2 Marks Questions

Question 12 Marks

How many kg of sugar costing ₹ 45 per kg must be mixed with 30 kg sugar costing ₹ 35 per kg so that there may be a gain of 12% by selling the mixture at ₹ 47.04 per kg?

Answer

Let the CP. of mixture be ₹ x per kg
Given S.P.= ₹ 47.04 and profit = 12%
$\begin{array}{l}\because \text { S.P. }=\text { C.P. }+ \text { profit } \\ \Rightarrow 47.04=x+12 \% \text { of } x \\ \Rightarrow x=\frac{4704}{112} \Rightarrow x=42\end{array}$
Given c = ₹ 35 per kg, d = ₹ 45 per kg, m = ₹ 42 per kg
and quantity of cheaper sugar = 30 kg
$\begin{array}{l}\text { So, } \frac{\text { quantity of cheaper sugar }}{\text { quantity of dearer sugar }}=\frac{45-42}{42-35} \\ \Rightarrow \frac{30}{\text { quantity of dearer sugar }}=\frac{3}{7}\end{array}$
$\Rightarrow$ quantity of dearer sugar $=70 \text{ kg}$
Hence, 70 kg of sugar costing ₹ 45 per kg should be mixed.

View full question & answer→

Question 22 Marks

A machine being used by a company is estimated to have a life of 15 years. At that time a new machine would cost ₹ $75000$ and the scrap of the old machine would yield ₹ $9600$ only. A sinking fund is created for replacing the machine at the end of its life. What sum should be retained by the company at the end of every year to accumulate at $6\%$ per annum?

Answer

Cost of new machine $=$ ₹ $75000$
Scrap value of old machine $=$ ₹ $9600$
Hence, the money required for new machine after 15 years
$=$ ₹ $75000 -$ ₹ $9600 =$ ₹ $65400$
So, we have $\text{A} =$ ₹ $65400, \text{r} = 6\%$ p.a. $\Rightarrow \text{i} =0.06$ and $\text{n} =15$ years.
Using formula,
$A = R \left[\frac{(1+i)^n-1}{i}\right]$
$\Rightarrow 65400=\left[\frac{(1.06)^{15}-1}{0.06}\right]$
$\Rightarrow R=\frac{65400\times0.06}{2.396-1}$
$\Rightarrow R=\frac{3924}{1.396}=$ ₹ $2810.89$
Let $\text{x}=(1.06)^{15}$ Taking logarithm on both sides, we get
log $\text{x}=15$ log $1.06$
$\Rightarrow \text{log x}=15\times 0.0253$
$\Rightarrow \text{log x}=0.3795$
$\Rightarrow \text{x}=$ antilog $0.3795$
$\Rightarrow \text{x}=2.396$

View full question & answer→

Question 32 Marks

Amrita bought a car worth ₹ $12,50,000$ and makes a down payment of ₹ $3,00,000.$ The balance amount is to be paid in 4 years by equal monthly instalments at an interest rate of $15\%$ p.a. Find the EMI that Amrita has to pay for the car. $\left\{\right.$Given $\left.\left.(1.0125)^{-48}=0.5508565\right)\right\}$

Answer

Here $\text{P} =$ ₹ $9,50,000,$ $\text{i} =\frac{15}{1200}=0.0125$
$\text{n} = 48$ months
Using the reducing balancing method,
$\text{E} =\frac{P_i}{1-(1+i)^{-n}}=\frac{9,5,0000 \times 0.0125}{1-(1+0.0125)^{-48}}$
$=\frac{11875}{1-(1.0125)^{-48}}=\frac{11875}{1-0.5508565}$
$=$ ₹ $26,439.21$

View full question & answer→

Question 42 Marks

Evaluate the definite integral:
$\int_2^e\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right) d x$

