CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A-E.
Section A has 20 MCQs carrying 1 mark each
Section B has 5 questions carrying 02 marks each.
Section C has 6 questions carrying 03 marks each.
Section D has 4 questions carrying 05 marks each.
Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section – A
(Section A consists of 20 questions of 1 mark each.)

Question 1.
Let a and b be two positive integers such that a = p³q⁴ and b = p²q³ , where p and q are prime numbers. If HCF(a, b) = p^mqⁿ and LCM(a, b) = pr(f, then (m + n)(r + s) = (1)
(a) 15
(b) 30
(c) 35
(d) 72 1
Answer:
(c) 35

Explanation:
Given,
a = p³q⁴
and b = p²q³
HCF of (ab) = p²q³
p^mqⁿ = p²q³ .
⇒ m = 2
⇒ n = 3

LCM of (a, b) = p³q⁴
p^rq^s = p³q⁴
⇒ r= 3
⇒ s = 4
(m + n)(r + s) = (2 + 3)(3 + 4)
⇒ 5 × 7
= 35

Question 2.
Let p be a prime number. The quadratic equation having its roots as factors of p is: (1)
(a) x² – px + p = 0
(b) x² – (p + 1)x + p = 0
(c) x² + (p + 1)x + p = 0
(d) x² – px + p + 1 = 0
Answer:
(b) x² – (p + 1)x + p = 0

Explanation:
If p is prime number, then it’s factors are 1 and p itself.
And it’s roots, a and p are it’s factors
Hence α = 1, β = p
x = α or β
0 = (x – α)(x – β)
Which means = (x – 1)(x – p)
= x²+ (-1- p)x + (-1 × -p)
⇒ x² – (p + 1)x + p is the required quadratic equation

Question 3.
If a and p are the zeros of a polynomial f(x) = px² – 2x + 3p and α + p = αp, then p is (1)
(a) \(\frac{-2}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{-1}{3}\)
Answer:
(b) \(\frac{2}{3}\)

Explanation:
Given, f(x) = px² – 2x + 3p
Compare it with standard equation ax² + bx + c = 0, we get
a = p, b = -2, c = 3p
Sum of zeros = α + β = –\(\frac{b}{3}\)
α + β = \(\frac{2}{p}\)
Product of zeros = αβ = \(\frac{c}{a}\)
αβ = \(\frac{c}{a}\)
αβ = \(\frac{3p}{p}\) = 3

It is given that, α + β = αβ
\(\frac{2}{p}\) = 3
p = \(\frac{2}{3}\)

Question 4.
If the system of equations 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1 is inconsistent, then k = (1)
(a) -1
(b) 0
(c) 1
(d) 2
Answer:
(d) 2

Explanation:
Given,
3x + y = 1 ……..(i)
(2k – 1)x + (k – 1)y = 2k + 1 …….(ii)

Both the equation are in the from of ax + by + c = 0
a₁ = 3, b₁ = 1, c₁ = 1
a₂ = (2k – 1), b₂ = (k – 1), c₂ = 2k + 1

The system of equation is inconsistent, if
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
∴ \(\frac{3}{2k – 1}=\frac{1}{k – 1} \neq \frac{1}{2k + 1}\)
\(\frac{3}{2k – 1}=\frac{1}{k – 1}\)
3k – 3 = 2k – 1
k = 2

Caution: While comparing the given equations with standard equation, we should also consider the signs of constant

Question 5.
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are: (1)
(a) (-2, -1)
(b) (-2, -3)
(c) (2, -1)
(d) (1,2)
Answer:
(c) (2, -1)

Explanation:
Since, the diagonals of a parallelogram bisect each other.
Let the coordinate of S is (x, y)
∴ Coordinates of the mid point of PR = coordinates of the mid point of QS
⇒ \(\left(\frac{3+(-3)}{2}, \frac{4-2}{2}\right)=\left(\frac{-2+x}{2}, \frac{3+y}{2}\right)\)
(0, 1) = \(\left(\frac{x-2}{2}, \frac{3+y}{2}\right)\)
\(\frac{x-2}{2}\) = 0 and \(\frac{3+y}{2}\) = 1
⇒ x – 2 = 0
x= 2
and 3 + y = 2
y = -1
Hence, the fourth vertex of the parallelogram is (2, -1)

