5 Marks Questions

Question 15 Marks

A firm bought a machinery for ₹ 7,40,000 on 1st April, 2020 and ₹ 60,000 is spent on its installation. Its useful life is estimated to be of 5 years. Its scrap value at the end of 5 years is estimated to be ₹ 40,000. Find the amount of annual depreciation and the rate of depreciation.

Answer

Cost = 740000 + 60,000 = ₹ 8,00,000
Determination of the amount of annual Depreciation and Rate of Depreciation = $\frac{\text { Cost of Asset - Estimated Realisable or Scrap }}{\text { No. of year of estimated useful life }}$
$=\frac{(740000+₹ 60,000)-₹ 40000}{5 .}$
= ₹ 1,52,000
Rate of Depreciation $=\frac{\text { Annual depreciation }}{\cos 1 \text { of Asset }} \times 100$.
$=\frac{1,52,000}{800,000} \times 100=19 \%$ p.a.

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Question 25 Marks

The diameter of shafts produced in a factory confirms to normal distribution. 31% of the shafts have a diameter less than 45 mm and 8% have more than 64 mm. Find the mean and standard deviation of the diameter of shafts.

Answer

Let X denote the diameter of shafts. Let be mean and be the standard deviation. It is given that X follows a normal distribution.
Let Z be the standard normal variate. Then, $Z =\frac{X-\mu}{\sigma}$
$X =45 \Rightarrow Z =\frac{45-\mu}{\sigma}= Z _1$ (Say) and, $X =64 \Rightarrow Z =\frac{64-\mu}{\sigma}= Z _2$ (Say).
Now,
P (X < 45) = 0.31
$= P \left( Z < Z _1\right)=0.31$
$= P ( Z \leq 0)- P \left( Z _1 \leq Z \leq 0\right)=0.31\left[\because P \left( Z < Z _1\right)\right.$ is less than $\left.0.5 \therefore Z _1<0\right]$
$\begin{array}{l}=0.5- P \left( Z _1 \leq Z \leq 0\right)=0.31 \\ = P \left( Z _1 \leq Z \leq 0\right)=0.19 \\ = P \left(0 \leq Z \leq\left| Z _1\right|\right)=0.19\end{array}$
$\begin{array}{l}=\left| Z _1\right|=0.5[\text { See Table }] \\ =- Z _1=0.5\left[\because Z _1<0 \therefore\left| Z _1\right|=- Z _1\right] \\ = Z _1=-0.5\end{array}$
$\begin{array}{l}=\frac{45-\mu}{\sigma}=-0.5\left[\because Z _1=\frac{45-\mu}{\sigma}\right] \\ =\mu-0.5 \sigma=4.5 \ldots \text { (i) }\end{array}$
And
P(X > 64) = 0.08
$= P \left( Z > Z _2\right)=0.08$

$\begin{array}{l}= P ( Z \geq 0)- P \left(0 \leq Z \leq Z _2\right)=0.08 \\ =0.5- P \left(0 \leq Z \leq Z _2\right)=0.08 \\ = P \left(0 \leq Z \leq Z _2\right)=0.42 \\ = Z _2=1.41 \\ =\frac{64-\mu}{\sigma}=1.41 \\ =\mu+1.41 \sigma=64 \ldots \text { (ii) }\end{array}$

Solving (i) and (ii), we get $\sigma=10$ and $\mu=49.9$
Hence the mean and standard deviation of the diameter of shafts are 49.9 mm and 10 mm. respectively.

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Question 35 Marks

If X denotes the number of heads in a single toss of 4 fair coins, then find
i. $P(X=3)$
ii. $P ( X <2)$
iii. $P ( X \leq 2)$
iv. $P (1< X \leq 3)$.

