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1 PROBLEMS ON DIGITS / NUMBERS / FRACTIONS Ex-1: The sum of five consecutive positive integers is 55. The sum of the squares of the extreme terms is (1) 308 (2) 240 (3) 250 (4) 180 Let the five consecutive in integers be x, x + 1, x + 2, x + 3 and x + 4 (x) + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 55 5x + 10 = 55 5x= 45 x = 9 The numbers are 9, 10, 11, 12 &13 Sum of squares of extremes = = = 250 Choice (3) Ex-2: The sum of two numbers is 18. The difference of the numbers is 4. Find the numbers. (1) 11, 7 (2) 12, 8 (3) 16, 2 (4) 8, 10 Let the two numbers be x and y. x + y = 18 x y = 4 (i) (ii) Add equations (i) and (ii), we get x+ y = 18 x y = 4 2x = 22 x = 11 Substitute x = 11 in equation (i), (11) + y = 18 y = 7 You can solve these types of questions by observation. Choice (1)
2 Ex-3: If two is subtracted from the denominator of a fraction, the fraction becomes 1/3. Instead, if two is added to the numerator and one is subtracted from the denominator, the fraction becomes 1/2. Find the fraction. (1) 6/17 (2) 7/18 (3) 4/15 (4) 3/11 Let the fraction be N D N 1 From statement one = D 2 3 3N D = -2 From 2 nd statement (i) N = D 1 2 2N D = -5 (ii) (i) (ii), we get 3N D = -2 2N D = N = 3 Substitute N = 3 in equation (i) 3 (3) D = -2 9 D = -2 D = 11 N 3 = D 11 Choice (4) Ex- 4: The sum of a two- digit number and the number formed by reversing the digits is 55. Find the number, if one of the digits is one more than the other. (1) 32 (2) 23 (3) 32 (or) 23 (4) None of these
3 Let the two-digit number be xy. Its numerical value is 10x + y. Number formed by reversing the digits is yx. Its numerical value is 10y + x 10x + y + 10y + x = 55 (given) 11x + 11y = 55 11(x + y) = 55 x+ y = 5 (i) As one digit is one more than the other Let x = y + 1 (or) Let y = x + 1 From equation (i) From equation (i) y y = 5 x + x + 1 = 5 2y = 4 2x = 4 x = 3 x = 2 xy = 32 xy = 23 The number can be either 32 (or) 23. Choice (3) Ex- 5: A two-digit number is such that twice the ten s digit added to eleven times the units digit is equal to the number itself. Find the number. (1) 48 (2) 86 (3) 73 (4) 54 Let the two-digit number be xy Its numerical value is 10x + y 2x + 11y = 10x + y (given) 8x = 10y x = = x: y= 5:4 y As 54 is the only number where the above condition is satisfied, with the given ratio, the number must be 54 uniquely. Choice (4) Ex-6: In a three digit number, the middle digit equals the average of its extreme digits. The sum of its digits is 9. How many possibilities can it take? (1) 8 (2) 6 (3) 7 (4) 5
4 Let the number be xyz. x+ z y= x+ z= 2 y ( i) 2 And x + y + z = 9 (ii) Substitute equation (i) in equation (ii) 2y + y = 9 3y = 9 y = 3 x + z = 6 Various possibilities for (x, z) are (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) and (6, 0) The number has 6 possibilities. {x 0; so (0,6) is not possible} Choice (2) Ex-7: A three digit number is equal to 17 times the sum of the digits. If 198 is added to the number, the digits get reversed; also the sum of the extreme digits of the original number is less than the middle digit by unity. Find the sum of digits of the original number. (1) 9 (2) 8 (3) 7 (4) 10 Let the 3-digit number be xyz Its numerical value is 100x+ 10y+ z Number formed by reversing the digits is xyz Its numerical value is 100z + 10y + x 100x + 10y + z = 17 (x+ y + z) (given) 83x 7y 16z = 0 (i) Also (100 x + 10y +z) = 100z + 10y + x 99z - 99x = 198 z - x = 2 (ii)
5 The sum of the extreme digits of the original number is less than the middle digit by unity, x + z = y - 1 (iii) Hence, Z = x + 2 & y = 2x + 3 [from (ii) and (iii)] Substitute y and z values in equation (i) 83x 7 (2x + 3) 16 (x + 2) = 0 83x 14x 21 16x 32 = 0 53x = 53 x= 1 y = 2(1) + 3 = 5 z = (1) + 2 = 3 The sum of the digits of the original number would be x + y + z = = 9 Choice (1) Ex-8: A three digit number is such that the sum of its digits is 17. The middle digit of the number is 6. If 297 is subtracted from the number, the hundred s digit and the unit s digit of the number are interchanged. Find the number. (1) 863 (2) 962 (3) 562 (4) 764 Let the three digit number be x6y. (As middle digit is given) Its numerical value is 100x y. Number formed by reversing its digits is y6x. Its numerical value is 100y x x y = 17 (given) x + y = 11 (i) Also, 100x y 297 = 100y x 99x - 99y = 297 x - y = 3 (ii)
6 Add equations (i) and (ii), we get x + y = 11 x y = 3 2x = 14 x = 7 From equation (i), (7) + y = 11 y = 4. The number is 764. Choice (4) NOTE: This problem can be solved by observation. Ex-9: Manish was asked to find 5/6 times a number and he instead multiplied it by 6/5. As a result, he got an answer which was more than the correct answer by 649. What was the number? (1) 1870 (2) 1770 (3) 1860 (4) 1760 Let the number be x. 6 5 x x= x = 649 x= x= The number was Choice (2) Ex-10: A two-digit number is formed by either subtracting 17 from nine times the sum of the digits or by adding 21 to 13 times the difference of the digits. Find the number. (1) 37 (2) 73 (3) 71 (4) cannot be determined Let the two-digit number be xy. Its numerical value is 10x+ y
7 10x + y = 9 (x + y) 17 x 8y = -17 (i) Also, 10x + y = 13 (x - y) + 21 (Here we taken as only (x-y ), because (y-x) gives factional values.) 3x - 14y = -21 (ii) (i) 3 (ii) 1, we get (i) 3 3x - 24y = - 51 (ii) 1 3x - 14y = y = -30 y = 3 Substitute y value in equation (i), x 8(3) = - 17 x = x = 7 The required number is 73. Choice (2)
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