RRB NTPC 18 January 2017 Paper

Question : 46 of 120

Marks: +1, -0

K borrowed ₹ P at a compound interest of 20% p a. for 2 years. Interest amount payable was ₹ 5,280. What was the value of P?

Solution:

\ { C } = \ { P } \[ \( 1 + \ { \ { R } } /{ 100 } \) ^ { 1 } - 1 \]

\⇒ 5280 = \ { P } \[ \( 1 + \ { 20 } /{ 100 } \) ^ { 2 } - 1 \]

\⇒ 5280 = P \[ \ { 36 }/ { 25 } - 1 \]

\⇒ 5280 = P \[ \ { 11 } /{ 25 } \]

\⇒ P = \ { 5280 } /{ 11 } \× 25 = 12000

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