Question 6.
∆ABC – ∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and AB²: PQ² = 4: 9, then AM : PN = (1)
(a) 3 : 2
(b) 16 : 81
(c) 4: 9
(d) 2 : 3
Answer:
(d) 2 : 3

Explanation:
We Know that the ratio the sides of the two similar triangles is equal to the ratio of their corresponding sides.
Therefore, (1)
\(\frac{AB}{PN}\) = \(\frac{AB}{PQ}\)
\(\frac{AB}{PN}\) = \(\sqrt{\frac{4}{9}}\)
\(\frac{AB}{PN}\) = \(\frac{2}{3}\)

Find AB: PQ,
So that,
AM : PN = 2 : 3
Hence, the ratio of AM to PN is 2 : 3.

Question 7.
If x tan 60°cos 60° = sin60°cot 60°, then x = (1)
(a) cos 30°
(b) tan 30°
(c) sin 30°
(d) cot 30°
Answer:
(b) tan30°

Explanation:
x tan 60° cos 60° = sin 60° cot 60°
\((x) \times(\sqrt{3}) \times \frac{1}{2}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}\)
\(\frac{\sqrt{3} x}{2}=\frac{1}{2}\)
\(x=\frac{1}{\sqrt{3}}\) = tan30°

Question 8.
If sin θ + cos θ = \(\sqrt{2}\) , then tan θ + cot θ = (1)
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Explanation:
Given, sin θ + cos θ = \(\sqrt{2}\)
On squaring both the sides, we get

From (i) and (ii) we get,
tan θ + cot θ = 2

Question 9.
In the given figure, DE || BC, AE = a units, EC = b units, DE = x unit and BC = y units. Which of the following is true? (1)

(a) \(x=\frac{a + b}{ay}\)
(b) \(y=\frac{ax}{a+b}\)
(c) \(x=\frac{ay}{a+b}\)
(d) \(\frac{x}{y}=\frac{a}{b}\)
Answer:
(c) \(x=\frac{ay}{a+b}\)

Explanation:
Given AE = a, EC = b, DE = x and BC = y.
As DE || BC
∠ADE = ∠ABC
∠AED = ∠ACB
So by AAA property,
∆ADE ~ ∆ABC
\(\frac{DE}{BC} = \frac{AE}{AC}\)
\(\frac{x}{y} = \frac{a}{a+b}\)
\(x = \frac{ay}{a+b}\)

Question 10.
ABCD is a trapezium with AD || BC and AD = 4cm. If the diagonals AC and BD intersect each other at O such that = \(\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}\) then BC = (1)
(a) 6 cm
(b) 7 cm
(c) 8 cm
(d) 9 cm
Answer:
(c) 8 cm

Explanation:
Given ABCD is a trapezium in which BC || AD and AD = 4 cm.
Also, the diagonals AC and BD intersect at O such that \(\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}\)

In ∆AOD and ∆COB,
∠OAD = ∠OCB [alternate angles]
∠ODA = ∠OBC [alternate angles]
∠AOD = ∠BOC [vertically opposite angles]
We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
∴ ∆AOD ~ ∆COB
We know that two triangles are similar if their corresponding sides are proportional.
\(\frac{AO}{OC} = \frac{DO}{OB} = \frac{AD}{BC}\)
⇒ \(\frac{1}{2} = \frac{AD}{BC}\)
⇒ \(\frac{1}{2} = \frac{4}{BC}\)
∴ BC = 8 cm

Question 11.
If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is equal to (1)
(a) \(\frac{3 \sqrt{3}}{2}\) cm
(b) 3 cm
(c) 6 cm
(d) \(3 \sqrt{3} \) cm
Answer:
(d) \(3 \sqrt{3} \) cm

Explanation:
Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°

Radius of the circle = 3 cm
Join OA and OP.
Also, OP is bisector line of ∠APC
∴ ∠APO = ∠CPO = 30°
OA ⊥ AP
Also, tangents at any point of a circle is perpendicular to the radius through the point of contact.
In right angled ∆OAP, we have
tan 30° = \(\frac{OA}{AP}\) = \(\frac{3}{AP}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{3}{AP}\)
⇒ AP = \(3 \sqrt{3} \)
AP =CP
= \(3 \sqrt{3} \) cm
[Tangents drawn from an external point are equal]
Hence, the length of each tangent is \(3 \sqrt{3} \) cm.

