Quantitative Aptitude

General Questions

Quantitative Aptitude Exercise Mode

General Questions

Practice and master this topic with our carefully crafted questions.

10 Questions
15 Minutes
0% Completed
QUEST ? ✓ !
Question 11
The difference between a two-digit number and the number obtained by interchanging the two digits is 63. Which is the smaller of the two numbers ?

A
29
B
70
C
92
D
Cannot be determined
E
None of these
Correct Answer: Option D

Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 63
<=> 9(x - y) = 63
<=> x - y = 7.
Thus, none of the numbers can be determined.

Question 12
The sum of the digits of a two-digit number is 1/5 of the difference between the number and the number obtained by interchanging the position of the digits. What is definitely the difference between the digits of that number ?

A
5
B
7
C
9
D
Data inadequate
E
None of these
Correct Answer: Option D

Let the ten's digit be x and unit's digit be y.
Then, x + y = 1/5 [(10x + y) - (10y + x)]
⇒ 5x + 5y = 9x - 9y
⇒ 4x = 14y.
Thus, the value of (x - y) cannot be determined from the given data.

Question 13
The difference between a two-digit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

A
4
B
16
C
8
D
None of these
Correct Answer: Option C

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

9x = 36

x = 4.

Required difference = (2x + x) - (2x - x) = 2x = 8.

Question 14
In a two-digit number, 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
Correct Answer: Option A

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

(x - 2)(11x + 35) = 0

x = 2.

Hence, required number = 11x + 2 = 24.

Question 15
If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true ?

A
Sum of the digits is a two-digit number.
B
Digit in the unit's place is twice the digit in the ten's place.
C
Digits in the unit's place and the ten's place are equal.
D
Digit in the unit's place is half of the digit in the ten's place.
E
None of these.
Correct Answer: Option B

Let the ten's digit be X and unit's digit be Y.

Then, 10 x 2X + Y/2 = 10Y + X

⇒ 20X - X = 10Y - Y/2

⇒ 19X = 19Y/2

⇒ Y = 2X

Thus, the unit's digit is twice the ten's digit.

Question 16
If the number obtained on interchanging the digits of a two-digit number is 18 more than the original number and the sum of the digits is 8, then what is the original number ?

A
26
B
35
C
53
D
Cannot be determined
E
None of these
Correct Answer: Option B

Let ten's digit = x Then, unit's digit = (8 - x).
∴ [10(8 - x) + x] - [10x + (8-x)] = 18
<=> 18x = 54
<=> x = 3.
So, ten's digit = 3 and unit's digit = 5.
Hence, original number = 35.

Question 17
A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is :

A
145
B
253
C
370
D
352
Correct Answer: Option B

Let the middle digit be x. Then, 2x = 10 or x = 5.
So, the n umber is either 253 or 352.
Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.
Hence, required number = 253.

Question 18
Of the three numbers, the sum of the first two is 45; the sum of the second and the third is 55 and the sum of the third and thrice the first is 90. The third number is :

A
3
B
20
C
30
D
25
Correct Answer: Option C

Let the number be x, y and z.

Then, x + y = 45, y + z = 55 and 3x + z =90

∴ y = 45 - x,
⇒ z = 55 - y = 55 - (45 - x) = 10 + x.
⇒ 3x + 10 + x = 90 or x = 20.
⇒ y = (45 - 20) = 25 and
∴ z = (10 + 20) = 30.
Third number = 30.

Question 19
The sum of three numbers is 136. If the ratio between first and second be 2 : 3 and that between second and third is 5 : 3, then the second number is :

A
40
B
48
C
72
D
60
Correct Answer: Option D

A : B = 2 : 3 and

B : C = 5 : 3 = 3/5 x 5 : 3/5 x 3 = 3 : 9/5.

So, A : B : C = 2 : 3 : 9/5 = 10 : 15 : 9.

∴ Second number = (136 x 15/34) = 60.

Question 20
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
Correct Answer: Option A

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) = 400 = 20

Previous 1 2 3 Next
Page 2 of 3