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SIMPLE AND COMPOUND INTEREST INTEREST
2. Find the amount to be paid back bac k on a loan of
EXAMPLE
Interest is the fixed amount paid on borrowed money. The sum lent is called the Principal. The sum of the principal and interest is called the Amount. Interest is of two two kinds : (i) Simpl implee int inter eres estt (ii) Comp Compou ound nd intere interest st (i) (i) Simp imple Inte Interrest est : When interest is calculated on the original principal for any length of time, it is called called simple simple interest. Simple interest S.I.
i.e.
P r in incipal Time Rate
18,000 at 5.5% per annum annum for 3 years Sol. P = 18000, R = 5.5%, T = 3 years `
S.I.
`
3. In how many many years will wil l a sum of money triple
EXAMPLE
itself, at 25% per annum simple simple interest. Sol. Let the sum of money be be P. So, A = 3P `
and S.I. = A – P = 3P – P = 2P
100
R = 25%
P R T 100
A= P+ I P
Principal (P) Rate (R) Time (T)
RT P 1 100 100
PRT
T
EXAMPLE
P R
1 00 2 P P 25
= 8 years
4. What rate per cent per annum will produce
EXAMPLE
250 as simple interest on
6000 in 2.5 years
`
R T
`
Rate
100 S.I. TP
S.I. 100 250 100 10 5 1 2 % 6000 2.5 6 3 3 PT
5. To buy furniture for a new n ew apartment, Sylvia
EXAMPLE
Chang borrowed borrowed 5000 at 11 % simple interest interest for 11 months. months.
100 S.I.
How much much interest will w ill she pay?
P R
(R1 R 2 R 3 .....) 100
Sol. From the formula, I = Prt, with P = 5000, r = .11, and t = 11/12 (in years). The total interest she s he will pay is
I = 5000 (.11) (11/12) = 504.17 or
1. Find the interest to be paid on a loan of of 6000
at 5% per year for 5 years ye ars Sol. P = 6000, R = 5% and T = 5 years `
S.I.
100 S.I.
Sol. P = 6000; Time Tim e (T) = 2.5 year years; s; S.I. = 250
100 S.I.
If rate of simple interest diffe differs rs from year year to year, year, then S.I. P
`
Amount = P + I = 18000 + 2970 = 20970
Amount = Principal + Interest i.e.
P R T 18000 5.5 3 = 2970 100 100
PR T 6000 5 5 = 1500 100 100 `
(ii) (ii)
504.17
`
Comp Compou ound nd Int Inter eres estt : Money is said to be lent at a t compound
interest when at the end of a year or other fixed period, the interest that t hat has become due is not paid to the lender, but is added to the sum lent, and a nd the amount am ount thus obtained becomes becomes the principal in the next year or period. The process is repeated until the amount for the last period has been found. Hence, Hence, When the interest charged after a certain specified time
2
period is added to form new principal for the next time period, the interest is said to be compounded and the total interest accrued is compound interest.
n
10920.25
r Amount (A) P 1 100
n
20.9 400 20
2
n
`
9. Suppose 1000 is deposited for 6 years in an
account paying 8.31% per year compounded annually.
6. Find the compound interest on 70000 for 4 years at the rate of 14% per annum compounded annually. EXAMPLE
Sol. P = 70000, n = 4, r = 14% `
n
n
`
EXAMPLE
r r r Amount P 1 1 1 2 1 3 ....... 100 100 100
n
20.9 20.9 20 20 Hence 10000 will become 10920.25 in 2 years at 4.5%. 436.81
If rate of compound interest differs from year to year, then
4
r 14 A P 1 70000 1 = 118227.20 100 100 `
C.I. = A – P= 118227.20 – 70000 = 48227.20 `
7. If
n
0.9 20.9 1 10000 20 20
n r ; –1 C.I. P 1 100
EXAMPLE
10920.25 10000 1 4.5 100
60000 amounts to 68694 in 2 years then
find the rate of interest.
