Practice MCQ Questions and Answer on Problems on Numbers
1.
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
2.
A number consists of two digits. If the digits interchange place and the new number is added to the original number, then the resulting number will be divisible by :
(A) 3
(B) 5
(C) 9
(D) 11
Solution:
Let the ten's digit be x and unit's digit be y Then, number = 10x + y Number obtained by interchanging the digits = 10y + x ∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11
3.
Out of six consecutive natural numbers if the sum of first three is 27, what is the sum of the other three ?
(A) 24
(B) 25
(C) 35
(D) 36
Solution:
Let the six numbers be, x , (x + 1), (x + 2), (x + 3), (x + 4) and (x + 5) Then, ⇔ x + (x + 1) + (x + 2) = 27 ⇔ 3x + 3 = 27 Required sum : = (x + 3) + (x + 4) + (x + 5) = 3x + 12 = (3x + 3) + 9 = 27 + 9 = 36
4.
What is the sum of two consecutive even numbers, the difference of whose squares is 84?
(A) 34
(B) 38
(C) 42
(D) 46
Solution:
Let the numbers be x and x + 2. Then, (x + 2)2 - x2 = 84 ⇒ 4x + 4 = 84 ⇒ 4x = 80 ⇒ x = 20. ∴ The required sum = x + (x + 2) = 2x + 2 = 42
5.
The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are :
(A) 25, 36
(B) 25, 46
(C) 35, 46
(D) None of these
Solution:
Let the numbers be x and y. Then, x2 + y2 = 3341..... (i) And, x2 - y2 = 891..... (ii) Adding (i) and (ii), we get : 2x2 = 4232 or x2 = 2116 or x = 46 Subtracting (ii) from (i), we get : 2y2 = 2450 or y2 = 1225 or y = 35 So, the numbers are 35 and 46
6.
If the numerator of a fraction is increased by and the denominator is decreased byy , the new fraction obtained is . What was the original fraction ?
(A)
(B)
(C)
(D) Cannot be determined
Solution:
Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{{x + \frac{1}{4}}}{{y - \frac{1}{3}}} = \frac{{33}}{{64}} \cr & \Leftrightarrow \frac{{3\left( {4x + 1} \right)}}{{4\left( {3y - 1} \right)}} = \frac{{33}}{{64}} \cr & \Leftrightarrow \frac{{4x + 1}}{{3y - 1}} = \frac{{33}}{{64}} \times \frac{4}{3} \cr & \Leftrightarrow \frac{{4x + 1}}{{3y - 1}} = \frac{{11}}{{16}} \cr & \Leftrightarrow 16\left( {4x + 1} \right) = 11\left( {3y - 1} \right) \cr & \Leftrightarrow 64x + 16 = 33y - 11 \cr & \Leftrightarrow 64x - 33y = - 27 \cr} $$ Which cannot be solved to find $$\frac{x}{y}$$ Hence, the original fraction cannot be determined from the given data.
7.
The difference between two integers is 5. Their product is 500. Find the numbers.
(A) 15, 20
(B) 20, 25
(C) 30, 25
(D) 21, 26
Solution:
Let the integers be x and (x + 5) Then, $$\eqalign{ & \Leftrightarrow x\left( {x + 5} \right) = 500 \cr & \Leftrightarrow {x^2} + 5x - 500 = 0 \cr & \Leftrightarrow \left( {x + 25} \right)\left( {x - 20} \right) = 0 \cr & \Leftrightarrow x = 20 \cr} $$ So, the numbers are 20 and 25
8.
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
9.
There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number ?
(A) 12
(B) 14
(C) 32
(D) 35
Solution:
Let the numbers be x and y Then, 2x + 3y = 100 ..... (i) And, 3x + 2y = 120 ..... (ii) Adding (i) and (ii), we get : 5x + 5y = 220 or x + y = 44 ..... (iii) Subtracting (i) from (ii): we get : x - y = 20 ..... (iv) Adding (iii) and (iv), we get : 2x = 64 or x = 32 Putting x = 32 in (iii), we get : y = 12 Hence, larger number = 32
10.
A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?
(A) 23
(B) 27
(C) 32
(D) 72
Solution:
Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Rightarrow 10x + y = 3\left( {x + y} \right) \cr & \Rightarrow 7x - 2y = 0.....(i) \cr & 10x + y + 45 = 10y + x \cr & \Rightarrow y - x = 5.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 2 and y = 7 ∴ Required number = 27