9&999999\\9&81\\\hline 189& 1899 \\9&1701\\\hline 1989&~~19899\\9&~~17901\\ \hline &~~~~1998 \end{array}$

We can see that square of 999 is less than 999999 by 1998

∴Required perfect square=999999-1998

=998001

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Question 7

A gardener arranges his plants in rows to form a perfect square. He finds that in doing so, 25 plants are left out.If the total number of plants be 2234 , find the numbers of plants in each row.

Sol :

Prefect square number=Total plants-Left out plants

=2234-25=2209

Number of plants in each row

$=\sqrt{2209}$

=47

$\begin{array}{r|l} &47 \\\hline 4&2209\\4&16\\\hline 87& 609 \\7&609\\\hline &~~0 \end{array}$

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Question 8

There are 800 children in a school. For a PT drill they have to stand in a square formation, such that the number of rows is equal to number of columns. Find the greatest number of children needed to complete the formation.

Sol :

Total number of student is 800

Given : Number of rows = Number of column=x

Required student t form a square=Number of rows ×Number of column

$=x^{2}=\sqrt{800}$

$\begin{array}{r|l} &28 \\\hline 2&800\\2&4\\\hline 48 & 400 \\8&384\\\hline &~~16 \end{array}$

We can see that square of 28 is less than 800 by 16

∴Required student=800-16=784

MCQS

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Question 9

Find the correct one:

The square root of 1585081 is

(a) 1259

(b) 2159

(c) 1251

(d) 1291

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Question 10

Find the value of

$ \sqrt{240.25}+\sqrt{2.4025}+\sqrt{0.024025}+\sqrt{0.00024025} $

(a) 1602205

(b) 16.2402

(c) 17.2205

(d) 155.2205

RS Aggarwal solution class 8 chapter 3 Squares and Square roots Exercise 3B

Exercise 3B

Page-45

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Question 1:

Give reason to show that none of the numbers given below is a perfect square:
(i) 5372
(ii) 5963
(iii) 8457
(iv) 9468
(v) 360
(vi) 64000
(viii) 2500000

Answer 1:

By observing the properties of square numbers, we can determine whether a given number is a square or not.

(i) 5372
A number that ends with 2 is not a perfect square.
Thus, the given number is not a perfect square.

(ii) 5963
A number that ends with 3 is not a perfect square.
Thus, the given number is not a perfect square.

(iii) 8457
A number that ends with 7 is not a perfect square.
Thus, the given number is not a perfect square.

(iv) 9468
A number ending with 8 is not a perfect square.
Thus, the given number is not a perfect square.

(v) 360
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

(vi) 64000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

(vii) 2500000
Any number ending with an odd number of zeroes is not a perfect square.
Hence, the given number is not a perfect square.

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Question 2:

Which of the following are squares of even numbers?
(i) 196
(ii) 441
(iii) 900
(iv) 625
(v) 324

Answer 2:

The square of an even number is always even.
Thus, even numbers in the given list of squares will be squares of even numbers.

(i) 196
This is an even number. Thus, it must be a square of an even number.

(ii) 441
This is an odd number. Thus, it is not a square of an even number.

(iii) 900
This is an even number. Thus, it must be a square of an even number.

(iv) 625
This is an odd number. Thus, it is not a square of an even number.

(v) 324
This is an even number. Thus, it is a square of an even number.

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Question 3:

Which of the following are squares of odd numbers?
(i) 484
(ii) 961
(iii) 7396
(iv) 8649
(v) 4225

Answer 3:

According to the property of squares, the square of an odd number is also an odd number.
Using this property, we will determine which of the numbers in the given list of squares is a square of an odd number.

(i) 484.
This is an even number. Thus, it is not a square of an odd number.

(ii) 961
This is an odd number. Thus, it is a square of an odd number.

(iii) 7396
This is an even number. Thus, it is not a square of an odd number.

(iv) 8649
This is an odd number. Thus, it is a square of an odd number.

(v) 4225
This is an odd number. Thus, it is a square of an odd number.

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Question 4:

Without adding, find the sum:
(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)
(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)
(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)

Answer 4:

Sum of first n odd numbers $= n^{2}$

(i) $(1 + 3 + 5 + 7 + 9 + 11 + 13)$

$= 7^{2} = 49$

(ii) $(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)$

$= 10^{2} = 100$

(iii) (1+3+5+7+9+11+13+15+17+19+21+23)

$= 12^2$ = 144

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Question 5:

(i) Express 81 as the sum of 9 odd numbers.
(ii) Express 100 as the sum of 10 odd numbers.

Answer 5:

Sum of first n odd natural numbers $= n^{2}$

(i) Expressing 81 as a sum of 9 odd numbers:
$81 = {(9)}^{2} n = 9 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$

(ii) Expressing 100 as a sum of 10 odd numbers:
$100 = {(10)}^{2} n = 10 100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19$

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Question 6:

Write a pythagorean triplet whose smallest member is
(i) 6
(ii) 14
(iii) 16
(iv) 20

Answer 6:

For every number m > 1, the Pythagorean triplet is $(2 m, m^{2} - 1, m^{2} + 1)$ .

Using the above result:

(i)
$2 m = 6 m = 3, m^{2} = 9 m^{2} - 1 = 9 - 1 = 8 m^{2} + 1 = 9 + 1 = 10$

Thus, the Pythagorean triplet is $[6, 8, 10]$ .

