Basics of Financial Mathematics Test 8

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Basics of Financial Mathematics Test 8

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Weekly Quiz Competition

Question 1
1 / -0

What is the principal amount which earns Rs. 1320 as compound interest for the second year at 10% per annum?

A
Rs. 10000

B
Rs. 12000

C
Rs. 13200

D
Rs. 11880

Solution

Let the principal at the end of first year be Rs. $$x$$.
Then $$\cfrac{x\times 10\times 1}{100} = 1320$$
or $$x = 13200$$
Now, let the original principal be Rs. P.
Then, the amount after 1 year $$= P+\cfrac{P\times 10\times 1}{100} = \cfrac{11P}{10}$$
$$\cfrac{11P}{10} = 13200$$
or $$P = (\cfrac{13200\times 10}{11}) = Rs. 12000$$
Question 2
1 / -0

A tree increases annually by one-fourth of its height. What will be it's height after $$2$$ years, if it stands $$64$$ cm high today?

A
$$76$$ cm

B
$$80$$ cm

C
$$100$$ cm

D
$$105$$ cm

Solution

Initial height $$=64\ cm$$
Every year it increases one-fourth of it's height.
Increase $$\%= (\cfrac{1}{4}\times 100\%)=25\%$$

Height after $$2$$ years $$= 64\times \left(1+\cfrac{25}{100}\right)^2\ cm$$
$$= 64\times \left(1+\cfrac{1}{4}\right)^2\ cm$$
$$= 64\times \left(\cfrac{5}{4}\right)^2\ cm$$
$$= (64\times \cfrac{5}{4}\times \cfrac{5}{4})\ cm$$
$$= 100\ cm$$

Hence, the height of the tree after $$2$$ years will be $$100\ cm$$
Question 3
1 / -0

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 12000, the compound interest on the same sum for the same period at the same rate, is

A
Rs. 12600

B
Rs. 12610

C
Rs. 12640

D
Rs. 12650

Solution

Clearly, Rate = $$5\; \%$$, Time = $$3$$ years and S.I. = $$\text{Rs.}\;12000$$
Sum $$= \text{Rs.} \left (\cfrac{100\times 12000}{3\times 5}\right ) = \text{Rs.}\; 80000$$
Amount $$= \text{Rs. } \left [80000\times \left (1+\cfrac{5}{100}\right )^3\right ]$$
$$= \text{Rs. } 92610$$
We know that C.I. = Amount - Sum
So,
C.I. $$ = 92610-80000$$
$$= \text{Rs. } 12610$$
Question 4
1 / -0

The current population of a town is 10,000. If the population increases by $$10\%$$ every year, then the population of the town after three years will be

A
13,310

B
13,500

C
14,000

D
14,500

Solution

$$A=P(1+\cfrac{R}{100})^n$$
$$A=10000(1+0.1)^3=13310$$
Question 5
1 / -0

If Rs.3,750 amount to Rs.4,620 in 3 years at simple interest, then find the rate of interest.

A
$$\displaystyle 7\frac{11}{15}$$%

B
$$\displaystyle 8\frac{11}{15}$$%

C
$$\displaystyle 6\frac{11}{15}$$%

D
$$\displaystyle 5\frac{11}{15}$$%

Solution

Simple Interest $$ I = \cfrac {PNR}{100} $$
Given,
Amount $$A = Rs. 4620 $$
$$ P = Rs. 3750 $$
$$ I = A - P = 870 $$
$$ N = 3$$

So, $$ 870 = \cfrac {3750 \times 3 \times R}{100} $$
$$ => R = 7 \cfrac {11}{15} \%$$
Question 6
1 / -0

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. Find the sum.

A
Rs. $$698$$

B
Rs. $$798$$

C
Rs. $$658$$

D
Rs. $$758$$

Solution

$$\textbf{Step 1: Find the simple interest in one year and 3 years.}$$

$$\text{Given,}$$
$$\text{Sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years.}$$
$$\text{So, Amount after 4 years= Amount after 3 years }+\text{ Simple interest in 1 year} $$
$$\therefore \text{Simple interest in one year}=854-815$$
$$=\text{Rs. } 39$$
$$\text{Simple interest for 3 years}=39\times 3$$
$$=\text{Rs. }117$$

$$\textbf{Step 2: Find the principle value.}$$

$$\text{Principle}=815-117$$ $$[\because\textbf{Principle}=\textbf{Amount}-\textbf{Interest}]$$
$$=\text{Rs. }698$$

$$\textbf{Hence, Option A is correct.}$$
Question 7
1 / -0

Calculate the compound interest(I) on :
Rs. $$8,000$$ in $$2\displaystyle \frac{1}{2}$$ years at $$15\%$$ per annum.

