Simple Interest Quiz Set 009

Shortcut is required here. The addition would be same as if R = 2%, T = 3 years and P = Rs. 200, which is ${200 × 2 × 3}/100$ = 12. So new amount is 254 + 12 = Rs. 266.

P = $(I × 100)/(R × T)$. Solving, we get P = $(768 × 100)/(4 × 4)$ = Rs. 4800.

Let the amount x be lent at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = ${x × r_1 × 6}/100$ + ${(P - x) × r_2 × 6}/100$, which simplifies to 100I = 6 × ($x × (r_1 - r_2) + P × r_2$). Putting r₁ = 8, r₂ = 3, P = 1100, I = 258, we get x = Rs. 200.

We should use the shortcut technique here. If r₁, t₁, r₂, t₂ and r₃, t₃ be the rates and times for three parts with same interest amount, then the three parts must be in the ratio $1/{r_1 t_1} : 1/{r_2 t_2} : 1/{r_3 t_3}$. In our case r₁ = r₂ = r₃ = 10, which cancels, so the ratio is $1/t_1 : 1/t_2 : 1/t_3$. The product of denominators is 3 × 17 × 15 = 765. Thus, the three parts are in the ratio $255 : 45 : 51$. The parts are: 56160 × $51/{255 + 45 + 51}$, 56160 × $45/{255 + 45 + 51}$ and 56160 × $255/{255 + 45 + 51}$, which are 40800, 7200 and 8160. The smaller is Rs. 7200.

Let the amount x be invested at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = ${x × r_1 × 4}/100$ + ${(P - x) × r_2 × 4}/100$, which simplifies to 100I = 4 × ($x × (r_1 - r_2) + P × r_2$). Putting r₁ = 10, r₂ = 3, x = 700, I = 220, we get P = Rs. 200.