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(i) 75,32,105

Seventy-Five Lakhs Thirty-Two Thousand One Hundred and Five

(ii) 9,75,63,453

Nine crores Seventy Five Lakhs Sixty Three Thousand Four Hundred and Fifty-Three.

(i) 47,38,561 

Place value of 5 is 5 x 100 = 500 (Five Hundred) 

(iii) 9,82,50,241 

Place value of 5 is 5 x 10000 = 50,000 (Fifty Thousand) 

(iv) 19,57,60,370 

Place value of 5 is 5 x 10,00,000 = 50,00,000 (Fifty Lakhs)

Since all the numbers to be multiplied are odd number, the digit at the units' place will be 5. Hence the number of zeros at the right end of the product is zero.

Arranging the numbers in place value chart.

(i) The highest mountain is Anaimudi [Comparing left-most digits] 

(ii) From the above chart 

In thousands place, Doddabetta and Anaimudi have greater value 2.

Comparing digits of 2637 and 2695 

2 = 2, 6 = 6, 3 < 9. 

2637 < 2695 

Again comparing the digits of 1647 and 1778 

1 = 1, 6 < 7 

1647 < 1778. 

The required order is 2695 > 2637 > 1778 > 1647. 

Anaimudi > Doddabetta > Veliangiri > Mahendragiri

(iii) The height of Anaimudi mountain = 2695 m 

The height of Mahendragiri mountain = 1647 m 

The Difference = 1048 m

(i) (10 + 17) ÷ 3 (Given) 

= 27 ÷ 3 (Bracket completed first) 

= 9 (÷ completed) 

∴ (10 + 17) ÷ 3 = 9

(ii) 12 – [3 – {6 – (5 – 1)}] (Given) 

= 12 – [3 – {6 – 4}] (Innermost bracket completed first) 

= 12 – [3 – 2] (Again Inner bracket completed second) 

= 12 – 1 (Bracket completed third) 

= 11 (- completed) 

∴ 12 – [3 – {6 – (5 – 1)}] = 11

(iii) 100 + 8 ÷ 2 + {(3 x 2) – 6 ÷ 2} (Given) 

= 100 + 8 ÷ 2 + {6 – 6 ÷ 2} (Innermost bracket completed first) 

= 100 + 8 ÷ 2 + {6 – 3} (To remove the next bracket ÷ within the bar completed second) 

= 100 + 8 ÷ 2 + 3 (bar completed third) 

= 100 + 4 + 3 (÷ completed fourth) 

= 107 (+ completed) 

∴ 100 + 8 ÷ 2 + {(3 x 2) – 6 ÷ 2} = 107

(i) Canopus star’s diameter is 25941900 miles 

Indian System: Two crores Fifty-Nine Lakh Fortyone thousand Nine Hundred 

International System: Twenty-Five Million Nine Hundred Forty-One Thousand Nine Hundred.

(ii) Sirus star’s diameter = 1556500 miles 

Sum of place values of 5 is 5 x 100000 + 5 x 10000 + 5 x 100 

= 500000 + 50000 + 500 

= 5,50,500

(iii) Given value is 864,000,730 

In Indian System 86,40,00,730 

Eighty-six crore forty lakhs seven hundred and thirty.

(iv) The diameter of the Arcturus Star is 19,888,800 miles 

Nineteen Million Eight Hundred and Eighty-Eight Thousand Eight Hundred.

(v) The diameter of Canopus = 25941900 

Diameter of Arcturus = 19888800 

Difference = 6053100 

In Indian System 60,53,100 

Sixty lakh fifty-three thousand one hundred. 

In International System 6,053,100 

Six Million fifty-three thousand one hundred.

