Problem Analysis:

• We are given the Highest Common Factor (HCF) of two numbers is 4.
• We are also told that 5 and 7 are two factors of the Least Common Multiple (LCM) of these numbers.
• The goal is to find the smaller of the two numbers.

Calculating the LCM

The HCF represents the common prime factors raised to the lowest power. The LCM includes all prime factors from both numbers, raised to their highest power. Given HCF = 4 and additional factors of the LCM are 5 and 7, the LCM must contain the prime factors of the HCF and these additional factors.

Since $HCF = 4 = 2^2$, the prime factors of the LCM must include $2^2$, 5, and 7.

Therefore, the $LCM = HCF \times (\text{product of other unique factors}) = 4 \times 5 \times 7 = 140$.

Finding the Two Numbers

Let the two numbers be $N_1$ and $N_2$. We know that:

• $N_1 = HCF \times a = 4a$
• $N_2 = HCF \times b = 4b$
• Here, $a$ and $b$ must be coprime integers (their HCF is 1).
• We also know the relationship $LCM = HCF \times a \times b$.

Substitute the known values:

$140 = 4 \times a \times b$

Divide both sides by 4:

$a \times b = \frac{140}{4}$

$a \times b = 35$

Now, we need to find pairs of coprime integers ($a, b$) whose product is 35.

• Possible pairs ($a, b$) are: (1, 35) and (5, 7).

Determining the Smaller Number

Let's calculate the two numbers ($N_1, N_2$) for each coprime pair:

• Case 1: Using ($a=1, b=35$)
  • $N_1 = 4 \times 1 = 4$
  • $N_2 = 4 \times 35 = 140$
  • The pair of numbers is (4, 140). The smaller number is 4.
• Case 2: Using ($a=5, b=7$)
  • $N_1 = 4 \times 5 = 20$
  • $N_2 = 4 \times 7 = 28$
  • The pair of numbers is (20, 28). The smaller number is 20.

Comparing the results with the given options (10, 14, 20, 28), the smaller number derived from Case 2, which is 20, matches option C.

Conclusion

The pair of numbers is (20, 28). The HCF(20, 28) is 4, and the LCM(20, 28) is 140 (which has factors 5 and 7). The smaller of these two numbers is 20.