Correct Answer: 20

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

In the above figure, there are a total of 20 triangles. They are ACG, AGH, AHF, AFB, HFE, DHE, DGH, CGD, CAH, CAF, AFG, AEF, CHE, CDA, AEC, AEB, CFE, ADH, ADE, CDH.

Hence, the

Correct Answer: gold

Solution : The correct option is gold.

Gold holds the distinguished title of being called the " king of metals" due to its exceptional rarity and consequent high value. Its scarcity makes it one of the most expensive metals to acquire. Beyond its monetary worth, gold stands

Correct Answer: Philistine

Solution : The correct choice is the third option.

The term "philistine" refers to a person who is hostile or indifferent to culture and the arts, often focusing on material or practical concerns. In the context given, someone interested in material gains and hostile to art or

Correct Answer: $\frac{3}{20}$

Solution : $=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=(\frac{1}{20}+\frac{1}{30})+(\frac{1}{42}+\frac{1}{56})+(\frac{1}{72}+\frac{1}{90})$
$=(\frac{3+2}{60})+( \frac{4+3}{168})+( \frac{5+4}{360})$
$=(\frac{1}{12})+( \frac{1}{24})+( \frac{1}{40})$
$=\frac{10+5+3}{120}$
$=\frac{18}{120}$
$=\frac{3}{20}$
Hence, the correct answer is $\frac{3}{20}$.

Correct Answer: 203

Solution : Given:
63 ÷ 3 + 14 × 7 – 16 = ?

After interchanging + and –, × and ÷, the equation becomes –
= 63 × 3 – 14 ÷ 7 + 16
= 63 × 3 – 2 + 16
= 189 –

Correct Answer: Rs. 2,160

Solution : Given:
The cost of 16 kg of wheat = Rs. 384

According to the question,
The cost of 1 kg wheat = 384 ÷ 16 
= 24 Rs.
Therefore, the cost of 90 kg of wheat = 90 × 24
= 2160 Rs.

So,

Correct Answer: 9 : 16

Solution : The ratio of the diameters of two right circular cones of equal height = 3 : 4
Let diameter of 1^st cone be $3x$.
Diameter of 2^nd cone $= 4x$
Radius of the 1^st cone, $r_1 = \frac{3x}{2}$
Radius of

Correct Answer: 9.23

Solution :
Given, In $\triangle \mathrm{CAB}, \angle \mathrm{CAB}=90^{\circ}$ and $\mathrm{AD} \perp \mathrm{BC}$.
$\mathrm{AC}=24 \mathrm{~cm}, \mathrm{AB}=10 \mathrm{~cm}$
Applying Pythagoras theorem,
$BC^2=AC^2+AB^2$
⇒ $BC^2 =24^2+10^2$
⇒ $BC = 26$ cm
Now, area of triangle = $\frac{1}{2}\times \text{base}\times \text{height}$
So, $\frac{1}{2}\times AC\times AB=\frac{1}{2}\times AD\times BC$
⇒ $24\times 10 = AD

Correct Answer: $\frac{28}{135}$

Solution : Consider, $\left(2 \frac{1}{2}÷ 1 \frac{7}{8}\right) ÷\left(9 \frac{3}{8}÷ 11 \frac{2}{3} \text { of } \frac{1}{8}\right)$
= $(\frac{5}{2}÷\frac{15}{8})÷\{\frac{75}{8}÷(\frac{35}{3}\times\frac{1}{8})\}$
= $(\frac{5}{2}\times\frac{8}{15})÷(\frac{75}{8}\times\frac{3}{35}\times\frac81)$
= $(\frac{4}{3})÷(\frac{15\times 3}{ 7})$
= $(\frac{4}{3})÷(\frac{15\times 3}{ 7})$
​​= $\frac{4}{3}\times\frac{7}{15\times 3}$
= $\frac{28}{135}$
Hence, the correct answer is $\frac{28}{135}$.

Correct Answer: Banabhatta

Solution : The correct option is Banabhatta.

The "Harshacharita" was written by the Indian poet Banabhatta. Banabhatta was a Sanskrit scholar and a court poet in the 7th century. The "Harshacharita" is a biography of King Harshavardhana, who ruled the Indian subcontinent from 606 to 647 CE.