{"id":5901,"date":"2025-06-09T07:14:51","date_gmt":"2025-06-09T07:14:51","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5901"},"modified":"2025-06-09T07:14:51","modified_gmt":"2025-06-09T07:14:51","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-problems-numbers-on","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-problems-numbers-on\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free download: PROBLEMS ON NUMBERS"},"content":{"rendered":"

PROBLEMS <\/strong>ON NUMBERS<\/strong><\/h1>\n

\u00a0<\/strong>In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.<\/p>\n

\u00a0SOLVED EXAMPLES <\/u><\/strong><\/p>\n

Ex.1.\u00a0 A number is as much greater than 36 as is less than 86. Find the number<\/strong>.<\/em><\/p>\n

Sol.<\/strong>\u00a0\u00a0\u00a0 Let the number be x. Then, x – 36 = 86 – x\u00a0 => 2x = 86 + 36 = 122 => x = 61. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0Hence, the required number is 61.<\/p>\n

Ex. 2. Find a number such that when 15 is subtracted from 7 times the number, the<\/h1>\n

Result is 10 more than twice the number<\/strong>.<\/em>\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Hotel Management, 2002)<\/p>\n

Sol.<\/strong>\u00a0\u00a0\u00a0 Let the number be x. Then, 7x – 15 = 2x + 10 => 5x = 25 =>x = 5.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 Hence, the required number is 5.<\/p>\n

Ex. 3. The sum of a rational number and its reciprocal is 13\/6. Find the number.<\/strong><\/p>\n

(S.S.C. 2000)<\/p>\n

Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let the number be x.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, x + (1\/x) = 13\/6 => (x2<\/sup> + 1)\/x = 13\/6 => 6x2<\/sup> \u2013 13x + 6 = 0<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 => 6x2 <\/sup>\u2013 9x \u2013 4x + 6 = 0 => (3x \u2013 2) (2x \u2013 3) = 0<\/p>\n

    \n
  • x = 2\/3 or x = 3\/2<\/li>\n<\/ul>\n

    Hence the required number is 2\/3 or 3\/2.<\/p>\n

    Ex. 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh<\/strong><\/p>\n

    of the other by 8, find the smaller number.<\/strong><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let the numbers be x and (184 – x). <\/em>Then,<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (X\/3) – ((184 \u2013 x)\/7) = 8 => 7x \u2013 3(184 \u2013 x) = 168 => 10x = 720 => x = 72.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0 So, the numbers are 72 and 112. Hence, smaller number = 72.<\/p>\n

    Ex. 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers<\/strong>.<\/em><\/p>\n

    Sol<\/strong>.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let the number be x and y. Then,<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x \u2013 y = 11\u00a0\u00a0\u00a0\u00a0\u00a0 —-(i)\u00a0\u00a0\u00a0\u00a0 and 1\/5 (x <\/em>+ y) <\/em>= 9\u00a0 => x <\/em>+ y <\/em>= 45\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —-(ii)<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, the numbers are 28 and 17.<\/p>\n

    Ex. 6. If the sum of two numbers is 42 and their product is 437, then find the<\/strong><\/p>\n

    absolute\u00a0 difference between the numbers<\/strong>.<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (S.S.C. 2003)<\/p>\n

    Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Let the numbers be x and y. Then, x + y = 42 and xy = 437<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x <\/em>– y <\/em>= sqrt[(x + y)2<\/sup> <\/em>– 4xy] = sqrt[(42)2<\/sup> – 4 x 437 ] = sqrt[1764 \u2013 1748] = sqrt[16] = 4.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 Required difference = 4.<\/p>\n

    Ex. 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the <\/strong><\/p>\n

    numbers.<\/strong><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0 Let the numbers be x and (15 – x).<\/em><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, x2<\/sup> + (15 – x)2<\/sup> = 113\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x2<\/sup> + 225 + X2<\/sup> – 30x = 113\u00a0\u00a0<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2x2<\/sup> <\/em>– 30x + 112 = 0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 x2<\/sup> – 15x <\/em>+ 56 = 0<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (x <\/em>– 7) (x <\/em>– 8) = 0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0 x = 7\u00a0 or\u00a0 x = 8.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, <\/em>the numbers are 7 and 8.<\/p>\n

