Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 5 with Solutions
Time : 2 Hr.
Max. Marks : 40
General Instructions:
- The question paper consists of 14 questions divided into three 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
Question 1.
If α and ß are the roots of the equation x² + px + q = 0, then what is value of α² + ß²? (2)
OR
For which value of k does the pair of equation x² – y² = 0 and (x – k)² + y² = 1 yield a unique positive solution of x?
Solution:
We know that a and p are the roots of the equation x² + px + q = 0, a + P = -p and αß = q
Now, α² + ß² = (α + ß)² – 2αß
= (- P)² – 2 (q)
= p² – 2q
OR
x² – y² = 0 … (i)
and (x – k)² + y² = 1
x² + k² – 2 kx + y² – 1 = 0 … (iii)
From equations (i) and (ii)
2x² – 2kx + k² – 1 = 0
For unique solution b² – 4ac = 0 must satisfy
(- 2k)² – 4 x 2 x (k² – 1) = 0
⇒ 4k² = 8
⇒ k = \(\sqrt{2}\)
Question 2.
Find how many integers between 200 and 500 are divisible by 8. (2)
Solution:
Smallest number divisible by 8 in the given range = 208
Last number divisible by 8 in the given range = 496
So, a = 208, d = 8, n = ?, Tn = 496
We know, Tn = a + (n – 1 )d
⇒ 496 = 208 + (n – 1) 8
or 8 n + 208 – 8 = 496
or 8 n = 496 – 200
⇒ 8 n = 296
⇒ n = \(\frac { 296 }{ 8 }\) = 37
So number of terms between 200 and 500 divisible by 8 are 37.
Question 3.
Draw a line segment of length 7 cm and divide it internally in the ratio 2 : 3. (2)
Solution:
Step I: Draw AB = 7 cm
Step II: At A, draw an acute angle with 5 equidistant marks A1, A2, A3, A4, A5
Step III: Join A5B
Step IV : Draw A2C || A5B to get point C on AB.
Step V : Thus, AC : CB = 2 : 3.
Question 4.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? (2)
Solution:
Let radius of hemisphere be r units.
Volume of hemisphere = S.A. of hemisphere (Given)
\(\frac { 2 }{ 3 }\)πr³ =3nr2 3
⇒ r = \(\frac { 9 }{ 2 }\) or diameter = 9 units
Question 5.
For the following distribution: (2)
Below | 10 | 20 | 30 | 40 | 50 | 60 |
Number of students | 3 | 12 | 27 | 57 | 75 | 80 |
Find the modal class.
Solution:
Below | Class interval | Cumulative frequency | Frequency |
10 | 0 – 10 | 3 | 3 |
20 | 10 – 20 | 12 | 9 |
30 | 20 – 30 | 27 | 15 |
40 | 30 – 40 | 57 | 30 |
50 | 40 – 50 | 75 | 18 |
60 | 50 – 60 | 80 | 5 |
Here, modal class is 30 – 40, with maximum frequency 30.
Question 6.
The sum of two numbers is 15 and their reciprocals is \(\frac { 3 }{ 10 }\). Find the numbers. (2)
Solution:
Let the numbers be x and (15 – x). So, their reciprocals are \(\frac { 1 }{ x }\) and \(\frac { 1 }{ 15-x }\) respectivley.
Now, \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\)
⇒ \(\frac{15-x+x}{x(15-x)}=\frac{3}{10}\)
⇒ \(\frac{5}{x(15-x)}=\frac{1}{10}\)
⇒ x(15 – x) = 50
⇒ 15x – x² = 50
⇒ x² – 15x + 50 = 0
⇒ x² – 10x – 5x + 50 = 0
⇒ x(x – 10) – 5(x – 10) = 0
⇒ (x – 5) (x -10) = 0
⇒ x = 5 and 10.
∴ Numbers are 5 and 10.
Section – B
Question 7.
