Question
Person 'P' invested ₹4y at a simple interest rate of
35% per annum and ₹3y at a compound interest rate of 20% per annum (compounded annually), both for a duration of 2 years. The simple interest earned exceeds the compound interest by ₹3,700. Determine the total amount invested by 'P'.Solution
Simple interest = (Principal X time X rate) ÷ 100 = (4y X 2 X 35) ÷ 100 = Rs. '2.8y' CI = P X (1 + r/100)t - 1) , where 'P' is the sum invested, 'R' is the annual rate of interest and 'T' is the time period. CI = 3y X (1 + 20/100)2 - 1) Or, CI = 3y X [(1 + 0.2) 2 - 1] Or, CI =3y X [(1.2) 2 - 1] So, CI = 3y X 0.44 = Rs. '1.32y' ATQ, 2.8y - 1.32y = 3,700 Or, 1.48y = 3,700 So, 'y' = 2,500 Therefore, total sum invested by 'P' = 3y + 4y = 7y = 7 X 2,500 = Rs. 17,500
√? = 120 - 102 + ∛125
350% of (450 / 1.5) = ?% of 4200
The value of (43-3× (4×6+12/6 ×6-4×5) +4)?
? = 6.25% of 240 + 25 2 + 17 2 – 16 × 17
What value should come in place of (?) question mark.
(5/9 of 396) − (35% of 320) + 7.2 = ?
What will come in the place of question mark (?) in the given expression?
[{(24) 2 ÷ (6) 2 } ÷ 8]4 = ? ÷ 5
The value of 42 ÷ 9 of 6 - [64 ÷ 48 x 3 – 15 ÷ 8 x (11 – 17) ÷ 9] ÷ 14 is:
3.3 Times 2/27 of 40% of 364=?
What value should come in place of the question mark (?) in the following question?
2000 ÷ 10 + 250 × 2 �...
Simplify the following expressions and choose the correct option.
(3/4 of 528) − (2/3 of 135) = ?