Question
Rs. 2,800 is invested in three parts all at simple
interest of 5% p.a. These three parts are invested randomly for 1 year, 2 years and 4 years such that the interest received from the given three parts after desired period of investment is same. Find the difference between the smallest and the largest part of the investment.Solution
Let the amount invested in three parts be Rs. ‘p’, Rs. ‘q’ and Rs. ‘r’ ATQ; (p × 0.05 × 1) = (q × 0.05 × 2) = (r × 0.05 × 4) Or, p = 2q = 4r Let p = 2q = 4r = ‘4t’ So, p = 4t
q = 2t
r = t So, p + q + r = 2800 Or, 4t + 2t + t = 2800 Or, 7t = 2800 Or, t = 400 So, required difference = 4t − t = 3t = 3 × 400 = Rs. 1,200
More Simple and compound interest Questions
(2197)1/3 + (18)2 − 121 = ? − 69 × 5
Solve the following equation.
143 + 14.3 + 1.43 + 0.143 + 0.0143 =?
Simplify the following expression:
(√121 + √196) × 7 =? × 5
- What will come in the place of question mark (?) in the given expression?
62.5% of 120 + ? = (720 + 90) ÷ √36 (54/6) × 5 + 12 × (17/2) = ?% of 700
What is the value of 7/9-11/12+12/16-1/8?
- What will come in the place of question mark (?) in the given expression?
40% of (320 ÷ 4) + 2² X 25 = ? + 42 (13)2 - 3127 ÷ 59 = ? x 4
(2 ÷ 3) × (4 ÷ 12) × (? ÷ 10) × 45 × (1 ÷ 5) = (? ÷ 6) + (2 ÷ 5)
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