We will learn how to solve different types of examples on digits and numbers.
1. The sum of a 2digit number & the number formed by interchanging the digits of the original (2digit number) number is divisible by
(a) 11
(b) 9
(c) 5
(d) 3
Solution:
(10a + b) + (10b + a) = 11(a + b)
Therefore, 11(a + b) must be divisible by 11.
Answer: (a)
Note: Any twodigit number and the number obtained by interchanging its digit:
⟹ Their sum is divisible by 11.
⟹ Their difference is divisible by 9.
2. The product of two positive integers is 24. The greatest number is 1 ½ times the smaller number. The difference of the numbers is
(a) 6
(b) 4
(c) 2
(d) 1
Solution:
Ratio of greater to smaller number = 3/2 = 3 : 2
Therefore, 3x × 2x = 24
or, 6x\(^{2}\) = 24
or, x\(^{2}\) 4
or, x = 2
Therefore, the required difference = (3x  2x) = 2
Answer: (c)
3. Find the sum of all 4digit numbers which are formed by the digits 1, 2, 3 and 4 only once?
(a) 66666
(b) 66662
(c) 66661
(d) 66660
Solution:
The required sum = 6666 × (1 + 2 + 3 + 4) = 66660
Answer: (d)
Note: Sum of all four digit numbers by using four difference digits (other than zero) = 6666 × Sum of digits
4. The number of digits in (125\(^{10}\) × 8\(^{9}\)) is:
(a) 19
(b) 28
(c) 29
(d) 30
Solution:
(125\(^{10}\) × 8\(^{9}\))
= 125(125 × 8)\(^{9}\)
= 125 × (1000)\(^{9}\)
= 125 × (10^3)\(^{9}\)
= 125 × (10)\(^{27}\)
Therefore, the required number of digits = 3 + 27 = 30
Answer: (d)
5. There are three consecutive positive integers. The difference of the squares of the extreme integers is 88. What is the average of three integers?
(a) 11
(b) 22
(c) 44
(d) None of these
Solution:
Of three consecutive positive integers, the difference of squares of two extreme integers = 88
Therefore, the average of three numbers = 88 ÷ 4 = 22
Answer: (b)
Note: If a, b and c are three consecutive integers, then average of the three numbers b =(c\(^{2}\)  a\(^{2}\)) ÷ 4.
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