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Question 1 of 5
1. Question
Abhi invested some amount on scheme ‘P’ which offer CI at the rate of 15% p.a.. After 2 years he got Rs. 1032 as interest. Abhi invest the amount he got from scheme ‘P’ in scheme ‘Q’ which offer 15% p.a. at SI for 4 years. Find the total interest he earned from scheme P and scheme Q together?
Correct
Solution: Option (a)
Explanation:
Let, sum Abhi have initially = 100x
100x × (115/100) × (115/100) -100x = 1032
132.25x – 100x = 1032
⇒ x = 1032/32.25 = 32
Amount initially Abhi have = 3200
Interest earned from scheme Q = (4232×15×4)/100 = 2539.2
Required amount = 2539.2 + 1032 = 3571.2
Incorrect
Solution: Option (a)
Explanation:
Let, sum Abhi have initially = 100x
100x × (115/100) × (115/100) -100x = 1032
132.25x – 100x = 1032
⇒ x = 1032/32.25 = 32
Amount initially Abhi have = 3200
Interest earned from scheme Q = (4232×15×4)/100 = 2539.2
Required amount = 2539.2 + 1032 = 3571.2
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Question 2 of 5
2. Question
Raheem invested Rs.x in simple interest at the rate of 12% per annum for 5 years and Sameer invested Rs.(x + 1000) in compound interest at the rate of 10% per annum for two years. If Raheem received the interest amount is Rs.1350 more than that of Sameer, then find the value of x.
Correct
Solution: Option (a)
Explanation:
SI = P * N * R/100
CI = P * (1 + R/100)n – P
SI = x * 12 * 5/100 = 0.6x
CI = (x + 1000) * (1 + 10/100)2 – (x + 1000)
= 21/100 * (x + 1000)
0.6x – 21x/100 – 210 = 1350
60x – 21x – 21000 = 135000
x = Rs.4000
Incorrect
Solution: Option (a)
Explanation:
SI = P * N * R/100
CI = P * (1 + R/100)n – P
SI = x * 12 * 5/100 = 0.6x
CI = (x + 1000) * (1 + 10/100)2 – (x + 1000)
= 21/100 * (x + 1000)
0.6x – 21x/100 – 210 = 1350
60x – 21x – 21000 = 135000
x = Rs.4000
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Question 3 of 5
3. Question
If a certain sum of money quadruples (becomes 4 times its original value) in 4 years under compound interest, how many years will it take for the amount to become 64 times the original value?
Correct
Solution: Option (c)
Explanation:
P * (1 + R/100)4 = 4P
(1 + R/100)4 = 4
((1 + R/100)4)3 = 43
P * (1 + R/100)12 = 64P
n = 12 years
Incorrect
Solution: Option (c)
Explanation:
P * (1 + R/100)4 = 4P
(1 + R/100)4 = 4
((1 + R/100)4)3 = 43
P * (1 + R/100)12 = 64P
n = 12 years
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Question 4 of 5
4. Question
Divide Rs. 11000 into two parts such that the amount of simple interest earned on the first part over 3 years at an annual interest rate of 5% equals the amount of simple interest earned on the second part over 2 years at an annual interest rate of 9%. Find the values of the two parts.
Correct
Solution: Option (b)
Explanation:
Let two parts be x and 11000 – x
(x*3*5)/100 = [(11000 – x)*9*2]/100
15x = 198000 – 18x
33x = 198000
x = 198000/33 = 6000
Two parts are Rs. 6000 and Rs. 5000
Incorrect
Solution: Option (b)
Explanation:
Let two parts be x and 11000 – x
(x*3*5)/100 = [(11000 – x)*9*2]/100
15x = 198000 – 18x
33x = 198000
x = 198000/33 = 6000
Two parts are Rs. 6000 and Rs. 5000
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Question 5 of 5
5. Question
Usha invests Rs.x in a simple interest scheme at the rate of 19% per annum for 4 years. After 4 years, he received the interest is Rs.(1872 + x/2). If Sujay invests Rs.x in a compound interest scheme at the rate of 20% per annum for 2 years, then find the total amount earned by Sujay after 2 years?
Correct
Solution: Option (d)
Explanation:
(1872 + x/2) = x * 19 * 4/100
46800 + 12.5x = 19x
x = 7200
CA received by Sujay = 7200 * (1 + 20/100)2 = Rs.10368
Incorrect
Solution: Option (d)
Explanation:
(1872 + x/2) = x * 19 * 4/100
46800 + 12.5x = 19x
x = 7200
CA received by Sujay = 7200 * (1 + 20/100)2 = Rs.10368