Caution: Remember that the point where tangent touches the circle is perpendicular to the radius.

Question 12.
The area of the circle that can be inscribed in a square of 6cm is: (1)
(a) 36π cm²
(b) 18π cm²
(c) 12π cm²
(d) 9π cm²
Answer:
(d) 9π cm²

Explanation:
Diameter of circle = side of square = 6 cm
Radius = \(\frac{6}{2}\)
Radius = 3 cm
Area of circle = πr²
π(3)²
= 9π
Therefore, the area of the circle is 9π square cm.

Question 13.
The sum of the length, breadth and height of a cuboid is \(6 \sqrt{3} \) cm and the length of its diagonal is \(2 \sqrt{3} \) cm. The total surface area Qf the cuboid is: (1)
(a) 48 cm²
(b) 72 cm²
(c) 96 cm²
(d) 108 cm²
Answer:
(c) 96 cm²

Explanation:
According to condition,
\(\sqrt{\left(l^2+b^2+h^2\right)}=2 \sqrt{3}\)
⇒ l² + b² + h² = 12
Again, 1 + b + h = 6V3
⇒ (l + b + h)² = 108
⇒ l² + b² + h² + 2 (lb + bh + h[) = 108
⇒ 2 (lb + bh + hl)= 108
⇒ 2 (lb + bh + hl) = 96
∴ The total surface area of cuboid = 96 cm2

Question 14.
If the difference of Mode and Median of a data is 24, then the difference of median and mean is: (1)
(a) 8
(b) 12
(c) 24
(d) 36 1
Answer:
(b) 12

Explanation:
Since, Mode = 3 Median – 2 Mean
⇒ Mode = 2 Median + Median – 2 Mean
⇒ Mode – Median = 2 Median – 2 Mean
⇒ Mode – Median = 2 (Median – Mean)
Given, Difference of mode and median of a data is 24
24 = 2 (Median – Mean)
⇒ Median – Mean = 12

Question 15.
The number of revolutions made by a circular wheel of radius 0.25m in rolling a distance of 11km is:
(a) 2800
(b) 4000
(c) 5500
(d) 7000
Answer:
(d) 7000

Explanation:
Given radius = 0.25m.
Distance covered in 1 revolution = 2πr
= 2 × \(\frac{22}{7}\) × 0.25
= \(\frac{11}{7}\)
Given total distance = 11km = 11000m.
Then the number of revolutions = 11000 × \(\frac{11}{7}\)
= \(\frac{77000}{11}\)
= 7000
Therefore the number of revolutions = 7000.

Caution: Remember that circumference of a wheel is the distance covered by the wheel in one round.

Question 16.
For the following distributions, (1)

Class	Frequency
0-5	10
5-10	15
10-15	12
15-20	20
20-25	9

the sum of the lower limits of the median and model class is:
(a) 15
(b) 25
(c) 30
(d) 35
Answer:
(b) 25
Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Here N = 66
∴ \(\frac{N}{2}\) = 33, which lies in the interval 10 – 15.
So, the lower limit of the median class is 10. The highest frequency is 20, which lies in the interval 15 – 20.
Therefore, the lower limit of modal class is 15. So, the required sum is 10 + 15 = 25.

