Basics of Financial Mathematics Test 18

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Basics of Financial Mathematics Test 18

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Weekly Quiz Competition

Question 1
1 / -0

Oscar earned Rs. $2400$ as simple interest on Rs. ` $4500$ for $3$ months. What is the annual rate of interest?

A
$16.66\%$

B
$17.77\%$

C
$18.88\%$

D
$19.99\%$

Solution

Given, principal $=4500$ , Interest $=$ Rs. $2400$ , period $=3$ months
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{2400\times 100}{4500\times 3}$
$\Rightarrow R = 17.77\%$
Therefore, the annual rate of interest is $17.77\%$ .
Question 2
1 / -0

You borrow Rs. $4,000$ from a loan shark. If you owe Rs. $7,200$ in $4$ years, what would be the simple interest rate?

A
$10\%$

B
$20\%$

C
$30\%$

D
$40\%$

Solution

Given:
Principal $=$ Rs. $4,000$
Interest $= 7200 - 4000 = 3200$
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{3200\times 100}{4000\times 4}$
$\Rightarrow R = 20\%$
Thus, the simple interest rate is $20\%$ .
Question 3
1 / -0

Sharmila got a Rs. $1300$ loan for $5$ years. She paid Rs. $100$ in interest. What was the interest rate?

A
$1.1\%$

B
$1.3\%$

C
$1.5\%$

D
$1.7\%$

Solution

Given, principal $=$ Rs. $1300$ , Interest $=$ Rs. $100$ , period $=5$ years
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{100\times 100}{1300\times 5}$
$\Rightarrow R = 1.5\%$
Therefore, the interest rate is $1.5\%$ .
Question 4
1 / -0

You invested Rs. $1500$ and received Rs. $5000$ after three years. What had been the interest rate?

A
$111.11\%$

B
$222.22\%$

C
$333.33\%$

D
$77.77\%$

Solution

Amount invested $= 1500$
Amount received after three years $= 5000$

$\Rightarrow$ Interest earned $= 5000-1500 = 3500$
Let interest rate $\%$ be $r$
$\Rightarrow$ $3500 = \dfrac{1500 \times 3 \times r}{100}$
$\Rightarrow$ $r = 77.77\%$
Question 5
1 / -0

To start a grocery shop, a woman borrowed Rs. $1,500$ . If the loan was for four years and the amount of interest was Rs. $150$ , what simple interest rate was she charged?

A
$1.5\%$

B
$2.5\%$

C
$3.5\%$

D
$4.5\%$

Solution

Given, principal $=$ Rs. $1500$ , Interest $=$ Rs. $150$ , period $=4$ years
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{150\times 100}{1500\times 4}$
$\Rightarrow R = 2.5\%$
Therefore, the simple interest is $2.5\%$ .
Question 6
1 / -0

Jenn borrowed Rs. $5,000$ for $5$ years and had to pay Rs. $1,500$ simple interest at the end of that time. What rate of interest did she pay?

A
$4\%$

B
$5\%$

C
$6\%$

D
$7\%$

Solution

Given, $I=1500,P=5000, T=5$ years
We need to find rate of interest $R$ .
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{1500\times 100}{5000\times 5}$
$\Rightarrow R = 6\%$
Question 7
1 / -0

The present worth of Rs. $450$ due in $2\dfrac{3}{2}$ years at $10\%$ per annum compound interest is:

A
$322.36$

B
$222.36$

C
$122.36$

D
$422.36$

Solution

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $450$ , $r = 10\%$ , $n = 1$ and $t =$ $2\dfrac{3}{2}$ years
Therefore, we have $450 = P\left (1+\dfrac{0.1}{1}\right)^{1\times 3.5}$
$\Rightarrow 450 = P\times 1.1^{3.5}$
$\Rightarrow 450 = 3500\times 1.395965$
$\Rightarrow P= \dfrac{450}{1.395965}$
$\Rightarrow P =$ Rs. $322.36$
Question 8
1 / -0

A machine was purchased $3$ years ago. Its value decreases by $5\%$ every year. Its present value is $Rs.23000$ . For how much money was the machine purchased?

