correct answer is: c) 22 years

Explanation

Here,
A’s principal = ₹ 8000
A’s Rate = 12 % p.a.
B’s principal = ₹ 9100
B’s Rate = 10 % p.a.
Let, Time = ‘t’ years
We know, Amount = (Principal + S.I)
Or, Amount =\left(p+\frac{p\times t\times r}{100}\right) _{[Use S.I. formula]}
According to the question,
A’s amount = B’s amount
\therefore\left(8000+\frac{8000\times t\times12}{100}\right)=\left(9100+\frac{9100\times t\times10}{100}\right)
\rightarrow\left(\frac{800000+96000\times t}{100}\right)=\left(\frac{910000+91000\times t}{100}\right) _[L.C.M=100]
\rightarrow\ 800000+96000t=910000+91000t _{[same denominator]}
\rightarrow96000t-91000t=910000-800000 _{[interchange]}
\rightarrow5000t=110000
\rightarrow t=\frac{110000}{5000}
\rightarrow t=22
\therefore Their amounts will equal in 22 years.
Ans: Their amounts will equal in 22 years.

Another way to ask this same question is:
(1) In how many years will the amounts borrowed by A and B, ₹8000 at 12% per annum and ₹9100 at 10% per annum respectively, be the same?
(2) Determine the time it takes for the borrowed sums of ₹8000 at 12% per annum and ₹9100 at 10% per annum to become equal.
(3) Find the number of years needed for A’s loan of ₹8000 at 12% per annum and B’s loan of ₹9100 at 10% per annum to reach an equal amount.
(4) Calculate the time required for the borrowed amounts of ₹8000 at 12% per annum and ₹9100 at 10% per annum to be equal.
(5) How many years will it take for the amounts borrowed by A and B to be equal, with A borrowing ₹8000 at 12% per annum and B borrowing ₹9100 at 10% per annum simple interest?