MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 5

A sum of Rs. 100 at CI becomes Rs. 300 in 4 years. What will it amount to in 12 years?

0%
Rs. 2,700
0%
Rs. 2,400
0%
Rs. 2,200
0%
Rs. 3,700

Explanation

Principal(P)=100
After 4 years amount=

$3\times P=3\times 100=300 Rs.$
After 8 years amount =

$3\times P=3\times 300=900 Rs$
After 12 years amount=

$3\times P=3\times 900=2700 Rs.$

The C.I. on a sum of money for 2 years is Rs. 832 and the S.I. on the same sum for the same period is Rs. The difference between the C.I and S.I. for 3 years will be

0%
Rs. 48
0%
Rs. 66.56
0%
Rs. 98.56
0%
None of these

Explanation

Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400 and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400 and compound interest for 2nd year will be

$832 - 400 = 432$
We can see that compound interest for 2nd year is more than simple interest for 2nd year by

$432 - 400 = Rs.32$
Rs. 32 is the interest obtained for Rs.400 for 1 year

$R = \frac{100\times SI}{PT} = \frac{100 \times 32}{400\times1} = 8$ %
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

$=\frac{PRT}{100} = \frac{832\times 8\times 1}{100} = 66.56$
Total difference between the compound and simple interest for 3 years

$= 32 + 66.56 = Rs.98.56$

A sum of money invested at C.I. amounts to Rs. 800 in 3 years and to Rs. 840 in 4 years. The rate of interest per annum is

0%
$2\, \displaystyle \frac{1}{2}$ %
0%
$4$ %
0%
$5$ %
0%
$6\, \displaystyle \frac{2}{3}$ %

Explanation

Simple interest for one year

$=840-800=40 Rs.$

Rate of interest=

$\dfrac{40\times 100}{800\times 1}=5$ %

What will be the difference between simple and compound interest at 10% per annum on a sum of Rs. 1000 after 4 years?

0%
Rs. 31
0%
Rs. 32.10
0%
Rs. 40.40
0%
Rs. 64.10

Explanation

Principal=1000 Rs.
Time=4 years
Rate of interes=10%
Then Simple interest

$=\frac{1000\times 10\times 4}{100}=400 Rs.$
Compound interest

$=$ Amount

$-$ Principal

Amount

$=1000(1+\frac{10}{100})^4$

$\Rightarrow 1000\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}\times$

$\Rightarrow 1464.10 Rs.$

$C.I.=1464.10-4000=464.10 Rs.$
Difference between C.I and S.I.=

$464.10-400=64.10 Rs.$

Madhav lent out Rs. 7953 for 2 years and Rs. 1800 for 3 years at the same rate of simple interest. If he got Rs.66 as total, then find the percent rate of interest.

0%
11%
0%
12%
0%
12.5%
0%
5%

Explanation

We know that Simple Interest

$= \cfrac {PNR}{100}$
Given,

$\cfrac {7953 \times 2 \times R}{100} + \cfrac {1800 \times 3 \times R}{100} = Rs 2343.66$

$=> 159.06R + 54R = Rs 2343.66$

$=> 213.06R = 2343.66$

$=> R = 11 \%$

A scooter was bought at

$\text{Rs. } 42000$ . Its value depreciated at the rate of

$8\%$ per annum, find its value after one year.

0%
$\text{Rs. } 38640$
0%
$\text{Rs. } 37740$
0%
$\text{Rs. } 38145$
0%
$\text{Rs. } 39640$

