Practice MCQ Questions and Answer on Problems on Numbers
31.
The digit in the unit's place of a numbers is equal to the digit in the ten's place of half of that number and the digit in the ten's place of that number is less than the digit in unit's place of half of the number by 1. If the sum of the digits of the number is 7, then what is the number ?
(A) 34
(B) 52
(C) 162
(D) Data inadequate
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \frac{{10x + y}}{2} = 10y + \left( {x + 1} \right) \cr & \Leftrightarrow 10x + y = 20y + 2x + 2 \cr & \Leftrightarrow 8x - 19y = 2.....(i) \cr & {\text{And}} \Leftrightarrow x + y = 7.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 5, y = 2 Hence, required number = 52
32.
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
(A) 20
(B) 30
(C) 40
(D) None of these
Solution:
Let the numbers be a, b and c Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131 (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400 ⇒ (a + b + c) = $$\sqrt {400} $$ = 20
33.
The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number ?
(A) 74
(B) 82
(C) 83
(D) Cannot be determined
Solution:
Answer & Solution Answer: Option E Solution: Let the three odd numbers be x, (x + 2), (x + 4) and The three even numbers be (x + 11), (x + 13) and (x + 15) Then, ⇔ x + (x + 2) + (x + 4) + (x + 11) + (x + 13) + (x + 15) = 231 ⇔ 6x + 45 = 231 ⇔ 6x = 186 ⇔ x = 31 ∴ Required sum : = (x + 4) + (x + 15) = 2x + 19 = 2 × 31 + 19 = 62 + 19 = 81
34.
The sum of the squares of three numbers is 138, while the sum of their product taken two at a time is 131. Their sum is :
(A) 20
(B) 30
(C) 40
(D) None of these
Solution:
Let the numbers be a, b and c. Then, a2 + b2 + c2 = 138 And (ab + bc + ca) = 131 ∴ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 × 131 = 400 ⇒ (a + b + c) = $$\sqrt {400} $$ = 20
35.
The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :
(A) 380
(B) 395
(C) 400
(D) 425
Solution:
Let the numbers be x and y Then, $$xy = 9375{\text{ and }}\frac{x}{y} = 15$$ $$\eqalign{ & \Leftrightarrow \frac{{xy}}{{\left( {\frac{x}{y}} \right)}} = \frac{{9375}}{{15}} \cr & \Leftrightarrow {y^2} = 625 \cr & \Leftrightarrow y = 25 \cr & \therefore x = 15y \cr & \Rightarrow x = 15 \times 25 \cr & \Rightarrow x = 375 \cr} $$ ∴ Sum of the numbers : = 375 + 25 = 400
36.
The sum of seven consecutive numbers is 175. What is the difference between twice the largest number and thrice the smallest number ?
The sum of four numbers is 64. If you add 3 to the first number, 3 is subtracted from the second number, the third is multiplied by 3 and the fourth is divided by 3, then all the results are equal. What is the difference between the largest and the smallest of the original numbers ?
(A) 21
(B) 27
(C) 32
(D) Cannot be determined
Solution:
Let the four numbers be , A, B, C and D Let A + 3 = B - 3 = 3C = $$\frac{D}{3}$$ = x Then, A = x - 3 B = x + 3 C = $$\frac{x}{3}$$ D = 3x $$\eqalign{ & \Leftrightarrow A + B + C + D = 64 \cr & \Leftrightarrow \left( {x - 3} \right) + \left( {x + 3} \right) + \frac{x}{3} + 3x = 64 \cr & \Leftrightarrow 5x + \frac{x}{3} = 64 \cr & \Leftrightarrow 16x = 192 \cr & \Leftrightarrow x = 12 \cr} $$ Thus, the numbers are 9, 15, 4 and 36 ∴ Required difference : = (36 - 4) = 32
38.
Three-forth of a number is 60 more than its one-third. The number is :
(A) 84
(B) 108
(C) 144
(D) None of these
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{3}{4}x - \frac{1}{3}x = 60 \cr & \Leftrightarrow \frac{{5x}}{{12}} = 60 \cr & \Leftrightarrow x = \left( {\frac{{60 \times 12}}{5}} \right) \cr & \Leftrightarrow x = 144 \cr} $$
39.
In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
(A) 24
(B) 26
(C) 42
(D) 46
Solution:
Let the ten's digit be x Then, unit's digit = x + 2 Number = 10x + (x + 2) = 11x + 2 Sum of digits = x + (x + 2) = 2x + 2 ∴ (11x + 2)(2x + 2) = 144 ⇒ 22x2 + 26x - 140 = 0 ⇒ 11x2 + 13x - 70 = 0 ⇒ 11x2 + (35 - 22)x - 70 = 0 ⇒ 11x2 + 35x - 22x - 70 = 0 ⇒ (x - 2)(11x + 35) = 0 ⇒ x = 2 Hence, required number = 11x + 2 = 24
40.
The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:
(A) 20
(B) 23
(C) 169
(D) None of these
Solution:
Let the numbers be x and y Then, xy = 120 and x2 + y2 = 289 ∴ (x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529 ∴ x + y = $$\sqrt {529} $$ = 23