Answer

A fair coin is tossed 4 times
$\begin{array}{l}\therefore n =4, \text { probability of getting a head }=p=\frac{1}{2}, \text { so } q =1-\frac{1}{2}=\frac{1}{2} . \\ P ( r )={ }^4 C_r p^r q^{4-r}={ }^4 C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{4-r}={ }^4 C_r\left(\frac{1}{2}\right)^4\end{array}$
i. $P ( X =3)= P (3)={ }^4 C_3\left(\frac{1}{2}\right)^4=4 \times \frac{1}{16}=\frac{1}{4}$.
$\begin{array}{l}\text { ii. } P ( X <2)=P(0)+P(1)={ }^4 C_0\left(\frac{1}{2}\right)^4+{ }^4 C_1\left(\frac{1}{2}\right)^4 \\ \quad=(1+4) \times\left(\frac{1}{2}\right)^4=\frac{5}{16} .\end{array}$
$\begin{array}{l}\text { iii. } P ( X \leq 2)= P (0)+ P (1)+ P (2)=\left({ }^4 C_0+{ }^4 C_1+{ }^4 C_2\right)\left(\frac{1}{2}\right)^4 \\ \quad=(1+4+6) \times \frac{1}{6}=\frac{11}{16}\end{array}$
$\begin{array}{l}\text { iv. } P (1< X \leq 3)= P (2)+ P (3)=\left({ }^4 C_2+{ }^4 C_3\right)\left(\frac{1}{2}\right)^4 \\ \quad=(6+4) \times \frac{1}{16}=\frac{10}{16}=\frac{5}{8}\end{array}$

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Question 45 Marks

A person can row a boat at 5 km/h in still water. It takes him thrice as long to row upstream as to row downstream. Find the rate at which the stream is flowing.

Answer

Let the distance covered be d km. and y be speed of stream
speed of boat = 5 km/h
speed of stream = y km/h
speed of boat in upstram(u): x - y km/h
= 5 - y km/h
speed of boat in downstream (v) = x + y km/h
= 5 + y km/h
ATQ.
$\begin{array}{l}\frac{d}{5-y}=3\left(\frac{d}{5+y}\right) \quad\left[\because T=\frac{D}{S}\right] \\ \frac{1}{5-y}=\frac{3}{5+y}\end{array}$
5 + y = 3(5 - y)
5 + y = 15 - 3y
y + 3y = 15 - 5
4y = 10
$\begin{array}{l} y =\frac{10}{4} \\ y =\frac{5}{2} km / h \\ y =2 \frac{1}{2} km / h \end{array}$
speed of stream is 2.5 km/h

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Question 55 Marks

The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result, we get 7. By adding second and third numbers to three times the first number, we get 12. Using matrices find the numbers.

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Question 65 Marks

By using determinants, solve the following system of linear equations:
x + y + z = 1
x + 2y + 3z = 4
x + 3y + 5z = 7

Answer

Here
$\begin{array}{l} D =\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5\end{array}\right|=1(10-9)-1(5-3)+1(3-2) \\ =1-2+1=0\end{array}$
$\begin{array}{l} D _1=\left|\begin{array}{lll}1 & 1 & 1 \\ 4 & 2 & 3 \\ 7 & 3 & 5\end{array}\right|=1(10-9)-1(20-21)+1(12-14) \\ =1+1-2=0\end{array}$
$\begin{array}{l} D _2=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 4 & 3 \\ 1 & 7 & 5\end{array}\right|=1(20-21)-1(5-3)+1(7-4) \\ =-1-2+3=0 \text { and }\end{array}$
$\begin{array}{l} D _3=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 7\end{array}\right|=1(14-12)-1(7-4)+1(3-2) \\ =2-3+1=0\end{array}$
Thus, $D = D _1= D _2= D _3=0$, therefore, the given system may or may not be consistent. Let us solve the first two equations for x and y in terms of z. These equations can be written as:
x + y = 1 - z
x + 2y = 4 - 3z
To solve these equations, we use Cramer's rule.
$\begin{array}{l} D =\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|=2-1=1, \\ D _1=\left|\begin{array}{cc}1-z & 1 \\ 4-3 z & 2\end{array}\right|=2-2 z -4+3 z = z -2, \text { and } \\ D _2=\left|\begin{array}{cc}1 & 1-z \\ 1 & 4-3 z\end{array}\right|=4-3 z -1+ z =3-2 z \end{array}$
By Cramer's rule,
$x =\frac{ D _1}{ D }=\frac{z-2}{1}= z -2, y =\frac{ D _2}{ D }=\frac{3-2 z}{1}=3-2 z$
Let
z = k where k is arbitrary number, then we get
x = k - 2, y = 3 - 2k, z = k, where k is any number.
Note that these values satisfy the third equation i.e. x + 3y + 5z = 7 of the given system. Hence, the system is consistent and it has infinitely many solutions given by x = k - 2, y = 3 - 2k, z = k where k is any number.

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Applied Maths 5 Marks Questions

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