The present population of a town is

$5000$ . If the population decreases at the rate of

$10\%$ per year, the population after

$2$ years will be

0%
$4820$
0%
$4050$
0%
$4980$
0%
$4500$

Explanation

Present population,

$P_o = 5000$
Rate

$= 10\%$
Time

$= 2$ years
Let after

$2$ years population

$= P$
Thus,

$P = P_o(1 - \frac{R}{100})^T$

$P = 5000 (1 - \frac{10}{100})^2$

$P = 4050$
Thus after two years the population will be

$4050$

The present population of a town is

$35,000$ . If the population becomes

$35,700$ next year, the rate of growth is

0%
$2\%$
0%
$20\%$
0%
$70\%$
0%
$7\%$

Explanation

Population next year,

$P = 35700$
Current population,

$P_o = 35000$

$P = P_o \left(1 + \cfrac{R}{100}\right)^T$

$35700 = 35000 \left(1+ \cfrac{R}{100}\right)^1$

$35700 - 35000 = \cfrac{35000 R}{100}$

$R = \cfrac{700 \times 100}{35000}$

$R = 2\%$

The population of a town increases from

$10000$ to

$10400$ . The growth rate is

0%
$4\%$
0%
$40\%$
0%
$2\%$
0%
$20\%$

Explanation

Population an year ago,

$P_o = 100000$

Population after one year

$P = 10400$

Let the growth rate be R,

then

$P = P_o \left(1 + \cfrac{R}{100}\right)^T$

$10400 = 10000\left(1 + \cfrac{R}{100}\right)$

$10400 = 10000 + 100 R$

$R = \cfrac{400}{100} = 4\%$

Thus, growth rate

$= 4\%$

A student purchased a pen for RsAt the end of 8 years it was valued at Rs 20 Assuming the yearly depreciation is a constant amount find the annual depreciation

0%
8
0%
9
0%
10
0%
11

Explanation

Cost Price of the pen

$=$

$Rs.100$

After

$8$ years, Value of the pen

$=Rs. 20$

Since, depreciation is yearly and constant

So, Annual depreciation

$=\dfrac{Rs.100-Rs.20}{8}=\dfrac{Rs.80}{8}=Rs.10$

Hence, C is the correct option.

Calculate compound interest on Rs

$2000$ for

$2$ years at the rate

$5\%$

0%
$105$
0%
$200$
0%
$205$
0%
$2205$

Explanation

For the first year,

$P = Rs. 2000, R = 5\%$

so, I =

$\cfrac { P \times R \times T }{ 100 } =\cfrac { Rs.\quad 2100 \times 5 \times 1 }{ 100 } =Rs.\quad 105$

The amount after the second year =

$Rs. 2100 + Rs. 105$

$=Rs. 2205$

Final amount =

$Rs. 2205$

Hence, Compound Interest = final amount - original amount

$= Rs. 2205 - Rs. 2000 = Rs. 205$

The difference between simple interest and compound interest is that

0%
Simple interest is easier to compute.
0%
Simple interest yields a lot of money.
0%
Simple interest is the interest on original principal only.
0%
None of these

Explanation

$\Rightarrow Simple Interest = \dfrac {P \times N\times R} {100}$

$where, P = Principal \:or\: the\: sum$

$N = Number \:of\: period \:or \:Time$

$R = Rate\:of\:interest$

$\Rightarrow Easier\: to\: compute.$

$\Rightarrow Simple\: interest\: is \:calculated \:on \: original \: principal.$

There is a question followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read, the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to be repaid if he is repaying the entire amount at the end of

$2$ years?

$I$ . The rate of interest is

$5$ p.c.p.a.

$II$ . Simple interest fetched on the same amount in one year is

$Rs. 600$ .

$III$ . The amount borrowed is

$10$ times the simple interest in

$2$ years.

0%
$I$ only
0%
$III$ only
0%
$I$ or $II$ only
0%
None of these

Explanation

Given
I. The rate of interest is 5 p.c.p.a.
II. Simple interest fetched on the same amount in one year is Rs. 600.
III. The amount borrowed is 10 times the simple interest in 2 years.
Now | and || give the sum,
For this sum , C.I and hence amount can be obtained
Thus , ||| is redundant
Again || gives S.I for 2 years =

$600\times 2=1200$
Now from ||| ,sum =

$10\times 1200=12000$
Thus Rate=

$\frac{100\times 1200}{2\times 1200}=5%$ p.a
Thus ,C.I for 2 years and therefore ,amount can be obtained.
Thus ,| is redundent
Hence either | and II or II and III are sufficient to answer the question

An emergency lamp costs Rs.Every year, its value depreciates by 10%. What is its value at the end of 3 years?