Question 17.
Two dice are rolled simultaneously. What is the probability that 6 will come up at least once? (1)
(a) \(\frac{1}{6}\)
(b) \(\frac{7}{36}\)
(c) \(\frac{11}{36}\)
(d) \(\frac{13}{36}\)
Answer:
(c) \(\frac{11}{36}\)
Explanation: Outcomes when 5 will come up at least once are:
(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4) and (6,5)
Probability that 5 will come up at least once = \(\frac{11}{36}\)

Question 18.
If 5 tan β = 4, then \(\frac{5 \sin \beta-2 \cos \beta}{5 \sin \beta+2 \cos \beta}\) = (1)
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) 6
Answer:
(a) \(\frac{1}{3}\)

Explanation:
Given 5 tan β = 4

Direction: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Statement A (Assertion): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340
Statement R (Reason): HCF is always a factor of LCM (1)
Answer:
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

Explanation:
We know that,
HCF × LCM = Product of the numbers
⇒ 17 × LCM = 5780
⇒ LCM = \(\frac{5780}{17}\)
⇒ LCM = 340
The LCM of the two numbers = 340

Question 20.
Statement A (Assertion): If the co-ordinates of the mid-points of the sides AB and AC of ∆ABC are D(3, 5) and E(-3, -3) respectively, then BC = 20 units
Statement R (Reason): The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it. (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
M and N are the mid-point of AB and AC.
∴ MN || BC (mid-point theorem)
∴ In ∆AMN and ∆ABC
∠A = ∠A (common)
∠AMN = ∠ABC (corresponding angles)
∠ANM = ∠ACB (corresponding angles)
∴ ∆AMN ~ ∆ABC (AAA)
∴ \(\frac{AM}{AB} = \frac{MN}{BC} = \frac{AN}{NC} = \frac{1}{2}\)
∴ \(\frac{MN}{BC}\) = \(\frac{1}{2}\)
BC = 2MN
∴ MN = \(\sqrt{(-3-3)^2+(-3-5)^2}\)
= \( \sqrt{36 + 64} \) = 10
∴ BC = 2 × 10 = 20

SECTION – B
(Section B consists of 5 questions of 2 marks each.)

Question 21.
If 49x + 51y = 499, 51x + 49y = 501, then find the value of x and y (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Add and subtract equations to simplify (1m) Calculate x (0.5m) Calculate y (0.5m)”

Answer:
Adding the two equations and dividing by 10,
we get: x + y = 10
Subtracting the two equations and dividing by -2,
we get: x – y =1
Solving these two new equations, we get.
x = \(\frac{11}{2}\)
y = \(\frac{9}{2}\)

Detailed Answer:
The given equations are:
49x + 51y = 499 …(i)
51x + 49y = 501 …(ii)
On adding the equations, we get,
100x + 100y = 1000
Dividing by 100
⇒ x + y = 10 ….(iii)
On subtracting the equations, we get,
-2x + 2y = -2
Dividing by 2
⇒ -x + y = -1 …(iv)
Now, solving equation (iii) and equation (iv),
x + y = 10
-x + y = -1
On adding,
2y = 9
⇒ y = \(\frac{9}{2}\)
Substituting y = \(\frac{9}{2}\) in equation (iii)
⇒ x + \(\frac{9}{2}\) = 10
⇒ x = 10 – \(\frac{9}{2}\)
⇒ x = \(\frac{11}{2}\)
Therefore, the value of x is \(\frac{11}{2}\) and that of y is \(\frac{9}{2}\).

Caution: When we make the coefficients equal then check their signs carefully to decide whether to add or subtract the two equations.

Question 22.
In the given figure below, \(\frac{AD}{AE} = \frac{AC}{BD}\) and M a AE BD ∠1 = ∠2. Show that ∆BAE ~ ∆CAD. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Use AD/AE=AC/BD to prove AC/AB = AD/AE (1m) Use SAS similarity criterion to prove ∆BAE ~ ∆CAD (1m)

Answer:
In ∆ABC,
∠1 = ∠2
∴ AB = BD ……(i)
Given \(\frac{AD}{AE} = \frac{AC}{BD}\)
Using equation 0, we get \(\frac{AD}{AE} = \frac{AC}{AB}\) ……(ii)
AE AB „(ii)
In ∆BAE and ∆CAD, by equation (ii),
\(\frac{AC}{AB} = \frac{AD}{AE}\)
∠A = ∠A (common)
∴ ∆BAE ~ ∆CAD [By SAS similarity criterion]

