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Question: Find the third proportional to: (i) 36 and 18. (ii) 5.25 and 7. (iii) Rs. 1.60 and Rs.0.40....

Find the third proportional to:
(i) 36 and 18.
(ii) 5.25 and 7.
(iii) Rs. 1.60 and Rs.0.40.

Explanation

Solution

Assume a variable to be the third proportional to each of these ratios and equate the fractions thus obtained and solve further to get the result. Remember third proportional is nothing but a common term in two fractions.

Complete step-by-step answer :
Let two ratios be a:b and b:c. Then in these two ratios ‘a’ is the third proportional. So the third proportional is the uncommon term.
The third proportional to a ratio a : b is the number x such that ab=bx or ba=xb\dfrac{a}{b}=\dfrac{b}{x}\ or\ \dfrac{b}{a}=\dfrac{x}{b}.
(i) 36 and 18
Let us assume the third proportional to be x. Then,
1836=x18\dfrac{18}{36}=\dfrac{x}{18}
By simplifying this, we get
12=x18 x=182 x=9 \begin{aligned} & \Rightarrow \dfrac{1}{2}=\dfrac{x}{18} \\\ & \Rightarrow x=\dfrac{18}{2} \\\ & \Rightarrow x=9 \\\ \end{aligned}
Hence, 9 is the third proportional to 36 and 18.

(ii) 5.25 and 7
Let us assume the third proportional to be x. Then,
75.25=x7\dfrac{7}{5.25}=\dfrac{x}{7}
On simplifying this, we get
43=x7 x=283 \begin{aligned} & \Rightarrow \dfrac{4}{3}=\dfrac{x}{7} \\\ & \Rightarrow x=\dfrac{28}{3} \\\ \end{aligned}
Hence, the third proportional to 5.25 and 7 is 283\dfrac{28}{3}.

(iii) Rs. 1.60 and Rs.0.40
Let us assume the third proportional to be Rs. x. Then,
0.401.60=x0.4\dfrac{0.40}{1.60}=\dfrac{x}{0.4}
On simplifying this, we get
x0.4=14 x=0.1 \begin{aligned} & \Rightarrow \dfrac{x}{0.4}=\dfrac{1}{4} \\\ & \Rightarrow x=0.1 \\\ \end{aligned}
Hence, the third proportional to Rs. 1.60 and Rs.0.40 is Rs.0.10.1.

Note :In the third part, make sure to maintain the unit i.e Rs.
Another approach is using the definition of the third proportional, that is, third proportional is equal to the square of equal terms and is divided by unequal terms.
For example, in the ratio ab=bx \dfrac{a}{b}=\dfrac{b}{x}\
The third proportional can be written as, x=b2ax=\dfrac{{{b}^{2}}}{a}.