- AMedian
- ✓Standard deviation
- CMode
- DMean
54 questions · 42 auto-graded MCQ + 12 self-marked written.

Head | Rent | Clothing | Education | Saving | Food | Miscellaneous |
Expenditure | Rs.2400 | Rs.1800 | Rs.1200 | Rs.1200 | Rs. 4800 | Rs.600 |
Blood group | O | A | B | AB |
% Of persons | 60 | 20 | 15 | 5 |
Length ( in cm) | 2.0 – 2.5 | 2.5 – 3.0 | 3.0 – 3.5 | 3.5 – 4.0 | 4.0 – 4.5 |
No. of nails | 5 | 18 | 7 | 11 | 9 |
\frac{3}{10}
$
Explanation:
$
\begin{aligned}
S=\{1,2, \ldots ., 50\}\\
\therefore \mathrm{n}(S)=50
\end{aligned}
$
Let $\mathrm{A}$ be the event of getting a prime number.
$
\begin{aligned}
& \therefore A=\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\} \\
& \therefore \mathrm{n}(\mathrm{A})=15 \\
& \therefore P(A)=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{15}{50}=\frac{3}{10}
\end{aligned}
$
Explanation:
Let the first four terms of an A.P are $a, a+d, a+2 d$ and $a+3 d$
Given that the first term is -2 and difference is also -2 , then the A.P would be:
-2,(-2-2),[-2+2(-2)],[-2+3(-2)] \
= -2,-4,-6,-8
Explanation:
A.P. 9, 15, 21, 27, ...(Given)
$t_1=9, t_2=15, t_3=21, t_4=27$
$\therefore \mathrm{t}_3=21$
\
Explanation:
\
The roots of the quadratic equation $x^2+8 x+15=0$ are
\[
\begin{aligned}
& x^2+5 x+3 x+15=0 \quad \ldots\left(\begin{array}{c}
5 \times 3=+15 \\
5+3=+8
\end{array}\right) \\
& x(x+5)+3(x+5)=0 \\
& (x+5)(x+3)=0 \\
& \therefore x+5=0 \\
& x=-5 \\
& \therefore x+3=0 \\
& x=-3
\end{aligned}
\]
\
Explanation:
\[
\begin{aligned}
& 3 x+6 y=3(3)+6(1) \\
& =9+6 \\
& =15 \neq 12
\end{aligned}
\]
\
Explanation:
\[
\begin{aligned}
& 49 x-57 y=172 \\
& 57 x-49 y=252 \\
&\frac{-\quad +\quad- \\}{}
\\& -8 x-8 y=-80 \\
& \therefore x+y=\frac{-80}{-8}=10
\end{aligned}
\]
\[
D x=26, D y=-39 \text { and } D=13
\]
Using Cramer's rule,
\[
\begin{aligned}
& x=\frac{D x}{x}=\frac{26}{13}=2 \\
& x=2
\end{aligned}
\]
\
Explanation:
\[
\begin{aligned}
& \text { Here, } a_1=4, b_1=3, c_1=19 \\
& a_2=4, b_2=-3, c_2=-11 \\
& \therefore D_x=\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|=\left|\begin{array}{cc}
19 & 3 \\
-11 & -3
\end{array}\right| \\
& =-57+33 \\
& =-24
\end{aligned}
\]