Basics of Financial Mathematics Test 22

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Basics of Financial Mathematics Test 22

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Weekly Quiz Competition

Question 1
1 / -0

Aman gave Rs. $9500$ as a loan to Megh at the rate of $4\%$ p.a. compounded yearly. Megh returned the amount after two years. Calculate the interest on the first year's interest.

A
$380$

B
$270$

C
$15.2$

D
$40$

Solution

$P=9500,R=4,T=2$ years

Interest in $1$ year $=\cfrac { P\times R\times 1 }{ 100 }$ $=\cfrac { 9500\times 4 }{ 100 }$ $=380$

Interest on interest $=\cfrac { 380\times 4\times 1 }{ 100 }$ $=15.2$
Question 2
1 / -0

The bank offers senior citizens intrest at the rate $9\%$ p.a compounded yearly on a fixed deposit. What interest will be received at the end of $5$ years if Rs. $10,000$ are deposited. Calculate by the method of simple interest.

A
$5386.23$

B
$5500.23$

C
$6523.65$

D
$4367.78$

Solution

$P=10000,R=9,T=5$ yrs
Interest in $1$ st year $=\cfrac { P\times R\times T }{ 100 }$
$=\cfrac { 10000\times 9\times 1 }{ 100 }$
$=900$
$P'=(900+10000)$
Interest in $2$ nd year $=\cfrac { 10900\times 9\times 1 }{ 100 } =981$
Interest in $3$ rd year $=\cfrac { 11881\times 9\times 1 }{ 100 }$
$=1069.29$
Interest in $4$ th year $=\cfrac { 12950.29\times 9\times 1 }{ 100 }$
$=1165.5261$
Interest in $5$ th year $=\cfrac { 14115.81\times 9\times 1 }{ 100 }$
$=1270.42$
Total interest $=900+981+1069.29+1165.52+1270.42$
$=5386.23$
Question 3
1 / -0

A bank pays interest at the rate of $8\%$ per annum compounded yearly. Find the interest by simple interest method if $Rs.\ 8000$ was deposited for $3$ years.

A
$Rs.\ 1000$

B
$Rs.\ 2077.7$

C
$Rs.\ 2000$

D
$Rs.\ 3456$

Solution

$P=Rs.\ 8000,R=8$ %, Time period $=3$ years

Interest in $1$ st year $=\cfrac { P\times R\times T }{ 100 }$
$=\cfrac { 8000\times 8\times 1 }{ 100 }$
$=Rs.\ 640$

Principal amount for $2$ nd year will be,
$P'=P+640$
$=Rs.\ 8000+Rs.\ 640$
$=Rs.\ 8640$

Interest in $2$ nd year $=\cfrac { 8640\times 8 }{ 100 }$
$=Rs.\ 691.2$

Principal amount for $2$ nd year will be,
$P''=P'+691.2$
$=Rs.\ 8640+Rs.\ 691.2$
$=Rs.\ 9331.2$

Interest in $3$ rd year $=\cfrac { 9331.2\times 8 }{ 100 }$
$=Rs.\ 746.5$

So, total interest $=640+691.2+746.5$
$=Rs.\ 2077.69$
$\approx Rs.\ 2077.7$
Question 4
1 / -0

A businessman borrowed the Rs. $45$ lakh for two years at the rate $2.5\%$ p.a. compounded annually. Calculate the compound interest by simple interest method.

A
$2,23000$

B
$2,27,812.5$

C
$5,65,563$

D
$2,00,000$

Solution

$\Rightarrow$ Here, Principal amount for first year is $Rs.45,00,000$ and $R=2.5\%$ .
$\Rightarrow$ Interest for first year = $\dfrac{4500000\times 2.5}{100}=Rs.1,12,500$

$\Rightarrow$ Principal amount for second year = $Rs.4500000+Rs.112500=Rs.46,12,500$

$\Rightarrow$ Second year interest = $\dfrac{4612500\times 2.5}{100}=Rs.1,15,312.5$

$\Rightarrow$ Total interest for 2 yeras = $Rs.1,12,500+Rs.1,15,312.5=Rs.2,27,812.5$
Question 5
1 / -0

The population of a town $3$ years ago was Rs. $50,000$ .If the population in these three years has increased at the rate of $10\%,15\%$ and $8\%$ respectively, find the present population.

A
Rs. $68,210$

B
Rs. $68,310$

C
Rs. $68,410$

D
Rs. $68,510$

Solution

Let $P=.50,000$ and
$R_1=10\%,\\ R_2=15\%,\\ R_3=8\%$ .
Present population = $P\times \left (1+\dfrac{R_1}{100}\right)\left (1+\dfrac{R_2}{100}\right)\left (1+\dfrac{R_3}{100}\right)$
On substituiting the given values,
Present population $=$ $50000\times \left (1+\dfrac{10}{100}\right)\times \left (1+\dfrac{15}{100}\right)\times \left (1+\dfrac{8}{100}\right)$
$=$ $50000\times \dfrac{11}{10}\times \dfrac {23}{20}\times \dfrac{27}{25} =68310$
Therefore, present population $=$ $68310$ .
Question 6
1 / -0

A man lends Rs. $12,500$ at $12$ % for the first year, at $15$ % for the second year and at $18$ % for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. for the first year and the compound interest for the third year.

A
$\text{Rs }1,498$

B
$\text{Rs }1,598$

C
$\text{Rs }1,298$

D
$\text{Rs }1,398$
Question 7
1 / -0

In what time will Rs. 4000 amount to Rs. 5324 at 10% per annum compound interest?

