It's a Question of your Career!

Correct Answer: 4523

Solution : Given:
6341, 5432, ____, 3614

Subtract 909 from the previous term to obtain the missing term –
6341 – 909 = 5432; 5432 – 909 = 4523; 4523 – 909 = 3614

So, 4523 is the missing term in the series. Hence, the fourth option

Correct Answer: 3

Solution : The sentence can be improved using the third option.

It means that something will last for a long time, or forever. It is often used to describe something resistant to decay or destruction. In the context of diamonds, imperishable is a more precise term, as

Correct Answer: 3

Solution : Given:
(I) C scored less than 2 persons. B scored more than C but less than A.
A > B > C > __ > __ > __ > __
(II) E scored more than F but less than D. F has not scored the

Correct Answer: 40

Solution : Given:
5 – 9 × 8 ÷ 12 + 4 = ?

After interchanging + and ÷, and × and –, the equation will be as follows –
= 5 × 9 – 8 + 12 ÷ 4
= 5 × 9 – 8 +

Correct Answer: (1)

Solution : The error lies in the first part of the sentence.

Books fair is the incorrect usage. When talking about multiple fairs, we use book fairs as we pluralise the main noun in the case of compound words. This makes the sentence grammatically accurate and

Correct Answer:

Solution : As per the mirror image properties, closer things appear close to the mirror in the reflection.
Here, according to the information provided, the mirror is placed on the right side of the figure (on line AB). So, the left side of the reflected image will appear

Correct Answer: $\frac{k^{32}-\frac{1}{k^{32}}}{k+\frac{1}{k}}$

Solution : Consider, $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)$
Multiplying and dividing by $(k+\frac{1}{k})$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k-\frac{1}{k})(k+\frac{1}{k})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
Now using, $(a+b)(a-b)=a^2 - b^2$
⇒ $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right)=\frac{(k^2-\frac{1}{k^2})(k^2+\frac{1}{k^2})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^4-\frac{1}{k^4})(k^4+\frac{1}{k^4})(k^8+\frac{1}{k^8})(k^{16}+\frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^8-\frac{1}{k^8})(k^8+\frac{1}{k^8})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{(k^{16} - \frac{1}{k^{16}})(k^{16} + \frac{1}{k^{16}})}{k+\frac{1}{k}}$
$=\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$
Hence, the correct answer is $\frac{k^{32} - \frac{1}{k^{32}}}{k+\frac{1}{k}}$.

Correct Answer: had worked hard

Solution : The correct choice is the fourth option. 

"Had worked hard" should be used to fill in the blank since it uses the past perfect tense, which is appropriate for describing an action that was completed before another past action. In this sentence, the

Correct Answer: P

Solution : Subtract 3 from the place value of the letters of each row to obtain the next letter.
Row I: K – 3 = H; H – 3 = E
Row II: B – 3 = Y; Y – 3 = V
Row III: S –

Correct Answer: 342

Solution : Given:
5 : 124 :: 7 : ?

Like, (5)³ – 1 = 125 – 1 = 124
Similarly, (7)³ – 1 = 343 – 1 = 342

342 is related to 7. Hence, the third option is correct.