243 has been divided into three parts such that half of the first part, one-third of the second part and one-fourth of the third part are equal. The largest part is :
(A) 74
(B) 86
(C) 92
(D) 108
Solution:
Let the three parts be A, B and C $$\eqalign{ & {\text{Let }}\frac{A}{2} = \frac{B}{3} = \frac{C}{4} = x \cr & {\text{Then, }}A = 2x,B = 3x{\text{ and }}C = 4x \cr & {\text{So, }}A:B:C = 2:3:4 \cr & \therefore {\text{Largest part :}} \cr & = \left( {243 \times \frac{4}{9}} \right) \cr & = 108 \cr} $$
43.
In a Mathematics examination the number scored by 5 candidates are 5 successive odd integers. If their total marks are 185, the highest score is :
(A) 39
(B) 43
(C) 41
(D) 47
Solution:
Let the five successive odd number be, x, x + 2, x + 4, x + 6, x + 8 Then, according to given information, 185 = x + x + 2 + x + 4 + x + 6 + x + 8 $$\eqalign{ & \Leftrightarrow 185 = 5x + 20 \cr & \Leftrightarrow 5x = 185 - 20 \cr & \Leftrightarrow 5x = 165 \cr & \Leftrightarrow x = 33 \cr} $$ Highest number = 33 + 8 = 41
44.
A number whose fifth part increased by 4 is equal to its fourth part diminished by 10, is :
(A) 240
(B) 260
(C) 270
(D) 280
Solution:
Let the number be x Then, $$\eqalign{ & \Leftrightarrow \left( {\frac{1}{5}x + 4} \right) = \left( {\frac{1}{4}x - 10} \right) \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$
45.
A number of two digits has 3 for its unit's digit, and the sum of digits is of the itself. The number is :
(A) 43
(B) 63
(C) 53
(D) 73
Solution:
Let the ten's digit be x Then, number = 10x + 3 and Sum of digits = (x + 3) So, (x + 3) = $$\frac{1}{7}$$(10x + 3) ⇔ 7x + 21 = 10x + 3 ⇔ 3x = 18 ⇔ x = 6 Hence, the number is : = (10x + 3) = (10 × 6 + 3) = 63
46.
In a two-digit number, if it is known that its unit's digits 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 :
The sum of three numbers is 264. If the first number be twice the second and third number be be one-third of the first, then the second number is :
(A) 48
(B) 54
(C) 72
(D) 84
Solution:
Let the second number be x Then, first number = 2x, And third number = $$\frac{2x}{3}$$ $$\eqalign{ & \therefore 2x + x + \frac{{2x}}{3} = 264 \cr & \Leftrightarrow \frac{{11x}}{3} = 264 \cr & \Leftrightarrow x = \left( {\frac{{264 \times 3}}{{11}}} \right) \cr & \Leftrightarrow x = 72 \cr} $$
48.
The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
(A) 3
(B) 4
(C) 9
(D) Cannot be determined
Solution:
Let the ten's digit be x and unit's digit be y. Then, (10x + y) - (10y + x) = 36 ⇒ 9(x - y) = 36 ⇒ x - y = 4
49.
If doubling a number and adding 20 to the result gives the same answer as multiplying the number by 8 and taking away 4 from the product, the number is :
(A) 2
(B) 3
(C) 4
(D) 6
Solution:
Let the number be x Then, ⇔ 2x + 20 = 8x - 4 ⇔ 6x = 24 ⇔ x = 4
50.
The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be :
(A)
(B)
(C)
(D)
Solution:
Let the numbers be x and y Then, $$\eqalign{ & x + y = 12\,\, \& \,\, xy = 35 \cr & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{12}}{{35}} \cr} $$