Question
Rahul placed Rs. 40,000 between two investment options,
‘E’ and ‘F’, for 6 years and 3 years, respectively. Option ‘E’ accrues simple interest at 10% per annum, and option ‘F’ grows by compound interest (compounded annually) at 20% per annum. What amount was put into option ‘F’ if the interest from ‘E’ exceeds that from ‘F’ by Rs. 1,200?Solution
ATQ, Let the investment in option ‘F’ be Rs. ‘z’. Investment in ‘E’ = Rs. (40000 - z). Simple interest from ‘E’ = (40000 - z) × 10% × 6. Compound interest from ‘F’ = z × [{1 + (20/100)}3 - 1]. Solving the equation for ‘z’ based on the interest difference, z = Rs. 15,000.
105   107   111   114   ?   127
8 6 2 ? -1 9.5
...64Â Â Â 48Â Â Â 36Â Â Â 22Â Â Â Â 16Â Â Â Â 8
(45)2 ÷ ∛729 + (35)2 ÷ 1.4 =?
24, 33, 50, ?, 108, 149
16 4 2 1.5 1 1.875
...Find the wrong number in the given series.
3 1.5 1.5 ? 4.5 11.25
...18Â Â Â Â Â Â Â Â Â Â 434Â Â Â Â Â Â Â Â 642Â Â Â Â Â Â Â Â 746Â Â Â Â Â Â Â Â 798Â Â Â Â Â Â Â Â ?
...720    ?     240     180      144     120
...If  204    196       223   x  284
Then, what is the average of the numbers of the above series?
...
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