0%
Rs.908.90
0%
Rs. 809.10
0%
Rs. 801.90
0%
Rs. 980.90

Explanation

Cost of Lamp = Rs 1100
Depreciation rate = 10 %
Now after 3 years, let the cost be C
then,

$C = 1100(1 - \frac{R}{100})^T$

$C = 1100(1 - \frac{10}{100})^3$

$C = 801.9$ Rs

Rakesh took a loan for

$7$ years at the rate of

$6\%$ p.a.

$S.I$ . If the total interest paid was Rs.

$2100$ , the principal was

0%
Rs. $4400$
0%
Rs. $4800$
0%
Rs. $5000$
0%
Rs. $5200$

Explanation

Simple interest

$S.I=\dfrac{P.R.T}{100}$

$\Rightarrow 2100 = \dfrac{P\times 6\times 7}{100}$

$\Rightarrow P=\dfrac{2100\times 100}{6\times 7}$

$\Rightarrow P=\dfrac{2100\times 100}{42}$

$\Rightarrow P=50\times 100$

$\Rightarrow P=5000$

Hence, the principal amount

$P=Rs.5000$

The cost of a car purchased 2 years ago depreciates at the rate of 20% per annum. If its present value is Rs 3, 15,600, find the value of the car after 2 years

0%
Rs 2,01,994
0%
Rs 50,496
0%
Rs 2,01,984
0%
Rs 10,09,920

Explanation

$Depreciation = A=P\left (1-\frac {r}{100}\right )^n=3,15,600\left (1-\frac {20}{100}\right )^2$

$=3,15,600\left (\frac {4}{5}\right )^2=3,15,600\times \frac {16}{25}=Rs 2,01,984$

Find the compound interest on Rs 1,200 for 2 years at 10% per annum

0%
Rs 252
0%
Rs 225
0%
Rs 300
0%
Rs 350

Explanation

Compound Interest

$C.I=A-P$
Formula for amount when compound interest is applied :

$A=P\left (1+\dfrac {r}{100}\right )^n$
Given

$P=1,200; n=2; r=10$

$\therefore A=1,200\left (1+\dfrac {10}{100}\right )^2=1,200\left (\dfrac {11}{10}\right )^2=\dfrac {1,200\times 121}{100}=Rs\ 1,452$

$CI=1,452-1,200=Rs\ 252$

Depreciation refers to

0%
Closure of a plant due to lock out
0%
Closure of a plant due to labour issues
0%
Loss of equipment over time due to wear to tear
0%
Destruction of plant in a fire accident

In what period will Rs 3,200 amount to Rs 3,528 at 5% per annum compounded annually?

0%
$3$ years
0%
$4$ years
0%
$2$ years
0%
$2\frac {1}{2}$ years

Explanation

$P = 3,200; A = 3,528; r = 5\%; n =?$

$A=P\left (1+\dfrac {r}{100}\right )^n$

$3,528=3,200\left (1+\dfrac {5}{100}\right )^n$

$\dfrac {3,528}{3,200}=\left (\dfrac {105}{100}\right )^n$

$\left (\dfrac {21}{20}\right )^2=\left (\dfrac {21}{20}\right )^n$

$\therefore n=2$ years.

Ram had Rs 2 lakh part of which he lent at 15% p.a. and rest at 12% p.a. Yearly interest on both the parts was Rs 27,600 How much did he lend at 15%?