Hypotenuse = \(\sqrt{13;}\sqrt{2^2+5^2}=\sqrt{4+9}\)

Perimeter = 2 + 3 + \(\sqrt{13}=5+\sqrt{13}\)

Let one side of the reactable be ‘a’

Area = a²

a² =10 sq. cm a = \(\sqrt{10}\)

Perimeter = 4 × a = 4 × \(\sqrt{10}\) cm

x = \(\frac{1}{\sqrt{2}}\)

\(\frac{1}{x}=\sqrt{2};(x+\frac{1}{x})^2=(\sqrt{2}+\frac{1}{\sqrt{2}})^2\)

2 + \(\frac{1}{2}\) + 2 = \(4\frac{1}{2}\) = 4.5

Perimeter = \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}=6\sqrt{2}\)

= 6 × 1.414 = 8.484 = 8.48 cm

\(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}\)

= 5 × 1.414 = 7.070 = 7.07

Hypotenuse = \(\sqrt{\sqrt{3}^2}+\sqrt{2}^2\)

\(=\sqrt{3}+2=\sqrt{5}\) cm

Sum of perpendicular sides = \(\sqrt{3}+\sqrt{2}\) Difference with the hypotenuse

\(=(\sqrt{3}+\sqrt{2})-\sqrt{5}\)

= (1.73 + 1.41) – 2.24 = 3.14 – 2.24 cm = 0.9

\(\sqrt{2}\) = 1.41; \(\sqrt{3}\) = 1.73

Numbers in between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.5, 1.6, 1.65 So fractions are \(\frac{15}{10},\frac{16}{10},\frac{165}{100}\)

\(\sqrt{3}+\sqrt{2}\)

\(\sqrt{3}+\sqrt{2}\) = 3.146

\(\sqrt{5}\) = 2.236;

\(\sqrt{3}+\sqrt{2}+\sqrt{3}+\sqrt{2}>\sqrt{5}.\)

Square of the third side is =

\((1\frac{1}{2})^2-(\frac{1}{2})^2\) \(=(1\frac{1}{2}+\frac{1}{2})(1\frac{1}{2}-\frac{1}{2})\) = 2 × 1 = 2

∴ Third side is \(\sqrt{2}\)

Perimeter = \(1\frac{1}{2}+\frac{1}{2}+\sqrt{2}\) = 2 + \(\sqrt{2}\)

= 2 + 1.41 = 3.412 m = 341.2 cm

43758, 45378, 47538, 48735

(i) 79009, 78999, 77890, 74077

(ii) 56109, 55499, 52888, 50000

If the numbers are written as largest to smallest order known as descending order.

(i) 25975, 26886, 30840, 37725, 40021

(ii) 53907, 57039, 57903, 59307, 59703

(iii) 74344, 74434, 74443, 77444, 77555

(i) 51425, 50925, 42325, 41525, 34152

(ii) 86067, 85032, 82511, 81316, 81154

(iii) 76543, 76435, 74653, 74356, 73456

Number of digit in 5 -digit Number 5.

Symbol falls between the numbers 12199 < 21200

5-digit numbers using digits 7,1,5 ,8 and 3 are-
71583, 17538, 75813, 81573, 57381

Many more 5-digit numbers can be formed using given digits.

(i) 2979 > 2932

(ii) 5423 < 5432

(iii) 8952 = 8952

(iv) 6850 < 6852

(v) 3675 = 3675

(vi) 9821 > 9799

5 × 2 × 6 = 2 × 5 × 6 = 60

(i) Yes, they are the same

(ii) They can be arranged as 2 × 6 × 5 = 6 × 5 × 2 = 5 × 6 × 2 = 6 × 2 × 5.

7 – 5 ≠ 5 – 7.

Because subtraction is not commutative 

[∵ 7 – 5 = 2; 5 – 7 = -2]

(15 – 8) – 6 = 7 – 6 = 1

(15 – 8) – 6 = 1

It is not same as 15 – (8 – 6).

15 – (8 – 6) = 15 – 2 = 13

(15 – 8) – 6 ≠ 15 – (8 – 6)

(i) (100 ÷ 10) ÷ 5 = 10 ÷ 5 = 2

(ii) 100 ÷ (10 ÷ 5) ≠ (100 ÷ 10) ÷ 5

(iii) Because division of whole numbers are not associative.

