model 3 combination of si & ci Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on compound interest topic of quantitative aptitude

Questions : The compound interest on a certain sum of money at a certain rate for 2 years is Rs.40.80 and the simple interest on the same sum is Rs.40 at the same rate and for the same time. The rate of interest is

(a) 5% per annum

(b) 2% per annum

(c) 4% per annum

(d) 3% per annum

The correct answers to the above question in:

Answer: (c)

Let the principal be P and rate of interest be r per cent per annum. Then,

C. I = P$[(1 + r/100)^2 - 1]$

40.80 = P$[(1 + r/100)^2 - 1]$...(i)

S.I. = ${P.r.t}/100 ⇒ 40 = {Pr × 2}/100$ ...(ii)

${40.80}/40 = P[(1 + r/100)^2 - 1]/{{2Pr}/100}$ ⇒ 1.02

= $100/{2r}[1 + r^2/10000 + {2r}/100 - 1]$

1.02 = $r/200$ +1

$r/200$ = 1.02 - 1

r = 0.02 × 200 = 4% per annum.

Using Rule 10,

Here, C.I. = Rs.40.80, S.I. = Rs.40, R = ?

C.I.= S.I.$(1 + R/200)$

40.80 = 40$(1 + R/200)$

$4080/4000 = 1 + R/200$

$408/400 = {200 + R}/200$

408 = 400 + 2R

2R = 8 ⇒ R = 4%

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Read more combination of si and ci Based Quantitative Aptitude Questions and Answers

Question : 1

The compound interest on a certain sum of money at a certain rate per annum for two years is Rs.2,050, and the simple interest on the same amount of money at the same rate for 3 years is Rs.3, 000. Then the sum of money is

a) Rs.25, 000

b) Rs.20,000

c) Rs.21,000

d) Rs.18,000

Answer: (b)

Using Rule 6,
The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will be
C.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$
For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$

S.I. for 3 years = Rs.3000

S.I. for 2 years = $3000/3 × 2$ = Rs.2000

C.I. - S.I. = 2050 - 2000 = Rs.50

S.I. = ${PR × 3}/100$

PR = ${3000 × 100}/3$ = Rs.100000

Difference = ${P × R^2}/10000$

50 = ${P × (100000)^2}/{10000 × P^2}$

P = $1000000/50$ = Rs.20000

Question : 2

The compound interest on a certain sum of money invested for 2 years at 5% per annum is Rs.328. The simple interest on the sum, at the same rate and for the same period will be

a) Rs.287

b) Rs.320

c) Rs.300

d) Rs.308

Answer: (b)

Let the principal be P.

C.I. = P$[(1 + R/100)^T - 1]$

328 = P$[(1 + 5/100)^2 - 1]$

328 = P$[441/400 -1]$

328 = P$[{441 - 400}/400]$

P = ${328 × 400}/41$ = Rs.3200

S.I. = ${PRT}/100 = {3200 × 5 × 2}/100$ = Rs.320

Using Rule 10,

Here, C.I. = Rs.328, R = 5%, S.I. = ?

C.I.= S.I.$(1 + R/200)$

328 = S.I.$(1 + 5/200)$

328 = S.I.$(1 + 1/40)$

S.I. = ${328 × 40}/41$

S.I. = 8 x 40 = Rs.320

Question : 3

If the compound interest on a certain sum for 2 years at 3% per annum is Rs.101.50, then the simple interest on the same sum at the same rate and for the same time will be

a) Rs.98.25

b) Rs.90.00

c) Rs.100.00

d) Rs.95.50

Answer: (c)

Let the sum be P.

101.50 = P$[(1 + 3/100)^2 - 1]$

[Since, C.I. = P$[(1 + r/100)^n - 1]$]

101.50 = P$[(103/100)^2 - 1]$

=P$({10609 - 10000}/10000)$

P = Rs.${101.50 × 10000}/609 = Rs.1015000/609$

S.I. = ${1015000 × 2 × 3}/{609 × 100}$ = Rs.100

Using Rule 10,
The simple interest for a certain sum for 2 years at an annual rate interest R% is S.I., then
C.I. = S.I.$(1 + R/200)$

Here, C.I. = Rs.101.50, R = 3%, S.I. = ?

C.I. = S.I.$(1 + R/200)$

101.50 = S.I.$(1 + 3/200)$

S.I. = ${101.50 × 200}/203$ = Rs.100

Question : 4

On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. If it was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned ?

a) Rs.35

b) Rs.3.50

c) Rs.14

d) Rs.7

Answer: (d)

Principal = $\text"S.I. × 100"/\text"Time × Rate"$

= ${350 × 100}/{2 × 4}$ = Rs.4375

Difference = ${PR^2}/10000$

= ${4375 × 4 × 4}/10000$ = Rs.7

Question : 5

If the compound interest on a sum of money for 3 years at the rate of 5% per annum is Rs.252.20, the simple interest on the same sum at the same rate and for the same time is

a) Rs.250

b) Rs.220

c) Rs.245

d) Rs.240

Answer: (d)

Using Rule 1,
If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,
A=P$(1 + r/100)^n$, C.I. = A - P
C.I. = P$[(1 + r/100)^n - 1]$

Suppose principal be x

$x((1 + 5/100)^3 - 1) = 252.20$

$x((21/20)^3 - 1)$ = 252.20

$x({21 × 21 × 21 - 20 × 20 × 20}/{20 × 20 × 20})$ = 252.20

$x 1261/8000$ = 252.20

x = ${252.20 × 8000}/1261$ = 1600

SI = ${1600 × 5 × 3}/100$ = Rs.240

Question : 6

If the compound interest on a sum for 2 years at 12$1/2$ p.a is Rs.510, the simple interest on the same sum at the same rate for the same period of time is

a) Rs.480

b) Rs.400

c) Rs.460

d) Rs.450

Answer: (a)

Principal = Rs.P (let)

C.I. = P$[(1 + R/100)^T - 1]$

510 = P$[(1 + 25/200)^2 - 1]$

510 = P$[(1 + 1/8)^2 - 1]$

510 = P$[(9/8)^2 - 1]$

510 = P$(81/64 - 1)$

510 = P$({81 - 64}/64)$

510 = ${17P}/64$

P = ${510 × 64}/17$ = Rs.1920

S.I. = $\text"Principal × Time × Rate"/100$

= ${1920 × 2 × 25}/{100 × 2}$ = Rs.480

Using Rule 10,

Here, C.I. = Rs.510, R = 12$1/2$%, S.I. = ?

C.I. = S.I.$(1 + R/200)$

510 = S.I.$(1 + 25/400)$

510 = S.I.$(425/400)$

S.I. = ${510 × 400}/425$ = Rs.480

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GET compound interest PRACTICE TEST 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
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