Given:
a=50, d= -4 and Sn = 0
Find:
value of n
Sn= $\dfrac{n}{2}[2a+(n-1)d] $= 0
[2(50)+(n-1)(-4)]=0 $\times \dfrac{2}{n}$
[100+(n-1)(-4)] = 0
(n-1)(-4) = -100
(n-1) = $\dfrac{-100}{-4}$
(n-1) =25
n = 25+1
n= 26
Given:
a= 14, d = 7, n= 12
Sn= $\dfrac{n}{2}[2a+(n-1)d] $
$S_{12} = \dfrac{12}{2}[2(14)+(12-1)7] $
=6 [28+11$\times$7]
=6 [28+77] = 6 [105]
$S_{12}$ =630
Given
Radius of Sphere = 3 cm
Radius of cylinder = 2 cm
Volume of sphere = Volume of cylinder
$\dfrac{4}{3} \times \pi \times r^3 = \pi × r^2 \times h $
$\dfrac{4}{3} \times \pi \times 3 \times 3 \times3 = \pi \times 2 \times 2 \times h$
⇒ h=9 cm
Given:
quadratic equation $x^2-5x + 9 = 0$
The general form of the above equation is
$ax^2 + bx + c = 0$
a= 1, b=(-5), c= 9
The roots are founded on determining the discriminant value.
The formula to find the discriminant is ∆ = $b^{2}-4ac$
∆ = $(-5)^{2}-4(1)(9)$
∆ = 25-36
∆ = -11 < 0
Therefore , equation has no real roots.
Given:
$\alpha = -3$
$\beta = 5$
$(x-\alpha)\times (x-\beta) = 0$
$(x+3)\times(x-5) = 0$
$x^{2}-5x+3x-15 = 0$
$x^{2}-2x-15 = 0$
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| Frequency | 8 | 7 | 12 | 5 | 3 |
Given:
l = 40
h = 20
$f_0$ = 7
$f_1$ = 12
$f_2$ = 5
mode =$ l+(\dfrac{(f_1-f_0)}{(2f_1-f_0-f_2)})\times h$
mode =$ 40+(\dfrac{(12-7)}{(2(12)-7-5)})\times 20$
mode =$ 40+(\dfrac{5}{12})\times 20$
mode =40+8.3
mode =48.3
Given :
a = 2
b = -5
c = -1
Quadratic Equation =$ ax^{2}+bx=c $
x = $ \dfrac{-b \pm\sqrt{b^{2}-4ac}}{2a} $
x =$ \dfrac{-(-5) \pm\sqrt{(-5)^{2}-(4\times2\times(-1))}}{2\times2} $
x = $\dfrac{5\pm\sqrt{25+8}}{4} $
x = $\dfrac{5\pm\sqrt{33}}{4}$
x =$\dfrac{5+\sqrt{33}}{4}$, $\dfrac{5-\sqrt{33}}{4}$
PA = PB (equal tangente)
In $ \triangle AOP \: and \: \triangle BOP,$
PA = PB (equal tangents)
OP = OP (common)
OA=OB (r)
SSS criterian,$ \angle AOP \cong \angle ABOP$
by Corresponding parts of Congruent Triangle,
$ \angle APO = \angle BPO $
$ \angle APO = \dfrac{1}{2} \angle APB $
$ \angle APO = \dfrac{1}{2}\times 70 = 35^o $
$ \angle PAO = \angle PBO = 90 $
$ in \triangle OAP$,
$ \angle OAP + \angle APO + \angle POA = 180^o $
$ 90^o + 35^o + \angle POA = 180^o $
$\angle POA = 55 ^o $
SECTION B
Question numbers 7 to 10 carry 3 marks each
Question 7
| Weight (in kg) | Number of Students |
|---|---|
| 40-45 | 9 |
| 45-50 | 5 |
| 50-55 | 8 |
| 55-60 | 9 |
| 6065 | 6 |
| 65-70 | 3 |
| Weight (in kg) | Number of Students | C.f |
|---|---|---|
| 40 - 45 | 9 | 9 |
| 45 - 50 | 5 | 14 |
| 50 - 55 | 8 | 22 |
| 55 - 60 | 9 | 31 |
| 60 - 65 | 6 | 37 |
| 65 -70 | 3 | 40 |
n = 40
$\dfrac{n}{2}=\dfrac{40}{2} = 20$
from the table cf just greater than 20 is 22
h = 5
l = 50
f = 8
cf = 14
median weight of the students = = l+$ (\dfrac{\dfrac{n}{2}-cf}{f}) h$
= 50 + $(\dfrac{20-14}{8}) \times 5$
= 50 + $(\dfrac{6\times5}{8})$
= 50 + 3.75
= 53.75
Steps:
1. Draw a circle of radius (r) = 4 cm.
2. Mark a from the centre at a distance of 6 cm from the pount 0.
3. Draw a perpendicular bisector of OP
4.Taking Q as a center and radius OP/PQ , Draw a circle to intrsect the given circle at T and T'
5. Joint PT and PT'
PT and PT' are the required tangents
