Please refer to Class 10 Maths Sample Paper Term 2 With Solutions Set C provided below. The Sample Papers for Class 10 Maths have been prepared based on the latest pattern issued by CBSE. Students should practice these guess papers for class 10 Maths to gain more practice and get better marks in examinations. The Term 2 Sample Papers for Maths Standard 10 will help you to understand the type of questions which can be asked in upcoming examinations.
Term 2 Sample Paper for Class 10 Maths With Solutions Set C
SECTION – A
1. If a point P is 13 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 12 cm. Then find the radius of the circle.
Answer : 1. If a point P is 13 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 12 cm. Then find the radius of the circle.
In ΔPTO, OP2 = PT2 + OT2
⇒ 132 = 122 + OT2 ⇒ 169 – 144 = OT2 ⇒ 25 = OT2 ⇒ OT = 5 cm
2. For what value of k will k + 11, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?
Answer : Let k + 11, 2k – 1 and 2k + 7 are in A.P.
∴ (2k – 1) – (k + 11) = (2k + 7) – (2k – 1)
⇒ 2k – 1 – k – 11 = 2k + 7 – 2k + 1
⇒ k – 12 = 8 ⇒ k = 20
3. Find the roots of the quadratic equation a2b2x2 + b2x – a2x – 1 = 0.
Answer : The given quadratic equation is a2b2x2 + b2x – a2x – 1 = 0
⇒ b2x (a2x + 1) – 1 (a2x + 1) = 0
⇒ (a2x + 1)(b2x – 1) = 0 ⇒ a2x + 1 = 0 or b2x – 1 = 0
⇒ x = −1 / a2 or x = 1 / b2
Find the value of k, for which the quadratic equation x2 – kx + 4 = 0 has equal roots.
Answer : The given equation is, x2 – kx + 4 = 0
For equal roots, D = b2 – 4ac = 0
⇒ (–k)2 – 4(1)(4) = 0 ⇒ k2 = 16 ⇒ k = ± 4
4. Find two consecutive positive integers, the sum of whose squares is 61.
Answer : Let the two consecutive positive integers be x and x + 1.
According to question, x2 + (x + 1)2 = 61
⇒ x2 + x2 + 2x + 1 = 61
⇒ 2x2 + 2x = 60 ⇒ x2 + x – 30 = 0
⇒ x2 + 6x – 5x – 30 = 0 ⇒ x(x + 6) – 5(x + 6) = 0 ⇒ (x + 6)(x – 5) = 0
⇒ x = -6 or x = 5 ⇒ x = 5 [Since x is a positive integer]
And x + 1 = 6
∴ The two consecutive positive integers are 5 and 6.
5. The mode of the following frequency distribution is 38. Find the value of x.
6. A cone and a hemisphere have equal bases and equal volumes. Find the sum of the numerator and denominator of ratio of height of cone and radius of hemisphere.
Answer : Let r be the radius of both the cone and hemisphere and h be the height of the cone.
∴ Volume of cone = Volume of hemisphere
∴ Required sum = 2 + 1 = 3
Three cubes each of edge 3 cm are joined end to end. Find the surface area of the resulting cuboid.
Answer : Length of the resulting cuboid = 3 + 3 + 3 = 9 cm
Breadth of the resulting cuboid = 3 cm
Height of the resulting cuboid = 3 cm
∴ Surface area of the resulting cuboid = 2(lb + bh + hl)
= 2(9 × 3 + 3 × 3 + 3 × 9) = 2 × 63 = 126 cm2
SECTION – B
7. Calculate the mean and median for the following frequency distribution:
8. Draw a circle of radius 5 cm. From a point P, 8 cm away from its centre, construct a pair of tangents to the circle. Measure the length of each one of the tangents.
Answer : Steps of construction :
Step-I : Draw a circle with O as centre and radius 5 cm.
Step-II : Mark a point P outside the circle such that OP = 8 cm.
Step-III : Join OP and draw its perpendicular bisector, which cuts OP at M.