0%
Rs 1,20,000
0%
Rs 1,00,000
0%
Rs 80,000
0%
Rs 60,000

Explanation

Given 200000 Rs lend @ 15% and 12% recd Rs 27600

So average rate of interest

$\frac{27600}{200000}\times 100=13.8$ %

Let x and y amount lent @ 15% and 12%

We have x:y =(13.8-12):(15-13.8)=2:3

So money lend at @ 15% =

$\frac{3}{5}\times 200000=120000$ Rs

X, Y and Z are the three sums of money. X is the simple interest on Y, and Y is the simple interest on Z. The rate percent per year and the time in years being the same in each case. Which one of the following relations in X, Y, Z is correct?

0%
$XYZ=100$
0%
$Y^2=XZ$
0%
$Z^2=XY$
0%
$X^2=YZ$

Explanation

$\displaystyle X=\frac{Y\times r\times t}{100}$ and

$\displaystyle Y=\frac{Z\times r\times t}{100}$

$\displaystyle\frac{X}{Y}=\frac{Y}{Z}$ i.e.,

$Y^2=XZ$ '

A sum of money at compound interest is doubled in

$20$ years. Then the number of years required to tripple itself are

0%
$32$ years
0%
$35$ years
0%
$40$ years
0%
$45$ years

Explanation

We know that

$A=P\left(1+\dfrac{r}{100}\right)^T$

According to the question,

$2P=P\left(1+\dfrac{r}{100}\right)^{20}$

$2=\left(1+\dfrac{r}{100}\right)^{20}$

$.............(1)$

And

$3P=P\left(1+\dfrac{r}{100}\right)^T$

$3=\left(1+\dfrac{r}{100}\right)^T$

$...........(2)$

From equation (1) and (2), we get

$3^{1/T}=2^{1/20}$

On taking

$log$ both sides, we get

$\dfrac{1}{T}\log\ 3=\dfrac{1}{20}\log\ 2$

$T=\dfrac{20log\ 3}{\log\ 2}$

$T=31.7$

$T\approx 32\ years$

Hence, this is the answer.

Find the rate, when

$Rs.\ 1800$ earns an interest of

$Rs.\ 432$ in

$3$ years

0%
$7\%$
0%
$6\%$
0%
$5\%$
0%
$8\%$

Explanation

Given:

Principal,

$P=Rs.\ 1800$

Interest,

$I=Rs.\ 432$

Times,

$T=3$ years

$I=\dfrac{P\times R\times T}{100}$

$R=\dfrac {I\times 100}{P\times T}$

$=\dfrac {432\times 100}{1800\times 3}$

$=8\%$

Rs.

$2000$ amounts to Rs.

$2226.05$ in two years at compound interest. The rate of interest is

0%
$5.0$ per cent
0%
$5.25$ per cent
0%
$5.5$ per cent
0%
$6.0$ per cent

Explanation

We know $A=P{ \left[ 1+\dfrac { R }{ 100 } \right] }^{ n }$

$2226.05= 2000{ \left[ 1+\dfrac { R }{ 100 } \right] }^{ 2 }\\ \dfrac { 2226.05 }{ 2000 } = { \left[ 1+\dfrac { R }{ 100 } \right] }^{ 2 }\\1.113025 ={ \left[ 1+\dfrac { R }{ 100 } \right] }^{ 2 }\\ 1.055= { \left[ 1+\dfrac { R }{ 100 } \right] }\\ 1.055 =\dfrac { 100+R }{ 100 } \\ 105.5= 100+R\\ R= 5.5\%$

A sewing machine was purchased

$2$ years ago. Its value decreases by

$10\%$ every year. Its present value is

$Rs.\ 19,083.60$ . For how much was the machine purchased?

