44 questions · self-marked practice — reveal the answer and mark yourself.
Difference in ages(in years) | No. of couples (𝑓𝑖) |
0 – 2 | 1 |
2 – 4 | 2 |
4– 6 | 8 |
6 – 8 | 5 |
8 – 10 | 3 |
10 – 12 | 1 |
Class | 1 – 10 | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 |
Frequency | 2 | 3 | 5 | 7 | 1 |
Use formula: Mode =𝐿 =$\frac{f_0-f_1}{2 f_0-f_1-f_2} \times h$
Marks | 20 - 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 -90 | 90-100 |
No. of students | 8 | 10 | 15 | 25 | 30 | 38 | 24 | 10 |
How many students are getting above 80 marks.
Marks | 10 – 19 | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 |
Number of students | 3 | 8 | 16 | 12 | 10 | 8 |
Class interval | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
Frequency | 30 | 35 | 40 | 75 | 15 |
$=5(3)-3(-2)$
$=15+6$
$=21$
$\neq \text { R.H.S }$
$\therefore$ The point $(3,-2)$ does not lie on the graph of $5 m-3 n=-21$.
$=2(2)-1$
$=4-1$
$=3$
$=\text {R.H.S}$
$\therefore x=2$ and $y=-1$ is the solution of the equation $2 x+y=3$.
$a+2(4)=7$
$\therefore a+8=7$
$\therefore a=7-8$
$=-1$
Substituting $x=2, y=-5$ in $2 x-k y=14$, we get
$2(2)-k(-5)=14$
$\therefore 4+5 k=14$
$\therefore 5 k=14-4=10$
$\therefore k=\frac{10}{5}=2$
we get $2(3)-y=2$
$\therefore 6-y=2$
$\therefore y=6-2$
$\therefore y=4$