The marks attained by 40 students in a short assessment is given below where a and b are missing. If the mean of the distribution is 7.2, find a and b. (3)
Marks | 5 | 6 | 7 | 8 | 9 |
No. of students | 6 | a | 16 | 13 | b |
Solution:
Class Interval (xi) | Frequency (fi) | (fi × xi) |
5 | 6 | 30 |
6 | a | 6a |
7 | 16 | 112 |
8 | 13 | 104 |
9 | b | 9b |
∑fi = 35 + a + b = 40 | ∑(fi × xi) = 246 + 6a + 9b |
We know that,
Mean = \(\frac{\Sigma\left(f_{i} \times x_{i}\right)}{\Sigma f_{i}}\)
7.2 = \(\frac{246+6 a+9 b}{40}\)
⇒ 246 + 6a + 9b= 40(7.2)
⇒ 246 + 6a + 9b= 288
⇒ 6a + 9b =42
⇒ 2a + 3b = 14 … (i)
⇒ 35 + a + b = 40
⇒ a + b =5
⇒ 2a + 2b = 10 … (ii)
Subtracting equation (ii) from equation (i),
b = 4
and a = 5 – 4 = 1.
Question 8.
If AB, AC, PQ are the tangents is the figure, and AB = 5 cm, find the perimeter of ∆APQ. (3)
Solution:
Given, AB, AC, PQ are tangents and AB = 5 cm
Perimeter of ∆APQ,
Perimeter = AP + AQ + PQ
= AP + AQ + (PX + QX)
We know that, the two tangents drawn from external point to the equal in length from point A,
So, AB = AC = 5 cm
From point P, PX = PB [Since, tangents from an external point to the circle are equal.
From point Q, QX = QC [Since, tangents from an external point to the circle are equal.
∴ Perimeter = AP + AQ + PQ
= AP + AQ + (PX + XQ)
= AP + AQ + (PB + QC)
= (AP + PB) + (AQ + QC)
= AB + AC = 5 + 5
= 10 cm.
Question 9.
The following table provides data about the weekly wages (in ?) of workers in a factory. Calculate the Mean and the Modal Class. (3)
Class Interval | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 |
Weekly wages (₹) | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
Solution:
Class Interval | Frequency (fi) | Class Mark (xi) | (fi×xi) |
50 – 55 | 5 | 52.5 | 262.5 |
55-60 | 20 | 57.5 | 1150 |
60-65 | 10 | 62.5 | 625 |
65-70 | 10 | 67.5 | 675 |
70-75 | 9 | 72.5 | 652.5 |
75-80 | 6 | 77.5 | 465 |
80-85 | 12 | 82.5 | 990 |
85-90 | 8 | 87.5 | 700 |
N = ∑fi = 80 | ∑(fi x Xi) = 5520 |
Thus Mean = \(\frac{\Sigma\left(f_{i} \times x_{i}\right)}{\Sigma f_{i}}=\frac{5520}{80}\) = ₹ 69
Since 55 – 60 has the highest frequency 20, so it is the Modal Class.
Question 10.
A kite is flying at a height of 30 m from the ground. The length of the string from the kite to the ground is 60 m. Assuming that there is no slack in the string, then find the angle of elevation of the kite at the ground. (3)
OR
A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45°. Then find the height (in meters) of the tower.
Solution:
Given, perpendicular height = 30 m and hypotenuse = 60 m.
Section – C
Question 11.