0%
$Rs.\ 23,940$
0%
$Rs.\ 23,560$
0%
$Rs.\ 24,500$
0%
$Rs.\ 24,000$

Explanation

Given,

Let, principal

$=P$

Rate of depreciation

$R=10\%$

Present price

$A=Rs.\ 19,083.60$

Time

$T=2$ years

So,

$n=2$

Amount

$A = P\left( 1-\dfrac { R }{ 100 } \right) ^{ n }$

$\Rightarrow 19,083.60= P \times \left( 1-\dfrac { 10 }{ 100 } \right) ^{ 2 }$

$\Rightarrow 19,083.60= P \times \left( 1-\dfrac { 1 }{ 10 } \right) ^{ 2 }$

$\Rightarrow 19,083.60= P \times \left(\dfrac { 9 }{ 10 } \right) ^{ 2 }$

$\Rightarrow 19,083.60= P \times \left(\dfrac { 81 }{ 100 } \right)$

$\Rightarrow P= 19,083.60\times \left(\dfrac { 100 }{ 81 } \right)$

$\Rightarrow P= 23,560$

Hence, the machine was purchased for

$Rs.\ 23,560$ .

Rajan borrowed Rs.

$5000$ from Rakesh at simple interest. After

$3$ years, Rakesh got Rs.

$300$ more than what he had given to Rajan. What was the rate of interest per annum?

0%
$2\%$
0%
$5\%$
0%
$8\%$
0%
$10\%$

Explanation

Given: Rajan borrowed

$Rs.5000$ from Rakesh

Let the rate of interest be

$R\ \%$

Then, simple interest after

$3$ years

$=\dfrac{5000\times 3 \times R}{100}$

Given: Rakesh get

$Rs.300$ more than the interest

$\Rightarrow \dfrac{5000\times 3\times R}{100}=300$

$\Rightarrow R=\dfrac{300\times 100}{5000\times 3}$

$\Rightarrow R=2\%$

Mr. Mittal finds that an increase in the rate of interest from

$4 \displaystyle \dfrac{7}{8}\%$ to

$5 \displaystyle \dfrac{1}{8}\%$ per annum increases his yearly income by Rs.

$250$ . His investment is

0%
Rs. $1,00,000$
0%
Rs. $1,20,000$
0%
Rs. $1,50,000$
0%
Rs. $2,00,000$

Explanation

Let the investment be

$Rs. x$ . Then

$x \times \displaystyle \dfrac{41}{8} \times \dfrac{1}{100} - x \dfrac{39}{8} \times \dfrac{1}{100} = 250$

$2x = 200000$

$x = 100000$

The compound interest on

$Rs.\ 5600$ for

$1 \dfrac{1}{2}$ years at

$10\%$ per annum, compounded annually, is

0%
$Rs.\ 882.70$
0%
$Rs.\ 873.50$
0%
$Rs.\ 868$
0%
$Rs.\ 840$

A sum of Rs 16,400 is borrowed to be paid back in 2 years by two equal annual payments, allowing 5% compound interest. The annual payment is

0%
Rs 8,000
0%
Rs 8,050
0%
Rs 8,820
0%
Rs 9,000

A man had

$\text{Rs. } 2,000$ . He lent a part of this at

$5\%$ interest and the rest at

$4\%$ interest. The total interest he received in one year was

$\text{Rs. } 92$ . The money he lent at

$5\%$ interest was

0%
$\text{Rs. } 1,200$
0%
$\text{Rs. } 1,250$
0%
$\text{Rs. } 1,300$
0%
$\text{Rs. } 1,350$

Explanation

Let the money lent at the rate of

$5\%$ be

$x.$ So, the money lent at

$4\%$ will be

$2000-x.$

It is given that the total interest obtained in

$1$ year is

$\text{Rs. } 92.$ So,

$\begin{aligned}{}\frac{{x \times 5 \times 1}}{{100}} + \frac{{\left( {2000 - x} \right) \times 4 \times 1}}{{100}} &= 92\\5x + 4\left( {2000 - x} \right) &= 9200\\5x + 8000 - 4x& = 9200\\x &=\text{Rs. } 1200\end{aligned}$

Hence, the money lent at

$5\%$ rate is

$\text{Rs. } 1200.$

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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