Basics of Financial Mathematics Test 26

Result

Basics of Financial Mathematics Test 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$1000$$ dollars is deposited in a bank account for $$4$$ years at $$8$$% per annum.
Calculate interest using compound interest formula.

A
$$340.5$$ dollars

B
$$350.5$$ dollars

C
$$360.5$$ dollars

D
$$370.5$$ dollars

Solution

$$CI=P[(1+\cfrac{r}{100})^n-1]$$
$$\implies CI=1000[(1+\cfrac{8}{100})^4-1]=360.48\approx 360.5$$
$$\implies CI=360.5$$dollars.
Question 2
1 / -0

A sum of Rs.$$15,000$$ is invested for $$3$$ years at $$13$$ % per annum compound interest. Calculate the compound interest.

A
$$6500.435$$

B
$$6689.245$$

C
$$6643.455$$

D
$$6276.585$$

Solution

We have,
$$P=Rs.\ 15000$$
$$T=3\ years$$
$$R=13\%$$
$$C.I=?$$

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

$$A=15000\left(1+\dfrac{13}{100}\right)^3$$

$$A=15000\left(\dfrac{113}{100}\right)^3$$

$$A=15000\left(\dfrac{113\times 113\times 113}{100\times 100\times 100}\right)$$

$$A=15\left(\dfrac{113\times 113\times 113}{1000}\right)$$

$$A=Rs.\ 21,643.455$$

So, the compound interest
$$=21.643.455-15000=Rs.\ 6643.455$$
Question 3
1 / -0

Raman borrowed Rs.$$1,20,000$$ for $$4$$ years at $$8$$ % per year compound interest. Calculate the final amount at the end of four years.

A
$$1,63,250$$

B
$$1,53,250$$

C
$$1,63,700$$

D
$$1,66,250$$

Solution

We have,
$$P=Rs.\ 120, 000$$
$$T=4\ years$$
$$R=8\%$$
$$A=?$$

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

So,
$$A=120000\left(1+\dfrac{8}{100}\right)^4$$

$$A=120000\left(\dfrac{108}{100}\right)^4$$

$$A=120000\left(\dfrac{108\times 108\times 108\times 108}{100\times 100\times 100\times 100}\right)$$

$$A=12\left(\dfrac{11664\times 11664}{10000}\right)$$

$$A=Rs.\ 163258.675\approx Rs.\ 163250$$

Hence, this is the answer.
Question 4
1 / -0

Find the compound interest on Rs. $$2,000$$ for $$3$$ years, compounded annually at $$12\%$$ per annum.

A
$$764$$

B
$$810$$

C
$$820$$

D
$$850$$

Solution

Given: $$P=Rs\ 2000$$

$$T=3$$ years

$$R=12\%$$ p.a.

So, Interst for the first year $$=\cfrac{2000\times 12\times 1}{100}$$

$$=Rs\ 240$$

Amount after the first year $$=P+$$ Interest for the first year

$$=Rs.\ 2000+Rs.\ 240=Rs.\ 2240$$

Interst for the second year $$=\cfrac{2240\times 12\times 1}{100}$$

$$=Rs.\ 268.8$$

Hence, the amount after the second year $$=Rs.\ 2240+Rs.\ 268.8=Rs.\ 2508.8$$

Interst for the third year $$=\cfrac{2508.8\times 12\times 1}{100}$$

$$=Rs.\ 301.056$$

Amount after the third year $$=Rs.\ 2508.8+Rs.\ 301.056=Rs.\ 2809.856$$

Compound interest $$=$$ Final amount $$-$$ original amount

$$=Rs.\ 2809.856-Rs.\ 2000$$

$$=Rs.\ 809.856$$

$$\approx Rs.\ 810$$
Question 5
1 / -0

A sum of Rs.$$100,000$$ is invested for $$4$$ years at $$13$$ % per annum compound interest. Calculate the interest.

A
$$61000$$

B
$$62000$$

C
$$63000$$

D
$$64000$$

Solution

$$CI=P[(1+\cfrac{r}{100})^n-1]$$
$$\implies CI=100,000[(1+\cfrac{13}{100})^4-1]\\ \implies CI=63,047.36\approx 63,000$$.
Hence, interest$$=Rs.63000$$.
Question 6
1 / -0

Find the rate which will amount to Rs. $$12,100$$ in two years, if the principle amount was Rs $$10,000$$.

