Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

Let the positive number be $x$.
According to the problem statement:

$$x + 17 = 60 \times \frac{1}{x}$$ $$x + 17 = \frac{60}{x}$$

Multiply both sides by $x$ (since $x$ is a positive number, $x \neq 0$):

$$x(x + 17) = 60$$ $$x^2 + 17x = 60$$ $$x^2 + 17x - 60 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to -60 and add up to 17. These numbers are 20 and -3.

$$(x + 20)(x - 3) = 0$$

This gives two possible solutions for $x$:

$$x + 20 = 0 \implies x = -20$$ $$x - 3 = 0 \implies x = 3$$

Since the problem asks for a positive number, we choose $x = 3$.

The correct option is (a).
The final answer is $\boxed{\text{3}}$.

52. The average monthly income of P and Q is ₹ 5,050. The average monthly income of Q and R is ₹ 6,250 and the average monthly income of P and R is ₹ 5,200. The monthly income of P is:

Let P, Q, and R be the monthly incomes of persons P, Q, and R respectively.

1. Average income of P and Q:

$$\frac{P + Q}{2} = 5050$$ $$P + Q = 2 \times 5050$$ $$P + Q = 10100 \quad \text{(Equation 1)}$$

2. Average income of Q and R:

$$\frac{Q + R}{2} = 6250$$ $$Q + R = 2 \times 6250$$ $$Q + R = 12500 \quad \text{(Equation 2)}$$

3. Average income of P and R:

$$\frac{P + R}{2} = 5200$$ $$P + R = 2 \times 5200$$ $$P + R = 10400 \quad \text{(Equation 3)}$$

$$(P + Q) + (Q + R) + (P + R) = 10100 + 12500 + 10400$$ $$2P + 2Q + 2R = 33000$$ $$2(P + Q + R) = 33000$$ $$P + Q + R = \frac{33000}{2}$$ $$P + Q + R = 16500 \quad \text{(Equation 4)}$$

Subtract Equation 2 from Equation 4:

$$(P + Q + R) - (Q + R) = 16500 - 12500$$ $$P = 4000$$

The correct option is (b).
The final answer is $\boxed{\text{₹ 4,000}}$.

53. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Total number of men = 7
Total number of women = 6
Total persons to be selected for the committee = 5
Condition: At least 3 men must be on the committee.

This means the committee can have the following combinations of men and women:

• 3 men and 2 women
• 4 men and 1 woman
• 5 men and 0 women

Number of ways to choose 3 men from 7 = $\binom{7}{3}$
Number of ways to choose 2 women from 6 = $\binom{6}{2}$

$$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ $$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$$

Ways for this case = $35 \times 15 = 525$

Number of ways to choose 4 men from 7 = $\binom{7}{4}$
Number of ways to choose 1 woman from 6 = $\binom{6}{1}$

$$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ $$\binom{6}{1} = 6$$

Ways for this case = $35 \times 6 = 210$

Number of ways to choose 5 men from 7 = $\binom{7}{5}$
Number of ways to choose 0 women from 6 = $\binom{6}{0}$

$$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21$$ $$\binom{6}{0} = 1$$

Ways for this case = $21 \times 1 = 21$

Total ways = (Ways for 3 men, 2 women) + (Ways for 4 men, 1 woman) + (Ways for 5 men, 0 women)
Total ways = $525 + 210 + 21 = 756$

The correct option is (d).
The final answer is $\boxed{\text{756}}$.