(a) Find the compound amount. In the formula above, P = 1000, i = .0831, and n = 6. The compound amount is A = P (1 + i) n A = 1000 (1.0831) 6 A = 1614.40. (b) Find the amount of interest earned. Subtract the initial deposit from the compound amount. `
Amount of interest = 1614.40 – 1000 = 614.40. `
`
`
Compound interest – when interest is compounded annually
Sol. Given : A = 68694 P = 60000 n = 2 years r =? `
but time is in fraction
`
A P 1
r 100
If time = t
n
68694 60000 1
r 100 2
1
1
100
r
100
r 1 10000 100 11449
years, then
EXAMPLE
p r q 1 100
10. Find the compound interest on 8000 at 15%
per annum for 2 years 4 months, compound annually. Sol. Time = 2 years 4 months = 2
1 years = 2 years 3 12 4
1.07
1.07 – 1 0.07
Amount
`
r = 0.07 × 100 = 7%
EXAMPLE
t
2
r 11449 1.1449 100 10000 r
q
r A P 1 100
2
r 1 60000 100 68694
p
8. In how many years, the sum of
10000 will
become 10920.25 if the rate of compound interest is 4.5% per annum? Sol. A = 10920.25 P = 10000 Rate of interest = 4.5% Time (n) = ? `
`
1 2 15 15 1 3 8000 1 100 100 23 23 21 11109 8000 20 20 20 `
C.I. = (11109 – 8000) = 3109. `
`
`
A P 1
r 100
n
Compound interest – when interest is calculated half-yearly
Since r is calculated half-yearly therefore the rate per cent will become half and the time period will become twice, i.e.,
3
Rate per cent when interest is paid half-yearly
r 2
When T = 3 %
and time = 2 × time given in years Hence,
C.I. – S.I. =
104
100
S.I. R R 3 100 (ii) C.I. – S.I. = 3 100 2
A P 1
2 100 r
2n
11. What will be the compound interest on 4000 in 4 years at 8 per cent annum. If the interest is calculated halfyearly. Sol. Given : P = 4000, r = 8%, n = 4 years Since interest is calculated half-yearly, therefore, `
8 2
13. The difference between compound interest and
EXAMPLE
EXAMPLE
r
(i)
PR 2 300 R
simple interest on a certain amount of money at 5% per annum for 2 years is 15. Find the sum : (a)
4500
(b)
7500
(c)
5000
(d)
6000
Sol. (d) Let the sum be 100. `
Therefore, SI 100 5 2 100
% 4% and n = 4 × 2 = 8 half years 8
4 A 4000 1 100
26 4000 25
`
10
8 2
5 and CI 1001 100 100
= 4000 × 1.3685 = 5474.2762 Amount = 5474.28 Interest = Amount – Principal = 5474.28 – 4000 = 1474.28 Compound interest – when interest is calculated quarterly Since 1 year has 4 quarters, therefore rate of interest will become `
`
1 4
`
21 21 100 20 20
100
`
41 4
`
Difference of CI and SI
th of the rate of interest per annum, and the time period will be
1
If the difference is
4 times the time given in years Hence, for quarterly interest
4
41 4
10
1 4
, the sum = 100
If the difference is 15, the sum `
r/4 A P 1 100
4n
= 400 × 15 = 6000
4n
r P 1 400
12. Find the compound interest on 25625 for 12 months at 16% per annum, compounded quarterly. Sol. Principal (P) = 25625 EXAMPLE
`
16 Rate (r) = 16% % 4% 4
`
and the compound interest compounded annually at the rate of 12% per annum on 5000 for two years will be : (a)
47.50
(b)
63
(c)
45
(d)
72
Sol. (d) Required difference
Time = 12 months = 4 quarters
4
4 26 A 256251 25625 100 25
= 50001
4
5000 12 2 5000 100 100 12
2
28 28 1 1200 25 25
26 26 26 26 = 29977.62 25 25 25 25 C.I. = A – P = 29977.62 – 25625 = 4352.62 25625
14. The difference between the simple interest
EXAMPLE
= 5000
`
`
784 625 1200 72 625
= 5000
Difference between Compound Interest and Simple Interest When T = 2
(i)
R C.I. – S.I. = P 100
(ii) C.I. – S.I. =
R S.I. 2 100
2
`
EFFECTIVE RATE
If 1 is deposited at 4% compounded quarterly, a calculator can be used to find that at the end of one year, the compound amount is 1.0406, an increase of 4.06% over the original 1. The actual increase of 4.06% in the money is somewhat `
`
`
4
higher than the stated increase of 4%. To differentiate between these two numbers, 4% is called the nominal or stated rate of interest, while 4.06% is called the effective rate. To avoid confusion between stated rates and effective rates, we shall continue to use r for the stated rate and we will use r e for the effective rate. EXAMPLE
`
15. Find the effective rate corresponding to a
stated rate of 6% compounded semiannually. Sol. A calculator shows that semiannually will grow to
`
100 at 6% compounded
2
Thus, the actual amount of compound interest is 106.09 – 100 = 6.09. Now if you earn 6.09 interest on 100 in 1 year with annual compounding, your rate is 6.09/100 = .0609 = 6.09%. Thus, the effective rate is r e = 6.09%. `
`
`
`
r e
(a) (c)
=
P
P
r
r m
m
1
m
or x
or y
Sol. Use the formula given above with r = .049 and m = 12. 12
1
= 1.050115575 – 1 .0501 or 5.01%
Present worth of P due n years hence `
P r 1 100
100
1020 100 110
927.27 `
1003 20 20 22 22
`
2
828.92
990 20 20 20
743.80 22 22 22 Hence, CP = 1500 + 927.27 + 828.92 + 743.80 = 3999.99 or 4000. `
`
18. The difference between the interest received
EXAMPLE
Sol. I 1 =
monthly. Find the effective rate.
