If a sum of money at simple interest doubles In 6 years, it will become 4 times in:

Correct option is D

Given:

A sum doubles in 6 years at simple interest.

Formula Used:
In Simple Interest (SI):

SI = $$\frac{P \times R \times T}{100}$$​​

Solution:

Let Principal P = 100

If it doubles in 6 years, Amount = 200

So, Interest = 200 - 100 = 100

​$$100 = \frac{100 \times R \times 6}{100}\\ \ \\R = \frac{100 \times 100}{100 \times 6} = \frac{100}{6}\%$$​​

Now, to become 4 times

Amount = 400

Interest = 400 - 100 = 300

Use formula again:

​$$300 = \frac{100 \times R \times T}{100} = R \times T \\ \ \\T = \frac{300}{R} = \frac{300}{100/6} = 18 \text{ years}$$​​

Alternate Solution:

If sum doubles in 6 years, then:

Interest = 1P in 6 years

To make amount = 4P (i.e., 3P interest)

Time = $$\frac{3P}{1P} \times 6 = 3 \times 6 = 18 \text{ years}$$​​