5 Marks Questions

Question 15 Marks

In 4 years, a mobile costing ₹ 36,000 will have a salvage value of ₹ 7200.
The following graph shows the depreciation of a mobile’s value over 4 years.

A new mobile at that time (i.e., after 4 years) is expected to cost for ₹ 55,200. In order to provide funds for the difference between the replacement cost and the salvage cost, a sinking fund is set up into which equal payments are placed at the end of each year. If the fund earns interest at the rate 7% compounded annually, how much should each payment be? Also find the amount of Annual Depreciation of the mobile’s value over 4 years and find the rate of depreciation (under straight line method).
Use (1.07)⁴=1.3107.

Answer

Amount needed after 4 years
= Replacement Cost - Salvage Cost = ₹ (55,200 – 7200) = ₹ 48,000
The payments into sinking fund consisting of 10 annual payments at the rate 7% per year is given by

$\Rightarrow R=\frac{48000}{4.4385} $ =₹10814.5
Amount of Annual Depreciation $=\frac{36000-7200}{4}=\frac{28800}{4}$ = ₹ 7200
and rate of Depreciation $=\frac{7200}{36000-7200} \times 100=25 \%$

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Question 25 Marks

Supply and demand curves of a tyre manufacturer company is given below:

The above graph showing the demand and supply curves of a tyre manufacturer company which are linear. ‘ABC’ tyre manufacturer sold 25 units every month when the price of a tyre was ₹ 20000 per units and ‘ABC’ tyre manufacturer sold 125 units every month when the price dropped to ₹ 15000 per unit. When the price was ₹ 25000 per unit, 180 tyres were available per month for sale and when the price was only ₹ 15000 per unit, 80 tyres remained. Find the demand function. Also find the consumer surplus if the supply function is given to be 𝑺(𝒙) = 𝟏𝟎𝟎 𝒙 + 𝟕𝟎𝟎𝟎

Answer

Let us consider demand function be $p=D(x)=a x+b \ldots \ldots . .(i)$
When $x = 2 5$ then $p = 2 0 0 0$
From equation (i), we have $2 0 0 0 0 = 2 5 a + b . \ldots . . . . . .(i i)$
And when $x = 1 2 5$ then $p = 1 5 0 0 0$
From equation ( $i$ ), we have $15000=125 a+b . . . . . . . . .(i i)$
On solving equations (i) and (ii), we get $a=- 5 0$ and $b = 2 1 2 5 0$
Therefore, demand function, $p=D(x)=-50 x+21250$
For equilibrium point $D\left(x_0\right)=S\left(x_0\right)$
$
\Rightarrow-50 x_0+21250=100 x_0+7000
$
$
\Rightarrow-150 x_0=-14250
$
$
\Rightarrow x_0=95
$
On putting value of $x_0$ in demand function and supply function, we get
$
p_0=16500
$
$\therefore$ Consumer surplus $(C S)$
$
=\int_0^{x_0} D(x) d x-p_0 x_0
$
$
=\int_0^{95}(-50 x+21250) d x-16500 \times 95
$
$
=\left(-50 \frac{x^2}{2}+2150 x\right)_0^{96}-1567500
$
= 1793125-1567500
=₹ 225625

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Question 35 Marks

A toy rocket is fired, from a platform, vertically into the air, its height above the ground after $t$ seconds is given by $s(t)= a t ^2+ b t + c$, where $a , b , c \in R ; a \neq 0$ and $s(t)$ is measured in
metres. After $1 0$ second, the rocket is $1 6 ~ m$ above the ground; after $2 0$ seconds, $2 2 ~ m$; after 30 seconds, $2 5$ m.
(i) Write down a system of three linear equations in terms of $a , b$ and $c$.
(ii) Hence find the values of $a , b$ and $c$, using matrix method.

Answer

(i) $s(t)=a t^2+b t+c ; t \geq 0$
Clearly, $(10,16),(20,22),(30,25)$ lie on the curve of $s(t)$.
Then, 100a+10b+c=16
$
\begin{aligned}
400 a+20 b+c & =22 \\
900 a+30 b+c & =25
\end{aligned}
$
(ii) Let, $A=\left(\begin{array}{ccc}100 & 10 & 1 \\ 400 & 20 & 1 \\ 900 & 30 & 1\end{array}\right) ; X=\left(\begin{array}{l}a \\ b \\ c\end{array}\right) ; B=\left(\begin{array}{l}16 \\ 22 \\ 25\end{array}\right)$
Then, the system becomes, $A X=B$
$
\begin{aligned}
|A|= & 100(-10)-400(-20)+900(-10) \\
& =-1000+8000-9000 \\
& =-2000 \neq 0
\end{aligned}
$
Now, adj $A=\left(\begin{array}{ccc}-10 & 500 & -6000 \\ 20 & -800 & 6000 \\ -10 & 300 & -2000\end{array}\right)^T=\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)$
Therefore, $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-2000}\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)$
Then, $X=A^{-1} B=\frac{1}{-2000}\left(\begin{array}{ccc}-10 & 20 & -10 \\ 500 & -800 & 300 \\ -6000 & 6000 & -2000\end{array}\right)\left(\begin{array}{l}16 \\ 22 \\ 25\end{array}\right)$
$
=\frac{1}{-2000}\left(\begin{array}{c}
30 \\
-2100 \\
-14000
\end{array}\right)
$
$
=\left(\begin{array}{c}
-\frac{3}{200} \\
\frac{21}{20} \\
7
\end{array}\right)
$
Therefore, $a=-\frac{3}{200}, b=\frac{21}{20}, c=7$.

