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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
TOTAL:
SUBJECT: MATHEMATICS
I. Choose the best alternative from following options
1) If Tn = 2n + 1 ,The first sum of first three terms --(A) 3
(B) 7 (C) 10 (D)
2) If n! =24 then the value of n’ is ---(A) 24
(B) 6
(C) 4
(D) 1
3) Probability of sure event is--(A)
0
(B)
(C)
(D) 1
4) The formula to find coefficient of variation--(A)
x100
(B
x 100
(C)
.
50
1x7=7
(D)
.
5) The slope of the equation 3x + 2y +1 is ----(A)
(B)
(C)
(D)
6) If the standard deviation of a certain data is 4 , its variance is-------(A) 2
(B)
4
(C) 8
(D)
7) The zero of the polynomial f(x) = 3x – 6 is ---(A) 2
(B) -2
(C)
(D)
II. Answer the following
8) Express 210 in prime factors
1x5=5
2 x 3x5x7
9) If A = { 1, 3, 4, 5 }, B = { 1, 4, 9 } ,find A – B
A-B = { 3, 5 }
10) State converse of Thales Theorem.
If a straight line divides two sides of a triangle proportionally,then the straight
line is parallel to the third side.
11) The two circles are intersecting, what is the maximum number of common tangents to
be drawn?
2
12) Find the curved surface area of a cone, if its radius is 7cm and slant height is 10cm.
A=
A=
7x10
A=
.
.
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III. Answer the following
2x 10 = 20
13) In a circle of radius 3cm draw two radii such that the angle between them is60 0
Construct tangents at the non-centre ends of the radii.
14) Rationalize the denominator and simplify
√
=
=
=
√
(√
x
√
√
√ √
√ )
√
(√
√
√ )
(√
√ )
=√ +√
15) Simplify: 4√63 + 5√7 - 8√28
4√63 + 5√7 - 8√28
= 4√9x7 + 5√7 - 8√4x7
= 4x3√7 + 5√7 – 8x2√7
= 12√7 + 5√7 – 16√7
= √7 (12+ 5 – 16)
=√
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16) There are 50 students in a school. Out o them 30 students are participating in
Dance,17 are in drama, 5 members are participating in both in a school day . Find the
number of students who are not participating in any one of these.
A-{Students Participating in Dance}
B-{Students participating in drama}
n(AUB) = n(A) + n(B) – n(A∩B)
n(AUB) = 30 + 17 – 5
n(AUB) = 42
∴ The number of students who are not participating in any = 50 – 42 = 8
17) Classify the following as permutation and combination.
a). Five different subject books to be arranged in a shelf.
Permutation
b).There is 8 chairs and 8 people to occupy them.
Permutation
c). A collection of 10 toys are to be divided equally between two children.
Combination
d).Five keys are to be arranged in a circular key ring.
permuation
OR
How many committees of five with a given chairperson can be selected from 12
persons?
!
Formula: nCr = ( )!
−
12
C1x 11C4
= 12 x
= 12 x 330
!
= 3960
18) Solve by using the formula x2 + x = 12
a = 1, b = 1 , c = -12
−1 ± √1 − 4x1x − 12
=
2x1
−1 ± √1 + 48
=
2
−1 ± √49
=
2
−1 ± 7
2
=
Or
=
=
19) =
Or
=
=
Or
=−
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20) Prove that(1 +
= sec .cos
=
).cos
= 1.
.cos
=
21) Find the distance between the points (2, 3) and (6, 6).
d = (6 − 2) + (6 − 3)
d = √4 + 3
d = √16 + 9
d = √25
d = units
22) Verify Euler’s formula for the given graph
N = 4, R = 4, A = 6
N+R = 4+ 4 = 8
A+2 = 6 + 2 = 8
∴ N+R = A+2
22) Sketch out the field to the following notes from the field book
(1 cm = 25m.)
To D
75To E
225
150
100
50
75 To C
50 To B
From A
1cm = 25m,
50m=2cm,75m=3cm,100m=4cm,150m=6cm,225m=9cm
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IV. Answer the following
23) Calculate the standard deviation of the following data
X
0-10
10-20
20-30
f
1
2
3
3 x 2= 6
30-40
4
CI
X
f
d=x-25
fd
d2
fd2
0-10
10-20
20-30
30-40
5
15
25
35
1
2
3
4
-2
-1
0
1
-2
-2
0
4
4
1
0
1
4
2
0
4
10
=
=
∑
−
0
10
∑
−
= √
=
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23) Prove that the tangents drawn from an external point to a circle are equal.
Given:A is the centre.B is an external point..
BP and BQ are the tangents.AP, AQ and AB are joined
To prove : (a).
BP = BQ
Proof: In ∆APB and ∆AQB,
[ ∵ Radius of the same circle
AP = AQ
∠APB = ∠AQB =900
[ ∵ Radius drawn at the point of contact is
perpendicular to the tangent
AB =
∴
∴
∆APB ≡ ∆AQB
(a).
AB
[ ∵ RHS postulates
BP = BQ
V) Answer the following
4 x 3= 12
25) Draw direct common tangents to two circles of radii 5cm and 2.5cm having their
Centers 10cm apart and measure their lengths.
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26) Draw the graph of y = x2 - 6x - 7
x
3
1
5
y
-16
-12
-12
0
-7
6
-7
-2
9
8
9
Roots are x= -1 and x=7
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27) State and prove converse of Pythagoras theorem.
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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