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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
TOTAL:
50
1x7=7
SUBJECT: MATHEMATICS
I . Choose the best alternative from following options
1)
The Arithmetic Mean between 12 and 30.
(A) 12
(B) 30 (C)
(D)
42
n
2) If P2 = 90,then the value of n ‘ is ---(A) 10
(B) 9
(C) 90
(D) 45
3) Three coins are tossed at a time, the probability of getting all heads.
(A)
(B)
(C)
(D)
4) If The Standard deviation of 10 data is 8, then the variance is ---(A) 16
(B) 4
(C) 80
(D)
5) If 2sin
= 1, then he value of θ is ---(A) 300
(B) 450 (C) 600
(D) 900
6) If in = , then, ec = -----
(B)
(A)
3
(C)
(D)
2
7) If f(x) = 3x + 3x – 11x - 6 ,then f(1) is ---(A) -11
(B) 22
(C) 0
(D)
-6
II. Answer the following
8) Express 120 as the product of prime factors...
1x5=5
2 120
2
60
2
30
3
15
5
∴ 120 = 2x2x2x3x5
9) If A and B are disjoint sets then what is n (A∩B)?
n(AUB) = n(A) + n (B)
10) The corresponding heights of two similar triangles are 3cm and 5cm. Find
The ratio of their areas.
Ratio of their areas are 32 : 52 ⇒ 9 : 25
11) What are concentric circles?
Circles having the same centre but different radii are called concentric
circles.
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12) Write a formula to find the total surface area of a hemisphere.
A =
III. Answer the following
2x 10 = 20
13) Cons ruct a pair of tangents to a circle of radius 5cm, and the angle between them
Is 500
14) Rationalize the denominator and simplify
=
=
=
=
=
√
√
√
√
√
√
x
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
− √
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15) Find the product. √2 x √3
√2
= 2 = 2
= √2
√3 = 3 = 3
√2 x3
√16x27
= √3
√432
16) In a group of passengers, 100 know Kannada, 50 know English and 25 know both. If
passengers know either Kanada or English. Drawing Venn diagram find the total
number of passengers are in a group.
A – { Know Kannada}, B – { Know English}
n(AUB) = ?, n(A) = 100, n(B) = 50 , n(A∩B) = 25
n(AUB) = n(A) + n(B) – n(A∩B)
= 100 + 50 - 25
= 125
17) How many triangles to be drawn through 12 points on a circle
The number of triangles to be drawn = 12C3
!
n
Cr = ( )!
12
C3 = (
12
C2 =
12
!
!
)! !
C2 =
18) Solve by using the formula. x2 - 2x + 4 =0
a = 1, b = -2 , c = 4
−(−2) ± (−2) − 4x1x4
=
2x1
2 ± √4 − 16
=
2
2 ± √−12
=
2
2 ± 2√−3
=
2
= 1 ± √−3
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19) If tan = ,then find sin
In the Figure,
AC = √
+
AC = √7 + 24
AC = √49 + 576
AC = √625
AC = 25
∴
and cos
=
=
20) Find the slope of a line forming by joining the points (-4, 1) and (-5, 2).
m=
−
−
m=
(
)
m=
m=
m= −
21) Verify Euler’s formula for the given graph
N = 4, R = 6, A = 8
N+R = 4+ 6 = 10
A+2 = 8 + 2 = 10
∴ N+R = A+2
22) Sketch out the field to the following notes from the field book
(1 cm = 25m.)
To D
100 To E
200
125
75
25
75 To C
50 To B
From A
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Scale 1cm = 25m
⇒75m = 3cm, 125m=5cm, 50m=2cm,100m=4cm,200m=8cm
IV. Answer the following
23) Calculate the standard deviation of the following data
X
5-15
15-25
25-35
35-45
f
8
12
CI
f
5-15
15-25
25-35
35-45
45-55
55-65
8
12
20
10
7
3
20
10
Assumed Mean Methdo
d=x-30
fd
X
10
20
30
40
50
60
-20
-10
0
10
20
30
-160
-120
0
100
140
90
60
50
=
∑
45-55
55-65
7
3
d2
fd2
400
100
0
100
400
900
3200
1200
0
1000
2800
2700
10900
∑
−
=
3 x 2= 6
−
=
.
= √
– .
.
=
.
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24) prove that “If two circles touch each other internally, the centers and the point of
Contact Collinear”
If two circles touch each other internally the centres and the point of contact are
collinear.
Given:A and B are centres of touching circles. P is point of contact.
To prove : A,P,and B are collinear
Construction: Draw the common tangent XPY . Join AP and BP.
Proof:In the figure
∠APX = 900……………..(1)
∵Radius drawn at the point of contact
∠BPX = 900 ………… ..(2)
is perpendicular to the tangent.
∠APX = ∠BPX = 900
[ From (1) and (2)
AP and BP lie on the same line
∴ APB is a straight line
∴ A, P and B are collinear.
V) Answer the following
4 x 3= 12
25) Draw transverse common tangents to two circles of radii 5cm and 2cm having
Their centers 9cm apart and measure their lengths.
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26) Solve graphically: x2 +3 x + 2 =0
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27) Prove that in a right angled triangle, the square on the hypotenuse is equal to the sum
of the squares on the other two sides.
Theorem:Phythagoras Therem
In a right angled triangle,the square of the hypotenuse is equal to the sum of the square
of the other two sides.
Given: ∆ABC In which ∠ABC = 900
To Prove : AB2 + BC2 = CA2
Construction: Draw BD ⟘ AC .
Proof: In ∆ABC and ∆ADB ,
∠ABC = ∠ADB = 900 [ ∵ Given and Construction
∠BAD =∠BAD [∵ Common angle
∴ ∆ABC ~ ∆ADB [∵ AA criteria
⇒
AB
AD
=
AC
AB
⇒ AB2 = AC.AD……..(1)
In ∆ABC and ∆BDC ,
∠ABC = ∠BDC = 900 [ ∵ Given and construction
∠ACB = ∠ACB [∵ Common angle
∴ ∆ABC ~ ∆BDC [∵ AA criteria
⇒
BC
DC
=
AC
BC
⇒ BC2 = AC.DC……..(2)
(1) + (2)
AB2+ BC2 = (AC.AD) + (AC.DC)
AB2+ BC2 = AC.(AD + DC)
AB2+ BC2 = AC.AC
AB2+ BC2 = AC2 [ ∵AD + DC = AC]
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