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Class 10 - Mathematics
Term-2 Sample Paper - 10 (Standard)

Maximum Marks: 40
Time Allowed: 2 hours

General Instructions:

1. The question paper consists of 14 questions divided into 3 sections A, B, C.

2. All questions are compulsory.
3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
4. Section B comprises of 4questions of 3 marks each. Internal choice has been provided in one question.
5. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question.
It contains two case study-based questions.

Section A
1. If mth term of an AP is and nth term is then find the sum of its first mn terms.
2. Find that non-zero value of k, for which the quadratic equation kx2 + 1 - 2(k - 1)x + x2 = 0 has equal
roots. Hence find the roots of the equation.
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that
OQ = 13 cm. Find the length of PQ.
4. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of
diameter 7 cm and height 3 cm. Find the number of cones so formed.
5. The mode of the following data is 36. Find the missing frequency in it.
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 8 10 16 12 6 7

6. Solve the quadratic equation (x-1)2- 5 (x - 1) - 6 = 0.

OR

Determine whether the given values are roots of the equation or not.

Section B
7. The percentage of marks obtained by 100 students in an examination are given below:
Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-65

Frequency 14 16 28 23 18 8 3

Determine the median percentage of marks.

8. Divide a line segment of length 14 cm internally in the ratio 2: 5. Also, justify your construction.
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9. The following data gives the information observed life times (in hours) of 225 electrical components.
Determine the modal life times of the components.

Life time (in hours) 0-20 20-40 40-60 60-80 80-100 100-200

Frequency 10 35 52 61 38 29
10. As observed from the top of a light-house, 100 m high above sea level, the angle of depression of a ship,
sailing directly towards it, changes from 30° to 60o . Determine the distance travelled by the ship during
the period of observation. (Use = 1.732)

OR

The angle of elevation of an aeroplane from a point on the ground is . After a flight of15 seconds,
the angle of elevation changes to . If the aeroplane is flying at a constant height of 1500 m, find
the speed of the plane in km/hr.

Section C

11. A copper wire 3 mm in diameter is wound about a cylinder whose length is 1.2 m, and diameter 10 cm,
so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the
density of the copper wire to be 8.88 gram per cm.
12. In a right triangle ABC in which∠B = 90°,a circle is drawn with AB as diameter intersecting the
hypotenuse AC at P. Prove that the tangent to the circle at P bisect BC.

OR

In fig. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents drawn at P and Q intersect at
T. Find the length of TP.

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13. The houses of Ajay and Sooraj are at 100 m distance and the height of their houses is the same as
approx 150m. One big tower was situated near their house. Once both friends decided to measure the
height of the tower. They measure the angle of elevation of the top of the tower from the roof of their
houses. The angle of elevation of ajay's house to the tower and sooraj's house to the tower are 45o and
30o respectively as shown in the figure.

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By using the above given information answer the following questions:

i. Find the height of the tower.
ii. What is the distance between the tower and the house of Sooraj?
14. In a school garden, Dinesh was given two types of plants viz. sunflower and rose flower as shown in the
following figure.

The distance between two plants is to be 5m, a basket filled with plants is kept at point A which is 10
m from the first plant. Dinesh has to take one plant from the basket and then he will have to plant it in
a row as shown in the figure and then he has to return to the basket to collect another plant. He
continues in the same way until all the flower plants in the basket. Dinesh has to plant ten numbers of
flower plants.
Now answer the following questions:
i. Find the distance covered by Dinesh to plant the first 5 plants and return to basket. (2)
ii. Find the distance covered by Dinesh to plant all 10 plants and return to basket. (1)
iii. If the speed of Dinesh is 10 m/min and he takes15 minutes to plant a flower plant then find the total
time taken by Dinesh to plant 10 plants. (1)

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Class 10 - Mathematics
Term-2 Sample Paper - 10 (Standard)

Section A

1. Let the first term = a, and common difference = d

Given , am= and an =
So,
.........(1)
and,
.........(2)
Subtracting (2) from (1), we get :-
(m-n)d =
Or, d=
Now putting value of d in equation (2), we have,

Now,

Smn = [ 2 + (mn - 1)] [taking (1/mn) common ]

Smn = [1 + mn ]

2. we have to find that non-zero value of k, for which the quadratic equation kx2 +1 - 2(k - 1)x + x2 = 0 has
equal roots.
Given, kx2 + 1 - 2(k - 1)x + x2 = 0
(k + 1)x2 - 2(k - 1)x + 1 = 0
For equal roots D = b2 - 4ac = 0
Here, a = k + 1, b = -2(k - 1), c = 1
4(k - 1)2 - 4(k + 1) 1 = 0
or, 4k2 - 8k + 4 - 4k - 4 = 0
or, 4k2 - 12k = 0
or, 4k (k - 3) = 0
k = 3 [k = 0 is rejected as it is coefficient of x2]
3. Since tangent at a point is perpendicular to the radius through that point. Therefore, OP is
perpendicular to PQ.
In right triangle OPQ, we have

OQ2 = OP2 + PQ2 [Using Pythagoras theorem]

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132 = 52 + PQ2
PQ2 = 169 - 25 = 144
PQ = 12 cm.
4. Radius of the solid metallic sphere = .

Volume =

Volume of a smaller cone with radius and height 3 cm

[Using volume of cone =

Since a solid metallic sphere is melted and recast into a number of smaller cones, therefore

Number of cones so formed =

Hence, the number of cones so formed is 126.