A hemispherical depression is cut out from one face of a cubical block of side 7 cm such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid. [Use π = \(\frac { 22 }{ 7 }\)]
OR
Sushant has a vessel in the shape of an inverted cone that is open at the top. Its height is 11 cm and the radius of the top is 2.5 cm. It is full of water and metallic spherical balls of diameter 0.5 cm are put in the vessel such that \(\frac { 2 }{ 5 }\) th of the water flows out. Find the number of balls that were put in the vessel. (4)
Solution:
Given, Side of the cube = 7 cm
Diameter of the hemisphere = 7 cm
Thus, Radius of the hemisphere = 3.5 cm
Total surface area of the cube = 6(7)² cm²
Curved surface area of the hemisphere = 2π(3.5)² cm²
Thus, total surface area of the remaining solid = Total surface area of cube + Curved surface area of the hemisphere – Area of the top face of the hemisphere
= [6(7)² + 2π(3.5)² – π(3.5)²] cm²
= \(\left[6(7)^{2}+\left(\frac{22}{7}\right)(3.5)^{2}\right]\)
= [294 + \(\frac { 77 }{ 2 }\)] cm²
= 332.5 cm²
OR
Given, Height of the cone = 11 cm
Radius of the top of the cone = 2.5 cm
Diameter of each ball = 0.5 cm
Thus, Radius of each ball = 0.25 cm
Volume of water that flows out = \(\frac { 2 }{ 5 }\) (Volume of the cone)
Now, Volume of the water in the cone = \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { π }{ 3 }\) (2.5)² (11) cm³
= \(\frac { 11π }{ 3 }\)(2.5 x 2.5) cm³
Thus, the volume of water that flows out
= \(\frac { 11 }{ 3 }\) π(2.5 x 2.5) (\(\frac { 2 }{ 5 }\))cm³
= Total volume of all the spherical balls
Now, volume of 1 spherical ball = (\(\frac { 4 }{ 3 }\))π(0.25)³ cm³
= (\(\frac { 4 }{ 3 }\))π(0.25 x 0.25 x 0.25) cm³
Hence, the number of spherical balls = \(\frac{\left(\frac{1}{3}\right) 11 \pi(2.5 \times 2.5)\left(\frac{2}{5}\right)}{\left(\frac{4}{3}\right) \pi(0.25 \times 0.25 \times 0.25)}\)
= \(\frac{11(2.5 \times 2.5)}{10(0.25 \times 0.25 \times 0.25)}\)
Thus, number of balls = 440.
Question 12.
In a circle C(0, r), OP is equal to diameter of the circle. PA and PB are two tangents. Prove that ΔABP is an equilateral triangle. (4)
Solution:
Let OP meet the circle at Q
It is given that,
OP = Diameter = 2r
⇒ OP = OQ + QP
⇒ 2r = r + QP
⇒ QP = r
∴ Q is the mid-point of OP.
Now AP is a tangent at A and OA is radius
⇒ OA ⊥ AP
∴ ∆OAP is right triangle with A = 90° and Q is mid-point of hypotenuse OP.
∴ AQ = OQ = OA
∴ ∆OAQ is equilateral.
⇒ ∠AOQ = 60°
∴ ∠APO = 180° – 90° – 60° = 30°
∴ ∠APB = 2 ∠APO
= 2 x 30° = 60°
Also, PA = PB
⇒ ∠PAB = ∠PBA
Now in AAPB, we have
∠APB + ∠PAB + ∠PBA = 180°
⇒ 60° + 2 ∠PAB = 180°
⇒ ∠PAB = 60°
∠PAB = ∠PBA = ∠APB = 60°
∴ ∆APB is equilateral.
Question 13.
Two hoardings are put on two poles of equal heights standing on either side of the road. From a point between them on the road (not the mid point) the angle of elevation of the top of poles are 60° and 30° respectively. Height of the each pole is 20 m. (4)
Answer the following questions. (Take \(\sqrt{3}\) = 1.73)
(i) Find the length of PO.
(ii) The width of the road.
Solution:
Question 14.
Aadita is celebrating her birthday. She invited her friends. She bought a packet of toffees/candies. She arranged the candies such that in the first row there are 3 candies, in second there are 5 candies, in third there are 7 candies and so on. (4)
(i) (a) Find the first term and common difference of A.P.
(b) How many candies are placed in the 9th row?
(ii) (a) Find the difference in number of candies placed in 7th and 3rd row.
(b) Find the number of candies in 12th row.
Solution:
(i) (a) As A.P. = 3,5,7…
∴ a = 3, d = 2
(b) Since, an = a + (n – 1 )d
a9 = 3 + 8 x 2 = 19
(ii) (a) a7 – a3 = a + 6d – a – 2d = 4d
= 4 x 2 = 8
(b) Number of candies in 12th row,
a12 = d + 11d
= 3 + 11 x 2 = 25.