Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 2 with Solutions
Time: 2 Hours
Maximum Marks: 40
General Instructions:
- The question paper consists of 14 questions divided into 3 sections A, B, C.
- All questions are compulsory.
- Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
- 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 [2 Marks Each]
Question 1.
Which term of the A.E 27,24,21,………. is zero ?
OR
In an Arithmetic Progression, if d = – 4, n = 7, an = 4, then find a.
Answer:
Since, nth term (an) = a + (n – 1)d
Let an be zero an = 0
0 = 27 + (n – 1)(- 3)
30 = 3 n
n = 10
10th term of the given A.P, is zero.
Detailed Solution:
Given A.P. = 27, 24, 21,…………….. .
Here, a = 27 and d = 24 – 27 = -3
and, an = 0
∴ an = a + (n – 1)d
⇒ 0 = 27 + (n – 1)(- 3)
⇒ – 3n + 3 = – 27
⇒ – 3n = – 27 – 3 = – 30
⇒ n = 10.
OR
Since,
an = a + (n – 1 )d
4 = a + 6 × (- 4)
a = 28
Detailed Solution:
We have, d = – 4, n = 7 and an = 4
∴ an = a + (n – 1 )d
⇒ 4 = a + (7 – 1)(- 4)
⇒ 4 = a + 6 (- 4)
= a – 24
⇒ a = 4 + 24
⇒ a = 28.
Question 2.
Find the roots of the equation x2 + 7x + 10 = 0 by using quadratic formula.
Answer:
Given, x2 + 7x + 10 = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = 7 and c = 10
Hence, the roots of the given equation are – 2 and – 5.
Question 3.
In the following, if PQ = 28 cm, then find the perimeter of ∆PLM.
Answer:
PQ = PT
PL + LQ = PM + MT
PL + LN = PM + MN (LQ = LN, MT = MN)
(Tangents to a circle from a common point)
Perimeter (∆ PLM) = PL + LM + PM
= PL + LN + MN + PM
= 2 (PL + LN)
= 2 (PL + LQ)
= 2 × 28 = 56 cm
Detailed Solution:
∴ Given, PQ = 28 cm
PQ = PT
(Length of tangents from an external point are equal)
i.e., PQ = PT = 28 cm
According to figure,
Let LQ = x, then
PL = (28 – x) cm and let MT = y, then
PM = (28 – y) cm
and LM = LN + NM
= x + y
Now, the perimeter of ∆ PLM = PL + LM + PM
= (28 – x) + (x + y) + (28 – y)
= 28 + 28 = 56 cm.
Question 4.
From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base removed. Find the volume of the remaining solid.
Answer:
Given, Height (h) = 14 cm
Base radius (r) = 6 cm
Volume of the remaining solid = Volume of a right circular cylinder – Volume of a right circular cone
= πr2h – \(\frac{1}{3}\)πr2h
= \(\frac{2}{3}\)πr2h
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 14
= 1056 cm3
Commonly Made Error:
Sometimes, students use the formula for curved surface.
Answering Tip:
They should read the question clearly and use right formula and correct calculations for which good practice is necessary.
Question 5.
Compute the mode for the following frequency distribution:
Answer:
Here, modal class = 12-16
∴ l1 = 12, f1= 17, f0 = 9, f2 = 12 and h = 4
Mode = l1 + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 12 + + \(\left(\frac{17-9}{2 \times 17-9-12}\right)\) × h
= 12 + + \(\frac{8 \times 4}{13}\)
= 12 + 2.46 = 14.46.(Approx)
Question 6.
For what values of k, the given quadratic equation 9x2 + 6kx + 4 = 0 has equal roots ?
OR
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.
Answer:
9x2 + 6kx + 4 = 0
For equal roots, b2 – 4ac = 0
Discriminant, D = 0
So, (6k)2 – 4 × 9 × 4 = 0
36 k2 = 144
k2 = 4
k = ±2
Detailed Solution:
Given, 9x2 + 6kx + 4 = 0.
