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 : There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs.8000 after 2 years at the same rate of interest.

(a) Rs.2125

(b) Rs.2500

(c) Rs.2250

(d) Rs.2000

The correct answers to the above question in:

Answer: (a)

Using Rule 1,

Let S.I. = Rs.100, & Principal = Rs.100

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

= ${100 × 100}/{100 × 8} = 25/2$%

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

= 8000$[(1 + 25/200)^2 - 1]$

= 8000 $(81/64 - 1)$

= ${8000 × 17}/64$ = Rs.2125

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

Question : 1

The simple interest on a certain sum of money for 2 years at 5% is Rs.1600. The compound interest at the same rate after 3 years interest compound annually, is

a) Rs.2535

b) Rs.2520

c) Rs.2555

d) Rs.2522

Answer: (d)

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

= ${1600 × 100}/{5 × 2}$ = Rs.16000

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

= 16000$[(1 + 5/100)^3 –1]$

= 16000$[(21/20)^3 - 1]$

= $16000(9261/8000 - 1)$

= ${16000 × 1261}/8000$ = Rs.2522

Question : 2

The simple interest on a sum of money for 3 years is Rs. 240 and the compound interest on the same sum, at the same rate for 2 years is Rs. 170. The rate of interest is :

a) 5$5/17$%

b) 8%

c) 12$1/2$

d) 29$1/6$%

Answer: (c)

Since, S.I. for 3 years = Rs.240

S.I. for 2 years = $240/3$× 2 = Rs.160

${PR × 2}/100$ = 160

PR = 160 × 50 = 8000...(i)

Again, C.I. - S.I. = 170 - 160 = Rs.10

${PR^2}/10000$ = 10

${8000 × R}/10000$ = 10

R = $100/8 = 25/2 = 12{1}/2$%

Question : 3

If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, the compound interest on the same at the same rate and for the same time is :

a) Rs.50.05

b) Rs.50.50

c) Rs.51.50

d) Rs.51.25

Answer: (d)

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

= ${50 × 100}/{2 × 5}$ = Rs.500

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

= 500$[(1 + 5/100)^2 - 1]$

= 500$[(1 + 1/20)^2 - 1]$

= 500$[(21/20)^2 - 1]$

= 500$(441/400 - 1)$

= ${500 × 41}/400$ = Rs.51.25

Question : 4

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 : 5

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 : 6

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

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