Question
A certain sum of money is invested in two schemes:
• Scheme A: Compound interest at 10% per annum, compounded annually, for 3 years. • Scheme B: Simple interest at 12% per annum for 4 years, on the same principal. The interest earned from Scheme B exceeds the interest earned from Scheme A by Rs. 1,490. Find the principal and also the total interest earned from both schemes together.Solution
Let principal = P. Interest from Scheme A (CI @ 10% for 3 years): CI = P[(1.10)^3 − 1] = P(1.331 − 1) = 0.331P. Interest from Scheme B (SI @ 12% for 4 years): SI = P × (12 × 4)/100 = 0.48P. Given: 0.48P − 0.331P = 1,490 0.149P = 1,490 P = 1,490 / 0.149 = 10,000. Now: Interest from Scheme A = 0.331 × 10,000 = Rs. 3,310. Interest from Scheme B = 0.48 × 10,000 = Rs. 4,800. Total interest from both schemes = 3,310 + 4,800 = Rs. 8,110. Answer: Principal = Rs. 10,000; total interest = Rs. 8,110.
- What will come in the place of question mark (?) in the given expression?
(198/13) X (52/11) - ? ÷ 5 = 13 + 68 ÷ 4 What will come in the place of question mark (?) in the given expression?
√4096 + √3249 = (?)2
Simplify the following expressions and choose the correct option.
(96 ÷ 8 + 72 ÷ 9) * 3
((12+12+12+12)÷4)/((8+8+8+8+8+8)÷16) = ?
{(5/8) + (4/5)} × (?/19) = 33
(72 × 52 + 1555 )/(79+60) = 2000 ÷ ?
What will come in the place of question mark (?) in the given expression?
25% of 1280 + (41 × 4) = ?2
- What will come in place of (?) in the given expression.
(14)² – (12)² = ? What value should come in the place of (?) in the following questions?
√(60 + 82 + 101) * 5 = ?
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