It's a Question of your Career!

Correct Answer: Onam

Solution : The correct option is Onam.

The Kumarakom boat race is the most admired boat race to take place on the Vembanad Lake in September during the Onam festival. The snake boat race is a major highlight of the Onam festivities, attracting locals and tourists alike.

Correct Answer: Hush

Solution : The correct choice is the fourth option.

"Alas" is typically used to express sorrow, regret, or disappointment. In the given sentence, it doesn't convey the right sentiment for the context, which is about maintaining silence in the library. "Hush" is a more suitable choice as

Correct Answer:

Solution : According to the given question figure –
1. In each box the small circle and small triangle interchange their position.
2. In each box the number of the arrow is increased by one and decreased by one in the alternative figure.
So, following the above pattern,

Correct Answer: monocarpic

Solution : The correct option is Monocarpic.

Monocarpic plants are those that only blossom and set seed once throughout their lives. Monocarpic plants have a unique life cycle in which they normally develop vegetatively for several years, collecting energy and nutrients, and then produce a single, spectacular

Correct Answer: 1872

Solution : Given: The expression is 32% of 75% of 7800.
$=\frac{32}{100}\times \frac{75}{100}\times 7800$
$=1872$
Hence, the correct answer is 1872.

Correct Answer: 50%

Solution : Given that the article was bought for Rs. 600 and sold for Rs. 900,
⇒ Profit = Selling price − Cost price = Rs. 900 − Rs. 600 = Rs.300
We know that, Gain percentage = $\frac{\text{Profit}}{\text{Cost price}}$ × 100
= $\frac{300}{600}$ × 100 =

Correct Answer: $\frac{\sqrt{41}}{5}$

Solution : Given: $\tan \theta=\frac{4}{5}$
We know, $\sec^2 \theta-\tan^2 \theta=1$
$⇒\sec^2 \theta-\frac{16}{25}=1$
$⇒\sec^2 \theta=\frac{41}{25}$
$\therefore\sec \theta=\frac{\sqrt{41}}{5}$
Hence, the correct answer is $\frac{\sqrt{41}}{5}$.

Correct Answer: 22

Solution : Given: $a + b + c = 0$ and $ab + bc + ca = -11$
We know,
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$⇒0^2 = a^2 + b^2 + c^2 + 2(-11)$
$\therefore

Correct Answer: VKOG

Solution : Given:
NOTE is coded as PQVG:
N + 2 = P ; O + 2 = Q ; T + 2 = V ; E + 2 = G

Similarly, TIME will be coded as:
T + 2 = V ; I + 2 =

Correct Answer: JEOMSC

Solution : Given:
GRANTED is written as HPBLUCE and FRAGMENT is written as GPBENCOR.

Add 1 and subtract 2 alternately from the place value of the letters of GRANTED to obtain the required code –
G + 1 = H, R – 2 = P, A +