Comparing with ax2 + bx + c = 0,
we get a = 9, b = 6k, c = 4
Since, Discriminant, D = b2 – 4ac
For equal roots, D = 0
b2 – 4ac = 0
⇒ (6k)2 – 4 × 9 × 4 = 0
⇒ 36k2 – 144 = 0
⇒ 36k2 = 144
⇒ k2 = 4
⇒ k = ±2
OR
Let the three consecutive natural numbers be x, x + 1 and x + 2.
∴ (x + 1)2 = (x + 2)2 – (x)2 + 60
⇒ x2 + 2x + 1 = x2 + 4x + 4 – x2 + 60
⇒ x2 – 2x – 63 = 0
⇒ x2 – 9x + 7x – 63 = 0
⇒ x(x – 9) + 7(x – 9) = 0
⇒ (x – 9)(x + 7) = 0
Thus, x = 9 or x = – 7
Rejecting – 7, we get x = 9
Hence, three numbers are 9, 10 and 11
Section – B [3 Marks Each]
Question 7.
Find the arithmetic mean of the following frequency distribution :
Answer:
Arithmetic Mean (x) = \(\frac{\Sigma f x}{\Sigma f}\)
= \(\frac{1002.5}{35}\)
= 28.6 (Approx)
Question 8.
Draw a line segment AB of length 9 cm. With A and B as centres, draw circles of radius 5 cm and 3 cm respectively. Construct tangents to each circle from the centre of the other circle.
Answer:
Detailed Solution:
Steps of Construction:
- Draw a line segment AB of 9 cm.
- Taking A and B as centres draw two circles of radii 5 cm and 3 cm respectively.
- Perpendicular bisect the line AB. Let mid-point of AB be C.
- Taking C as centre draw a circle of radius AC which intersects the two circles at point P, Q, R and S.
- Join BP, BQ, AS and AR.
BP, BQ and AR, AS are the required tangents.
Question 9.
The median of the following data is 525. Find the values of x and y, if total frequency is 100.
Answer:
Also, 76 + x + y = 100
⇒ x + y = 100 – 76 = 24 …..(i)
Given, Median = 525, which lies between class 500 – 600.
⇒ Median class = 500 – 600
Now, Median = l1 + \(\frac{\frac{N}{2}-c . f .}{f}\) × h
⇒ 525 = 500 + \(\left[\frac{\frac{100}{2}-(36+x)}{20}\right]\) × 100
⇒ 25 = (50 – 36 – x)5
⇒ 14 – x = \(\frac{25}{5}\) = 5
⇒ x = 14 – 5 = 9
Putting the value of x in eq. (j), we get
y = 24 – 9 = 15
Hence, x = 9 and y = 15.
Question 10.
The two palm trees are of equal heights and are standing opposite to each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O
from the trees, (use √3 = 1.73)
OR
The ratio of the length of a vertical rod and the length of its shadow is 1 : V3 . Find the angle of elevation of the Sun at that moment ?
Answer:
Let BD = width of river = 80 m
AB = CD = height of both trees = h
BO = x
OD = 80 – x
In ∆ABO, ∠B = 90°
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
h = √3x …….. (i)
In ∆CDO, ∠D = 90°
tan 30° = \(\frac{h}{(80-x)}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{(80-x)}\)
Solving (i) and (ii), we get
x = 20 m
h = √3x [From eqn. (i)]
= 1.73 × 20 m
= 34.6 m
The height of the trees = h = 34.6 m
BO = x = 20 m
DO = 80 – x
= 80 – 20
= 60 m
∴ The distances of the point O from the trees are 20 m and 60 m respectively.
OR
Let AB be a vertical rod and BC be its shadow.
From the figure, ∠ACB = 9.
In ∆ABC, ∠B = 90°
⇒ tan θ = \(\frac{A B}{B C}\)
⇒ tan θ = \(\frac{1}{\sqrt{3}}\) [∵ \(\frac{A B}{B C}\) = \(\frac{1}{\sqrt{3}}\)(Given)]
⇒ tan θ = tan 30°
⇒ θ = 30°
Hence, the angle of elevation of the Sun is 30°.