5. Here , given mode is 36, which lies between 30-40 . Therefore, the modal class is 30-40.
l = 30 and f1 = 16, f0 = x, f2 = 12, h = 10

Now, Mode =

or, 36 =
or 36( 20 - x) = 600 - 30x + 160 - 10x
or, 720 - 36x = 760 - 40x
40x - 36x = 760 - 720
4x = 40

6. Given, ( x - 1 )2- 5 ( x - 1 ) - 6 = 0
or, x2 - 2x + 1 - 5x + 5 - 6 = 0
or,x2- 7x + 6 - 6 = 0
or,x2 - 7x = 0
or,x(x - 7) = 0
x = 0, 7

OR

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Hence, is not the root of the given equation.

Put in the given equation

which is not true

Hence, is not a root of the given equation.

Section B
7.
Marks Number of Students
Cumulative frequency
(Class) (Frequency)

30-35 14 14

35-40 16 30

40-45 18 48

45-50 23 71 (Median class)

50-55 18 89

55-60 8 97

60-65 3 100
Here, N = 100
Therefore, = 50, This observation lies in the class 45-50.
(the lower limit of the median class) = 45
cf (the cumulative frequency of the class preceding the median class) = 48
f (the frequency of the median class) = 23
h (the class size) = 5

Median

So, the median percentage of marks is 45.4.

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8. Steps of construction

i. Draw a line segment AB of 14 cm.

ii. Through the points A and B, draw two parallel lines AX and BY on the opposite side of AB.
iii. Starting from A. Cut 2 equal parts on AX and starting from B cut 5 equal parts on BY such that
AX1 = X1X2 and BY1 = Y1Y2 = Y2Y3 = Y3Y4 = Y4Y5
iv. Join X2Y5 which intersect AB at P

Justification
In APX2 and BPY5
(vertically opposite angles)
(alternate interior angles)
then (by AA similarity)
(by c.p.c.t)
9. Since the maximum frequency = 61 and it corresponds to the class 60-80,
∴ Modal class = 60-80
Here, l = 60, h = 20, f1 = 61, f0 = 52, f2 = 38
We know that mode Mo is given by

= 60 + 5.625
= 65.625 hour
Thus, modal life times = 65.625 hours

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10.

Height of the tower = 100 m

Let BC = x and BD = y
Consider the ,

Consider the ,

We know that ,
BD = BC + CD
y = x + CD
CD = y - x
=

OR

Let A and B be the two positions of the aeroplane.

Let and Then,

and AC = BD =
From right we have

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From right we have

Thus, the aeroplane covers 300m in 15 seconds.
speed of the aeroplance
= 720 km /hr.
Section C
11. Given,
Diameter of copper wire = 3 mm = cm.
One round of the wire covers cm of the height of the cylinder
Also, given,
Length of the cylinder = 1.2 m = 120 cm.

Now, wire will cover entire length of cylinder

Let's find number of rounds.
Number of rounds taken by the wire to cover the curved surface of the cylinder =
rounds

Length of wire used in taking one round = cm = 31.4 cm

Total length of wire used in covering the curved surface of the cylinder = = 12560 cm =
125.6 m
Mass of the wire = Length Density = gram = 111532.8 gm 111.533 kg
12.

ΔABC is a right angled triangle.

∠ABC = 90°.
A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which
intersects BC at Q.
Join BP.
PQ and BQ are tangents drawn from an external point Q.
∴ PQ = BQ -------------- (1) (Length of tangents drawn from an external point to the circle are equal)
⇒ ∠PBQ = ∠BPQ (In a triangle, angles opposite to equal sides are equal)

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Given that, AB is the diameter of the circle.

∴ ∠APB = 90° (Angle in a semi-circle is a right angle)
∠APB + ∠BPC = 180° (Linear pair)
∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°
Consider ΔBPC,
∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)
∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90°................(2)
∠BPC = 90°
∴ ∠BPQ + ∠CPQ = 90° ...(3)
From equations (2) and (3), we get
∠PBC + ∠PCB = ∠BPQ + ∠CPQ
⇒ ∠PCQ = ∠CPQ (Since, ∠BPQ = ∠PBQ)
Consider ΔPQC,
∠PCQ = ∠CPQ
∴ PQ = QC.................(4)
From equations (1) and (4), we get
BQ = QC
Therefore, tangent at P bisects the side BC.

OR

PT = QT (Tangents)
is an isosceles and TO is the angle bisector of PTQ.
, So, OT also bisects PQ.
Since OPR is right angled isosecles triangle

= 3 cm
Now, Let TP = x and TR = y then
x2 = y2 + 16 ....(i)
Also in ....(ii)
Solving (i) and (ii) we get

Hence,
13. The above figure can be redrawn as shown below:

i. Let PQ = y
In ΔPQA,
tan 45 = =

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1 =
x = y .....(i)
In ΔPQB,

From the figure, Height of tower h = PQ + QR

= x + 150 = 136.61 + 150 = 286.61 m

ii. Distance of Sooraj's house from tower = QA + AB
= x + 100 = 136.61 + 100 = 236.61 m
14. i. The distance covered by Dinesh to pick up the first flower plant and the second flower plant,
= 2 10 + 2 (10 + 5) = 20 + 30
therefore, the distance covered for planting the first 5 plants
= 20 + 30 + 40 + ........ 5 terms
This is in AP where the first term a = 20
and common difference d = 30 - 20 = 10
We know that:
so, the sum of 5 terms

hence, Dinesh will cover 200 m to plant the first 5 plants.

ii. As a = 20 ,d = 10 and here n = 10
so,
hence Ramesh will cover 650 m to plant all 10 plants.
iii. Total distance covered by Ramesh = 650 m

Time taken to plant all 10 plants=15 10 = 150 minutes

Total time = 65 + 150 = 215 minutes = 3 hrs 35 minutes

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