Practice MCQ Questions and Answer on Problems on Numbers
61.
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
62.
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
63.
The difference between a number and its three-fifths is 50, What is the number ?
(A) 75
(B) 100
(C) 125
(D) None of these
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - \frac{3}{5}x = 50 \cr & \Leftrightarrow \frac{2}{5}x = 50 \cr & \Leftrightarrow x = \left( {\frac{{50 \times 5}}{2}} \right) \cr & \Leftrightarrow x = 125 \cr} $$
64.
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
(A) 18
(B) 24
(C) 42
(D) 81
Solution:
Let the ten's and unit digit be x and $$\frac{8}{x}$$ respectively $$\eqalign{ & {\text{Then,}} \cr & \left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Rightarrow 9{x^2} + 18x - 72 = 0 \cr & \Rightarrow {x^2} + 2x - 8 = 0 \cr & \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr} $$ ∴ first digit will be 2 and second digit will be 4. i.e digit is 24.
65.
The difference between th of rd of a number and th of th of the same number is 288. What is the number ?
(A) 850
(B) 895
(C) 955
(D) 960
Solution:
Let the number be x Then, $$\eqalign{ & \frac{3}{5}{\text{of }}\frac{2}{3}{\text{of }}x - \frac{2}{5}{\text{of }}\frac{1}{4}{\text{of }}x = 288 \cr & \Leftrightarrow \left( {x \times \frac{3}{5} \times \frac{2}{3}} \right) - \left( {x \times \frac{2}{5} \times \frac{1}{4}} \right) = 288 \cr & \Leftrightarrow \frac{2}{5}x - \frac{1}{{10}}x = 288 \cr & \Leftrightarrow \frac{{3x}}{{10}} = 288 \cr & \Leftrightarrow x = \left( {\frac{{288 \times 10}}{3}} \right) \cr & \Leftrightarrow x = 960 \cr} $$
66.
The sum of five consecutive odd numbers is 575. What is the sum of the next set of five consecutive odd numbers ?
Three numbers are in in the ratio of 3 : 4 : 6 and their product is 1944. The largest of these numbers is :
(A) 6
(B) 12
(C) 18
(D) None of these
Solution:
Let the numbers be 3x, 4x and 6x Then, $$\eqalign{ & \Leftrightarrow 3x \times 4x \times 6x = 1944 \cr & \Leftrightarrow 72{x^3} = 1944 \cr & \Leftrightarrow {x^3} = 27 \cr & \Leftrightarrow x = 3 \cr} $$ ∴ Largest number = 6x = 18
68.
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
(A) 9
(B) 11
(C) 13
(D) 15
Solution:
Let the three integers be x, x + 2 and x + 4 Then, 3x = 2(x + 4) + 3 ⇔ x = 11 ∴ Third integer = x + 4 = 15
69.
The difference between the numerator and the denominator of a fraction is 5. If 5 is added to its denominator, the fraction is decreased by . Find the value of the fraction.
(A)
(B)
(C)
(D)
Solution:
Let the denominator be x Then, numerator = x + 5 Now, $$\eqalign{ & \Leftrightarrow \frac{{x + 5}}{x} - \frac{{x + 5}}{{x + 5}} = \frac{5}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{5}{4} + 1 \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{9}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = 2\frac{1}{4} \cr} $$ So, the fraction is $$2\frac{1}{4}$$
70.
If the product of two numbers is 5 and one of the number is , then sum of two numbers is :
(A)
(B)
(C)
(D)
Solution:
Let two numbers be a and b Given ab = 5 and a = $$\frac{3}{2}$$ $$\eqalign{ & \Rightarrow b = \frac{5}{a} \cr & \Rightarrow b = \frac{5}{{\frac{3}{2}}} \cr & \Rightarrow b = \frac{{5 \times 2}}{3} \cr & \Rightarrow b = \frac{{10}}{3} \cr} $$ Required sum of : ⇒ a + b = $$\frac{{3}}{2}$$ + $$\frac{{10}}{3}$$ L.C.M. of 2 and 3 = 6 ⇒ a + b = $$\frac{{9 + 20}}{6}$$ ⇒ a + b = $$\frac{{29}}{6}$$ ⇒ a + b = $$4\frac{{5}}{6}$$