CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 5 with Solutions

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 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.
Solve the following quadratic equation for x:
4x² – 4a²x + (a⁴ – b⁴) = 0
OR
Find the value(s) of k for which the quadratic equation x² + 2√2kx + 18 = 0 has equal roots.
Answer:
Given, 4x² – 4a²x + (a⁴ – b⁴) = 0
Comparing with Ax² + Bx + C = 0, we get
Here, A = 4, B = -4a² and C = (a⁴ – b⁴)

OR
Since, x² + 2√2kx + 18 = 0 has equal roots then discriminant, D = 0
Comparing x² + 2√2kx + 18 = 0 with ax² + bx + c = 0,
Now, 0 = 1, b = 2√2 k and c = 18
D = 0
⇒ b² = 4ac
(2√2k)² = 4 × 1 × 18
4 × 2 × k² = 72
k² = \(\frac{72}{8}\)
k² = 9 ⇒ k = √9
k = ±3.

Question 2.
Show that (a – b)², (a² + b²) and (a + b)² are in A.E .
OR
Which term of the A.P 3, 15, 27, 39,… will be 120 more than its 21st term ?
Answer:
Given, (a – b)², (a² + b²) and (a + b)²
Common difference,
d₁ = (a² + b²) – (a – b)²
= a² + b² – (a² + b² – 2ab)
= a² + b² – a² – b² + 2ab = 2ab
and d₂ = (a + b)² – (a² + b²)
= a² + b² + 2ab – a² – b²
= 2ab
Since, d₁ = d₂
Hence, (a – b)², (a² + b²) and (a + b)² are in an
A.P. Hence Proved.
OR
a_n = a₂₁ + 120
= (3 + 20 × 12) + 120
= 363
∴ 363 = 3 + (n – 1) × 12
⇒ n = 31
or 31st term is 120 more than a₂₁.

Detailed Solution:
Given A.P. is: 3, 15, 27, 39
Here, first term, a = 3 and common difference,
d = 12
Now, 21st term of A.P. is
t₂₁ = a + (21 – 1) d
t₂₁ = 3 + 20 × 12 = 243

Therefore, 21st term is 243
We need to calculate term which is 120 more than 21st term
i.e., it should be 243 + 120 = 363
Therefore, t_n = 363
∴ t_n = a + (n -1 )d
⇒ 363 = 3 + (n -1)12
⇒ 360 = 12(n – 1)
⇒ n – 1 = 30
⇒ n = 31
So, 31st term is 120 more than 21st term.

Question 3.
In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If radius of in circie is 10 cm, tìn find the value of x.

Answer:
∠A = ∠OPA = ∠OSA = 90°
Hence, ∠SOP = 90°
Also, AP = AS
Hence, OSAP is a square.
AP = AS = 10 cm
CR = CQ = 27 cm
BQ = BC – CQ
= 38 – 27- 11 cm
BP = BQ = n cm
x = AB = AP + BP
= 10 + 11 = 21 cm

Detailed Solution:
With O as centre, draw a perpendicular OP on AB.
Now, in quadrilateral APOS,

∠SAP = 90° (Given)
∠APO = 90° (By construction) and
∠ASO = 90° (Angle between tangent and radius)
Finally ∠SOP = 360° – (90° + 90° + 90°)
= 90° ft
AP = AS (Tangents from external point A)
∴ OSAP is a square.
AP = AS = SO = 10 cm
∵ CR = CQ (Tangents from external point C)
CR = CQ = 27 cm
But BC = 38 cm (Given)
∴ BQ = BC – CQ = (38 – 27) cm
BQ = 11 cm
BP = BQ (Tangent from external point B)
∴ BP = 11 cm
So, x = AB = AP + PB
= (10 + 11) cm = 21 cm
Hence, the value of x is 21 cm.

Commonly Made Error:
Some students do not use appropriate figure to solve the question.

Answering Tip:
Carefully read the question and draw the figure as per the required condition.