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Question 45 Marks

An owl was sitting at $( 0 , k ) ; k > 0$. Then it starts flying along the path whose equation is given by $y = a x ^ { 2 } + b x + c$, where $a \in R -\{0\}, b , c \in R$. It passes through the points $( 1 , 2),(2,1)$ and (4,5). Using Cramer's Rule, find the values of $a, b, c$ and hence $k$

Answer

$
y=a x^2+b x+c
$
Owl passes through the points $( 1 , 2 ),( 2 , 1 )$ and $( 4 , 5 )$. So, it must satisfy the given equation
Therefore,
$
\begin{array}{l}
2=a+b+c \\
1=4 a+2 b+c \\
5=16 a+4 b+c
\end{array}
$
Now, $\quad D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1\end{array}\right|=1(2-4)-1(4-16)+1(16-32)=-6 \neq 0$
$
\begin{aligned}
D_a & =\left|\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
5 & 4 & 1
\end{array}\right|=2(2-4)-1(1-5)+1(4-10)=-6 \\
D_b & =\left|\begin{array}{ccc}
1 & 2 & 1 \\
4 & 1 & 1 \\
16 & 5 & 1
\end{array}\right|=1(1-5)-2(4-16)+1(20-16)=24
\end{aligned}
$
and
$D_c=\left|\begin{array}{ccc}1 & 1 & 2 \\ 4 & 2 & 1 \\ 16 & 4 & 5\end{array}\right|=1(10-4)-1(20-16)+2(16-32)=-30$
$
\therefore a=\frac{D_a}{D}=\frac{-6}{-6}=1 ;, b=\frac{D_b}{D}=\frac{24}{-6}=-4,, c=\frac{D_c}{D}=\frac{-30}{-6}=5
$
Therefore, equation of the curve is $y = x ^2- 4 x + 5$
When owl is sitting at $( 0 , k )$ then $x = 0 \Rightarrow k = 5$

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Question 55 Marks

The quarterly profits of a small-scale industry (₹ in thousands) are as follows.

Year	Quarter 1	Quarter 2	Quarter 3	Quarter 4
2020	39	47	20	56
2021	68	59	66	72
2022	88	60	60	67

Calculate 4-quarterly moving averages.

Answer

Yearly/ Quarterly		Small scale industry	4-quarterly moving total	4-quarterly moving average	4-year centered moving average
2020	I	39
	II	47	162	40.5
	III	20	191	47.75	44.125
	iv	56	203	50.75	49.25
2021	I	68	249	62.25	56.5
	II	59	265	66.25	64.25
	III	66	285	71.25	68.75
	iv	72	286	71.5	71.375
2022	I	88	280	70.00	70.75
	II	60	275	68.75	69.375
	III	60
	iv	67

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Question 65 Marks

Fit a straight-line trend by using the method of least squares for the following data and calculate the trend values.

Year	Production (in tonnes)
1962	2
1963	4
1964	3
1965	4
1966	4
1967	2
1968	4
1969	9
1970	7
1971	10
1972	8

Answer

Here, number of observations
n=11 (odd number)

Year (t)	Production (y)	𝑥=𝑡_𝑖−1967	$x^2$	xy
1962	2	-5	25	-10
1963	4	-4	16	-16
1964	3	-3	9	-9
1965	4	-2	4	-8
1966	4	-1	1	-4
1967	2	0	0	0
1968	4	1	1	4
1969	9	2	4	18
1970	7	3	9	21
1971	10	4	16	40
1972	8	5	25	40
Total	$\sum y=57$	$\sum x=0$	$\sum x^2=110$	$\sum xy=76$

Year $1 9 6 7$ is taken as year of origin.
The normal equations are $\sum y=n a+b \sum x$ and $\sum x y=a \sum x+b \sum x^2$
Since, $\sum x = 0$ i.e., deviation from actual mean is zero,
we have $a=\frac{\sum y}{n}=\frac{57}{11}=5.18, b=\frac{\sum x y}{\sum x^2}=\frac{76}{110}=0.69$
Therefore, the required equation of the trend line $y = 5 . 1 8 + 0 . 6 9 x$
The trend values are
$
1.73,2.42,3.11,3.8,4.49,5.18,5.87,6.56,7.25,7.94,8.63
$

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Applied Maths 5 Marks Questions

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