The difference between compound interest and simple interest at the same rate on Rs 5000 for 2 years is Rs 72. The rate of interest per annum is:___________?
A. 6%
B. 8%
C. 10%
D. 12%
A. 6%
B. 8%
C. 10%
D. 12%
A. 5%
B. 6%
C. 10%
D. 15%
Explanation:
P = Rs 20000, A =Rs 24200,
t = 2 years 20000 × (1 + R/100)2 = 24200
=> (1 + R/100)2 = 24200/20000 = 121/100 = (11/10) 2
=> 1 + R/100 = 11/10
=> R/100 = (11/10 – 1) = 1/10
=> R = (100 × 1/10) % p.a = 10 % p.a
Hence, Rate = 10 % p.a
A. Rs 6.06%
B. Rs 6.07%
C. Rs 6.08%
D. Rs 6.09%
Explanation:
Let the sum be Rs 100.
Then P = Rs 100, R = 3 %
per half – year, t = 2 half – years
Amount = Rs [100 × (1 + 3/100)2]
= Rs (100 × 103/100 × 103/100)
= Rs 10609/100 = Rs 106.09
Effective Annual Rate = 6.09%
A. Rs 6500
B. Rs 6565
C. Rs 65065
D. Rs 65650
Explanation:
Let the sum be Rs x.
Then [x × (1 +10/100)2 – x] – (x × 10/100 × 2) = 65
=> (x × 11/10 × 11/10 – x) – x/5 = 65
=> (121x/100 – x) – x/5 = 65
=> (21x/100 – x/5) = 65
=> (21x – 20x) = 6500
=> X = 6500
A. Rs 230
B. Rs 232
C. Rs 600
D. Rs 832
Explanation:
S.I = Rs (6000 × 5/100 × 2) = Rs 600
C.I = Rs [5000 × (1 + 8/100)2 – 5000]
= Rs (5000 × 27/25 × 27/25 – 5000)
= Rs(5832 – 5000)
= Rs 832
(C.I) – (S.I) = Rs (832 – 600) = Rs 232
A. Rs 4500
B. Rs 5000
C. Rs 5500
D. Rs 6000
Explanation:
Let the money borrowed be Rs x
Interest paid by the money lender = Rs (x × 4/100 × 1) =Rs x/25
Interest received by the money lender
= Rs [x × (1 + 3/100)2 – x]
= Rs(x × 103/100 × 103/100 – x)
Gain = Rs (609x/1000 – x /25)
= Rs 209x/1000 = Rs 609x/10000
Therefore 209x/10000 = 104.50
=> 209x = 1045000
=> x = 5000
Hence money borrowed = Rs 5000
A. Rs 1575.20
B. Rs 1600
C. Rs 1625.80
D. Rs 2000
Explanation:
Amount = Rs[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)]
= Rs(1000 × 26/25 × 21/20 × 53/50)
= Rs (57876/5) = Rs 11575.20
C.I = Rs (11575.20 – 10000) = Rs 1575.20
A. 7 years
B. 10 years
C. 15 years
D. 20 years
Explanation:
Let principal be Rs x and the rate is R % p.a.
Then X × (1 + R/100)5 = 2x
=> (1 + R/100)5 = 2
Let x × (1 + R/100)t = 8x
=> (1 + R/100)t = 8 = 23
={(1 + R/100)5 }3
=> (1 + R/100)t = (1 + R/100)15
=> T = 15 Years
A. Rs 100000
B. Rs 11000
C. Rs 120000
D. Rs 170000
Explanation:
Let the sum be Rs x.
Then S. I = Rs (x × 8/100 × 2) = Rs 4x/25
C. I = Rs [x × (1 + 8/100)2 – x]
= Rs (x × 27/25 × 27/25 – x)
=Rs 104x/625 (C.I) – (S.I)
= Rs(104x/625 – 4x /25)
= Rs 4x /625
Therefore 4x/625 = 768
=> x = ((768 ×625)/4) = 120000
Therefore sum = Rs 120000
A. Rs 900
B. Rs 1000
C. Rs 1050
D. Rs 1100
Explanation:
Let the principal be Rs x.
Then X × (1 + 10/100)3 – x = 331
=> (x × 11/10 × 11/10 × 11/10 – x) = 331
=> ((1331x-1000x)/1000) = 331
=> 331x = 331000
=> X = 1000
Hence the principal is Rs 1000