Question Detail

How many terms are there in 2,4,8,16,....,1024

Answer: Option D

Explanation:

It is a GP. with r = 2, ie \begin{aligned} 2^1, 2^2, 2^3... \end{aligned}
If number of terms is n. Then,
\begin{aligned} 2 \times 2^{n-1} = 1024 \end{aligned} =>
\begin{aligned} 2^{n-1} = 512 \end{aligned}
or \begin{aligned} 2 \times 2^{n-1} = 2^9 \end{aligned}
so n-1 = 9 => n = 10

1. A number when divided by 44, gives 432 as quotient and 0 as remainder. What will be the remainder when dividing the same number by 31

Answer: Option A

Explanation:

P ÷ 44 = 432

=> P = 432 * 44 = 19008

P / 31 = 19008 / 31 = 613, remainder = 5

2. Simplify (212 * 212 + 312 * 312 )

132288
142088
142188
142288

Answer: Option D

Explanation:

Trick: Above equation can be solved by using following formula
\begin{aligned}
(a^2 + b^2) = \frac{1}{2}((a+b)^2 + (a-b)^2)
\end{aligned}

3. What least value should be replaced by * in 223*431 so the number become divisible by 9

Answer: Option A

Explanation:

Trick: Number is divisible by 9, if sum of all digits is divisible by 9, so (2+2+3+*+4+3+1) = 15+* should be divisible by 9,
15+3 will be divisible by 9,
so that least number is 3.

4. Simplify 586645 * 9999

5865863355
5665863355
4865863355
4665863355

Answer: Option A

Explanation:

Although it is a simple question, but the trick is to save time in solving this.
Rather than multiplying it we can do as follows:
586645 * (10000 - 1) = 5866450000 - 586645 = 5865863355

5. What is the largest 4 digit number exactly divisible by 88

9900
9999
9988
9944

Answer: Option D

Explanation:

Largest 4 digit number is 9999
After doing 9999 ÷ 88 we get remainder 55
Hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944