Download App

Applied Maths 5 Marks Questions

Model Paper 6 · CBSE STD 12 Science (CBSE - English Medium). 6 questions.

Questions

5 Marks Questions

🎯

Test yourself on this topic

6 questions · timed · auto-graded

Question 15 Marks
Anil plans to send his daughter for higher studies abroad after 10 years. He expects the cost of the studies to be ₹ 2,00,000. How much must he set aside at the end of each quarter for 10 years to accumulate this amount, if money is worth 6% compounded quarterly? [Given: (1.015)40 = 1.8140]
Answer

$
\begin{array}{l}
FC=P \times\left(\frac{(1+r)^{n t}-1}{r}\right) \\
2,00,000=P \times\left(\frac{(1+0.015)^{4 \times 10}-1}{0.015}\right)
\end{array}
$
Now, calculate the value inside the parentheses:
$
\begin{array}{l}
(1.015)^{40}-1=1.8140-1=0.8140 \\
2,00,000=P \times\left(\frac{0.8140}{0.015}\right)
\end{array}
$
Now, calculate the value inside the second set of parentheses:
$
\frac{0.8140}{0.015} \approx 54.267
$
Now, solve for P:
$
P=\frac{2,00,000}{54.267}
$
$P \approx$ ₹ 3,684.81
So, Anil must set aside approximately ₹ 3,684.81 at the end of each quarter for 10 years to accumulate ₹ 2,00,000 with a 6% quarterly compounded interest rate.
View full question & answer→
Question 25 Marks
A die is thrown 5 times. Find the probability that an odd number will turn up
i. exactly 3 times
ii. atleast 4 times
iii. maximum 3 times
Answer
When a die is thrown, sample space $=\{1,2,3,4,5,6\}$. It has six equally likely outcomes.
$p =$ probability of an odd number $=\frac{3}{6}=\frac{1}{2}$, so $q =1-\frac{1}{2}=\frac{1}{2}$.
As the die is thrown 5 times, so there are 5 trials i.e. $n=5$.
$
P(r)={ }^5 C_r p^r q^{5-r}={ }^5 C_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}={ }^5 C_r\left(\frac{1}{2}\right)^5
$
i. Probability of an odd number exactly 3 times $= P (3)$
$
={ }^5 C_3\left(\frac{1}{2}\right)^5=\frac{10}{32}=\frac{5}{16}
$
ii. Probability of an odd number atleast 4 times $=P(X \geq 4)$
$
\begin{array}{l}
=P(4)+P(5)={ }^5 C_4\left(\frac{1}{2}\right)^5+{ }^5 C_5\left(\frac{1}{2}\right)^5 \\
=\left({ }^5 C_4+{ }^5 C_5\right)\left(\frac{1}{2}\right)^5=(5+1) \times \frac{1}{32}=\frac{3}{16}.
\end{array}
$
iii. Probability of an odd number maximum 3 times $=P(X \leq 3)$
$
\begin{array}{l}
=1-(P(4)+P(5))=1-\frac{3}{16}(\text { see part }(ii)) \\
=\frac{13}{16}
\end{array}
$
View full question & answer→
Question 35 Marks
The probability that Rohit will hit a shooting target is $\frac{2}{3}$. While preparing for an international shooting competition. Rohit aims to achieve the probability of hitting the target atleast once to be 0.99 . What is the minimum number of chances must he shoot to attain this probability?
Answer
Given probability of hitting a shooting target $=p=\frac{2}{3}$.
So, $q =1- p =1-\frac{2}{3}=\frac{1}{3}$.
Let the number of trials be $n$.
The probability of hitting target atleast once $= P ( X \geq 1)=1- P (0)$
$
=1-{ }^n C_0 q^n=1-\left(\frac{1}{3}\right)^n
$
According to given,
$
\begin{array}{l}
1-\left(\frac{1}{3}\right)^n>0.99 \Rightarrow 1-\frac{1}{3^n}>\frac{99}{100} \\
\Rightarrow 1-\frac{99}{100}>\frac{1}{3^n} \Rightarrow \frac{1}{100}>\frac{1}{3^n} \\
\Rightarrow 100<3^n
\end{array}
$
$\Rightarrow 3^n>100$, which is satisfied if n is atleast 5.
Hence, Rohit must shoot the target at 5 times.
View full question & answer→
Question 55 Marks
Find $A^{-1}$, where $A=\left[\begin{array}{rrr}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right]$. Hence solve the system of equations: $x+2 y-3 z=-4,2 x+3 y+2 z=2,3 x-$ $3 y-4 z=11$
Answer
We have,
$
\begin{array}{l}
A=\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right] \\
\therefore|A|=\left|\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right|=-6+28+45=67 \neq 0
\end{array}
$
So, A is invertible
Let $C _{ ij }$ be the co-factors of $a _{ ij }$ in $A =\left[ a _{ ij }\right]$. Then,
$
\begin{array}{l}
C_{11}=(-1)^{1+1}\left|\begin{array}{rr}
3 & 2 \\
-3 & -4
\end{array}\right|=-6, C_{12}=(-1)^{1+2}\left|\begin{array}{rr}
2 & 2 \\
3 & -4
\end{array}\right|=14, \\
C_{13}=(-1)^{1+3}\left|\begin{array}{rr}
2 & 3 \\
3 & -3
\end{array}\right|=-15, C_{21}=(-1)^{2+1}\left|\begin{array}{rr}
2 & -3 \\
-3 & -4
\end{array}\right|=17, \\
C_{22}=(-1)^{2+2}\left|\begin{array}{rr}
1 & -3 \\
3 & -4
\end{array}\right|=5, C_{23}=(-1)^{2+3}\left|\begin{array}{rr}
1 & 2 \\
3 & -3
\end{array}\right|=9, \\
C_{31}=(-1)^{3+1}\left|\begin{array}{rr}
2 & -3 \\
3 & 2
\end{array}\right|=13, C_{32}=(-1)^{3+2}\left|\begin{array}{rr}
1 & -3 \\
2 & 2
\end{array}\right|=-8, \\
\text { and } C_{33}=(-1)^{3+3}\left|\begin{array}{rr}
1 & 2 \\
2 & 3
\end{array}\right|=-1
\end{array}
$
$
\therefore \operatorname{adj} A=\left[\begin{array}{rrr}
-6 & 14 & -15 \\
17 & 5 & 9 \\
13 & -8 & -1
\end{array}\right]=\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]
$
So, $A ^{-1}=\frac{1}{|A|}$ adj $A =\frac{1}{67}\left[\begin{array}{rrr}-6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1\end{array}\right] \quad \ldots(i)$
The given system of equations is
$
\begin{array}{l}
x+2 y-3 z=-4 \\
2 x+3 y+2 z=-2 \\
3 x-3 y-4 z=11
\end{array}
$
$
\text { or, } AX=B \text {, where } A=\left[\begin{array}{rrr}
1 & 1 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right]
$
As discussed above A is non-singular and so invertible. The inverse of A is given by (i) The solution of the given system of equations is given by
$X = A ^{-1} B$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{67}\left[\begin{array}{rrr}-6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1\end{array}\right]\left[\begin{array}{r}-4 \\ 2 \\ 11\end{array}\right]=\frac{1}{67}\left[\begin{array}{rrr}24 & +34 & +143 \\ -56 & +10 & -88 \\ 60 & +18 & -11\end{array}\right]=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]$
$\Rightarrow x=3, y=-2$ and $z=1$ is the required solution.
View full question & answer→
Question 65 Marks
Express the matrix $A =\left[\begin{array}{rrr}4 & 2 & -1 \\ 3 & 5 & 7 \\ 1 & -2 & 1\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.
Answer

