1. No category

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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
I.
SUBJECT: MATHEMATICS
Choose the best alternative from following options
TOTAL:
50
1x7=7
1) If Tn = 5-4n , then T2 =
(A) 3
(B) −1 (C) −
(D) 8
2) The value of nP0 is--(A) n!
(B) 0
(C) 1
(D) n
3) An unbiased dice is thrown, what is the probability that a perfect square number
occurs
(A)
(B)
(C)
(D) 1
4) Mean =70 and coefficient of variation is 6 ,Find standard deviation
(A) 4
(B) .
(C)0.42
(D) 0.4
5) The distance between the points (2,3) and (6,6) is
(A) 5
(B)
4
(C) 6
(D) 3
6) If cos = , then the value equal to sec is
(A)
(B)
(C)
(D)
2
7) If
f(x) = x + x - 6 , the value of f(2) is
(A) 2
(B) 4
(C) 0
(D) 1
II. Answer the following
1x5=5
8) The Venn diagram shows the students who can speak English and Hindi. If the total
number of student in a class is 70 then, find the number of students speaks either
English or Hindi.
=60
9) Write the prime factors of 3825.
3x3x5x5x17
10) In ∆ABC If DE॥BC, AD = 7cm, CD = 5cm, BC = 18cm find CE.
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∆ABC, DE॥BC,
∴
=
= ⇒ AE =
∴ CE = 18-10.5
= 10.5
∴ CE = 7.5cm
11) What is the name of a straight line which intersects a circle at two distinct points?
secant
12) Write a formula to find the circumference of the base of the cylindrical vessel.
C=2 r
III. Answer the following
2x 10 = 20
13) In a medical examination of 150 people, it was found that 90 had eye problem, 50
had heart problem and 30 had both complaints. Draw a Venn diagram to show how
many people had either eye problem or heart problem.
A-{The people had eye problem}
B-{The people had Heart problem}
n(A) = 90 ,n(B) = 50, n(A∩B) = 30
n(AUB) = n(A) + n(B) –n(A∩B)
= 90 + 50 - 30
= 110
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
= √3
√2 x3
√16x27
√432
16) In a circle of radius 3.5cm draw two radii such that the angle between them is700
Construct tangents at the non-centre ends of the radii.
17) How many diagonals can be drawn in a pentagon
Formula - nC2 - n
5
C2 – 5
!
= ( )! - 5
!
=
!
( )! !
- 5
=
- 5
= 10 - 5
=
5
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18) Solve by using the formula x2 - 4x + 2 =0
a = 1, b = -4 , c = 2
−(−4) ± (−4) − 4x1x2
=
2x1
4 ± √16 − 8
=
2
4 ± √8
=
2
4 ± 2√2
=
2
= 2 ± √2
19) If cot = then find cos and cosec
cos
.
=
cosec
=
20) Find the slop of a line joining the points (4,-8) and (5,-2).
m=
m=
m=
−2− (−8)
5−4
−2+8
6
1
1
m=
21). Verify Euler’s formula for Pentahedron.
F = 5, E = 9, V = 6
F+V = 5+ 6 = 11
E+2 = 9 + 2 = 11
∴ F+V = E+2
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22) Sketch out the field to the following notes from the field book
(1 cm = 20m.)
To C
220
120 To D
200
120
80 To D
80
200 To B
From A
1cm = 20m ⇒80m=4cm, 120m=6cm, 200m=10cm, 220m=11cm,
IV.
Answer the following
23) Calculate the standard deviation of the following data
X
f
10
4
20
3
=
∑
40
5
X
f
d=x-30
fd
d2
fd2
10
20
30
40
50
4
3
6
5
2
-20
-10
0
10
20
-80
-30
0
50
40
400
100
0
100
400
1600
300
0
500
800
20
=
30
6
3 x 2= 6
−
−
-20
50
2
3200
∑
−
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=
–
= √
=
.
24) prove that “If two circles touch each other, the centers and the point of contact
Collinear”
If two circles touch each other, the centres and the point of contact are collinear.
Case-1). If two circles touch each other externally, thecentres and the point of
contact are collinear.
Given:A and B are the centres of touching circles. P is the point of contact.
To prove : A,P,and B are collinear.
Construction: Draw the tangent XPY.
Proof:In the figure
∠APX = 900……………..(1) ∵Radius drawn at the point of contact is
∠BPX = 900 ………… ..(2) perpendicular to the tangent
∠APX + ∠BPX = 900 +900 [ by adding (1) and (2)
∠APB = 1800
[ APB is a straight line
∴ APB is a straight line
∴ A, P andB are collinear.
Case-2 ). 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.
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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 4cm and 2cm having their
Centers 8cm apart and measure their lengths.
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26) Draw the graph of y = x2 – 8x + 7
x
y
0
7
8
7
1
0
7
0
3
-8
5
-8
4
-9
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27) Prove that “A straight line is drawn parallel to one side of the triangle, and then it
divides other two sides proportionally”.
∆ABC ಯ , DE॥BC
Given:
AD
To prove:
=
DB
AE
EC
Construction: 1. Join D,E and E,B.
2.Draw EL ⟘ AB and DN⟘ AC .
∆ABC
Proof: ∆BDE =
∆ABC
∆BDE
∆ADE
∆CDE
∆ADE
∆CDE
∴
=
=
=
1
2
1
2
xADxEL
xDBxEL
1
[∵ A = 2 xbxh
AD
DB
1
2
1
2
xAExDN
xDBxDN
1
[∵ A = xbxh
2
AE
EC
=
[∵∆BDE ≡ ∆
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