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 simple interest on a sum of money at 4% per annum for 2 years is Rs.80. The compound interest in the same sum for the same period is

(a) Rs.81.60

(b) Rs.82.60

(c) Rs.81.80

(d) Rs.82.20

The correct answers to the above question in:

Answer: (a)

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

= ${80 × 100}/{2 × 4}$ = Rs.1000

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

= 1000$[(1 + 4/100)^2 - 1]$

= 1000$[(25/26)^2 - 1]$

= 1000$(676/625 - 1)$

= 1000$({676 - 625}/625)$

= ${1000 × 51}/625$= Rs.81.60

Using Rule 10,

Here, S.I. = Rs.80, R = 4%, C.I. = ?

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

C.I.= 80$(1 + 4/200)$

= $80(1 + 1/50)$

= 80 × $51/50$ = Rs.81.60

Practice compound interest (model 3 combination of si & ci) Online Quiz

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

Question : 1

There is 40% increase in an amount in 8 years at simple interest. What will be the compound interest (in rupees) of Rs 30000 after 2 years at the same rate ?

a) 3075

b) 6150

c) 4612.5

d) 7687.5

Answer: (a)

According to the question,

If principal

= Rs.100 then interest = Rs.40.

Case I.

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

= ${40 × 100}/{100 × 8}$ = 5% per annum

Case II.

A = P$(1 + R/100)^T$

= 30000$(1 + 5/100)^2$

= 30000$(1 + 1/20)^2$

= 30000$({20 + 1}/20)^2$

= 30000$ × 21/20 × 21/20$

= Rs.33075

C. I. = Rs.(33075 - 30000) = Rs.3075

Question : 2

On a certain sum of money the compound interest for 2 years is Rs.282.15 and the simple interest for the same period of time is Rs.270. The rate of interest per annum is

a) 12.15%

b) 6.07%

c) 9%

d) 10%

Answer: (c)

Using Rule 10,

If SI on a certain sum for two years is x and CI is y, then

$y = x(r + /200)$

$282.15 = 270(1 + r/100)$

$1 + r/200 = 282.15/270$

$r/200 = 282.15/270$ - 1

$r/200 = {12.15}/270$

r = ${12.15 × 200}/270 = 9%$

Question : 3

A certain amount of money earns Rs.540 as Simple Interest in 3 years. If it earns a Compound Interest of Rs.376.20 at the same rate of interest in 2 years, find the amount (in Rupees).

a) 2100

b) 1600

c) 2000

d) 1800

Answer: (c)

S.I. for 2 years

= $2/3$ × 540 = Rs.360

C.I. - S.I.

= 376.20 - 360 = Rs.16.20

Rate of interest

= ${16.20}/180$ × 100 = 9% per annum

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

= ${180 × 100}/{1 × 9}$ = Rs.2000

Question : 4

A man borrowed some money from a private organisation at 5% simple interest per annum. He lended 50% of this money to another person at 10% compound interest per annum and thereby the man made a profit of Rs.3,205 in 4 years. The man borrowed.

a) Rs.1,50,000

b) Rs.80,000

c) Rs.1,20,000

d) Rs.1,00,000

Answer: (d)

Using Rule 1,

Sum borrowed = Rs.x

Simple interest after 4 years

= ${x × 4 × 5}/100$ = Rs.$x/5$

Amount lent of on compound interest = Rs.$x/2$

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

= $x/2[(1 + 10/100)^4 - 1]$

= $x/2[(1.1)^4 - 1]$

= $x/2$ (1.4641 - 1) = Rs.${0.4641x}/2$

${0.4641x}/2 - x/5$ = 3205

${2.3205x - 2x}/10$ = 3205

0.3205x = 32050

$x = 32050/{0.3205}$ = Rs.100000

Question : 5

If the compound interest on a certain sum for two years at 12% per annum is Rs.2,544, the simple interest on it at the same rate for 2 years will be

a) Rs.2,440

b) Rs.2,400

c) Rs.2,480

d) Rs.2,500

Answer: (b)

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

2544 = P$[(1 + 12/100)^2 - 1]$

2544 = P$[(28/25)^2 - 1]$

2544 = P$(784/625 - 1)$

2544 = P$({784 - 625}/625)$

2544 = ${P × 159}/625$

P = ${2544 × 625}/159$ = Rs.10000

S.I. = ${P × R × T}/100$

= ${10000 × 2 × 12}/100$ = Rs.2400

Using Rule 10,

Here, C.I. = Rs.2544, R = 12%, S.I. = ?

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

2544 = S.I.$(1 + 12/200)$

2544 = S.I.$(212/200)$

S.I. = ${2544 × 200}/212$ = Rs.2400

Question : 6

The compound interest on a certain sum of money for 2 years at 5% per annum is Rs.410. The simple interest on the same sum at the same rate and for the same time is

a) Rs.405

b) Rs.400

c) Rs.350

d) Rs.300

Answer: (b)

Compound interest = P $[(1 + R/100)^T - 1]$

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

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

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

410 = P$(441/400 - 1)$

410 = P$(41/400)$

P = ${410 × 400}/41$ = Rs.4000

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

= ${4000 × 2 × 5}/100$ = Rs.400

Using Rule 10,

Here, C.I. = Rs.410, R = 5%, S.I. = ?

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

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

410 = S.I.$(205/200)$

S.I. = ${410 × 200}/205$ = Rs.400

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