Correct Answer: 8

Solution :
Given, AB is a tangent, OB = 10 cm and OA = 6 cm
Since tangent is perpendicular to the radius at its point of contact, $\angle$OAB = 90$^\circ$
Using Pythagoras theorem,
OB² = AB² + OA²
⇒ 10² = AB² + 6²
⇒ AB² = 64
⇒ AB = 8 cm
Hence, the correct answer is 8.

Correct Answer: DO

Solution : Given:
VW, XU, ZS, BQ, ?

Add 2 to the positional value of the first letter and subtract 2 from the positional value of the second letter of the previous terms to get the first and second letters respectively of the next term.
First letter of the series→V + 2 = X; X + 2 = Z; Z + 2 = B; B + 2 = D
Second letter of the series→W – 2 = U; U – 2 = S; S – 2 = Q; Q – 2 = O

So, from the above, DO is the required term of the series. Hence, the second option is correct.

Correct Answer: 4

Solution :
Given: $\tan\theta=\frac{3}{4} =\frac{Perpendicular}{Base}$
Using Pythagoras' theorem,
$AC=\sqrt{AB^2+BC^2}$
⇒ $AC=\sqrt{3^2+4^2} =5$
$\sin\theta =\frac {\text {Perpendicular}}{\text {Hypotenuse} } = \frac{3}{5}$
Now,
$\frac{1+\sin\theta}{1-\sin\theta} = \frac{1+\frac{3}{5}}{1-\frac{3}{5}} =\frac{8}{2}=4$
Hence, the correct answer is 4.

Correct Answer: JGZSKZ

Solution : Given:
BOUCLE is coded as AJYSKZ.

Subtract 2 and 4 alternatively in the place value of each letter of BOUCLE and write the letters in reverse order, to get the required code. –

Thus, BOUCLE is coded as AJYSKZ.
Similarly, follow the same pattern for BOUDIN –

So, BOUDIN is coded as JGZSKZ. Hence, the third option is correct.

Correct Answer: 64 cm³

Solution : Given: Total surface area of cube = 96 cm²
Let the sides of the cube be $a$ cm.
⇒ $6a^2 = 96$
⇒ $a^2 = 16$
⇒ $a = 4$
Volume of the cube $=a^3 =4^3= 64$ cm³
Hence, the correct answer is 64 cm³.

Correct Answer: Drab

Solution : The correct choice is the second option.

Explanation:

Jejune is an adjective used to describe something that is dull, lacking interest, or insipid. It often refers to ideas, writing, or things that are simplistic and unstimulating.

Drab is a fitting synonym in this context. It denotes something dull, lacking brightness or interest.

The meaning of the other options are as follows:

• Poignant means evoking a keen sense of sadness or regret.
• Unchildlike focuses on the absence of childlike qualities.
• Cosmopolitan refers to someone or something that is sophisticated, worldly, or global.

Correct Answer: have seen

Solution : The correct choice is the second option.

The sentence structure in the passage is in the present perfect tense, indicating actions or events that have occurred at an unspecified time before now or have a connection to the present.

The phrase "as UK pension savers" in the passage suggests a connection to the present time. Therefore, the appropriate verb form to use in this context is have seen.

Correct Answer: ii, i, iii, iv

Solution : Given:
i. Worrisome ii. Wonderful iii. Worthless iv. Wrong

Step 1: Consider the first letter of each word. Since all the words start with the same letter W, move on to the next letter.
Step 2: The second letter of each word is either o or r. Based on the alphabetical order, we can arrange them as – Worrisome, Wonderful, Worthless, and Wrong.
Step 3: Consider the third letter of (Worrisome, Wonderful, Worthless). Wonderful will come before (Worrisome, Worthless) as n comes before r in the alphabet.
Step 4: Consider the fourth letter of (Worrisome, Worthless). Worrisome will come before Worthless as r comes before t in the alphabet.

So, the sequence is Wonderful, Worrisome, Worthless, Wrong or, ii, i, iii, iv. Hence, the second option is correct.

Correct Answer: 2

Solution : Given:
1. MINER ⇒ 56342
2. ANIME ⇒ 63457

After comparing both words and their codes, we get –
In both words, the common letters are M, I, N, and E, and the common codes are 5, 6, 3, 4.
The remaining letter and code in the word MINER are R and 2
The remaining letters and code in the word ANIM are A and 7

So, 2 is the code for R. Hence, the fourth option is correct.

Correct Answer: 10 cm

Solution :
Let O be the centre of the circle. OA = 13 cm and OM = 12 cm.
Let AB be the chord of the smaller circle which is tangent to a smaller circle.
Since the tangent is perpendicular to the radius at its point of contact, angle OMA is a right angle.
By Pythagoras theorem,
OA² = AM² + OM²
⇒ 13² = AM² + 12²
⇒ AM² = 25
⇒ AM = 5 cm
Also, perpendicular from the centre to a chord bisects the chord.
AB = 2 × AM = 2 × 5 = 10 cm
Hence, the correct answer is 10 cm.