A
$1$ years

B
$2$ years

C
$3$ years

D
$4$ years

Solution

Given that, $P=Rs.\ 4000,\,A=Rs.\ 5324$ and $R=10\%$
Let the time be $T$ years
We know that, for compound interest,
$A=P\times \left(1+\dfrac{R}{100}\right)^T$

$\therefore \ 5324=4000\times \left(1+\dfrac{10}{100}\right)^T$

$\Rightarrow$ $\dfrac{5324}{4000}=\left(1+\dfrac{1}{10}\right)^T$

$\Rightarrow$ $\dfrac{1331}{1000}=\left(\dfrac{11}{10}\right)^T$

$\Rightarrow$ $\left(\dfrac{11}{10}\right)^3=\left(\dfrac{11}{10}\right)^T$

$\therefore$ $T=3\,years$ .

Hence, $Rs.\ 4000$ will amount to $Rs.\ 5324$ at $10\%$ per annum in $3$ years.
Question 8
1 / -0

What amount will sum up to Rs. $6,655$ at $10$ % p.a. in CI in $3$ years?

A
$4,000$

B
$5,000$

C
$6,000$

D
$5,500$

Solution

$\Rightarrow$ We have, $A=Rs.6,655,\,R=10\%$ and $T=3\,years$
$\Rightarrow$ $A=P\times (1+\dfrac{R}{100})^T$

$\Rightarrow$ $6655=P\times (1+\dfrac{10}{100})^3$

$\Rightarrow$ $6655=P\times (\dfrac{11}{10})^3$

$\Rightarrow$ $P=\dfrac{6655\times 1000}{1331}=Rs.5000$
Question 9
1 / -0

What sum of money will amount to Rs. $31,944$ in three years at $10$ % per annum compounded yearly?

A
Rs. $22,000$

B
Rs. $27,000$

C
Rs. $24,000$

D
Rs. $25,000$

Solution

$\Rightarrow$ Here, we have $A=Rs.31,944,\,T=3\,years$ and $R=10\%$ .
$\Rightarrow$ $A=P\times (1+\dfrac{R}{100})^T$

$\Rightarrow$ $31,944=P\times (1+\dfrac{10}{100})^3$

$\Rightarrow$ $31,944=P\times (\dfrac{11}{10})^3$

$\Rightarrow$ $31,944=P\times \dfrac{1331}{1000}$

$\Rightarrow$ $P=\dfrac{31,944\times 1000}{1331}$

$\therefore$ $P=Rs.24,000$ .
Question 10
1 / -0

A sum of $Rs.\ 15,000$ is invested for $3$ years at $10 \%$ per annum compound interest. Calculate the interest for the second year.

A
$$Rs.\ 1,680$$

B
$$Rs.\ 1,650$$

C
$$Rs.\ 1,710$$

D
$$Rs.\ 1,640$$

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Basics of Financial Mathematics Test 22

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Basics of Financial Mathematics Test 22

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Question 1

380380380

270270270

15.215.215.2

404040

Solution

Question 2

5386.235386.235386.23

5500.235500.235500.23

6523.656523.656523.65

4367.784367.784367.78

Solution

Question 3

Rs. 1000Rs.\ 1000Rs. 1000

Rs. 2077.7Rs.\ 2077.7Rs. 2077.7

Rs. 2000Rs.\ 2000Rs. 2000

Rs. 3456Rs.\ 3456Rs. 3456

Solution

Question 4

2,230002,230002,23000

2,27,812.52,27,812.52,27,812.5

5,65,5635,65,5635,65,563

2,00,0002,00,0002,00,000

Solution

Question 5

Rs. 68,21068,21068,210

Rs. 68,31068,31068,310

Rs. 68,41068,41068,410

Rs. 68,51068,51068,510

Solution

Question 6

Rs 1,498\text{Rs }1,498Rs 1,498

Rs 1,598\text{Rs }1,598Rs 1,598

Rs 1,298\text{Rs }1,298Rs 1,298

Rs 1,398\text{Rs }1,398Rs 1,398

Question 7

111 years

222 years

333 years

444 years

Solution

Question 8

4,0004,0004,000

5,0005,0005,000

6,0006,0006,000

5,5005,5005,500

Solution

Question 9

Rs.22,00022,00022,000

Rs.27,00027,00027,000

Rs.24,00024,00024,000

Rs.25,00025,00025,000

Solution

Question 10

$$Rs.\ 1,680$$

$$Rs.\ 1,650$$

$$Rs.\ 1,710$$

$$Rs.\ 1,640$$

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$380$

$270$

$15.2$

$40$

$5386.23$

$5500.23$

$6523.65$

$4367.78$

$Rs.\ 1000$

$Rs.\ 2077.7$

$Rs.\ 2000$

$Rs.\ 3456$

$2,23000$

$2,27,812.5$

$5,65,563$

$2,00,000$

Rs. $68,210$

Rs. $68,310$

Rs. $68,410$

Rs. $68,510$

$\text{Rs }1,498$

$\text{Rs }1,598$

$\text{Rs }1,298$

$\text{Rs }1,398$

$1$ years

$2$ years

$3$ years

$4$ years

$4,000$

$5,000$

$6,000$

$5,500$

Rs. $22,000$

Rs. $27,000$

Rs. $24,000$

Rs. $25,000$