Question 4.
Solve for x:
x² + 5x – (a² + a – 6) = 0
Answer:
Given,
x² + 5x – (a² + a – 6) = 0

Detailed Solution:
x² + 5x – (a² + a – 6) =0
⇒ x² + 5x – (a² + 3a – 2a – 6) = 0
⇒ x² + 5x – [a(a + 3) – 2(a + 3)] = 0
⇒ x² + 5x – (a + 3)(a – 2) = 0
⇒ x² + [(a + 3) – (a – 2)]x – (a + 3){a – 2) = 0
⇒ x² + (a + 3)x -(a- 2)x – (a + 3 )(a – 2) = 0
⇒ x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0
⇒ [x + (a + 3)] [x – (a – 2)] = 0
⇒ x = -(a + 3) or x = a – 2
Hence, roots of given equation are – (a + 3) and a – 2.

Question 5.
The \(\frac{3}{4}\) th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer:
Radius of conical vessel = 5 cm and its height = 24 cm
Volume of cone vessel = \(\frac{1}{3}\)πR²H
= \(\frac{1}{3}\) × π × 5 × 5 × 24
= 200 πcm³
Internal radius of cylindrical vessel = 10 cm

Let the height of emptlèd water be h.
∴ Volume of water in cylmder
= \(\frac{3}{4}\) × Volume ofcone
⇒ πr²h = \(\frac{3}{4}\) × Vohime of cone
⇒ π × 10 × 10 × h = 150π
⇒ h = 15cm
Hence the height of water = 15cm

Question 6.
The mean of the following frequency distribution is 25. Find the value of p.

Answer:

Mean, x̄ = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
⇒ 25 = \(\frac{570+45 p}{26+p}\)
⇒ 650 + 25p = 570 + 45p
⇒ 650 – 570 = 45p – 25p
∴ p = 4

Section – B [3 Marks Each]

Question 7.
The marks obtained by 100 students in an examination are given below :

Find the median marks of the students.
Answer:

Marks	Number of students (f)	c.f
30-35	14	14
35-40	16	30
40-45	28	58
45-50	23	81
50-55	18	99
55-60	8	107
60-65	3	110
	N = Σf = 110

\(\frac{N}{2}\) = 55
The 55th term lies in class 40 – 45.
l = 40, h = 5, f = 28, \(\frac{N}{2}\) = 55, c.f = 30
We know,
Since,Median = l + \(\left(\frac{\frac{N}{2}-c . f}{f}\right)\) × h
= 40 + \(\left(\frac{55-30}{28}\right)\) × 5
= 40 + \(\frac{25}{28}\) × 5
= 40 + 4.46 = 44.46 (Approx)

Question 8.
Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm.
From P, draw two tangents to the circle.
Answer:
Steps of construction :

1. Draw a line segment OP = 6.5 cm.
2. Taking O as centre and radius 2 cm, draw a circle.
3. Draw perpendicular bisector of OP, intersect at OP at S.
4. Taking S as centre draw another circle which intersects the first circle at Q and R.
5. Join P to Q and P to R.
   Hence PQ and PR are two tangents.

Question 9.
Find the mode of the following distribution of marks obtained by the students in an examination :

Given the mean of the above distribution is 53, using empirical relationship estimate the value of its median.
Answer:
Modal class = 60 – 80
∴ Mode = l + \(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\) × h
Here, l = 60, f₁ = 29, f₀ = 21, f₂ = 17 and h = 20
Mode = 60+ \(\frac{29-21}{2 \times 29-21-17}\) × 20
= 60 + \(\frac{8}{58-38}\) × 20
= 60 + 8 = 68

Empirical relationship,
Mode = 3 median – 2 mean
Mode = 68 and mean = 53 (given)
∴ 3 Median = Mode + 2 Mean
3Median = 68 + 2 × 53
Median = \(\frac{174}{2}\) = 58
Hence, Median = 58.