Section – C [4 Marks Each]
Question 11.
A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, then find the ratio of the radius and the height of the conical part.
Answer:
Let ABC be a cone, which is mounted on a hemisphere.
Given, OC = OD = r cm
Curved surface area of the hemispherical part
= \(\frac{1}{2}\)(4pr2)
= 2pr2
Slant height of a cone,
l = \(\sqrt{r^{2}+h^{2}}\)
and curved surface area of a cone
= prl
= pr\(\sqrt{h^{2}+r^{2}}\)
According to the problem,
2pr2 = pr\(\sqrt{h^{2}+r^{2}}\)
⇒ 2r = \(\sqrt{h^{2}+r^{2}}\)
on squaring both of the sides, we get
⇒ 4r2 = h2 + r2
⇒ 4r2 – r2 = h2
⇒ 3r2 = h2
⇒ \(\frac{r^{2}}{h^{2}}\) = \(\frac{1}{3}\)
⇒ \(\frac{r}{h}\) = \(\frac{1}{\sqrt{3}}\)
Hence, the ratio of the radius and the height of conical part
= 1 : √3.
Question 12.
PQ is a tangent to a circle with centre O at point P. If AOPQ is an isosceles triangle, then find ZOQP.
OR
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.
Answer:
In ∆OPQ,
∠P + ∠Q + ZO = 180° (∠O = ∠Q, isosceles triangle)
2 ∠Q + ZP = 180°
2∠Q + 90° = 180°
2 ∠Q = 90°
∠Q = 45°
Detailed Solution:
Since, ∠OPQ = 90° (Angle between tangent and radius)
Let ∠PQO be x°, then
∠QOP = x°
(Since OPQ is an isosceles triangle) (OP = OQ) (given)
In ∆ OPQ,
∠OPQ + ∠PQO + ∠QOP = 180°
(Sum of the angles of a triangle)
∴ 90° + x° + x° = 180°
⇒ 2x° = 180° – 90° = 90°
⇒ x = \(\frac{90^{\circ}}{2}\) = 45°
Hence, ∠OQP is 45°
OR
∠APB = 90° (angle in semi-circle) and ∠ODB = 90° (radius is perpendicular to tangent)
∆ABP ~ ∆OBD
⇒ \(\frac{A B}{O B}=\frac{A P}{O D}\)
⇒ \(\frac{26}{13}=\frac{A P}{8}\)
Hence, AP = 16 cm
Case Study-1
Question 13.
‘Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively.
Based on the following figure related to sky sailing answer the questions:
(i) In the given figure, if tan θ = cot (30° + θ), where Q and 30° + θ are acute angles, then the value of θ.
Answer:
Given,
tan θ = cot(30° + θ)
= tan[90° – (30° + θ)]
= tan(90° – 30° – θ)
⇒ tan θ = tan(60° – θ)
⇒ θ = 60°- θ
⇒ 2θ = 60°
⇒ θ = 30°.
(ii) Find the value of tan 30°. cot 60°.
Answer:
tan 30° = \(\frac{1}{\sqrt{3}}\)
cot 60° = \(\frac{1}{\sqrt{3}}\)
Now, tan 30° × cot 60° = \(\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\)
= \(\frac{1}{3}\)
Case Study-2
Question 14.
A ladder has rungs 25cm apart. (see the below).
The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and the
bottom rungs are 2\(\frac{1}{2}\)m apart.
(i) Find the number of the rungs: [2]
Answer:
The distance between the two rungs is 25 cm.
Hence, the total number of rungs
= \(\frac{2 \frac{1}{2} \times 100}{25}\) + 1 = \(\frac{250}{25}\) + 1 = 11.
(ii) What is the length of the wood required for the rungs ? [2]
Answer:
Here, length of first rung be a and last rung be l, then
Sn = \(\frac{n}{2}\)[a + l]
Here, a = 25, l = 45
and n = 11
Then,
the required length of the wood,
Sn = \(\frac{11}{2}\) [25 + 45]
= \(\frac{11}{2}\) × 70
= 385 cm.