Answer

$\int_2^e\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right) d x=\int_2^e(\log x)^{-1} \cdot 1~d x-\int_2^e \frac{1}{(\log x)^2} d x$
(integrate the first integral by parts, taking $(\log x)^{-1}$ as the first function)
$\begin{array}{l}=\left[(\log x)^{-1} \cdot x\right]_2^e-\int_2^e(-1)(\log x)^{-2} \cdot \frac{1}{x} \cdot x ~d x-\int_2^e \frac{1}{(\log x)^2} d x \\ =\left[\frac{x}{\log x}\right]_2^e+\int_2^e \frac{1}{(\log x)^2} d x-\int_2^e \frac{1}{(\log x)^2} d x \\ =\frac{e}{\log e}-\frac{2}{\log 2}=\frac{e}{1}-\frac{2}{\log 2}=e-\frac{2}{\log 2}\end{array}$

View full question & answer→

Question 52 Marks

Find the effective rate of return equivalent to a nominal rate of $6\%$ per annum compounded
i. quarterly
ii. continuously.$\left(\right.$Given $(1.015)^4=1.06136$ and $\left.e ^{0.06}=1.06183\right)$

Answer

i. Given $r = 6\%$ p.a.
p = 4 quarters.
So, effective rate (per rupee) = $\left(1+\frac{6}{400}\right)^4-1=(1.015)^4-1$
$= 1.06136 - 1 = 0.06136 = 0.0614$
Hence, effective rate $=0.0614\times100\%=6.14\%$
ii. Given $r=6\%$ p.a.
So, $\text{i}=\frac{6}{100}=0.06$
When compounded continuously, then effective rate (per rupee) $=\text{e}^\text{i}-1=\text{e}^{0.06}-1$
$=1.06183-1=0.06183=0.0618$
Hence, effective rate $=0.0618\times 100\%=6.18\%$

View full question & answer→

Question 62 Marks

Surjeet purchased a new house, costing ₹ 40,00,000 and made a certain amount of down payment so that he can pay the balance by taking a home loan from XYZ Bank. If his equated monthly installment is ₹ 30,000 at $9\%$ interest compounded monthly (reducing balance method) and payable for 25 years, then what is the initial down payment made by him? $[$Use $(1.0075)^{-300}=0.1062$$]$

Answer

Let the initial down payment made by Surjeet be ₹ $x,$ then $P =$ ₹ $(4000000 - x)$
Given EMI $=$ ₹ $30000, i =\frac{9}{12 \times 100}=0.0075, n =25 \times 12=300$ months
Using formula EMI $=\text{P}\frac{i}{1-(1+i)^{-n}}$
$\Rightarrow 30000=(4000000-\text{x})\times \frac{0.0075}{1-(1.0075)^{-300}}$
$\Rightarrow 30000=(4000000-\text{x}) \times \frac{0.0075}{1-0.1062}$
$\Rightarrow 4000000-\text{x}=\frac{30000\times0.8938}{0.0075}$
$\Rightarrow 4000000-\text{x}=3575200$
$\text{x}=$ ₹ $424800$

View full question & answer→

Question 72 Marks

Construct 5-yearly moving averages from the following data of the number of industrial failures in a country during 2003-2018:

Year	No. of failures
2003	23
2004	26
2005	28
2006	32
2007	20
2008	12
2009	12
2010	10
2011	9
2012	13
2013	11
2014	14
2015	12
2016	9
2017	3
2018	1

Answer

Computation of moving averages

Year	No. of failures	5-yearly moving totals	5-yearly moving averages
2003	23	-	-
2004	26	-	-
2005	28	129	25.8
2006	32	118	23.6
2007	20	104	20.8
2008	12	86	17.2
2009	12	63	12.6
2010	10	56	11.2
2011	9	55	11.0
2012	13	57	11.4
2013	11	59	11.8
2014	14	59	11.8
2015	12	49	9.8
2016	9	39	7.8
2017	3	-	-
2018	1	-	-

View full question & answer→

Applied Maths 2 Marks Questions

Test yourself on this topic