exam Maths Puzzle

Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of 3 years and to $ 400 at the end of another 5 years. What was the rate of interest in which he invested his sum?

Read Solution (Total 2)

33.33%
If the Sum invested=p,Time=t,Rate in % =r
Given p +(p*3*r)/100 =300 or 100p+3pr=30000 ---(i) &
p+ (p*5*r)/100 =400 or 100p+5pr=40000 ---(ii)
Solving (i) & (ii), pr=5000
Substituting value of pr in (i),100p+3*5000=30000, p=150
Substituting value of p in (i),100*150+3*150*r=30000, r=300/9 =33.33 %
11 years agoHelpfull: Yes(0) No(0)

Initial amount invested = $ X

Amount at the end of year 3 = $ 300

Amount at the end of year 8 (another 5 years) = $ 400

Therefore, the interest earned for the 5 year period between the 3rd year and 8th year = $400 - $300 = $100

So, interest earned per year = 10051005 = $20.

Therefore, interest earned for 3 years = 3 * 20 = $ 60.

Hence, initial amount invested X = Amount after 3 years - interest for 3 years

= 300 - 60 = $ 240.

Rate of interest = interestperyearprincipalinvestedinterestperyearprincipalinvested * 100 = 2024020240 * 100 = 8.33%
7 years agoHelpfull: Yes(0) No(1)

exam Other Question

Ann invested a certain sum of money in a bank that paid simple interest. The amount grew to $240 at the end of 2 years. She waited for another 3 years and got a final amount of $300. What was the principal amount that she invested at the beginning? In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?