Correct Answer: Zaid

Solution : The correct option is Zaid.

The Zaid season typically falls between the Rabi and Kharif seasons. It usually begins in March and lasts until June. Zaid crops are typically short-duration and drought-resistant. Examples of Zaid crops include watermelon, cucumber, muskmelon, bitter gourd, and maize.

Correct Answer: 20%

Solution : As per the given chart,
The corresponding angle of expense on the Electricity head = 72°
The sum of all the central angles in a pie chart = 360°
$\therefore$ The total expenditure on Electricity = $\frac{72^\circ}{360^\circ}$ × 100 = 20%
Hence, the correct answer

Correct Answer: peeked

Solution : The correct choice is the first option.

In this context, peeked is the most appropriate antonym for stared. Stared implies a direct, intense gaze, while peeked suggests a brief, cautious look or appearance, which fits well with the sun emerging from behind the

Correct Answer: 8

Solution : The given figure can be labelled as shown below –

In the above-labelled figure, there are a total of 8 quadrilaterals. They are ABED, BDGF, AEGD, AEFB, BCEF, ECDG, FBAD, GDAB.

Hence, the fourth option is correct.

Correct Answer: managed currency

Solution : The correct answer is managed currency.

A managed currency is one in which the government or central bank of a country intervenes and modifies its market value or purchasing power. By issuing new money, controlling interest rates, and overseeing foreign exchange reserves, central

Correct Answer: $\frac{55}{28}$

Solution : $\frac{\sqrt{72}\times \sqrt{363}\times \sqrt{175}}{\sqrt{32}\times \sqrt{147}\times \sqrt{252}}$
$=\sqrt{\frac{72}{32}} \times \sqrt{\frac{363}{147}} \times \sqrt{\frac{175}{252}}$
Simplifying each square root,
$=\sqrt{\frac{9}{4}} \times \sqrt{\frac{121}{49}} \times \sqrt{\frac{25}{36}}$
$=\frac{3}{2} \times \frac{11}{7} \times \frac{5}{6} = \frac{55}{28}$
Hence, the correct answer is $\frac{55}{28}$.

Correct Answer: 1974

Solution : The correct answer is 1974.

On July 13, 1974, the Indian cricket team played their first-ever One-Day International (ODI) against England at Headingley, Leeds. Ajit Wadekar led the Indian team. England defeated India by four wickets. The England team's captain was Michael Denness. 

Correct Answer: bacc

Solution : Given:
ac_c_cb_acbcacbca_bc

To fill the series we have to divide the series→ac_c / _cb_ / acbc / acbc / a_bc
Let's check each option –
First option: abbb; acac / bcbb / acbc / acbc / abbc (No

Correct Answer: $0$

Solution : Given:
$\frac{\cos^{2}25^\circ-\sin^{2}65^\circ}{\cos^{2}25^\circ+\sin^{2}65^\circ}$
= $\frac{\cos^{2}25^\circ-\sin^{2}(90^\circ-25^\circ)}{\cos^{2}25^\circ+\sin^{2}(90^\circ-25^\circ)}$
= $\frac{\cos^{2}25°-\cos^{2}25°}{\cos^{2}25°+\cos^{2}25°}$
= $\frac{0}{2\cos^{2}25^\circ}$
= $0$
Hence, the correct answer is $0$.

Correct Answer: Neither I nor II

Solution : Statement I:
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\ldots \ldots \frac{1}{110}<\frac{5}{6}$
Expand LHS
$⇒\frac{1}{2}+\frac{1}{2\times{3}}+\frac{1}{3\times{4}}+\ldots \ldots \frac{1}{10\times{11}}<\frac{5}{6}$
$⇒\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots \ldots \frac{1}{10}-\frac{1}{11}<\frac{5}{6}$
$⇒1-\frac{1}{11}<\frac{5}{6}$
$⇒\frac{10}{11}<\frac{5}{6}$
which is wrong,
So, statement I is incorrect.
Statement II:
$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\ldots \ldots \frac{1}{143}>\frac{7}{13}$
$⇒\frac{1}{3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots \ldots \frac{1}{11\times13}>\frac{7}{13}$
$⇒\frac{1}{3}+\frac{2}{2}[\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots \ldots \frac{1}{11\times13}]>\frac{7}{13}$
$⇒\frac{1}{3}+\frac{1}{2}[\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots \ldots \frac{1}{11}-\frac{1}{13}]>\frac{7}{13}$
$⇒\frac{1}{3}+\frac{1}{2}[\frac{1}{3}-\frac{1}{13}]>\frac{7}{13}$
$⇒\frac{1}{3}+\frac{5}{39}>\frac{7}{13}$
$⇒\frac{6}{13}>\frac{7}{13}$
which is