Correct Answer: To promote equal justice and to provide free legal aid.

Solution : The correct option is To promote equal justice and provide free legal aid.

Article 39A of the Indian Constitution pertains to the promotion of equal justice and the provision of free legal aid. It states: The State shall secure that the operation of the legal system promotes justice on a basis of equal opportunity and shall, in particular, provide free legal aid, by suitable legislation or schemes or in any other way, to ensure that opportunities for securing justice are not denied to any citizen because of economic or other disabilities.

Correct Answer: JLMO

Solution : Given:
IKLN is related to FHIK –

Subtract 3 from the place value of each letter of IKLN to obtain the required code –
I – 3 = F; K – 3 = H; L – 3 = I; N – 3 = K
Thus, IKLN is coded as FHIK.
And, for GIJL –
G – 3 = D; I – 3 = F; J – 3 = G; L – 3 = I
Thus, GIJL is coded as DFGl.
Similarly, follow the same pattern for MOPR –
M – 3 = J; O – 3 = L; P – 3 = M; R – 3 = O 

So, MOPR is coded as JLMO. Hence, the third option is correct.

Correct Answer: $12 \sqrt{5} \;\mathrm{Sq} . \mathrm{cm}$

Solution : Using the formula:
$s = \frac{(a + b + c)}{2}$
$s = \frac{(7 + 9 + 8)}{2}$
$s = \frac{24}{2}$
$s = 12$ cm
Now, plug the values of a, b, c, and s into Heron's formula to find the area of the triangle:
$\sqrt{(s(s - a)(s - b)(s - c))}$
$=\sqrt{(12(12 - 7)(12 - 9)(12 - 8))}$
$=\sqrt{(12 × 5 × 3 × 4)}$
$=12\sqrt{5}$ Sq. cm
Therefore, the area of the triangle with side lengths 7 cm, 9 cm, and 8 cm is approximately $12\sqrt{5}$ Sq. cm
Hence, the correct answer is $12 \sqrt{5} \;\mathrm{Sq}. \mathrm{cm}$.

Correct Answer: Acute

Solution : The correct answer is Acute.

Acute diseases last for only shorter periods. The conditions within acute diseases typically have a rapid onset and short duration, often resolving without prolonged medical intervention. Examples of acute diseases include the common cold, flu, gastroenteritis, and some respiratory infections.

Correct Answer: 70

Solution : Given:
Number = 228, divisor = $a$, and remainder = 18
Subtract 18 from 228 = 210
Factors of 210 = 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, and 210
$\therefore$ The biggest two-digit value of $a$ will be 70.
Hence, the correct answer is 70.

Correct Answer: lighting

Solution : The correct option will be the first option.

Lighting the lamp is, in a way, an invocation of the waves of knowledge that are activated on stage. It also means keeping the waves of knowledge generated in the knowledge-imparting stage constantly active through the medium of the flame of the lamp.

Other words are used in different contexts. Their meanings are as follows:

• Lightening: It is used for thunder.
• Sparking: It is the act or appearance of something that sparks.
• Glittering: It is the appearance of something that glitters.

Therefore, according to the sentence, lighting is the correct answer.

Correct Answer: between 35 years and 40 years and between 55 years and 60 years

Solution : Total patients = 35
From the given graph,
About 11% of the patients were admitted.
⇒ 11% of 35 $=\frac{11}{100} \times 35=3.85 \approx 4$
Here, The number of patients between 35 and 40 years is 4.
Also, The number of patients between 55 and 60 years is 4.
Hence, the correct answer is between 35 and 40 years and between 55 and 60 years.

Correct Answer: The horse was frightened by a sudden noise.

Solution : The second option is the correct choice.

Passive voice is the voice in which the object experiences an action rather than the person who performs the action. Hence, the object in the active sentence becomes the subject in the passive voice.

The structure of the passive sentence in the simple past tense is: Singular object + was + past participle form of the verb + by followed by the agent of the action.

Therefore, the correct sentence is: The horse was frightened by a sudden noise.

Correct Answer: 52

Solution : Given: $p^2+\frac{1}{p^2}=14$
we know that,
$p^2+\frac{1}{p^2}=(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}$
$⇒(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}=14$
$⇒(p+\frac{1}{p})^2=14+2$
$⇒(p+\frac{1}{p})^2=16$
$⇒(p+\frac{1}{p})=4$
Now cubing both sides, we get:
$⇒(p+\frac{1}{p})^3=4^3$
$⇒p^3+\frac{1}{p^3}+3\times p\times \frac{1}{p}(p+ \frac{1}{p})=4^3$
​​​​​​​​​​​​​​$⇒​​​​​​​p^3+\frac{1}{p^3}+3\times4=64$
​​​​​​​$⇒​​​​​​​p^3+\frac{1}{p^3}=64-12$
​​​​​​​$⇒​​​​​​​p^3+\frac{1}{p^3}=52$
​​​​​​​Hence, the correct answer is 52.

Correct Answer: GMJZ

Solution : Given:
DPMW, EOLX, FNKY, ?

In the above-given letter clusters, add 1 to the place value of the first and fourth letters and subtract 1 from the place value of the second and third letters of the previous term to obtain the next term in the series.
DPMW→D + 1 = E; P – 1 = O; M – 1 = L; W + 1 = X→EOLX
EOLX→E + 1 = F; O – 1 = N; L – 1 = K; X + 1 = Y→FNKY
FNKY→F + 1 = G; N – 1 = M; K – 1 = J; Y + 1 = Z→GMJZ

So, GMJZ is the missing term in the series. Hence, the fourth option is correct.