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. 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}

2. What least number should be added to 1056, so that the sum is completely divisible by 23

Answer: Option C

Explanation:

(1056 / 23) gives remainder 21

21 + 2 = 23, So we need to add 2

3. 217 * 217 + 183 * 183

70578
82578
80578
80568

Answer: Option C

Explanation:

\begin{aligned}
\frac {1}{2}\times 2(a^2 + b^2) = \frac {1}{2} \times [(a + b)^2 + (a - b)^2]
\end{aligned}
\begin{aligned}
= \frac {1}{2} \times [(217 + 183)^2 + (217 - 183)^2] = 80578
\end{aligned}

4. 1939392 * 625

1212120010
1212120000
1212120011
1212121010

Answer: Option B

Explanation:

Trick: when multiplying with \begin{aligned} 5^n
\end{aligned} then put n zeros to the right of multiplicand and divide the number with \begin{aligned} 2^n
\end{aligned}
So using this we can solve this question in much less
time.
\begin{aligned} 1939392 \times 5^4 = \frac {19393920000}{16} = 1212120000
\end{aligned}

5. 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.