Question 10.
A man on the top of a vertical observation tower observes a car moving at uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point ?
OR
The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the hill is 30°. If tower is 50 metre high, find the height of the hill.
Answer:
Let the speed of car by x m/minute
In ∆ABC, ∠B = 90°
\(\frac{h}{y}\) = tan 45°
h = y ………(i)
∠B = 90°
\(\frac{1}{y+12 x}\) = tan 30°
⇒ h√3 = y + 12x

y = 6x(√3 + 1)
Hence, time taken froM C to B 6 (√3 + 1) minutes.

OR

Let AB = 50 m be the height of the tower and CD be the height of hill
Now, in ∆ABC,
∠ABC = 90°

Or, BC = 50√3 m
Again in ∆BCD, ∠BCD = 90°
tan 60°= \(\frac{D C}{B C}\)
DC = BC tan 60°
= 50√3 × √3 m
DC = 150m
∴ The height of hill is 150 m.

Commonly Made Error .
The concept of angle of depression is not dear to many students. That why they are not able to draw the diagram correctly.

Answering Tip:
The concept of angle of depression and angle of elevation must be clear to the students.

Section – C [4 Marks Each]

Question 11.
Rampal decided to donate canvas for 10 tents conical in shape with base diameter 14 m and height 24 m to a centre for handicapped person’s welfare. If the cost of 2 m wide canvas is ₹ 40 per metre, find the amount by which Rampal helped the centre.
Answer:
Diameter of tent = 14 m and height = 24 m
∴ radius of tent = 7 m
Slant height = \(\sqrt{h^{2}+r^{2}}=\sqrt{24^{2}+7^{2}}\)
= \(\sqrt{576+49}\) = 25 m

Surface area of the tent = πrl
= \(\frac{22}{7}\) × 7 × 25
= 550m²
Surface area of 10 tents = 550 × 10
Total = 5500 × \(\frac{40}{2}\)
= ₹ 1,10,000

Hence, the amount by which Rampal helped the centre = 1,10,000

Question 12.
In the given figure, two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

a, b and c are the sides of a right triangle, where c is the hypotenuse. A circle, of radius r, touches the sides of the triangle. Prove that r = \(\frac{a+b-c}{2}\)
Answer:
Let ∠OPQ be θ, then
∠TPQ = 90° – θ
Since, TP = TQ
∴ ∠TQP = 90° – θ (opposite angles of equal sides)

Now, ∠TPQ + ∠TQP + ∠PTQ = 180° (Angle sum property of a Triangle)
⇒ 90° – θ + 90° – θ + ∠PTQ = 180°
⇒ ∠PTQ= 180°- 180° + 2θ
⇒ ∠PTQ = 2θ
Hence, ∠PTQ = 2∠OPQ
Hence Proved.

OR

Let circle touches CB at MICA at N and AB at P.
Now OM ⊥ CB and ON ⊥ AC
(radius ⊥ tangent)
OM = ON (radii)
CM = CN (Tangents)
∴ OMCN is a square.
Let OM = r = CM = CN
AN = AP, CW= CM and BM = BP (tangent from external point)
AN = AP,

⇒ AC = CN = AB – BP
b – r = c – BM
b – r = c – (a – r)
17 – r = c – a + r
2r = a + b – c
r = \(\frac{a+b-c}{2}\)
Hence Proved.

Case Study-1

Question 13.
A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka ,them being Nanda Devi(height 7,816 m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

(i) Find the distance of the satellite from the top of Nanda Devi. [2]
Answer:

(ii) Find the total distance of the satellite from the top of Mullayanagiri. [2]
Answer:

Case Study-2

Question 14.
Your elder brother wants to buy a car and plans to take loan from a bank for his car. Fie repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month, answer the following:

(i) Find the amount paid by him in 30th installment. [2]
Answer:
First term a = 1000
Common difference d = 100
30th term, a₃₀ = a + (n – 1)d
= 1000 +(30 – 1)100
= 1000 + 2900
= ₹ 3900

(ii) The amount paid by him in the 30 installments. [2]
Answer:
Sum of 30 installments
=\(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{30}{2}\)[2 × 1000 + (30 – 1)100]
= 15[2000 + 2900]
= 15 × 4900
= ₹ 73500
Total amount paid in 30 installments = ₹ 73500