We will learn how to solve different types of problems on digits and numbers.
1. Numbers 1, 2, 3, 4, ........., 98, 99, 100 are multiplied together. The number of zeros at the end of the product on the right will be equal to
(a) 24
(b) 21
(c) 22
(d) 13
Solution:
The value of the ‘n’ = 100.
Therefore, number of zeros at the end of 1 × 2 × 3 × 4 × 5 × .............. × 99 × 100
= (100 ÷ 5) + (100 ÷ 5^2)
= 20 + 4
= 24
Answer: (a)
Note: Number of zeros at the end of the product of natural numbers = n/5 + n/5^2 + n/5^3 + ........... (upto n terms)
2. How many three digit natural numbers are possible?
(a) 999
(b) 990
(c) 890
(d) 900
Solution:
Number of three digit numbers = 9 × 10^(3  1) = 9 × 10^2 = 900
Answer: (d)
Note: Number of numbers with specific number of digits = 9 × 10^(d  1), where ‘d’ = number of digits.
3. The difference of the squares of two nos. is 135 & their difference is 5. The product of the numbers is
(a) 182
(b) 178
(c) 180
(d) 176
Solution: Let, two numbers be ‘a’ and ‘b’
According to the problem,
a  b = 5 and a^2  b^2 = 135
Therefore, a + b = 135 ÷ 5 = 27
Since, a = (27 + 5) ÷ 2 = 16 and b = 16  5 = 11
Therefore, required value of ab = 16 × 11 = 176
Answer: (d)
4. The sum of x and y is three times their difference. Find the ratio of x and y:
(a) 2 : 1
(b) 2 : 3
(c) 3 : 4
(d) 4 : 3
Solution:
a + b = 3(a  b)
or, 2a = 4b
Therefore, a : b = 4 : 2 = 2 : 1
Answer: (a)
5. The sum of the two numbers m and n is 5760, & the difference is onethird of the greater number. Which is the greater number?
(a) 3450
(b) 3456
(c) 3475
(d) 3500
Solution:
Let, two numbers be x and y.
Now according to the problem,
x/3 = x  y
or, 3x  3y = x
or, 2y = 3y
or, x : y = 3 : 2
Therefore, the greatest number = 5760 × 3/(3 + 2) =5760 × 3/5 = 3456
Answer: (b)
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