5 NUMBERS AND REAL NUMBERS. Find the largest 4-digit number which is exactly divisible by 459. Ans.The largest 4-digit natural number = 9999 We divide 9999 by 459 and find the remainder 459 9999 98 89 459 360 Hence, the required number = 9999 360 = 9639. Find the smallest 4-digit number which is exactly divisible by 76. Ans.The smallest 4-digit number = 000. We divide 000 by 76 and find the remainder 76 000 3 76 40 8 Hence, the least number which should be added to 000 so that the sum is exactly divisible by 76 = 76 = 64 Hence, the smallest 5-digit whole number which is exactly divisible by 76 = 000 + 64 = 064. 3. Simplify: (i) ( 47) + 8 + ( 53) + ( 37) + 69 (ii) 69 + ( 689) + ( 54) + 77 + ( 3) Ans.(i) ( 47) + 8 + ( 53) + ( 37) + 69 = 47 + 8 53 37 + 69 = 8 + 69 47 53 37 = (8 + 69) (47 + 53 + 37) = 50 437 = 87 Math Class VIII Question Bank
(ii) 69 + ( 689) + ( 54) + 77 + ( 3) = 69 689 54 + 77 3 = 69 + 77 689 54 3 = (69 + 77) (689 + 54 + 3) = 46 766 = 480 4. A shopkeeper bought a pen for Rs 75, a book for Rs 40 and a pencil box for Rs 46. He sold the pen for Rs 68, book for Rs 50 and the pencil box for Rs 5. What was his gain or loss? Ans. Cost price of pen = Rs 75 Cost price of book = Rs 40 and cost price of pencil box = Rs 46 Total cost price = Rs (75 + 40 + 46) = 36 Selling price of pen = Rs 68 Selling price of pen = Rs 50 and selling price of pecil box = Rs 5 Total selling price = Rs (68 + 50 + 5) = 370 Thus, total selling price > cost price Hence, gain = Rs 370 Rs 36 = Rs 9 5. A man starts from his home and drives 48 km to the North and then 7 km to the South. How far is he from his home finally? Ans. Goes to North = 48 km Then back to South = 7 km Hence, 7 48 = 4 km to South from his home. 6. From 38 subtract the sum of ( 93) and 67. Ans. 38 ( 93 + 67) = 38 + 93 67 = 38 67 + 93 = 709 + 93 = 04 7. Subtract the sum of 73 and ( 5) from the sum of ( 486) and 597. Ans.Sum of ( 486) and 597 = 486 + 597 = Sum of 73 and 5 = 73 5 = 39 Hence, ( 39) = + 39 = 350 Math Class VIII Question Bank
8. Simplify: (i) [ ( 54)] + [ 33 ( 97)] (ii) [9 ( )] + [45 ( 87)] Ans.(i) [ ( 54)] + [ 33 ( 97)] = [ + 54] + [ 33 + 97] = + 54 33 + 97 = 54 + 97 33 = (54 + 97) ( + 33) = 5 54 = 97 (ii) [9 ( )] + [45 ( 87)] = (9 + ) + (45 + 87) = 4 + 3 = 73 9. During a month, the average day temperature of a desert region was 37 C and the average night temperature was 4 C. Find the difference between the average day temperature and the average night temperature. Ans. Average day temperature = 37 C and average night temperature = 4 C Thus, difference between the average day and average night temperature = 37 C ( 4 C) = 37 C + 4 C = 4 C 0. Simplify: (i) 4 [6 {9 (4 8 4)}] (ii) 47 [9 {6 3 (6 9 3) 3)}] (iii) 53 [8 {6 (37 + 3 35)}] (iv) [67 ( 3) {6 (56 33)}] [3 + 4 {5 + ( 3) ( )} 4] Ans.(i) 4 [6 {9 (4 8 4)}] = 4 [6 {9 (4 8 + 4)}] = 4 [6 {9 4 + 8 4}] = 4 [6 9 + 4 8 + 4] = 4 6 + 9 4 + 8 4 Math Class VIII 3 Question Bank