Present worth =
10 Similarly, 1003 y1 100
16. A bank pays interest of 4.9% compounded
.049 The effective rate is re 1 12
10
from two different banks on 500 for 2 yrs is difference between their rates.
m
I2 =
Equal annual instalment to pay the borrowed amount
2.5. Find the
500 2 r1 = 10 r 1 100 500 2 r2
= 10 r 2 100 I1 – I2 = 10r 1 – 10r 2= 2.5 Or, I1 – I2=
2.5
= 0.25% 10 Examination method : When t1= t2, (r 1 –r 2) =
Id 100 sum t
2.5 100 = 0.25% 500 2
19. At what rate per cent compound interest does a sum of money becomes nine - fold in 2 years? Sol. Let the sum be x and the of compound interest be r% per annum; then EXAMPLE
n
n
2,000 5,000
Then, 1020 x1
1
P
r 1 100
`
and z
(b) (d)
`
principal P 1
2
x
`
compound amount principal
m
r 1 100
3,000 4,000
principal
= re 1 EXAMPLE
r 1 100
.....
Sol. (c) Cash down payment = 1500 Let x becomes 1020 at the end of first year.
compound interest
r P 1 m
x
that he is required to pay 1,500 cash down payment followed by 1,020 at the end of first year, 1,003 at the end of second year and 990 at the end of third year. Interest is charged at the rate of 10% per annum. Calculate the cash price :
In the preceding example we found the effective rate by dividing compound interest for 1 year by the original principal. The same thing can be done with any principal P and rate r compounded m times per year. Effective rate
x
17. Subash purchased a refrigerator on the terms
EXAMPLE
.06 A 100 1 100 (1.03) 2 $ 106.09 2
`
Let the value of each instalment = x Rate = r% and time = n years Then, Borrowed Amount
`
5
r 9x = x 1 100
r
or, 3 = 1+
100
2
r or, 9 = 1 100
=
1600 100 5 7000
32
=
7
%
ratio of two amounts = 2 : 5
r
; or,
2
100
= 2
6%
4%
r = 200% Examination method : The general formula of compound interest can be changed to the following form : If a certain sum becomes ‘m’ times in ‘t’ years, the rate of
32 7
%
1/ t compound interest r is equal to 100 (m) 1
In this case , r = 100 (9)
1
1/ t
4
= 100 (3 – 1) = 200%
7
20. The simple interest on a certain sum of money
EXAMPLE
at 4% per annum for 4 years is 80 more than the interest on the same sum for 3 years at 5% per annum. Find the sum. Sol. Let the sum be x, then at 4% rate for 4 years the simple
Interest =
100
`
x 53 100
=
`
Now, we have ,
or
16x 15x 100
2 years and upto 5324 in 3 years on compound interest. Find the rate percent Sol. We have,
P + CI of 3 yrs = 5324.......(1)
20
P + CI of 2yrs = 4840........(2)
25
`
`
3x 20
Subtracting (2) from (1), we get
80
CI of 3rd year = 5324 – 4840 = 484. `
Thus, the CI calculated in the third year which is 484 is basically the amount of interest on the amount generated after 2 years which is 4840. `
x = 8000
= 80
`
`
For this type of question Difference 100 [r2 t1 r2 t 2 ]
r =
80 100
4 4 3 5 = 8000
484 100 4840 1
=10%
Examination method :
`
Difference of amount after n yearsand(n 1) years 100
21. Some amount out of
EXAMPLE
4840 in
25
3x
4x
× 2 = 2000 `
7
22. As n amount of money grows upto
EXAMPLE
Examination Method :
Sum =
7000
%
4x
At 5%rate for 3 yrs the simple interest =
7
amount lent at 6 % =
`
x 44
10
%
7000 was lent at 6 %
per annum and the remaining at 4 % per a nnum. If the total simple interest from both the fractions in 5 years was 1600, find the sum lent at 6 % per annum.
Amountafter 2 years
In this, n = 2.
rate = Difference of
amount after 2 yearsand3years 100 Amountafter 2 years
Sol. Suppose x was lent at 6 % per annum. `
Thus,
or,
or,
3x 10
x 65 100
+
+
(7000 x) 4 5
7000 x 5
100
= 1600
3x 14, 000 2x 10
= 1600
=
(5324 4840) 4840
484 100 4840
= 10%
23. A certain amount of money at compound
EXAMPLE
interest grows upto 51168 in 15 yrs and upto years. Find the rate per cent per annum.
= 1600
x = 16000 – 14000 = 2000
× 100 =
Sol. Rate =
(51701 51168) 100 51168
`
By Method of Alligation : Overall rate of interest
=
100 96
=
25 24
=1
1 24
%
=
533 100 51168
51701 in 16
6 24. Find the compound interest on 18,750 in 2 years the rate of interest being 4% for the first year and 8%
EXAMPLE
for the second year. Sol. After first year the amount
= 18750 1
4 104 = 18750 100 100
104 108 100 100
After 2nd year the amount = 18750
26 25
= 18750
27 = 21060 25
CI = 21060 –18,750 = 2310. `
A computer gives the following results for various values of n.
n
n
1 1 n
Annually
1
1 1 2 1
Semiannually
2
1 1 2.25 2
Quarterly
4
1 1 4 2.4414
Monthly
12
1 1 12
Daily
365
1 1 365
Hourly
8760
1 1 8760
Every minute
525,600 2.7182792
1 1 525,600
31,536,000
1 1 31,536,000
Interest is compounded
1
2
4
Every second
12
2.6130 365
2.71457 8760
2.7182818
2.718127 525,600
31,536,600
7
EXERCI SE 1.
