The simple interest on a certain sum of money after x years is Rs. 247

1.	The simple interest on a certain sum of money after x years is Rs. 247.50 and after 2 more years is Rs. 412.50. Find the simple interest on the same sum at the same rate of interest after 10 years.1). Rs. 4952). Rs. 5773). Rs. 6604). Rs. 825
Answer» Let the sum of money be RS. ’P’ and the RATE of interest be ‘R%’ PER annum As we know, Simple interest ⇒ SI = (P × R × T)/100 Given, SI = Rs. 247.50 after T = x YEARS ⇒ 247.50 = (PR × x)/100 ⇒ PR = 24750/x----(1) Also, SI = Rs.412.50 after T = (x + 2) years ⇒ 412.50 = (PR × (x + 2))/100 ⇒ PR = 41250/(x + 2)----(2) Equating (1) and (2), we get, ⇒ 41250/(x + 2) = 24750/x ⇒ 41250x = 24750x + 49500 ⇒ 16500x = 49500 ⇒ x = 49500/16500 ⇒ x = 3 Substituting in (1), PR = 24750/3 = 8250 Hence, SI after T = 10 years, SI = (PR × 10)/100 ⇒ SI = (8250 × 10)/100 ⇒ SI = Rs. 825

The simple interest on a certain sum of money after x years is Rs. 247.50 and after 2 more years is Rs. 412.50. Find the simple interest on the same sum at the same rate of interest after 10 years.1). Rs. 4952). Rs. 5773). Rs. 6604). Rs. 825

Answer»

Let the sum of money be RS. ’P’ and the RATE of interest be ‘R%’ PER annum

As we know, Simple interest

⇒ SI = (P × R × T)/100

Given, SI = Rs. 247.50 after T = x YEARS

⇒ 247.50 = (PR × x)/100

⇒ PR = 24750/x----(1)

Also, SI = Rs.412.50 after T = (x + 2) years

⇒ 412.50 = (PR × (x + 2))/100

⇒ PR = 41250/(x + 2)----(2)

Equating (1) and (2), we get,

⇒ 41250/(x + 2) = 24750/x

⇒ 41250x = 24750x + 49500

⇒ 16500x = 49500

⇒ x = 49500/16500

⇒ x = 3

Substituting in (1),

PR = 24750/3 = 8250

Hence, SI after T = 10 years,

SI = (PR × 10)/100

⇒ SI = (8250 × 10)/100

⇒ SI = Rs. 825

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