Simple Interest Quiz Set 008

Shortcut is required here. The addition would be same as if R = 2%, T = 9 years and P = Rs. 3200, which is ${3200 × 2 × 9}/100$ = 576. So new amount is 4640 + 576 = Rs. 5216.

Let the amount x be invested at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = ${x × r_1 × 6}/100$ + ${(P - x) × r_2 × 6}/100$, which simplifies to 100I = 6 × ($x × (r_1 - r_2) + P × r_2$). Putting r₁ = 9, r₂ = 4, P = 1200, I = 558, we get x = Rs. 900.

His gain is ${(1500 + x) × 8 × 19}/100$ - ${1500 × 4 × 19}/100$ = 2052. We can solve this for x to get x = Rs. 600.

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 5 × 11 × 18 = 990. Thus, the three parts are in the ratio $198 : 90 : 55$. The parts are: 34300 × $55/{198 + 90 + 55}$, 34300 × $90/{198 + 90 + 55}$ and 34300 × $198/{198 + 90 + 55}$, which are 19800, 9000 and 5500. The smaller is Rs. 5500.

If I is the interest, and principal is P, time is P, and rate is P, then, I = $(P × P × P)/100$. Which gives P = $√^3{100 × I}$, which is $√^3{100 × 3430}$, which is $√^3{1000 × 7 × 7 × 7}$ = Rs. 70.