    Ex. 8. The average\u00a0 of\u00a0 four consecutive even numbers is 27. Find the largest of these<\/strong><\/p>\n

    numbers.<\/strong><\/p>\n

    Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, sum of these numbers = (27 x 4) = 108.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, <\/em>x + (x <\/em>+ 2) + (x <\/em>+ 4) + (x <\/em>+ 6) = 108\u00a0 or\u00a0 4x <\/em>= 96\u00a0 or\u00a0 x = 24.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 :. Largest number = (x <\/em>+ 6) = 30.<\/p>\n

    Ex. 9. The sum of the squares of three consecutive odd numbers is 2531.Find the<\/strong><\/p>\n

    numbers. <\/strong><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let the numbers be x, x + 2 and x + 4.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, X2<\/sup> + (x <\/em>+ 2)2<\/sup> + (x <\/em>+ 4)2<\/sup> = 2531 => 3x2<\/sup> <\/em>+ 12x <\/em>– 2511 = 0<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0 X2<\/sup> + 4x <\/em>– 837 = 0\u00a0\u00a0\u00a0\u00a0 => (x <\/em>– 27) (x <\/em>+ 31) = 0\u00a0\u00a0\u00a0 =>\u00a0\u00a0 x = 27.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, the required numbers are 27, 29 and 31.<\/p>\n

    Ex. 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of\u00a0 the numbers is larger than 6 times their difference by 6, find the two numbers<\/strong>.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0 Let the numbers be x and y, <\/em>such that x > y<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, 3x <\/em>– 4y = 5 …(i)\u00a0 and\u00a0 (x <\/em>+ y) <\/em>– 6 (x <\/em>– y) <\/em>= 6 =>\u00a0 –5x <\/em>+ 7y = 6\u00a0\u00a0\u00a0\u00a0 \u2026(ii)<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Solving (i) and (ii), <\/em>we get: x = 59 and\u00a0 y <\/em>= 43.<\/p>\n

    \u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Hence, the required numbers are 59 and 43.<\/p>\n

    Ex. 11. The ratio between a two-digit number and the sum of the digits of that<\/strong><\/p>\n

    number is 4 : 1.If the digit in the unit’s place is 3 more than the digit in the ten\u2019s place, what is the number?<\/strong><\/p>\n

    Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Let the ten’s digit be x. Then, unit’s digit = (x <\/em>+ 3).<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sum of the digits = x + (x <\/em>+ 3) = 2x <\/em>+ 3. Number = l0x + (x <\/em>+ 3) = llx + 3.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 11x+3 \/ 2x + 3\u00a0 = 4 \/ 1 => 1lx + 3 = 4 (2x + 3)\u00a0\u00a0\u00a0 =>\u00a0 3x = 9\u00a0\u00a0\u00a0\u00a0\u00a0 => x <\/em>= 3.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, required number = 11x <\/em>+ 3 = 36.<\/p>\n

    Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63\u00a0 is subtracted<\/strong><\/p>\n

    from the number, its digits are interchanged. Find the number.<\/strong><\/p>\n

    Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Let the ten’s digit be x. Then, unit’s digit = (9 – x).<\/em><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number = l0x + (9 – x) = 9x <\/em>+ 9.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number obtained by reversing the digits = 10 (9 – x) + x = 90 – 9x.<\/p>\n

    therefore, (9x <\/em>+ 9) – 63 = 90 – 9x<\/em>\u00a0 \u00a0\u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 18x <\/em>= 144\u00a0 \u00a0 =>\u00a0\u00a0\u00a0\u00a0\u00a0 x = 8.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, <\/em>ten’s digit = 8 and unit’s digit = 1.<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, the required number is 81.<\/p>\n

    Ex. 13. A fraction becomes 2\/3 when 1 is added to both, its numerator and denominator. <\/strong><\/p>\n

    And ,it becomes 1\/2 when 1 is subtracted from both the numerator and\u00a0 denominator. Find the fraction.<\/strong><\/p>\n

    Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 Let the required fraction be x\/y. Then,<\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x+1 \/ y+1 = 2 \/ 3\u00a0 =>\u00a0 3x \u2013 2y = – 1\u00a0\u00a0 \u2026(i) and\u00a0\u00a0 x \u2013 1 \/ y \u2013 1 = 1 \/ 2<\/p>\n