= 4 + 9 + 8 6 4 4 = (4 + 9 + 8) (6 + 4 + 4) = 5 34 = 7 (ii) 47 [9 {6 3 (6 9 3) 3)}] = 47 [9 { (6 3) 3}] = 47 [9 { 3 3}] = 47 [9 { }] = 47 [9 ] = 47 8 = 9 (iii) 53 [8 {6 (37 + 3 35)}] = 53 [8 {6 (37 + 6 35)}] = 53 [8 {6 8}] = 53 [8 ] = 53 6 = 7 (iv) [67 ( 3) {6 (56 33)}] [3 + 4 {5 + ( 3) ( )} 4] = [67 + 3{6 3}] [3 + 4 {5 + 6} 4] = [67 + ( ) ] [3 + 44 4] = [67 ] [3] = 46 3 =. Simplify: (i) 0 85 35 + 85 (ii) 7 {39 + (38 ) (04 9)} (iii) 9 (9 30 30) (iv) 56 [45 {56 (45 56 45)}] (v) 84 {7 {7 + (7 44 43)}] Ans.(i) 0 85 35 + 85 = 0 85 + 35 + 85 = 330 85 = 45 (ii) 7 {39 + (38 ) (04 9)} = 7 {39 + (6) (3)} = 7 {39 + 338} = 7 {377} = 6409 (iii) 9 (9 30 30) = 9 (9 30 + 30) = 9 (9) = 9 9 = 00 Math Class VIII 4 Question Bank
(iv) 56 [45 {56 (45 56 45)}] = 56 [45 {56 (45 56 + 45)}] = 56 [45 {56 (90 56)}] = 56 [45 {56 34}] = 56 [45 {}] = 56 [45 ] = 56 3 = 33 (v) 84 {7 {7 + (7 44 43)}] = 84 [7 {7 + (7 44 + 43)}] = 84 [7 {7 + (5 44)] = 84 [7 {7 + 7}] = 84 [7 {43}] = 84 [7 43] 84 = 84 9 = 84 = = 9 9 9. Simplify: (i) 9 [6 {7 + 3 (8 9 6 )}] (ii) [9 {4 (9 4 )}] (4 5 + ) (iii) 4 3 9 3 + (5 8 4) 6 (iv) 9 [48 6 7 {5 3 (5 3 8)}] Ans.(i) 9 [6 {7 + 3 (8 9 6 )}] = 9 [6 {7 + 3 (8 9 + 6)}] = 9 [6 {7 + 3 (34 9)}] = 9 [6 {7 + 45}] = 9 [6 {6}] = 9 [6 6] = 9 6 + 6 = 9 6 = 75 (ii) [9 {4 (9 4 )}] (4 5 + ) = [9 {4 (9 4 + )}] (4 5 ) = [9 {4 (9 )}] (4 7) = [9 {4 (7)}] ( 3) = [9 {4 7}] + 3 = [9 { 3}] + 3 = [9 + 3] + 3 = + 3 = 5 Math Class VIII 5 Question Bank
(iii) 4 3 9 3 + (5 8 4) 6 = 4 3 9 3 + 48 6 = 4 9 3 + 48 3 6 = 8 7 + 8 = 6 7 = (iv) 9 [48 6 7 {5 3 (5 3 8)}] = 9 [48 6 7 {5 3 (5 3 + 8)}] = 9 [48 6 7{5 3 (3 3)}] = 9 [48 6 7 {5 3 0}] = 9 48 6 7 5 0 3 = 9 [48 6 7 {0 0}] = 9 [48 6 7{0}] = 9 [48 6 0] = 9 48 0 6 = 9 [8 0] = 9 [8] = 9 8 = 3. Simplify: (i) 8 45 6 of 3 (ii) 57 9 64 3 + 96 6 of 4 Ans.(i) 8 45 6 of 3 = 8 45 8 = 8 45 = 8 45 8 8 = 0 = 8 (ii) 57 9 64 3 + 96 6 of 4 = 57 9 64 3 + 96 4 = 57 64 + 96 9 3 4 57 64 96 = + 9 3 4 = 6 4 + 4 = 0 4 = 6 Math Class VIII 6 Question Bank
4. (i) When a number is divided by 4; the quotient is 5 and the remainder is 7. Find the number. (ii) When 5 is divided by 3; the remainder is 4. Find the quotient. (iii) When 49 is divided by a certain number, the quotient is 8 and the remainder is 9. Find the number. Ans.(i) Let the required number be x. We know that Dividend = Quotient Divisor + Remainder x = 5 4 + 7 x = 360 + 7 x = 367 Hence, required number is 367. (ii) Let the quotient be x. We know that Quotient Divisor + Remainder = Dividend x 3 + 4 = 5 3x + 4 = 5 3x = 5 4 3x = x = 3 x = 7 Hence, required quotient is 7. (iii) Let the required number be x. We know that Quotient Divisor + Remainder = Dividend 8 x + 9 = 49 8x + 9 = 49 8x = 49 9 8x = 40 x = 40 x = 5 8 Hence, required number is 5. 5. Insert one rational number between (i) 3 and 6 (ii) 3 and 5 7 (iii) 8 9 and 5 9 Math Class VIII 7 Question Bank