A
$$10$$%

B
$$11$$%

C
$$12$$%

D
$$13$$%

Solution

We have,
$$P=Rs.\ 10000$$
$$T=2\ years$$
$$R=?$$
$$A=Rs.\ 12100$$

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

So,
$$12100=10000\left(1+\dfrac{R}{100}\right)^2$$

$$\left(1+\dfrac{R}{100}\right)^2=\dfrac{121}{100}$$

$$1+\dfrac{R}{100}=\dfrac{11}{10}$$

$$\dfrac{R}{100}=\dfrac{1}{10}$$

$$R=10\%$$

Hence, this is the answer.
Question 7
1 / -0

A car which costs $$\text{Rs. } 3,50,000$$ depreciates by $$10$$% every year. What will be the worth of the car after three years?

A
$$\text{Rs. } 2,55,150$$

B
$$\text{Rs. } 1,65,150$$

C
$$\text{Rs. } 2,25,350$$

D
$$\text{Rs. } 1,55,350$$

Solution

Given:
Principal, $$P =Rs.\ 350000$$
Rate, $$R= 10\%$$
Time, $$T=3$$ years

The worth of car after $$3$$ years will be,
$$\begin{aligned}{}350000{\left( {1 - \frac{{10}}{{100}}} \right)^3}& = 350000 \times \frac{9}{{10}} \times \frac{9}{{10}} \times \frac{9}{{10}}\\ &= \text{Rs. }2,55,150\end{aligned}$$

So, the price of the car after $$3$$ years will be equal to $$\text{Rs. } 2,55,150.$$
Question 8
1 / -0

Madhavi bought a scooter for $$Rs. 40,000$$ but she does not have enough money to pay all at once. The shopkeeper says that she can divide the cost into 4 equal monthly instalments but she would have to pay $$Rs.100$$ extra each month for 4 months. What is this additional amount paid by Madhavi to the shopkeeper called as and calculate the total extra amount at the end of 4 months paid by Madhavi?

A
Profit, $$Rs. 400$$

B
Principal, $$Rs. 400$$

C
Loss, $$Rs. 400$$

D
Interest, $$Rs. 400$$

Solution

$$Cost \space of \space scooter=Rs.40000$$
$$No. \space of \space installments=4$$
$$\therefore cost \space of \space each \space installments=\frac{40000}{4}$$
$$=10000$$
$$But \space 100 \space is \space paid \space extra \space in \space each \space instalment.$$
$$Total \space extra \space amt. \space paid=100\times4=400$$
The seller costs this extra $$400$$ as interest as the profit is included in $$Rs.40000$$
$$\therefore option \space D$$
Question 9
1 / -0

Belose Infrastructures just issued 10 million Rs100-par bonds payable carrying 8% coupon rate and maturing in 5 years. The bond indenture requires GI to set up a sinking up to pay off the bond at the maturity date. Semi-annual payments are to be made to the fund which is expected to earn 10% per annum. Find the amount of required periodic contributions.

A
7905155

B
7950515

C
8950515

D
6950515
Question 10
1 / -0

The simple interest at $$x\%$$ for $$x$$ years will be Rs. $$x$$ on a sum of _________.

A
Rs. $$x$$

B
Rs. $$100x$$

C
Rs. $$\left(\displaystyle\frac{100}{x}\right)$$

D
Rs. $$\left(\displaystyle\frac{100}{x^2}\right)$$

Solution

Let principal sum be $$y$$
Simple Interest $$=$$ Principal $$\times$$ Rate $$\times$$ Time

$$\therefore x = y\times \dfrac{x}{100} \times x$$

$$\therefore y =$$ Rs. $$\dfrac{100}{x}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Basics of Financial Mathematics Test 26

Result

Basics of Financial Mathematics Test 26

Score

-

Rank

-

SHARING IS CARING

Question 1

$$340.5$$ dollars

$$350.5$$ dollars

$$360.5$$ dollars

$$370.5$$ dollars

Solution

Question 2

$$6500.435$$

$$6689.245$$

$$6643.455$$

$$6276.585$$

Solution

Question 3

$$1,63,250$$

$$1,53,250$$

$$1,63,700$$

$$1,66,250$$

Solution

Question 4

$$764$$

$$810$$

$$820$$

$$850$$

Solution

Question 5

$$61000$$

$$62000$$

$$63000$$

$$64000$$

Solution

Question 6

$$10$$%

$$11$$%

$$12$$%

$$13$$%

Solution

Question 7

$$\text{Rs. } 2,55,150$$

$$\text{Rs. } 1,65,150$$

$$\text{Rs. } 2,25,350$$

$$\text{Rs. } 1,55,350$$

Solution

Question 8

Profit, $$Rs. 400$$

Principal, $$Rs. 400$$

Loss, $$Rs. 400$$

Interest, $$Rs. 400$$

Solution

Question 9

7905155

7950515

8950515

6950515

Question 10

Rs. $$x$$

Rs. $$100x$$

Rs. $$\left(\displaystyle\frac{100}{x}\right)$$

Rs. $$\left(\displaystyle\frac{100}{x^2}\right)$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.