Step-IV : Draw a circle with M as centre and radius equal to MP to intersect the given circle at the point T and T′. Join PT and PT′.
Hence, PT and PT′ are the required tangents.
9. A person walking 20 m towards a chimney in a horizontal line through its base observes that its angle of elevation changes from 30° to 45°. Find the height of chimney.
Answer : Suppose height of the chimney is h metres. Let A and B be the point of observation and BC be x m.
A man rowing a boat away from a lighthouse 150 m high takes 2 minutes to change the angle of elevation of the top of lighthouse from 45° to 30°. Find the speed of the boat. [Use √3 = 1.732]
Answer : Let AB be the lighthouse of height 150 m.
Let initially boat is at C and after 2 minutes it reaches at D.
10. If the median of the distribution given below is 28.5, then find the values of x and y.
SECTION – C
11. The sums of n, 2n, 3n terms of an A.P. are S1, S2, S3 respectively. Prove that S3 = 3(S2 – S1).
Answer : Let a be the first term and d be the common difference of the A.P.
12. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles
at the centre of the circle.
Answer : Let ABCD be the quadrilateral circumscribing a circle at the center O such that it touches the circle at the point P,Q,R,S. Let join the vertices of the quadrilateral ABCD to the center of the circle
In ΔOAP and ΔOAS
AP=AS ( Tangents from to same point A)
PO=OS ( Radii of the same circle)
OA=OA ( Common side)
so, ΔOAP=ΔOAS (SSS congruence criterion)
∴ ∠POA=∠AOS (CPCT)
Similarly, ∠2=∠3, ∠4=∠5 and ∠6=∠7
∠1+∠2 +∠3+∠4+∠5+∠6+∠7+∠8 = 360⁰
⇒ (∠1 +∠8) +(∠2 +∠3) + (∠4 +∠5) + (∠6 +∠7) = 360⁰
⇒ 2(∠1) + 2(∠2) + 2(∠5) + 2(∠6) = 360⁰
⇒ (∠1) + (∠2) + (∠5) + (∠6) = 180⁰
∴ ∠AOD + ∠COD=180⁰
Similarly, ∠BOC + ∠DOA = 180⁰
A circle is inscribed in a DABC having sides 16 cm, 20 cm and 24 cm as shown in figure. Find AD, BE and CF.
Answer : Since, tangents drawn from an external point to a circle are equal.
∴ AD = AF = x, BD = BE = y and CE = CF = z
According to the question,
AB = x + y = 24 cm …(i)
BC = y + z = 16 cm …(ii)
AC = x + z = 20 cm …(iii)
Subtracting (iii) from (i), we get
y – z = 4 …(iv)
Adding (ii) and (iv), we get
2y = 20 ⇒ y = 10 cm
Substituting the value of y in (i) and (ii), we get
x = 14 cm, z = 6 cm
∴ AD = 14 cm, BE = 10 cm and CF = 6 cm.
CASE STUDY QUESTION – 1
13. Anita is studying in X standard. While helping her mother in kitchen, she saw rolling pin made of steel and empty from inner side, with two small hemispherical ends as shown in the figure.
(i) Find the curved surface area of two identical cylindrical parts, if the diameter is 2.5 cm and length of each part is 5 cm.
(ii) Find the volume of big cylindrical part.
CASE STUDY QUESTION – 2
14. A boy 4 m tall spots a pigeon sitting on the top of a pole of height 54 m from the ground. The angle of elevation of the pigeon from the eyes of boy at any instant is 60°. The pigeon flies away horizontally in such a way that it remained at a constant height from the ground. After 8 seconds, the angle of elevation of the pigeon from the same point is 45°.
Based on the above information, answer the following questions. (Take √3 =1.73 )
(i) Find the distance of first position of the pigeon from the eyes of the boy.
(ii) How much distance the pigeon covers in 8 seconds?
Answer : (i) Distance of first position of pigeon from the eyes of boy = AC
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