$\begin{array}{l}\text { Given, } A=\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right] \text { Then } A^{T}=\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right] \\
X=\frac{1}{2}\left(A+A^{T}\right) \\
=\frac{1}{2}\left(\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]+\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right]\right) \\
=\frac{1}{2}\left[\begin{array}{ccc}
4+4 & 2+3 & -1+1 \\
3+2 & 5+5 & 7-2 \\
1-1 & -2+7 & 1+1
\end{array}\right] \\
=\frac{1}{2}\left[\begin{array}{ccc}
8 & 5 & 0 \\
5 & 10 & 5 \\
0 & 5 & 2
\end{array}\right] \\
X=\left[\begin{array}{ccc}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right] \\
X^{T}=\left[\begin{array}{lll}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right]^{T}=\left[\begin{array}{lll}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right]=X
\end{array}
$
$\therefore X$ is a symmetric matrix.
$
\begin{array}{l}
Y=\frac{1}{2}\left(A-A^{T}\right) \\
=\frac{1}{2}\left(\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]-\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right]\right) \\
=\frac{1}{2}\left[\begin{array}{ccc}
4-4 & 2-3 & -1-1 \\
3-2 & 5-5 & 7+2 \\
1+1 & -2-7 & 1-1
\end{array}\right] \\
=\frac{1}{2}\left[\begin{array}{ccc}
0 & -1 & -2 \\
1 & 0 & 9 \\
2 & -9 & 0
\end{array}\right] \\
Y=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right] \\
Y^{T}=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right]^{T}=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right]=Y
\end{array}
$
$\Rightarrow Y$ is a skew-symmetric matrix.
Now,
$X+Y=\left[\begin{array}{ccc}4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1\end{array}\right]+\left[\begin{array}{ccc}0 & -\frac{1}{2} & -1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & -\frac{9}{2} & 0\end{array}\right]$
$
\begin{array}{l}
=\left[\begin{array}{ccc}
4+0 & \frac{5}{2}-\frac{1}{2} & 0-1 \\
\frac{5}{2}+\frac{1}{2} & 5+0 & \frac{5}{2}+\frac{9}{2} \\
0+1 & \frac{5}{2}-\frac{9}{2} & 1+0
\end{array}\right] \\
=\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]=A
\end{array}
$
Hence, $X + Y = A$.
Thus matrix A is expressed as the sum of symmetric and skew-symmetric matrices.
View full question & answer→
Generate Paper FreeAll Model Paper 6 topics
5 Marks Questions - Applied Maths STD 12 Science Questions - Vidyadip