Question 12 Marks
If $A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]$, then find the value of $\alpha$ (if exists) for which $A^2=B$.
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Question 22 Marks
The lifetime of an item produced by a machine has a normal distribution with mean 12 months and standard deviation of 2 months. Find the probability of an item produced by this machine will last
(i) less than $7$ months
(ii) between $7$ and $1 4$ months.
(Given $P\left(Z<\frac{5}{2}\right)=0.9938$ and $\left.P(Z<1)=0.8413\right)$
(i) less than $7$ months
(ii) between $7$ and $1 4$ months.
(Given $P\left(Z<\frac{5}{2}\right)=0.9938$ and $\left.P(Z<1)=0.8413\right)$
Answer
View full question & answer→We have, mean $\mu=12$ and standard deviation $\sigma=2$, i.e., $X \sim N\left(\mu, \sigma^2\right)$
(i) Let $X$ denote the count of the months for which this machine lasts.
The probability of an item produced by this machine will last less than 7 months is
$
P(X<7)
$
For $X=7, Z=\frac{7-12}{2}=-\frac{5}{2}$
Now,
$P(X<7)=P\left(Z<-\frac{5}{2}\right)=P\left(Z>\frac{5}{2}\right)$
$=1-P\left(Z<\frac{5}{2}\right)=1-0.9938=0.0062$
(ii) The probability of an item produced by this machine will last more than 7 months and less than 14 months is $P(7 < X < 14)$
For $X=7, Z=\frac{7-12}{2}=-\frac{5}{2}$
and for $X=14, Z=\frac{14-12}{2}=1$
$P(7 < X < 14)=P\left(-\frac{5}{2} < Z < 1\right)$
$\begin{array}{l}=P(Z < 1)-P\left(Z < -\frac{5}{2}\right) \\
=0.8413-0.0062=0.8351\end{array}$
(i) Let $X$ denote the count of the months for which this machine lasts.
The probability of an item produced by this machine will last less than 7 months is
$
P(X<7)
$
For $X=7, Z=\frac{7-12}{2}=-\frac{5}{2}$
Now,
$P(X<7)=P\left(Z<-\frac{5}{2}\right)=P\left(Z>\frac{5}{2}\right)$
$=1-P\left(Z<\frac{5}{2}\right)=1-0.9938=0.0062$
(ii) The probability of an item produced by this machine will last more than 7 months and less than 14 months is $P(7 < X < 14)$
For $X=7, Z=\frac{7-12}{2}=-\frac{5}{2}$
and for $X=14, Z=\frac{14-12}{2}=1$
$P(7 < X < 14)=P\left(-\frac{5}{2} < Z < 1\right)$
$\begin{array}{l}=P(Z < 1)-P\left(Z < -\frac{5}{2}\right) \\
=0.8413-0.0062=0.8351\end{array}$
Question 32 Marks
The incidence of occupational disease in an industry is such that the workers have a 20% chance of suffering from it. What is the probability that out of six workers 4 or more will catch the disease?
Answer
View full question & answer→Let X be the random variable denoting the number of workers who catch the disease.
Given, $p=\frac{20}{100}=\frac{1}{5} \Rightarrow q=\frac{4}{5}$ and $n=6$
Now, $P(X=x)={ }^6 C_x\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{6-x}, x=0,1, \ldots ., 6$
So, the required probability that out of six workers 4 or more will catch the disease is
$P(X \geq 4)=P(X=4)+P(X=5)+P(X=6)$
$\begin{array}{l}={ }^6 C_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^2+{ }^6 C_5\left(\frac{1}{5}\right)^5\left(\frac{4}{5}\right)^1+{ }^6 C_6\left(\frac{1}{5}\right)^6\left(\frac{4}{5}\right)^0 \\ =\frac{265}{5^6} \text { or } 0.017 .\end{array}$
Given, $p=\frac{20}{100}=\frac{1}{5} \Rightarrow q=\frac{4}{5}$ and $n=6$
Now, $P(X=x)={ }^6 C_x\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{6-x}, x=0,1, \ldots ., 6$
So, the required probability that out of six workers 4 or more will catch the disease is
$P(X \geq 4)=P(X=4)+P(X=5)+P(X=6)$
$\begin{array}{l}={ }^6 C_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^2+{ }^6 C_5\left(\frac{1}{5}\right)^5\left(\frac{4}{5}\right)^1+{ }^6 C_6\left(\frac{1}{5}\right)^6\left(\frac{4}{5}\right)^0 \\ =\frac{265}{5^6} \text { or } 0.017 .\end{array}$
Question 42 Marks
A boat takes thrice as long to go upstream to a point as to return downstream to the starting point. If the speed of the stream is $5 km / h$, find the speed of the boat in still water.
