Problems on Numbers

Q1.

What largest number of five digits is divisible by 99?

9909	99981
99990	99999

Answer:

Option C

Solution:

Largest number of 5 digits = 99999. On dividing 99999 by 99, we get 9 as remainder.

.'.	Required number = (99999 - 9) = 99990.

Wrong Answer

Q2.

The sum of the smallest six digit number and the greatest five digit number is

201110	199999
211110	099999

Answer:

Option B

Solution:

Requires sum	=(100000 + 99999)
Requires sum	=199999.

Wrong Answer

Q3.

In a division sum, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, the divident is

4236	4306
4336	5336

Answer:

Option D

Solution:

Wrong Answer

Q4.

The difference between the squares of two consecutive odd integers is always divisible by:

3	6
7	8

Answer:

Option D

Solution:

Let the two consecutive odd integers be (2x + 1) and (2x + 3)

Then, (2x + 3)² - (2x + 1)²
= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)
= (4 + 4) x 2

= 8 (x + 1),
which is always divisible by 8

Wrong Answer

Q5.

A positive integer, which when added to 1000, gives a sum which is greater than when it is multiplied by 1000. This positive integer is:

1	3
5	7

Wrong Answer

Written

Thursday, November 13, 2014