2 Marks Questions

Question 12 Marks

A vessel contains a mixture of two liquids X and Y in the ratio 3 : 5. 8 litres of mixture are drawn off from the vessel and 8 liters of liquid X is filled in the vessel. If the ratio of liquids X and Y is now becomes 7: 10, how many litres of liquids X and Y were contained by the vessel initially?

Answer

Let initially liquids X and Y be 3x litres and 5x litres respectively in the vessel.
After drawing off 8 litres of mixture:
Quantity of liquid $X$ left in the mixture $=3 x-\frac{3}{8} \times 8=(3 x-3)$ litres
Quantity of liquid $Y$ left in the mixture $=5 x-\frac{5}{8} \times 8=(5 x-5)$ litres
Further 8 litres of liquid X are mixed in the mixture.
So, quantity of liquid X in the mixture = (3x - 3 + 8) litres = (3x + 5) litres
According to given, $\frac{7}{10}=\frac{3 x+5}{5 x-5}$.
$\begin{array}{l}\Rightarrow 35 x-35=30 x+50 \Rightarrow 5 x=85 \\ \Rightarrow x=17\end{array}$
Hence, the quantity of liquid X = 3 x 17 = 51 litres and the quantity of liquid Y = 5 x 17 = 85 litres initially.

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Question 22 Marks

Find the effective rate which is equivalent to nominal rate of 10% p.a. compounded monthly. [Given that: $\left.(1.00833)^{12}=1.1047\right]$

Answer

Here, r = 10% p.a., p = 12 months
So, effective rate (per rupee) $=\left(1+\frac{10}{1200}\right)^{12}-1=(1.00833)^{12}-1$
= 1.1047 - 1 = 0.1047
Hence, the effective rate = 0.1047 * 100% = 10.47%

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Question 32 Marks

The value of a car depreciates by 12.5 % every year. By what percent will the value of the car decrease after 3 years?

Answer

Let the present value of the car be ₹P, then
value of the car after 3 years $= P (1- i )^3= P \left(1-\frac{12.5}{100}\right)^3$
$= P \left(1-\frac{1}{8}\right)^3= P \left(\frac{7}{8}\right)^3$
Decrease in the value of car $= P - P \left(\frac{7}{8}\right)^3= P \left[1-\frac{343}{512}\right]= P \times \frac{169}{512}$
$\therefore$ Percentage decrease in the value of the car after 3 years
$\begin{array}{l}=\left(\frac{\text { Decrease in value }}{\text { Present value }} \times 100\right) \%=\left(\frac{ P \times \frac{169}{512}}{ P } \times 100\right) \% \\ =\frac{169 \times 25 \%}{128}=\frac{4225}{128} \%=33 \frac{1}{128} \%\end{array}$

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Question 42 Marks

Evaluate: $\int_0^3[x] d x$

Answer

$\begin{array}{l}\int_0^3[x] d x=\int_0^1[x] d x+\int_1^2[x] d x+\int_2^3[x] d x \\ =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x \\ =0+[x]_1^2+[2 x]_2^3=1+6-4=3\end{array}$

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Question 52 Marks

Mrs Vandana invested ₹ 35000 in a shares of a company and reinvested the earnings every year in buying the shares of the same company. At the end of 5 years, the value of shares increased to ₹ 56000. Calculate the compound annual growth rate of her investment.

Answer

Given P.V. = ₹ 35000, F.V. = ₹ 56000, n = 5 years
So, CAGR $=\left(\frac{\text { F.V. }}{\text { P.V. }}\right)^{\frac{1}{n}}-1=\left(\frac{56000}{35000}\right)^{\frac{1}{5}}-1=(1.6)(1.6)^{\frac{1}{2}}-1$
Let $x =(1.6)^{\frac{1}{5}}$
$\begin{array}{l}\Rightarrow \log x=\frac{1}{5} \log 1.6 \\ =\frac{1}{5} \times 0.2041=0.04082 \\ \Rightarrow x=\text { antilog } 0.04082=1.098\end{array}$
So, CAGR = 1.098 - 1 = 0.098
Hence, CAGR = 0.098 * 100 % = 9.8 %

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Question 62 Marks

A machine has a scrap value of ₹ 22500 after 15 years of its purchase. If the annual depreciation charge is ₹ 8500, find its original cost using linear method.

Answer

Scrap value = ₹ 22500, useful life = 15 years, annual depreciation = ₹ 8500
$\begin{array}{l}\therefore 8500=\frac{\text { original value }-22500}{15} \\ \Rightarrow \text { original value }-22500=127500 \\ \Rightarrow \text { original value }=₹ 127500+₹ 22500=₹ 150000\end{array}$

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Question 72 Marks

Assuming a four yearly cycle, calculate the trend by the method of moving averages from the following data:

Year	1984	1985	1986	1987	1988	1989	1990	1991	1992	1993
Value	12	25	39	54	70	87	105	100	82	65

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Applied Maths 2 Marks Questions

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