Answer
View full question & answer→Let the total distance be 𝑑 km and the speed of boat in still water be 𝑥 km/h
Speed of stream = 5 km/h
Speed upstream =(𝑥−5) km/h
Speed downstream =(𝑥+5) km/h
According to question, $\frac{d}{x-5}=3 \times \frac{d}{x+5}$
Solving, we get $x=10$
Hence, the speed of boat in still water is $10 km / h$
Speed of stream = 5 km/h
Speed upstream =(𝑥−5) km/h
Speed downstream =(𝑥+5) km/h
According to question, $\frac{d}{x-5}=3 \times \frac{d}{x+5}$
Solving, we get $x=10$
Hence, the speed of boat in still water is $10 km / h$
Question 52 Marks
In a 200 𝑚 race, A can give a start of 18 𝑚 to B and a start of 31 𝑚 to C. In a race of 350 𝑚, how much start can B give to C?
Answer
View full question & answer→In a 200 m race, when $A$ covers 200 m
then $B$ covers $( 2 0 0 - 1 8 )= 1 8 2 m$
and $C$ covers $( 2 0 0 - 3 1 )= 1 6 9 m$
$
\Rightarrow A: C=200: 169
$
$
\frac{B}{C}=\frac{A}{C} \times \frac{B}{A}=\frac{200}{169} \times \frac{182}{200}=\frac{182}{169}
$
When $B$ covers $1 8 2 m$ then $C$ covers $1 6 9 m$
When $B$ covers $3 5 0 m$ then $C$ covers $\frac{ 1 6 9 }{ 1 8 2 } \times 3 5 0 = 3 2 5 m$
Therefore, $B$ can give a start of $( 3 5 0 - 3 2 5 )= 2 5 m$ to $C$.
then $B$ covers $( 2 0 0 - 1 8 )= 1 8 2 m$
and $C$ covers $( 2 0 0 - 3 1 )= 1 6 9 m$
$
\Rightarrow A: C=200: 169
$
$
\frac{B}{C}=\frac{A}{C} \times \frac{B}{A}=\frac{200}{169} \times \frac{182}{200}=\frac{182}{169}
$
When $B$ covers $1 8 2 m$ then $C$ covers $1 6 9 m$
When $B$ covers $3 5 0 m$ then $C$ covers $\frac{ 1 6 9 }{ 1 8 2 } \times 3 5 0 = 3 2 5 m$
Therefore, $B$ can give a start of $( 3 5 0 - 3 2 5 )= 2 5 m$ to $C$.
Question 62 Marks
A pump can fill a tank with water in 2 hours. Because of leakage, it took $\frac{7}{3}$ hrs to fill the tank. How much time will it take for the leakage to drain all the water in the full tank?
Answer
View full question & answer→Time taken to drain full tank $= x$ hours i.e., the time rate of drain the tank $=\frac{ 1 }{ x }$ units per hour
Time taken to fill the full tank is 2 hours i.e., the time rate of filling the tank $=\frac{1}{2}$ units per hour
Again, with the leakage, the pipe takes $2 \frac{1}{3}=\frac{7}{3}$ hours to fill the full tank.
The rate of filling the tank along with the leakage will be $=\frac{ 3 }{ 7 }$ units per hour.
Now, according to question,
$
\left(\frac{1}{2}\right)-\left(\frac{1}{x}\right)=\left(\frac{3}{7}\right)
$
Solving, we get $x = 1 4$
Hence, 14 hours are required to drain the full tank.
Time taken to fill the full tank is 2 hours i.e., the time rate of filling the tank $=\frac{1}{2}$ units per hour
Again, with the leakage, the pipe takes $2 \frac{1}{3}=\frac{7}{3}$ hours to fill the full tank.
The rate of filling the tank along with the leakage will be $=\frac{ 3 }{ 7 }$ units per hour.
Now, according to question,
$
\left(\frac{1}{2}\right)-\left(\frac{1}{x}\right)=\left(\frac{3}{7}\right)
$
Solving, we get $x = 1 4$
Hence, 14 hours are required to drain the full tank.
Question 72 Marks
In what ratio water must be added in milk costing ₹ 60 per litre, so that the resulting mixture would be of worth ₹ 50 per litre?
Answer
C.P of 1 litre of water = ₹ 0
C.P of 1 litre of milk = ₹ 60
Mean Price = ₹ 50
Therefore, ratio of water and milk $=\frac{60-50}{50-0}=\frac{10}{50}=\frac{1}{5}$ or $1: 5$
View full question & answer→C.P of 1 litre of water = ₹ 0
C.P of 1 litre of milk = ₹ 60
Mean Price = ₹ 50
Therefore, ratio of water and milk $=\frac{60-50}{50-0}=\frac{10}{50}=\frac{1}{5}$ or $1: 5$