Arun borrowed a sum of money from Jayant at the rate of 8% per annum simple interest for the first four years, 10% per annum for the next six years and 12% per annum for the period beyond ten years. If he pays a total of 12,160 as interest only at the end of 15 years, how much money did he borrow? (a) 8000 (b) 10,000 (c) 12,000 (d) 9,000 (e) None of these A sum fetched total simple interest of 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum? (a) 8925 (b) 8032.50 (c) 4462.50 (d) 8900 (e) None of these At a simple interest 800 becomes 956 in three years. If the interest rate, is increased by 3%, how much would 800 become in three years? (a) 1020.80 (b) 1004 (c) 1028 (d) Data inadequate (e) None of these On 3,000 invested at a simple interest rate 6 p.c.p.a, 900 is obtained as interest in certain years. In order to earn 1,600 as interest on 4,000 in the same number of years, what should be the rate of simple interest? (a) 7 p.c.p.a. (b) 8 p.c.p.a. (c) 9 p.c.p.a. (d) Data inadequate (e) None of these A sum of money doubles itself in 10 years at simple interest. In how many years would it treble itself? (a) 10 (b) 15 (c) 20 (d) 25 (e) None of these A certain amount earns simple interest of 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned? (a) 35 (b) 350 (c) 245 (d) Cannot be determined (e) None of these What will be the difference in simple and compound interest on 2000 after three years at the rate of 10 percent per annum? (a) 160 (b) 42 (c) 62 (d) 20 (e) None of these Nikhilesh invested certain amount in three different schemes A, B and C with the rate of interest 10 p.c.p.a., 12 p.c.p.a.
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and 15 p.c.p.a. respectively. If the total interest accrued in one year was 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B? (a) 8000 (b) 5000 (c) 6500 (d) Cannot be determined (e) None of these Aniket deposited two parts of a sum of 25000 in different banks at the rates of 15% per annum and 18% per annum respectively. In one year he got 4050 as the total interest. What was the amount deposited at the rate of 18% per annum? (a) 9000 (b) 18000 (c) 15000 (d) Data inadequate (e) None of these Mr X invested an amount for 2 years @ 15 p.c.p.a at simple interest. Had the interest been compounded, he would have earned 450/- more as interest. What was the amount invested? (a) 22000 (b) 24000 (c) 25000 (d) Data inadequate (e) None of these Difference between the compound interest and the simple interest accrued on an amount of 18000, in two years was 405. What was the rate of interest p.c.p.a? (a) 16 (b) 12 (c) 15 (d) Cannot be determined (e) None of these Anish borrowed 15000 at the rate of 12% and an other amount at the rate of 15% for two years. The total interest paid by him was 9000. How much did he borrow? (a) 32,000 (b) 33,000 (c) 30,000 (d) 35,000 (e) None of these The compound interest on any sum at the rate of 5% for two years is 512.50. Find the sum. (a) 5200 (b) 4800 (c) 5000 (d) 5500 (e) None of these Mr Amin borrowed some money from Mr Vishwas. The rate of interest for first two years is 8% p.a., for the next three years is 11 % p.a. and for the period beyond 5 years 14% p.a. Mr Vishwas got an amount of 10920 as an interest at the end of eight years. Then what amount was borrowed by Mr Amin’? (a) 12000 (b) 15000
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(c) 1400 (d) Data inadequate (e) None of these The C.I. on a certain sum of money for the 4th year at 8% p.a. is 486. What was the compound interest for the third year on the same sum at the same rate? (a) 450 (b) 475 (c) 456 (d) 480 (e) None of these Seema invested an amount of 16000 for two years at compound interest and received an amount of 17640 on maturity. What is the rate of interest? (a) 8 pcpa (b) 5 pcpa (c) 4 pcpa (d) Data inadequate (e) None of these Amit Kumar invested an amount of 15,000 at compound interest rate of 10 pcpa for a period of two years. What amount will he receive at the end of two years? (a) 18,000 (b) 18,500 (c) 17,000 (d) 17,500 (e) None of these In a business A and C invested amounts in the ratio 2:1. Whereas the ratio between amounts invested by A and B was 3:2. If 1,57,300 was their profit, how much amount did B receive? (a) 72,600 (b) 48,400 (c) 36,300 (d) 24,200 (e) None of these Mr. Sane invested a total amount of 16,500 for two years in two schemes A and B with rate of simple interest 10 p.c.p.a. and 12 p.c.p.a. respectively. If the total amount of interest earned was 3,620, what was the amount invested in scheme B? (a) 8,000 (b) 8,600 (c) 8,150 (d) Data inadequate (e) None of these The difference between the simple and the compound interest compounded every six months at the rate of 10% p.a. at the end of two years is . 