BELTHANGADY TALUK MATHS TEACHERS

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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
TOTAL:
SUBJECT: MATHEMATICS
50
1x7=7
I . Choose the best alternative from following options
1) If Tn = 2n - 1 , then the value of S2 ----
(A)
1
(B) 2 (C) 3
(D)
2) The value of n+1P2 is --(
(A) n(n+1)
(B)
n(n-1)
(C) (n+1)(n-1)
(D)
3) The two coins are tossed at a time , the total number of outcomes n (S) ---(A)
2
(B)
(C)
6
(D) 8
4) The formula to find coefficient of variation ------(A)
5) If sin
(B)
=
x100
(C)
10
)
(D)
, then = -----
(A)
cos
(B)
(C)
6) The distance between the points (6, 5) and (4, 4).
sec
(A) √
(B)
1
7) If f(x) = 4x + 5, then f( 1 ) =
(A) 9
(B) -9
(C) 1
2
(C)
(D)
tan
(D)
(D)
−1
-1
II. Answer the following
8) Write 144 as product of prime factors.
1x5=5
2 144
2 72
2 36
2 18
3
9
3
144 = 2x2x2x2x3x3
9) Draw Venn diagram of A ∩ (B\C)
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10) State Thales Theorem.
Thales theorem :If a straight line is drawn parallel to a side of a triangle,
then it divides the other two sides proportionally
11) Write the formula to find the total surface area of a Sphere.
A = 4 r2
12) What are concentric circles?
Circles having the same centre but different radii are called concentric circles.
III. Answer the following .
2x 10 = 20
12) Draw a pair of tangents to a circle of radius 4cm from an external point 7cm apart
from the centre.
13) Rationalize the denominator and simplify
√
√
√
=
√
=
√
√
√
√
√
√
√
=
√
x
√
√
14) Find the product :√3 x √5
√3
= 3 = 3 = √3
√5 = 5 = 5 = √5
√3 x5
√27x25
√675
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15) If U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 2, 4, 6, 8 }, B= { 1, 3, 5, 7 } ,then Prove that
(AUB)1 = A1 ∩B1 .
U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8 }, B= {1, 3, 5, 7 } ,
LHS (AUB)1 = {1,2, 3,4, 5,6, 7,8 }1 = {0, 9 }
RHS = A1∩B1 ={ 2, 4, 6, 8 }1∩{ 1, 3, 5, 7 }1
= { 0, 1,3, 5, 7,9 }∩{0, 2, 4, 6, 8,9 }
= {0, 9 }
LHS = RHS
∴(AUB)1 = A1 ∩B1 .
16) How many 3 digit numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 without
repeating any digit.
The number of 3 digit numbers can be formed using the digits 1,2,3,4,5,6
Hundreds
Tens
ones
6
5
4
P1
P1
P1
6
5
4
=6x5x4 = 120
17. 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
17) Find the value of
60 + 2
60 + 2
45 .
45 .
= √3 +2x1
= 3 +2
=
5
18) Find the coordinates of point P which divides the line joining A (4,-5) and B (6, 3) in
the ratio 2:5.
P (x,y) =
,
P (x,y) =
P (x,y) =
,
,
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19) Verify Euler’s formula for the given graph
N = 3, R = 5, A = 6
N+R = 3+ 5 = 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 = 20m.)
To C
120To D
210
200
120
80
200To B
From A
1cm = 20m
⇒ 80m = 4cm, 120m = 6cm, 210m = 10.5cm, 200m = 10cm,
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IV. Answer the following
23) Calculate the standard deviation of the following data
X
0-4
5-9
10-14
f
1
4
3
CI
f
0-4
5-9
10-14
15-19
1
4
3
2
Assumed Mean Methdo
d=x-12
fd
X
2
7
12
17
-10
-5
0
5
10
=
∑
=
=
15-19
2
d2
fd2
100
25
0
25
100
100
0
50
-20
250
∑
−
−
-10
-20
0
10
3 x 2= 6
−
–
= √
= .
24) prove that “If two circles touch each other internally, the centers and the point of
Contact is 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.
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V. Answer the following
4 x 3= 12
25) Draw transverse common tangents to two circles of radii 6cm and 2cm having
their Centers 8 cm
26) Solve graphically: x2 + 4x – 5 = 0.
x
-6
-5
-3
-2
y
7
0
-8
-9
-1
-8
0
-5
1
0
2
7
--
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27) State and Prove Pythagoras Theorem.
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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