Correct Answer: I

Solution : Given:
A, C, E, G, ?

Like, A + 2→C; C + 2→E; E + 2→G
Similarly, G + 2→I

I is the missing term of the given series. Hence, the second option is correct.

Correct Answer: 6

Solution : Given that, On dividing a certain number by 363, we get 17 as the remainder.
Let the number be $363x+17$.
The number = $33x(11)+1(11) + 6$
The number = $11(33x+1) + 6$
So, the remainder when the given number is divided by 11 is 6.

Correct Answer: Arkose

Solution : The correct answer is Arkose.

Sedimentary rocks are the rocks which are formed by the deposition of sediments carried by rivers and winds. They also carry petroleum and mineral deposits. Some common sedimentary rocks are sandstone, limestone, arkose and shale etc.

While marble and charnockite

Correct Answer: 64 : 81

Solution : $1\frac{1}{2} : 1\frac{1}{3}$
$=\frac{3}{2}$ : $\frac{4}{3}$
After squaring the fractions, we get,
⇒ $\frac{9}{4}:\frac{16}{9}$
Their reciprocals are,
⇒ $\frac{4}{9}:\frac{9}{16}$
Multiply each ratio by 144 (LCM of 9 and 16)
$\therefore$ The required ratio = 64 : 81
Hence, the correct answer is 64

Correct Answer: 229

Solution : Given:
21, 25, 52, 68, 193, ?

Let's apply the pattern –
21 + (2)² = 21 + 4 = 25
25 + (3)³ = 25 + 27 = 52
52 + (4)² = 52 + 16 = 68
68 + (5)

Correct Answer: INLM

Solution : Given:
AVTE, CTRG, ERPI, GPNK, ?

Add 2 in the first and the fourth alphabet and subtract 2 from the second and the third alphabet to obtain the next term.
AVTE: A + 2 = C; V – 2 = T; T – 2 =

Correct Answer: $5$ cm

Solution : Given: PQRA is a rectangle, AP = 22 cm and PQ = 8 cm. $\triangle$ABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm.

Let D and E be the two

Correct Answer: 9390

Solution : The principal gives interest of Rs. 3130 in 10 years.
Hence, the principal gives interest in 5 years = Rs. $\frac{3130 \times 5}{10}$ = Rs. 1565
After 5 years, the principal is 5 times.
The interest for the remaining 5 years is also 5 times

Correct Answer: $\frac{53}{60}$

Solution : $3 \frac{1}{5} \div 4 \frac{1}{2}$ of $5 \frac{1}{3}+\frac{1}{8} \div \frac{1}{2}$ of $\frac{1}{4}-\frac{1}{4}\left(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4}\right)$
= $\frac{16}{5} \div \frac{9}{2}$ of $ \frac{16}{3}+\frac{1}{8}\div \frac{1}{8} -\frac{1}{4} (4×\frac{1}{4})$
= $\frac{16}{5} \div  24 + 1 - \frac{1}{4}$
= $\frac{16}{5} × \frac{1}{24}+ \frac{3}{4}$
= $\frac{2}{15} + \frac{3}{4}$
= $\frac{53}{60}$

Correct Answer: 16

Solution : Given:
27 (15) 33; 41 (13) 53

Like, 27 (15) 33 is (2 + 7) + (3 + 3) = 9 + 6 = 15
And, 41 (13) 53 is (4 + 1) + (5 + 3) = 5 + 8 = 13
Similarly, 26