124.05. What is the sum? (a) 10,000 (b) 6,000 (c) 12,000 (d) 8,000 (e) None of these Parameshwaran invested an amount of 12,000 at the simple interest rate of 10 pcpa and another amount at the simple interest rate of 20 pcpa. The total interest earned at t he end of one year on the total amount invested became 14 pcpa. Find the total amount invested. (a) 22,000 (b) 25,000 (c) 20,000 (d) 24,000 (e) None of these Raviraj invested an amount of 10,000 at compound interest rate of 10 pcpa for a period of three years. How much amount will Raviraj get after three years? (a) 12,310 (b) 13,210 (c) 13,320 (d) 13,120 (e) None of these A sum of money doubles itself in 6 years at a certain rate of compound interest. In how many years will be 16 times at the same rate of interest? `
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(a) 18 years (b) 24 years (c) 30 years (d) 12 years (e) None of these Amal borrowed a sum of money with simple interest as per the following rate structure: 1. 6 p.c. p.a. for the first three years 2. 8 p.c. p.a. for the next five years 3. 12 p.c. p.a. for the next eight years If he paid a total of 5,040 as interest at the end of twelve years, how much money did he borrow? (a) 8,000 (b) 10,000 (c) 12,000 (d) 6,000 (e) None of these The simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the r ate of 8 per cent in 8 months. How much amount was borrowed? (a) 15000 (b) 25000 (c) 7500 (d) 14500 (e) None of these On retirement, a person gets 1.53 lakhs of his provident fund which he invests in a scheme at 20% p.a. His monthly income from this scheme will be (a) 2, 450 (b) 2,500 (c) 2, 550 (d) 2, 600 (e) None of these A sum was put at simple interest at a certain rate for 4 years Had it been put at 2% higher rate, it would have fetched 56 more. Find the sum. (a) 500 (b) 600 (c) 700 (d) 800 (e) None of these Simple interest on a certain sum is 16 over 25 of the sum. Find the rate per cent and time, if both are equal. (a) 8% and 8 years (b) 6% and 6 years (c) 10% and 10 years (d) 12 % and 12 years (e) None of these The simple interest on 200 for 7 months at 5 paise per rupee per month is (a) 70 (b) 7 (c) 35 (d) 30.50 (e) None of these 1 A tree increases annually by th of its height. By how 8 1 much will it increase after 2 yearly, if it stands today 10ft 2 high? (a) 3 ft (b) 3.27 ft (c) 3.44 ft (d) 3.62 ft (e) None of these If there are three sum of money P,Q and R so that P is the simple interest on Q and Q is the simple interest of R, rate % and time are same in each case, then the relation of P, Q and R is given by
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(a) P2 = QR (b) Q2 = PR (c) R2 = PQ (d) PQR = 100 (e) None of these In how many minimum number of complete years, the interest on 212.50 at 3% per annum will be in exact number of rupees? (a) 6 (b) 8 (c) 9 (d) 7 (e) None of these A milk man borrowed 2,500 from two money lenders. For one loan, he paid 5% p.a. and for the other, he paid 7% p.a. The total interest paid for two years was 275. How much did he borrow at 7% rate? (a) 600 (b) 625 (c) 650 (d) 675 (e) None of these What annual instalment will discharge a debt of 4,200 due in 5 years at 10% simple interest? (a) 500 per year (b) 600 per year (c) 700 per year (d) 800 per year (e) None of these Aman borrows Rs 12,500 at 20% compound interest. At the end of every year he pays Rs 2000 as part repayment. How much does he still owe after three such instalments? (a) Rs 12,000 (b) Rs 12,864 (c) Rs 15,560 (d) None of these (e) None of these A person borrows 5000 for 2 years at 4% p.a. simple
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`
1 1 (d) 4 % % 5 5 (e) None of these An automobile financier claims to be lending money at simple 6
interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of
`
`
th of the 16 principal and the number of years is equal to the rate per cent per annum. The rate per cent annum is ________ . 1 1 (a) 6 % (b) 6 % 3 4
(c)
41.
1
The simple interest on a sum of money is
10%, the effective rate of interest becomes : (a) 10%
(b) 10.25%
(c) 10.5%
(d) None of these
(e) None of these
`
42. A lent 5000 to B for 2 years and 3000 to C for 4 years on `
`
`
`
simple interest at the same rate of interest and received
`
`
`
2200 in all from both of them as interest. The rate of interest
per annum is: (a) 5% 1 (c) 7 % 8 (e) None of these 43.