Ans.If a and b are two rational numbers, then (a + b) isa rational number between a and b. (i) A rational number between 3 and 6 is ( 3 + 6 ) = 9. We have, 3 < 9 < 6. Thus, a rational number between 3 and 6 is 9. (ii) A rational number between 3 and 5 7 is + 5 3 7 4 + 5 = = 9 4. We have, < 9 < 5. 3 4 7 Hence, a rational number between 3 and 5 7 is 9. 4 (iii) A rational number between 8 9 and 5 8 4 = and 9 9 9 is 8 4 8 4 + = =. We have, < < 9 9 9 9 9 9 9 Hence, a rational number between 8 9 and 5 is. 9 9 6. Insert two rational numbers between: (i) 4 and 6 (iii) and Math Class VIII 8 Question Bank (ii) 3 3 (iv) and 3 and 4 + 6 = 5. Ans. (i) A rational number between 4 and 6 is We have, 4 < 5 < 6. 5 + 6 =. A rational number between 5 and 6 is
Thus, we have 4 < 5 < < 6. Thus, two rational numbers between 4 and 6 are 5 and. (ii) A rational number between and 3 is 3 + 4 7 + = = 3 6 We have, < 7 <. 3 A rational number 7 and 3 is 7 7 + 8 5 5 + = = =. 3 4 Thus, we have < 7 < 5 <. 4 3 Hence, two rational numbers between and 3 are 7 (iii) A rational number between and 3 is 3 5 0 5 + 0 35. + = = 3 6 We have, 5 < 35 < 0. 3 A rational number between 35 and 0 3 is 35 0 35 + 40 75. + = = 3 4 and 5 4. Thus, we have 5 < 35 < 75 < 0. 4 3 Hence, two rational numbers between 5 and 0 3 35 75 are and 4 Math Class VIII 9 Question Bank
(iv) A rational number between and is +. + = = 4 We have, < <. 4 A rational number between and is 4 4 5 +. + = = 4 4 8 5 Thus, we have < < <. 4 8 Hence, two rational numbers between and are 4 and 5 8. 7. Rationalise the denominators of: (i) (iv) (vii) 6 3 5 3 6 3 Ans.(i) Given fraction is (ii) (v) 4 3 6 Math Class VIII 0 Question Bank 5 + 3 The rationalising factor of 6 is 6.. (iii) (vi) 3 3 Multiplying and dividing the given fraction by 6, we have, 6 = 6 6 6 (ii) Given fraction is 4 3 = 6 = 6 6 3
The rationalising factor of is. Multiplying and dividing the given fraction by, we have, 4 3 = 4 6 6 = (iii) Given fraction is 3 The rationalising factor of 3 is 3 Maltiplying and dividing the given fraction by 3, 3 3 3 we have, = = 3 3 3 6 (iv) Given fraction is 3 5 The rationalising factor of 3 5 is 3 + 5 Multiplying and dividing the given fraction by 3 + 5 ( 3 + 5 ) ( 3 5)( 3 + 5 ) ( 3) ( 5) (v) Given fraction is 3 + 5 3 + 5 3 + 5 = = = 9 5 4 5 + 3 The rationalising factor of ( 5 + 3) is ( 5 3) Multiplying and dividing the given fraction by ( 5 3 ) we have, ( 5 3) 5 3 5 3 5 3 = = ( 5 + 3) 5 3 5 3 5 3 = = 5 3, Math Class VIII Question Bank
(vi) Given fraction is 3 The rationalising factor of ( 3 ) is ( 3 + ) Multiplying and dividing the given fraction by ( 3 + ). We have, ( 3 + ) ( 3 )( 3 + ) (vii) Given fraction is = ( 3) ( + ) 3 + 3 + = 3 = = 3 + 3 6 3 The rationalising factor of ( 6 3) is ( 6 + 3) Multiplying and dividing the given fraction by ( 6 + 3). We have, 3( 6 + 3) ( 6 3)( 6 + 3) 3( 6 + 3) = 3( 6 + 3) ( 6 ) ( 3) = 6 3 3 6 + 3 8 + 3 = = 3 3 ( + ) 9 + 3 3 + 3 3 = = = = + 3 3 3 Math Class VIII Question Bank