Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 4 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.
How many two digits numbers are divisible by 3?
OR
Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer:
Lowest two digit number divisible by 3 is 12. and highest two digit number divisible by 3 is 99.
Hence, the sequence start with 12 and ends with 99 and common difference is 3.
So, the A.P. will be 12,15,18, …………, 96, 99
Here, a = 12, d = 3, l = 99
l = a + (n – 1 )d
∴ 99 = 12 + (n – 1)3
⇒ 99 – 12 = 3(n – 1)
⇒ n – 1 = \(\frac{87}{3}\)
⇒ n – 1 = 29
⇒ n = 30
Therefore, there are 30, two digit numbers which are divisible by 3.
OR
The number which ends with 0 is divisible by 2 and 5 both.
∴ Such numbers between 102 and 998 are:
110,120,130, ………….., 990.
Last term, an = 990
a + (n + 1)d = 990 1
110 + (n -1) × 10 = 990 (∵ a = first term = 110)
110+ 10n – 10 = 990
10n + 100 = 990
10n =990-100
10n = 890
n = \(\frac{890}{10}\) = 89
Question 2.
Solve the following quadratic equation for x:
9x2 – 9 (a + b)x + 2a2 + 5 ab + 2b2 = 0
Answer:
Given,
9x2 – 9 (a + b)x + 2a2 + 5ab + 2b2 = 0
First, we solve,
2a2 + 5 ab + 2b2 = 2a2 + 4 ab + ab + 2 b2
Here, = 2a[a + 2b] + b[a + 2b]
= (a + 2b) (2a + b)
Hence, the equation becomes 9x2 – 9 (a + b)x + (a + 2b)(2a + b) = 0
9x2 – 3[3a + 3b]x + (a + 2b)(2a + b) = 0
⇒ 9x2 – 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a +b) = 0
⇒ 9x2, – 3(a + 2b)x – 3(2a + b)x +(a + 2b) (2a + b)= 0
⇒ 3x[3x – (a + 2b)] – (2a + b) [3x – (a + 2b)] = 0
⇒ [3x – (a + 2b)][3x – (2a + b)] = 0
⇒ 3x – (a + 2b) = 0 or 3x – (2a + b) = 0
⇒ x = \(\frac{a+2 b}{3}\) or x = \(\frac{2 a+b}{3}\)
Hence, the roots = \(\frac{a+2 b}{3}, \frac{2 a+b}{3}\)
Question 3.
Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Answer:
Given, AP and BP are tangents of circle having centre O.
To Prove : AP = BP Construction : Join OP, AO and BO
Proof: In ∆OAF and ∆OBP
OA = OB (Radius of circle)
OP = OP (Common side)
∠OAP = ∠OBP = 90° (Radius ⊥ tangent)
∆OAP = ∆OBP (RHS congruency rule)
AP ≅ BP (By cpct)
Hence Proved,
Question 4.
The volume of a right circular cylinder with its height equal to the radius, is 25 \(\frac{1}{7}\) cm . Find the height of the cylinder. (Use π = \(\frac{22}{7}\))
Answer:
Given,
Volume of a right circular cylinder = 25 \(\frac{1}{7}\) cm3
i.e πr2h = \(\frac{176}{7}\)
Since, r = h…. given
\(\frac{22}{7}\) × h2 × h = \(\frac{176}{7}\)
⇒ h3 = \(\frac{176}{22}\) = 8 = 23
Hence, height of the cylinder = 2 cm
Question 5.
Given below is the distribution of weekly pocket money received by students of a class. Calculate the socket money that is received by most of the students.
Answer:
| Class Interval | Frequency |
| 0-20 | 2 |
| 20-40 | 2 |
| 40-60 | 3 |
| 60-80 | 12 |
| 80 -100 | 18 |
| 100 -120 | 5 |
| 120 -140 | 2 |
| Total | 44 |
Here, Modal Class = 80 -100
l = 80, f1 = 18, f2 = 5, f0 = 12 and h = 20
∴ Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 80 + \(\left(\frac{18-12}{36-12-5}\right)\) × 20
= 80+ \(\frac{6}{19}\) × 20
=80 + 6.31
= 86.31 (approx)
Hence, mode = 86.31
Question 6.
For what value(s) of ‘a’ quadratic equation 3ax2 – 6x + 1 = 0 has no real roots?
OR
If one root of the equation (k – 1)x2 – 10x + 3 = 0 is the reciprocal of the other, then find the value of k.
Answer:
Given that,
3ax2 – 6x + 1 =0
For no real roots b2 – 4ac < 0
Discriminant D < 0 so,
(- 6)2 – 4(3a) (1) < 0 12a > 36
a > 3
Detailed Solution:
Given, 3ax2 – 6x + 1 = 0
On Comparing with AX2 + BX + C = 0,
we get A = 3a, B = – 6 and C = 1
Discriminant, D = B2 – 4AC
= (- 6)2 – 4 × 3a × 1
= 36 – 12a
For condition of ‘no real roots’,
B2 – 4AC < 0
⇒ 36 – 12a < 0 ⇒ 12a > 36
⇒ a > 3.
OR
Let one root = α
Product of roots = α × \(\frac{1}{\alpha}\) = 1
Given equation, (k – 1)x2 – 10x + 3 = 0
Comparing it with standard quadratic equation
ax2 + bx + c = 0
we get, a = (k – 1), b = – 10 & c = 3
Product of roots = \(\frac{c}{a}=\frac{3}{(k-1)}\)
1 = \(\frac{3}{k-1}\)
or, 3 = k – 1
0r k = 4
Section – B [3 Marks Each]
Question 7.