(b) 7% (d) 10%
A sum of 725 is lent in the beginning of a year at a certain `
rate of interest. After 8 months, a sum of 362.50 more is `
`
interest. He immediately lends it to another person at 6
1 4
%
p.a. for 2 years. Find his gain in the transaction per year. (a) 112.50 (b) 125 (c) 150 (d) 167.50 (e) None of these 37. A certain amount earns simple interest of 1750 after 7 years Had the interest been 2% more, how much more interest would it have earned? (a) 35 (b) 245 (c) 350 (d) Cannot be determined (e) None of these 38. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years? (a) 1 : 3 (b) 1 : 4 (c) 2 : 3 (d) Data inadequate (e) None of these 39. Two equal sums of money were invested, one at 4% and the other at 4.5%. At the end of 7 years, the simple interest received from the latter exceeded to that received from the former by 31.50. Each sum was : (a) 1,200 (b) 600 (c) 750 (d) 900 (e) None of these `
`
`
`
lent but at the rate twice the former. At the end of the year, `
the original rate of interest? (a) 3.6%
(b) 4.5%
(c) 5%
(d) 3.46%
(e) None of these 44.
`
`
33.50 is earned as interest from both the loans. What was
The difference between the simple interest received from two different sources on 1500 for 3 years is 13.50. The `
`
difference between their rates of interest is:
`
`
(a) 0.1%
(b) 0.2%
(c) 0.3%
(d) 0.4%
(e) None of these 45.
The rates of simple interest in two banks A and B are in the ratio 5 : 4. A person wants to deposit his total savings in two banks in such a way that he received equal half-yearly interest from both. He should deposit the savings in banks A and B in the ratio. (a) 2 : 5
(b) 4 : 5
(c) 5 : 2
(d) 5 : 4
(e) None of these
`
`
`
`
`
46.
The price of a T.V. set worth 20,000 is to paid in 20 `
instalments of 1000 each. If the rate of interest be 6% per `
annum, and the first instalment be paid at the time of
10
purchase, then the value of the last instalment covering the
3 p.a. If the total interest at the end of one year is 9 % , then 4 the amount invested in each share was:
interest as well will be : (a) 1050 (b) 2050 (c) 3000 (d) None of these (e) 2020 A man buys a music system valued at 8000. He pays 3500 at once and the rest 18 months later, on which he is charged simple interest at the rate of 8% per annum. Find the total amount he pays for the music system. (a) 9260 (b) 8540 (c) 8720 (d) 9410 (e) None of these An amount of 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% `
`
(a)
`
`
47.
(b)
72,500: 27,500
(d)
`
`
`
62, 500; 37,500
`
`
82, 500; 17,500
`
`
`
`
48.
(c)
52,500; 47,500
`
`
`
`
`
`
(e) None of these 49. Find the compound interest on 12450 for 9 months at 12% per annum compounded quarterly. (a) 1154.45 (b) 1125.18 (c) 1198.72 (d) 1164.32 (e) None of these 50. A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is: (a) 200 (b) 600 (c) 800 (d) 1200 (e) None of these 51. Divide 2379 into 3 parts so that their amounts after 2, 3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is: (a) 759 (b) 792 (c) 818 (d) 828 (e) None of these `
`
`
`
`
`
`
`
`
`
ANSWER KEY 1 2 3 4 5 6
(a) (a) (c)
7
(b) (c) (d)
10
8 9 11 12
(c) (b) (e)
13
(e) (c) (b)
16
14 15 17 18
(c) (a) (a)
19
(b) (e) (b)
22
20 21 23 24
(a) (d) (c)
25
(e) (b) (e)
28
26 27 29 30
(a) (c) (c)
31
(a) (a) (c)
34
32 33 35 36
(c) (d) (b)
37
(c) (d) (a)
40
38 39 41 42
(d) (c) (d)
43
(a) (b) (d)
46
44 45 47 48
(d) (c) (b) (d) (b) (b)
49 50 51
(a) (d) (d)
11
An sw e r s & Ex p l a n a t i o n s 1.
(a)
Let the Principal = P P 8 4 P 10 6 P 12 5 Then 100 100 100 = 12160 152P = 12160 ×100
2.
12160 100 = 8000 152 (a) Let the sums be P. Now, 45% of P = 4016.25 or, P = 8925
3.
(c) Rate of interest =
or
4.
100 :
800 + 228 = 1028 900´ 100
or, or,
´ 100 = 6.50%
80k + 60k + 180k = 3200 × 100 320k = 3200 × 100 or, k = 1000
800´ 9.5 ´ 3
=
150 100 :150 8 : 5 :12 240
8k 10 5k 12 12k 15 3200 100 100 100
3´ 800
Amount = 800 +
`
Now, according to the question,
956 - 800
\
(b) Time =
(c)
8.
`
100
amount invested in scheme B willl be = 1000 × 5 = 5000 (e) Let the amount deposited at the rate of 15% per annum be x. 15% of x + 18% of (25000 – x) = 4050 or, 15% of x + 18% of 25000 – 18% of x = 4050 `
9.
`
= 5 years
3000´ 6
1600´ 100
3% of x = 4500 – 4050 = 450 x = 15000 Amount deposited at 18% = (25000 – 15000 =) 10000
or,
= 8% 5´ 4000 S.I. in I case = 2P – P = ` P Rate =
5.
