Q.1 Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$.

(OR)

Q.2 We have been given two numbers 726 and 275 and we have to find their 𝐻. 𝐶. 𝐹.

 

Here, we need to divide 27 into two parts. Let the parts be 𝑥 and 27 − 𝑥 Now, the sum of their reciprocals is $\frac{3}{20}$.

$\begin{array}{r}\Rightarrow \frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\\ \Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}\\ \Rightarrow \frac{27}{27x-x^2}=\frac{3}{20}\\ \Rightarrow x^2-27x+180=0\end{array}$

Factorising by splitting the middle term, we get

$x^2-15x-12x+180=0$

Grouping common terms,

𝑥(𝑥 − 15) − 12(𝑥 − 15) = 0 

⇒ (𝑥 − 15)(𝑥 − 12) = 0 

⇒ 𝑥 − 15 = 0 or, (𝑥 − 12) = 0 

⇒ 𝑥 = 15 or 12

Therefore, the two parts are 15 and 12.

(OR)

We have been given two numbers 726 and 275 and we have to find their 𝐻. 𝐶. 𝐹. 

By Euclid’s Division Lemma, we know that For any two positive integers 𝑎 and 𝑏, there exist unique whole numbers 𝑞 and 𝑟 such that 𝑎 = 𝑏𝑞 + 𝑟, where 0 ≤ 𝑟 < 𝑏. 𝑎 is called the dividend, 𝑏 is the divisor, 𝑞 is the quotient and 𝑟 is the remainder. 

Euclid’s division algorithm is based on Euclid’s division lemma. Now, using division lemma on 𝑎 = 726 with 275 as divisor, we get

726 = 𝑏𝑞 + 𝑟 

⇒ 726 = 275×2 + 176 

Here, the remainder obtained is not zero. So, the process is continued until we get remainder as zero. 

176 = 99×1 + 77 99 

= 77×1 + 22

77 = 22×3 + 11 22 

= 11×2 + 0 

𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 𝑏𝑞 + 𝑟 𝑤ℎ𝑒𝑟𝑒 𝑟 = 0 ⇒𝑏 = 11, 𝑞 = 2, 𝑟 = 0

Since for 𝑏 = 11, the value of remainder is 0, therefore, by Euclid’s division algorithm, the 𝐻. 𝐶. 𝐹 of two numbers 726 and 275 is 11.