On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table :
Find mean and median of the above data.
Answer:
Let a = Assumed mean = 12
Mean, x̄ = a + \(\frac{\Sigma d_{i} f_{i}}{\Sigma f_{i}}\) × h
Median class = \(\frac{N}{2}\)
= \(\frac{400}{2}\) = 200 which lie in class 09 – 11
Median = l + \(\left(\frac{\frac{N}{2}-c . f}{f}\right)\) × h
Median = 9 + \(\frac{200-190}{32}\) × 2 = 9 + \(\frac{10}{32}\) × 2 = 9 + 0.625 = 9.625
Question 8.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Answer:
Steps of construction :
- Draw a line segment AB = 7 cm.
- With A as centre and radius 3 cm draw a circle.
- With B as centre and radius 2 cm draw another circle.
- Taking AB as diameter and draw perpendicular bisector, which intersects first two circles at P and Q, R and S.
- Join B to P and Q, A to R and S.
Hence, BP, BQ, AR and AS are the required tangents.
Question 9.
The mean of the following distribution is 53. Find the missing frequency k?
Answer:
Given, Mean = 53
∴ Mean = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)
⇒ 53 = \(\frac{3340+70 k}{72+k}\)
⇒ 53(72 + k) = 3340 + 70k
⇒ 3816 + 53k = 3340 + 70k
⇒ 70k – 53k = 3816 – 3340
⇒ 17k = 476
⇒ k =28
Hence, value of k is 28.
Question 10.
Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points.
OR
A moving boat observed from the top of a 150 m high cliff, moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat.
Answer:
In ∆DCB, ∠C = 90°
\(\frac{D C}{C B}\) = tan 45° = \(\frac{15}{x+y}\) = 1
⇒ x + y =15
⇒ 5√3 + y = 15
⇒ y = 15 – 5√3
= 5(3 – √3) m
Hence, the distance between the points
= 5(3 – √3) m
OR
Let the speed of the boat be x m/mm.
∴ Distance covered in 2 minutes 2x
∴ CD = 2x
Let BC be y m.
In ∆ABC, ∠B=90°
⇒ y + 2x = 150 ………….(ii)
Substituting the value of y from (i) in (ii),
50√3 + 2x = 150
2x = 150 – 50√3
2x= 5o(3 – √3)
x = 25(3 – √3)m.
Speed of the boat = 25(3 – √3) m/mm.
= \(\frac{25(3-\sqrt{3}) \times 60}{1000}\) km/h
= \(\frac{3}{2}\)(3 – √3) km/h
= 19.02 km/h
Section – C [4 Marks Each]
Question 11.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in fig. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.
Answer:
Total surface Area of article
= CSA of cylinder + CSA of 2 hemispheres
CSA of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 10
= 220 cm2
Surface area of two hemispherical scoops
= 4πr
= 4 × \(\frac{22}{7}\) × 3.5 × 3.5
= 154 cm2
∴ Total surface area of article = 220 + 154
= 374 cm2
Commonly Made Error:
Sometimes, students subtract the area of hemisphere from T.S.A. of cylinder in place of adding these.
Answering Tip:
They should read the question clearly and use right formula and correct calculation for which ample practice is necessary.
Question 12.
If the radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of one circle which is tangent to the other circle.
OR
In Fig. O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Answer:
Here, radii of two concentric circles,
OC = 4 cm and OB = 5 cm
In right angled ∆OBC, ∠C = 90°
CO2 + BC2 = OB2
(Using Pythagoras theorem)
42 + BC2 = 52
16 + BC2 = 25
BC2 = 25 – 16
BC2 = 9
BC = 3 cm
Since, AB is the chord and OC ⊥ AB, that divides AB in two equal parts
Hence, AB = 2 × BC
= 2 × 3
= 6cm
OR
PT= \(\sqrt{169-25}\) = 12cm
TE = OT – OE = 13 – 5
= 8cm
Let PA = AE = x. (Tangents)
Then, TA2 = TE2 + EA2
or, (12 – x)2 = 82 + x2
24x = 80
or, x = 33 cm (Approx)
Thus AB = 2 × x = 2 × 3.3
= 6.6 cm. (Approx.)
Case Study-1
Question 13.
A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kings way), is about 138 feet (42 metres) in height.
(i) What is the angle of elevation if they are standing at a distance of 42 m away from the monument? [2]
(ii) They want to see the tower at an angle of 60°. So, they want to know the distance where they should stand and hence find the distance. [2]
Answer:
(i) Height of India gate = 42 m
Distance between students and Indian gate = 42 m
Now, ∆ABC, ∠B = 90°
tan θ = \(\frac{A B}{B C}\)
tan θ = \(\frac{42}{42}\)
tan θ = 1
tan θ = tan 45°
θ = 45°
Hence, angle of elevation = 45°.
(ii) Height of India gate = 42 m
Angle = 60°
Let the distance between students and India gate = x m.
Now, In ∆ABC, ∠B = 90°
Case Study-2
Question 14.
Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
(i) Write in AP for the given situation.
Answer:
a = 51
d = -2
For an AP) the common difference of each term will be same.
First term (a) = 51 seconds
Common difference (d) = 2 seconds
AP = 51, 49, 47……..
(ii) What is the minimum number of days he needs to practice till his goal is achieved?
Answer:
Goal = 31 second
n = number of days
∴ an = 31
a + (n – 1)d = 31
51 + (n – 1)(-2) =31
51 – 2n + 2 =31
-2n = 31 – 53
-2n = -22
n = 11.