2000 10 10 310 62 100 100 100 (b) Ratio of Nikhilesh’s investments in different schemes
`
`
P 100
R
P 10
10%
Now, S.I. in II case = 3P – P = 2P
10.
`
T
6.
2P 100 P 10
20 years
(d) Let p and r be the principal amount and rate of interest respectively. Then, or,
p r 7
100 pr = 25000
Now, SI =
1750
7.
`
(c) Rate %
12.
(b) Let x be the other amount
10
p r 7
13.
(100)3
15%
`
\ total borrowed sum = 33000 (c) Let the sum be x. `
éæ
\ x = 14.
3600 9000 x = 18000
=
100 100 M = SI when the rate of interest is 2% more. When we solve this equation, we find that we have two variables and one equation. Therefore, can’t be determined the correct answer. (c) For 3 years: Diff.
18000
512.50 x êêç 1+ ç ç è
= M – 1750
Sum (rate) 2 (300 rate)
405 100 100
11.
3 x
100 We have to find the value of
éæ ù ö2 15 ÷ ú 1+ p + 450 = êêp ç ÷ ç ú è 100 ÷ø 100 êë ç úû 30 p
p = 20,000.
p r 2 7
p ( r 2) 7
(e)
ê
û
512.50 ´ 400 41
ù
2
æ441- 400 ö÷ 5 ö ÷÷ - 1ú= x ç ÷ ç ú ÷ è 400 ø÷ 100 ø ú ç
= 5000 `
(a) Let ‘ x’ be the amount borrowed by Mr Amin.
x 2 8
100
x 3 11
100
x 3 14
100
10920
12
or, 15.
91 100
x = 10920 or x =
91
12000 ´
= 12000
(a) If ‘x’ be the interest of third year, then 108% of x= 486
x 486
16.
10920´ 100
(b)
100 108
100 x
or, 1200 +
450 or,
2
t
17640 r r 1 1 or, 16000 100 P 100
A
10
5
7
-
50
20
= (12000 + x ) ´
100
= 1680 +
14 100
7 x 50
x = 480
3
x = 480 50 \ x = 8000 Total amount invested = (12000 + 8000) = 20000
or, r 21 1 20 100 r = 5%
x
5
+ x ´
`
\
`
`
17.
10 (e) Amount = 15000 1 100 = 15000
18.
11
22.
= 10000´
= 18150 10 10 (b) Ratio A : B = 3 : 2 and A : C = 2 : 1 A : B : C = 6 : 4 : 3
(a)
æ 10 ö÷3 1+ ÷ (e) Amount = =10000ç ç ç è 100 ÷ø
11
`
23.
3620 100 16500
`
362 33
6
Now, use Alligation method. Scheme B 12%
33
24.
%
(e)
100 = 4754.71 106 (a) Let the amount be x. `
`
33 10
From the question,
362 362 10 17 :16 = 12 : 33 33 Hence, amount invested in B =
26.
`
27.
1200
1.53 105 20 100
(c) Difference in S.I. P T (R1 R 2 ) 100
56
x 10 2 x 124.05 100
P
`
`
28.
x 8 8
`
`
Solving the above eqns, we get x = 8,000. (c) Let the amount invested at 20% rate be x. According to the question,
1200
`
16 16500 = 8000 17 16
x 14 6
x = 15000 (c) Let S.I. = x
(d) Let the sum be x. 4 5 1 x Then, 100
21.
5040
x
362
Hence, ratio of amount invested in schemes A and B
24
r 1 100
P becomes 16 P in 24 years Let x be the amount Amal borrowed. 18% of x + 40% of x + 48% of x = 5040 or, 106% of x = 5040
25.
362 12 33
20.
64
r 2 1 100 4
Scheme A 10%
6
r 1 2 100
%
362
`
r 2P P 1 100
= 48400 13 % interest on total amount per annum
=
11 11 11 ´ ´ × = 13310 10 10 10
(b) P becomes 2P in 6 years at r% p.a.
4 157300
Profit share of B = 19.
2
(a)
16 25
P
P 4 2
( R 1 – R 2 = 2)
100
56 100 42
`
P R R 100
700
250
13 2 20 20 2000 1 2000 1 2000 100 100
1600 40 R 8% 25 5 Also, time = 8 years Rate = 5 paise per rupee = 5%
R2
29.
(a)
S.I. 30.
(c)
200 5 7
`
= 70 `
100
Increment
6 6 6 12500 5 5 5 6 6 6 2000 2000 2000 5 5 5
1 100 12 1 % 2 8
25 H 10 1 100 2
1
10 1 8
5/ 2
Hence, sum borrowed = 12,000. (a) Gain in 2 years `
5/ 2
36. 2
1/ 2
=
1 1 10 1 1 8 8
25 2 5000 4 2 5000 4 100 100
`
= (625 – 400) = 225. `
2
1 10 81 17 9 10 1 13.44ft 64 16 8 28
31.
(b)
P
Qr t
P
100
Q
and Q
R r t
rt
Q R 100 Q2 = PR.