(ii)
$2 m = 14 m = 7, m^{2} = 49 m^{2} - 1 = 49 - 1 = 48 m^{2} + 1 = 49 + 1 = 50$

Thus, the Pythagorean triplet is $[14, 48, 50]$ .

(iii)
$2 m = 16 m = 8, m^{2} = 64 m^{2} - 1 = 64 - 1 = 63 m^{2} + 1 = 64 + 1 = 65$

Thus, the Pythagorean triplet is: $[16, 63, 65]$

(iv)
$2 m = 20 m = 10, m^{2} = 100 m^{2} - 1 = 100 - 1 = 99 m^{2} + 1 = 100 + 1 = 101$

Thus, the Pythagorean triplet is $[20, 99, 101]$ .

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Question 7:

Evaluate:
(i) (38)² − (37)²
(ii) (75)² − (74)²
(iii) (92)² − (91)²
(iv) (105)² − (104)²
(v) (141)² − (140)²
(vi)(218)² − (217)²

Answer 7:

Given: $[{(n + 1)}^{2} - n^{2}] = (n + 1) + n$

(i) ${(38)}^{2} - {(37)}^{2} = 38 + 37 = 75$

(ii) ${(75)}^{2} - {(74)}^{2} = 75 + 74 = 149$

(iii) ${(92)}^{2} - {(91)}^{2} = 92 + 91 = 183$

(iv) ${(105)}^{2} - {(104)}^{2} = 105 + 104 = 209$

(v) ${(141)}^{2} - {(140)}^{2} = 141 + 140 = 281$

(vi) ${(218)}^{2} - {(217)}^{2} = 218 + 217 = 435$

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Question 8:

Using the formula (a + b)² = (a² + 2ab + b²), evaluate:
(i) (310)²
(ii) (508)²
(iii) (630)²

Answer 8:

(i) $310^{2} = {(300 + 10)}^{2}$

$= (300^{2} + 2 (300 \times 10) + 10^{2})$

$= 90000 + 6000 + 100 = 96100$

(ii) $508^{2} = {(500 + 8)}^{2}$

$= (500^{2} + 2 (500 \times 8) + 8^{2}) = 250000 + 8000 + 64 = 258064$

(iii) $630^{2} = {(600 + 30)}^{2}$

$= (600^{2} + 2 (600 \times 30) + 30^{2})$

$= 360000 + 36000 + 900$

$= 396900$

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Question 9:

Using the formula (a − b)² = (a² − 2ab + b²), evaluate:
(i) (196)²
(ii) (689)²
(iii) (891)²

Answer 9:

(i) ${(196)}^{2} = {(200 - 4)}^{2} = 200^{2} - 2 (200 \times 4) + 4^{2}$

$= 40000 - 1600 + 16 = 38416$

(ii) ${(689)}^{2} = {(700 - 11)}^{2} = 700^{2} - 2 (700 \times 11) + 11^{2}$

$= 490000 - 15400 + 121 = 474721$

(iii) ${(891)}^{2} = {(900 - 9)}^{2} = 900^{2} - 2 (900 \times 9) + 9^{2}$

$= 810000 - 16200 + 81 = 793881$

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Question 10:

Evaluate:
(i) 69 × 71
(ii) 94 × 106

Answer 10:

(i) $69 \times 71 = (70 - 1) \times (70 + 1) = (70^{2} - 1^{2})$

$= 4900 - 1 = 4899$

(ii) $94 \times 106 = (100 - 6) \times (100 + 6)$

$= (100^{2} - 6^{2}) = 10000 - 36 = 9964$

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Question 11:

Evaluate:
(i) 88 × 92
(ii) 78 × 82

Answer 11:

(i) $88 \times 92 = (90 - 2) \times (90 + 2)$

$= (90^{2} - 2^{2}) = 8100 - 4 = 8096$

(ii) $78 \times 82 = (80 - 2) \times (80 + 2)$

$= (80^{2} - 2^{2}) = 6400 - 4 = 6396$

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Question 12:

Fill in the blanks:
(i) The square of an even number is .........
(ii) The square of an odd number is .........
(iii) The square of a proper fraction is ......... than the given fraction.
(iv) n² = the sum of first n ......... natural numbers.

Answer 12:

(i) The square of an even number is even.

(ii) The square of an odd number is odd.

(iii) The square of a proper fraction is smaller than the given fraction.

(iv) $n^{2} =$ the sum of first n odd natural numbers.

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Question 13:

Write (T) for true and (F) for false for each of the statements given below:
(i) The number of digits in a perfect square is even.
(ii) The square of a prime number is prime.
(iii) The sum of two perfect squares is a perfect square.
(iv) The difference of two perfect squares is a perfect square.
(v) The product of two perfect squares is a perfect square.

Answer 13:

(i) F
     The number of digits in a square can also be odd. For example: 121

(ii) F
      A prime number is one that is not divisible by any other number, except by itself and 1. Thus, square of any number cannot be a prime number.

(iii) F
      Example: $4 + 9 = 13$
4 and 9 are perfect squares of 2 and 3, respectively. Their sum (13) is not a perfect square.

(iv) F
       Example: $36 - 25 = 11$
36 and 25 are perfect squares. Their difference is 11, which is not a perfect square.

(v) T