model 1 basic compound interest using formula Practice Questions Answers Test with Solutions & More Shortcuts
compound interest PRACTICE TEST [6 - EXERCISES]
model 1 basic compound interest using formula
model 2 at ci sum becomes ‘n’ times after ‘t’ years
model 3 combination of si & ci
model 4 difference in ci & si
model 5 ci with instalments
model 6 comparing sum in different years
Question : 21 [SSC CGL Prelim 2000]
The compound interest on Rs.10,000 in 2 years at 4% per annum, the interest being compounded half-yearly, is :
a) Rs.828. 82
b) Rs.636.80
c) Rs.912. 86
d) Rs.824.32
Answer »Answer: (d)
Using Rule 1,
A = 10,000$(1 + 2/100)^4$
=10,000$(51/50)^4$ =10824.3216
Interest = 10,824.3216 - 10,000
= Rs.824.32
Question : 22 [SSC CGL Tier-I 2013]
A man borrows Rs.21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years ?
a) Rs.12300
b) Rs.12000
c) Rs.12200
d) Rs.12100
Answer »Answer: (d)
If each instalment be x, then Present worth of first instalment
= $x/{1 + 10/100} = {10x}/11$
= Present worth of second instalment
= $x/(1 + 10/100)^2 = 100/121x$
$10/11x + 100/121x$ = 21000
${110x + 100x}/121 = 21000$
210x = 21000 × 121
$x = {21000 × 121}/210$ = Rs.12100
Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$
Here, n = 2, p = Rs.21000, r = 10%
Each annual instalment
= $p/{(100/{100 + r}) + (100/{100 + r})^2$
= $21000/{100/110 + (100/110)^2}$
= $21000/{100/110 + 10000/12100}$
= $21000/{10/11 + 100/121}$
= $21000/{110 + 100} × 121$
= $21000/210 × 121$ = 12100
Question : 23 [SSC CGL Prelim 2008]
A certain sum, invested at 4% per annum compound interest, compounded half yearly, amounts to Rs.7,803 at the end of one year. The sum is
a) Rs.7,700
b) Rs.7,000
c) Rs.7,500
d) Rs.7,200
Answer »Answer: (c)
Using Rule 1,
Let the sum be P.
As, the interest is compounded half-yearly,
R = 2%, T = 2 half years
A = P$(1 + R/100)^T$
7803 = P$(1 + 2/100)2$
7803 = $(1 + 1/50)^2$
7803 = P$× 51/50 × 51/50$
P = ${7803 × 50 × 50}/{51 × 51}$ = Rs.7500
Question : 24 [SSC CGL Prelim 2005]
In how many years will Rs.2,000 amounts to Rs.2,420 at 10% per annum compound interest?
a) 1$1/2$ years
b) 3 years
c) 2 years
d) 2$1/2$ years
Answer »Answer: (c)
Using Rule 1,
According to question,
2420 = 2000$(1 + 10/100)^t$
$2420/2000 = (11/10)^t$
or $(11/10)^t = 121/100$
or, $(11/10)^t = (11/10)^2$
t = 2 years
Question : 25 [SSC CGL Prelim 2005]
At what rate per cent per annum will Rs.2304 amount to Rs.2500 in 2 years at compound interest ?
a) 4$1/3$%
b) 4$1/2$%
c) 4$1/6$%
d) 4$1/5$%
Answer »Answer: (c)
Using Rule 1,
Let the rate per cent per annum be r. Then,
2500 = 2304$(1 + r/100)^2$
$(1 + r/100)^2 = 2500/2304 = (50/48)^2$
$1 + r/100 = 50/48 = 25/24$
$r/100 = 25/24 - 1 = 1/24$
r = $100/24 = 25/6 = 4{1}/6$%
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model 1 basic compound interest using formula Shortcuts »
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model 1 basic compound interest using formula
Defination & Shortcuts … -
model 2 at ci sum becomes ‘n’ times after ‘t’ years
Defination & Shortcuts … -
model 3 combination of si & ci
Defination & Shortcuts … -
model 4 difference in ci & si
Defination & Shortcuts … -
model 5 ci with instalments
Defination & Shortcuts … -
model 6 comparing sum in different years
Defination & Shortcuts …
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