32.
Gain in 1 year = 37.
100
38.
(b) Interest for one year 212.50 `
3
100 Thus in 8 years, the interest is 51. (b) Let he borrowed at 5% = x he borrowed at 7% = (2500 – x) Now I1 + I2 = 275
1
225 = 112.50 2 `
P R 6 100 6PR 6 2:3. Required ratio = P R 9 9PR 9 100
8
`
33.
`
(d) We need to know the S.I., principal and time to find the rate. Since the principal is not given, so data is inadequate. (c) Let the principal be P and rate of interest be R%.
51 `
`
`
`
x 5 2
39.
(2500 x) 7 2
275 100 100 10x + 14 (2500 – x) = 27500 4x = 35000 – 27500 = 7500 x = Rs 1875 Sum borrowed at 7% rate = 2500 – 1875 = 625 Shortcut method : If borrowed amount be M and it is to be paid in equal instalments, then
(d) Difference of S.I. = 31.50 Let each sum be x. Then `
`
1 x4 7 x47 2 31.50 100 100
`
34.
(c)
or
`
M na
ra
n(n 1)
100 Y 2 where Y = no. of instalments per annum a = annual instalment Here, M = 4200, y = 1, r = 10, n = 5, a = ? 4200 5a
10a 100
40.
7x 100
1 2
63 2
or
x = 900
(a)
Let the rate of interest = r % times = r years
`
Now,
S 16
S r r 100
5(5 1) 100 or, r2 = 16
2
4200 a 5 1 6a 4200 a = 700
41.
r=
25 4
1
6 % 4
(b) Let the sum be Rs 100. Then,
`
35.
( d)
Balance 3 20 12500 1 100
S.I. for first 6 months =
`
100 10 1 5. 100 2 `
`
S.I. for last 6 months =
`
105 10 1 5.25. 100 2 `
So, amount at the end of 1 year = (100 + 5 + 5.25) `
14
2x = (1100000 – 975000) = 125000 x = 62500. Sum invested at 9% = 62500.
= 110.25. Effective rate = (110.25 – 100) = 10.25%. (d) Let the rate be R% p.a. Then, `
`
42.
`
Sum invested at 11% = (100000 – 62500) = 37500. `
5000 R 2 3000 R 4 2200 100 100
49.
P = 12450, n = 9 months `
R A P 1 4 100
(d) Let the original rate be R%. Then, new rate = (2R)%.
725 R 1 362.50 2R 1 33.50 100 3 100
4n
44.
(c)
2900
`
`
`
45.
1
1
1
0.3%
X
100
4x
4
i.e., X : Y = 4 : 5. (d) Money paid in cash = 1000. Balance payment = (20000 – 1000) = 19000.
47.
(b)
(7800 5x) 8 3
52x (7800 8)
`
`
Cost of the music system = 8000 Money paid at once = 3500 Money left = (8000 – 3500) = 4500
100
7800 8 x 1200. 52 Money invested at 4% = 1200.
`
`
y 4 2 2 or y x. x 6 3 3
5 2600 x 8 x 4 1 3 So,
Y 4x , or Then, X 5x 2 100 2 100 Y 5 46.
(d) Let the parts be x, y and [2600 – (x + y)]. Then,
(b) Let the savings be X and Y and the rates of simple interest be 5x and 4x respectively. 1
`
x 4 1 y 6 1 [2600 (x y)] 8 1 100 100 100
1500 R1 3 1500 R 2 3 13.50 100 100
4500
3
`
50.
1350
103 100
= 13604.45 CI = (13604.45 – 12450) = 1154.45
3.46%
4500(R 1 R 2 ) 1350 R 1 R 2
yr,
9 4 12 12 4 12450 1 100
`
R
12
3 3 12450 1 = 12450 × 100
(2175 725)R 33.50 100 3 10050 10050
9
R = 12% per annum
2200 100R 120R 2200 R 10. 220 43.
(a)
`
51.
(d) Let the parts be x, y and z
`
x x 2
`
`
Time =
`
1 18 yr 1 yr 12 2
5 5 y y 3 100 100
5 z z4 100
Rate = 8% per annum SI
3 1 PTR 4500 8 540 2 100 100 `
`
Money to be paid at the end = (4500 + 540) = 5040 `
48.
`
Cost of music system = (3500 + 5040) = 8540 `
`
(b) Let the sum invested at 9% be x and that invested at 11% be (100000 – x). Then, `
`
x 9 1 (100000 x) 11 1 100 100 39 1 100000 4 100
11x 10
23y 20
6z 5
kx
9x 1100000 11x
10k
100
39000 4
9750
20k 23
,z
20k
5k
2379 11 23 6 1380 k + 1320 k + 1256 k = 2379 × 11 × 23 × 6 k
2379 11 23 6 3965
3 11 23 6 5
10 3 11 23 6 x 828. 5 11 Hence, the first part is 828.
11
,y
But x + y +z = 2379.
`
10k
5k 6