A
$Rs.16828.45$

B
$Rs.26826.07$

C
$Rs.36828.56$

D
$Rs.46828.80$

Solution

Given,
Depreciated value $A=Rs. 23,000$
Rate of depreciation $R=5\%$
Time $T=3$ years
So, $n=3$
let the machine was purchased in $Rs. P$

Depreciated value, $A = P\left( 1-\dfrac { R }{ 100 } \right) ^{ n }$
$\Rightarrow 23000 = P \times \left( 1-\dfrac { 5 }{ 100 } \right) ^{ 3 }$

$\Rightarrow 23000= P \times \left( 1-\dfrac { 1 }{ 20 } \right) ^{ 3 }$

$\Rightarrow 23000 = P \times \left(\dfrac { 19}{ 20 } \right) ^{ 3 }$

$\Rightarrow 23000 = P \times \dfrac { 19}{ 20 } \times\dfrac{19}{20}\times\dfrac{19}{20}$

$\Rightarrow P=23,000 \times \dfrac { 20}{ 19 } \times\dfrac{20}{19}\times\dfrac{20}{19}$

$\Rightarrow P=26,826.07$

Therefore, the machine was purchased for $Rs.26,826.07$ .
Question 9
1 / -0

Calculate the compound interest on a sum of Rs. $20000$ at the end of $3$ years at the rate of $10\%$ p.a. compounded annually.

A
$6620$

B
$5620$

C
$3410$

D
$2386$

Solution

Given, $P = 20000, r = 10\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 20000\left [\left (1+\dfrac{10}{100}\right)^3\right]$
$A = 20000\times\dfrac{11}{10}\times\dfrac{11}{10}\times\dfrac{11}{10}$
$A = \text{Rs.} 26620$
Compound interest $=$ Amount $-$ Principal
$C.I. = 26620 - 20000$
$C.I. = \text{Rs.} 6620$
Question 10
1 / -0

A sum of $Rs.1728$ becomes $Rs.3375$ in $3$ years at compound interest, compound annually. Find the rate of interest.

A
$10\%$

B
$15\%$

C
$20\%$

D
$25\%$

Solution

Given: $P = Rs.1728, A =Rs. 3375, n = 3$ years
We need to find rate of interest i.e. $r\%$
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$\Rightarrow 3375 = 1728\left [\left (1+\dfrac{r}{100}\right)^3\right]$
$\Rightarrow \dfrac{3375}{1728}=\left (1+\dfrac{r}{100}\right)^3$
$\Rightarrow \left (\dfrac{15}{12}\right)^3 = \left (1 +\dfrac{r}{100}\right)^3$
Taking cuberoot on both the sides, we get
$\dfrac{15}{12}=1+\dfrac{r}{100}$
Thus $r = 25$

The rate of interest is $25\%$

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Basics of Financial Mathematics Test 18

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Basics of Financial Mathematics Test 18

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Question 1

16.66%16.66\%16.66%

17.77%17.77\%17.77%

18.88%18.88\%18.88%

19.99%19.99\%19.99%

Solution

Question 2

10%10\%10%

20%20\%20%

30%30\%30%

40%40\%40%

Solution

Question 3

1.1%1.1\%1.1%

1.3%1.3\%1.3%

1.5%1.5\%1.5%

1.7%1.7\%1.7%

Solution

Question 4

111.11%111.11\%111.11%

222.22%222.22\%222.22%

333.33%333.33\%333.33%

77.77%77.77\%77.77%

Solution

Question 5

1.5%1.5\%1.5%

2.5%2.5\%2.5%

3.5%3.5\%3.5%

4.5%4.5\%4.5%

Solution

Question 6

4%4\%4%

5%5\%5%

6%6\%6%

7%7\%7%

Solution

Question 7

322.36322.36322.36

222.36222.36222.36

122.36122.36122.36

422.36422.36422.36

Solution

Question 8

Rs.16828.45Rs.16828.45Rs.16828.45

Rs.26826.07Rs.26826.07Rs.26826.07

Rs.36828.56Rs.36828.56Rs.36828.56

Rs.46828.80Rs.46828.80Rs.46828.80

Solution

Question 9

662066206620

562056205620

341034103410

238623862386

Solution

Question 10

10%10\%10%

15%15\%15%

20%20\%20%

25%25\%25%

Solution

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Question Analysis

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$16.66\%$

$17.77\%$

$18.88\%$

$19.99\%$

$10\%$

$20\%$

$30\%$

$40\%$

$1.1\%$

$1.3\%$

$1.5\%$

$1.7\%$

$111.11\%$

$222.22\%$

$333.33\%$

$77.77\%$

$1.5\%$

$2.5\%$

$3.5\%$

$4.5\%$

$4\%$

$5\%$

$6\%$

$7\%$

$322.36$

$222.36$

$122.36$

$422.36$

$Rs.16828.45$

$Rs.26826.07$

$Rs.36828.56$

$Rs.46828.80$

$6620$

$5620$

$3410$

$2386$

$10\%$

$15\%$

$20\%$

$25\%$