A
$$8000$$

B
$$11000$$

C
$$11373.50$$

D
$$3373.50$$

Solution

$$Amount =8000\times \left(1+\cfrac{15}{100} \right)^2\left(1+\cfrac{7.5}{100} \right)$$

$$\Rightarrow 8000\times \dfrac{115}{100}\times \dfrac{115}{100}\times \dfrac{107.5}{100}$$

$$\Rightarrow \dfrac{11373500}{1000}\Rightarrow 11373.50 Rs.$$

$$C.I=11373.50-8000=3373.50 Rs.$$
Question 8
1 / -0

A man borrows $$Rs.\ 1,200$$ at rate $$10$$ percent per annum compound interest. If he repays $$Rs.\ 250$$ at the end of each year, find the amount outstanding at the beginning of the fourth year.

A
$$Rs.\ 769.70$$

B
$$Rs. 796$$

C
$$Rs. 799.70$$

D
$$Rs.\ 777.70$$

Solution

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

$$A=1200\left ( 1+\dfrac{10}{100} \right )\Rightarrow 1200\times \dfrac{110}{100}\Rightarrow 1320$$
According to question, borrower paid $$Rs\ 250$$ per year
Then borrowed amount for second year is $$Rs.\ 1320-250=Rs.\ 1070$$
$$A=1070\left ( 1+\dfrac{10}{100} \right )\Rightarrow 1070\times \dfrac{110}{100}\Rightarrow 1177$$
Then borrowed amount for third year is $$Rs.\ 1177-250=Rs.\ 927$$
$$A=927\left ( 1+\dfrac{10}{100} \right )\Rightarrow 927\times \dfrac{110}{100}\Rightarrow Rs.\ 1019.70$$
Borrowed amount for fourth year is $$Rs.\ 1019.70-250=Rs.\ 769.70$$

Hence, option A.
Question 9
1 / -0

Calculate the amount and the compound interest on :
Rs. $$3,500$$ at $$10\%$$ per annum in $$2$$ years

A
Rs. $$4,935$$ and Rs. $$529$$.

B
Rs. $$4,235$$ and Rs. $$735$$.

C
Rs. $$3,506$$ and Rs. $$825$$.

D
Rs. $$3,120$$ and Rs. $$900$$.

Solution

$$A=P \left(1+\dfrac{R}{100} \right)^n$$
Here, $$P=3500, R=10\%, n=2$$ years
$$A=3500\left(1+\dfrac{10}{100} \right)^2$$
$$A=3500\times \dfrac{110}{100}\times \dfrac{110}{100}$$

$$A=35\times 121=4235 Rs.$$
$$C.I=A-P=4235-3500=735 Rs.$$
Question 10
1 / -0

Calculate the amount after five years on $$Rs. \ 10,000$$ invested for $$5$$ years at $$10 \%$$ per annum.

A
$$16105.1$$

B
$$15000$$

C
$$16288.95$$

D
None of these

Solution

Given P = Rs.$$10000$$, r $$= 10$$%, n $$= 1$$ and nt $$= 5$$ years

Formula for Amount with compound interest = $$P { \left( 1+\frac { r }{100} \right) }^{ nt }$$

A $$= 10000 { \left( 1+\frac {10}{ 100 } \right) }^{ 5}$$

$$= 10000 { \left( 1.1 \right) }^{ 5 }$$

$$= 10000 (1.61051)$$

A $$= 16105.1$$

Ans Amount after $$5$$ years = Rs. $$16105.1$$

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Basics of Financial Mathematics Test 8

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Question 1

Rs. 10000

Rs. 12000

Rs. 13200

Rs. 11880

Solution

Question 2

$$76$$ cm

$$80$$ cm

$$100$$ cm

$$105$$ cm

Solution

Question 3

Rs. 12600

Rs. 12610

Rs. 12640

Rs. 12650

Solution

Question 4

13,310

13,500

14,000

14,500

Solution

Question 5

$$\displaystyle 7\frac{11}{15}$$%

$$\displaystyle 8\frac{11}{15}$$%

$$\displaystyle 6\frac{11}{15}$$%

$$\displaystyle 5\frac{11}{15}$$%

Solution

Question 6

Rs. $$698$$

Rs. $$798$$

Rs. $$658$$

Rs. $$758$$

Solution

Question 7

$$8000$$

$$11000$$

$$11373.50$$

$$3373.50$$

Solution

Question 8

$$Rs.\ 769.70$$

$$Rs. 796$$

$$Rs. 799.70$$

$$Rs.\ 777.70$$

Solution

Question 9

Rs. $$4,935$$ and Rs. $$529$$.

Rs. $$4,235$$ and Rs. $$735$$.

Rs. $$3,506$$ and Rs. $$825$$.

Rs. $$3,120$$ and Rs. $$900$$.

Solution

Question 10

$$16105.1$$

$$15000$$

$$16288.95$$

None of these

Solution

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