Also 100 ÷ (10 ÷ 5) = 100 ÷ 2 = 50

But (100 ÷ 10) ÷ 5 = 10 ÷ 5 = 2 = 50 ≠ 2

(i. e) (100 ÷ 10) ÷ 5 ≠ 100 ÷ (10 ÷ 5)

(i) 15 ÷ 5 = 3

(ii) 15 ÷ 5 ≠ 5 ÷ 15

(iii) Division is not commutative for whole numbers.

Answer is (d) 1

(b) 12 

24 ÷ {8 – 3 x 2} = 24 ÷ {8 – 6} 

= 24 ÷ 2 

= 12

24 + 2 x 8 ÷ 2 – 1 

= 24 + 2 x 4 – 1 

= 24 + 8 – 1 

= 32 – 1 

= 31

(i) 50 x 102 = 50 x (100 + 2) 

= (50 x 100) + (50 x 2) 

= 5000 + 100 

= 5100 

(ii) 500 x 689 – 500 x 89 = 500 x (689 – 89) 

= 344500 – 44500

63785, 53781

(c) x

The correct operator is x 

\(\frac{198}{891}\) = \(\frac{1}{891}\times198\) = \(\frac{2}{9}\) = 0.2222........

\(\frac{\sqrt{1}}{\sqrt{2}}=\frac{\sqrt{1}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

Remaining part = \(\sqrt{27}-\sqrt{12}\)

= \(3\sqrt{3}-2\sqrt{3}\) = \(\sqrt{3}\) cm

\(\sqrt{12}-\frac{1}{\sqrt{3}}=\frac{2\sqrt{3}}1{}-\frac{1}{\sqrt{3}}\)\(=\frac{6}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}\)

= \(\frac{5}{1.73}\) = 2.89

N = 4¹¹ × 14⁵ × 11² = (2² )¹¹ × (2 × 7)⁵ × 11² = 2²² × 2⁵ × 7⁵ × 11² = 2²⁷ × 7⁵ × 11² 

\(\therefore\) Total number of prime factors of 

N = (27 + 1) × (5 + 1) × (2 + 1) = 28 × 6 × 3 = 504.

(a) \(\sqrt{25}\times2\times\sqrt{2}\) = \(\sqrt{25}\times\sqrt{2}\times\sqrt{2}\) = 5 × 2 = 10

(b) \(\sqrt{9}\times3\times\sqrt{3}\) = 3 × 3 = 9

(c) \(\sqrt{4}\times3\times6\) = 2 × \(\sqrt{3}\times6\) = \(12\sqrt{3}\)

(d) \(\sqrt{6}\times4\) × \(\sqrt{6}\) = 2 × 6 = 12

(e) \(\sqrt{4\times7}\times\sqrt{7}=2\times7=14\)

(f) \(\sqrt{16\times2}\times\sqrt{2}=4\times2=8\)

(g) \(\sqrt{5}\times\sqrt{5}\times\sqrt{2}\times\sqrt{2}=10\)

(h) \(\sqrt{4}\times\sqrt{2}\times\sqrt{5}\times\sqrt{2}\times\sqrt{5}\) = \(2\times2\times5=20\)

AB = \(\sqrt{50}\) = \(5\sqrt{2}\) cm

BC = \(\sqrt{98}\) = \(7\sqrt{2}\) cm

AC = \(\sqrt{288}\) = \(12\sqrt{2}\) cm

\(5\sqrt{2}+7\sqrt{2}=12\sqrt{2}\)

AB + BC = AC

The three points lie on a straight line.

If A, B, C are points on the same line then AB + BC = AC

AB + BC = \(\sqrt{50}+\sqrt{98}\)

\(=\sqrt{25\times2}+\sqrt{49\times2}\)

\(=5\sqrt{2}+7\sqrt{2}=12\sqrt{2}\) cm

AC = \(\sqrt{288}=\sqrt{144\times2}\)

= \(12\sqrt{2}\) cm

Here AB + BC = AC. So the three given points are on the same straight line.

The answer is: (d) 15

The answer is: (b) 1540