## Aptitude question and answer on Problems on numbers

 1 The sum of two numbers is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number. A.    69 B.    70 C.    71 D.    72 Answer : D.  72 Explanation: Let the numbers be x and (184 - x). Then, ${X \over3 }- {184-x \over7} = 8$ => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72. So, the numbers are 72 and 112. Hence, smaller number = 72.   Discuss 2 The ratio between a two-digit number and the sum of the digits of that number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?   A.    34 B.    35 C.    36 D.    37 Answer : C.  36 Explanation: Let the ten's digit be x. Then, unit's digit = (x + 3). Sum of the digits = x + (x + 3) = 2x + 3. Number = l0x + (x + 3) = llx + 3. ${{11x+3} \over {2x + 3 } }= {4 \over 1 }$ => $11x + 3 = 4 (2x + 3)$  =>  3x = 9      => x = 3. Hence, required number = 11x + 3 = 36.   Discuss 3 The sum of two numbers is 15 and the sum of their squares is 113. Find the numbers.   A.    6 and 9 B.    7 and 8 C.    4 and 11 D.    10 and 5 Answer : B.   7 and 8 Explanation: Let the numbers be x and (15 - x).           Then,$x^2 + (15 - x)^2 = 113?$           =>         $x^2 + 225 + x^2 - 30x = 113$    =>       $2x^2 - 30x + 112 = 0$         =>        $x^2 - 15x + 56 = 0$ =>       (x - 7) (x - 8) = 0               =>      x = 7  or  x = 8.            So, the numbers are 7 and 8.   Discuss 4 If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the numbers  A.    6, 4 and 15 B.    7, 4 and 14 C.    6, 2 and 17 D.    6, 5 and 13 Answer : A.  6, 4 and 15 Explanation: Let the numbers be x, y and z. Then, x+ y = 10      ...(i)      y + z = 19               ...(ii)       x + z = 21       …(iii) Adding (i) ,(ii) and (iii), we get:  2 (x + y + z ) = 50  or   (x + y + z) = 25. Thus, x= (25 - 19) = 6;  y = (25 - 21) = 4;  z = (25 - 10) = 15. Hence, the required numbers are 6, 4 and 15.   Discuss 5 50 is divided into two parts such that the sum of their reciprocals is $1\over 12$.Find the two parts. A.    10 and 40 B.    15 and 35 C.    45 and 5 D.    30 and 20 Answer : D.  30 and 20 Explanation: Let the two parts be x and  (50 - x). Then, 1 / x + 1 / (50 – x) =  $1 \over 12$ => (50 – x + x) / x ( 50 – x) = $1 \over 12$ => x2 – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20. So, the parts are 30 and 20.   Discuss 6 The average  of  four consecutive even numbers is 27. Find the largest of these numbers.   A.    29 B.    30 C.    31 D.    32 Answer : B.   30 Explanation: Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6. Then, sum of these numbers = (27 x 4) = 108. So, x + (x + 2) + (x + 4) + (x + 6) = 108  or  4x = 96  or  x = 24. $\dot{..}$Largest number = (x + 6) = 30.   Discuss 7 A number is as much greater than 36 as is less than 86. Find the number. A.    71 B.    61 C.    51 D.    41 Answer : B.   61 Explanation: Let the number be x. Then, x - 36 = 86 - x  => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.   Discuss 8 Find a number such that when 15 is subtracted from 7 times the number, the Result is 10 more than twice the number.  A.    5 B.    6 C.    7 D.    8 Answer : A.  5 Explanation: Let the number be x. Then, 7x - 15 = 2x + 10 => 5x = 25 =>x = 5. Hence, the required number is 5.   Discuss