Question 23.
In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB= 75°. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Apply angle sum property to find ∠AOB (1m) Use property of tangent to find ∆AQB (1m)

Answer:
∠PAO = ∠PBO = 90°
(angle b/w radius and tangent)
∠AOB = 105°
(By angle sum property of a triangle)
∠AQB = \(\frac{1}{2}\) × 105° = 52.5°
(Angle at the remaining part of the circle is half the angle subtended by the arc at the centre)

Question 24.
The length of the minute hand of a clock is 6cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Find the angle swept by the minute hand of the clock in 1 min (0.5m) Find the angle swept by the minute hand of the clock in 35 min (0.5m) Calculate the area swept by the minute hand of the clock (lm)

OR
In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Write the formula for required area with centre A, B, C and D (0.5m) Substitute the values (0.5m) Apply angle sum property of triangle (0.5m) Calculate the required area and get the desired results (0.5m)

Answer:
We know that, in 60 minutes, the tip of minute hand moves 360°

In 1 minute, it will move = \(\frac{360}{60}\) = 6°
∴ From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
∴ Area of swept by the minute hand in 35 min = Area of sector with sectorial angle 0 of 210° and radius of 6 cm
= \(\frac{210}{320}\) × π × 6²
= \(\frac{7}{12}\) × \(\frac{22}{7}\) × 6 × 6
= 66 cm²
OR
Let the measure of ∠A, ∠B, ∠C and ∠D be θ₁, θ₂, θ₃ and θ₄ respectively Required area = Area of sector with centre A + Area of sector with centre B + Area of sector with centre C + Area of sector with centre D

Question 25.
If sin(A + B) = 1 and cos(A – B) = \(\frac{\sqrt{3}}{2}\), 0°< A + B < 90° and A > B, then find the measures of angles A and B. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Form a pair of linear equations in two variables, namely A and B, from the given relations (0.5m) Add equations to simplify (0.5m) Calculate A (0.5m) Calculate B (0.5m)

OR
Find an acute angle 0 when
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Divide numerator and denominator by COS θ (1m) Simplify it and find an acute angle (1m)

Answer:
sin(A + B) = 1 = sin 90, so A + B = 90 ……. (i)
cos(A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30,
so A – B= 30 …(ii)
From (i) & (ii) ∠A = 60°
And ∠B = 30°
Detailed Answer:
We have sin(A + B) = 1
(A + B) = sin^-11
We know, sin^-1 1 = 90°
A + B = 90° ….(i)
cos(A – B) = \(\frac{\sqrt{3}}{2}\)
A – B = cos^-1 \(\frac{\sqrt{3}}{2}\)
we know cos^-1 \(\frac{\sqrt{3}}{2}\) = 30°
solve (i) and (ii)
A + B = 90°
A – B = 30°
add
2A = 120°
A = 60° and ∠B = 30°
OR
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Dividing the numerator and denominator of LHS by cos θ, we get
\(\frac{1-\tan \theta}{1+\tan \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Which on simplification (or comparison) gives
tan θ = \( \sqrt{3} \)
Or θ = 60°

SECTION – C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
Given that \( \sqrt{3} \) is irrational, prove that 5 + \(2 \sqrt{3} \) is irrational. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Suppose 5 + \(2 \sqrt{3} \) to be a rational number (0.5m) Write it in fraction form i.e., p/q, where p, q are co-primes. (1m) Find prime factors of p and q (0.5m) p and q will have common prime factors which contradict our assumption that p, q are co-primes and 5 + \(2 \sqrt{3} \) is an irrational number (1m)

Answer:
Let us assume 5 + \(2 \sqrt{3} \) is rational, then it must be in the form of \(\frac{p}{q}\) where p and q are coprime
integers and q ≠ 0
i.e 5 + \(2 \sqrt{3} \) = \(\frac{p}{q}\)
So \( \sqrt{3} \) = \(\frac{p-5q}{2q}\) ….(i)
Since p, q, 5 and 2 are integers and q ≠ oR,
HS of equation (i) is rational. But LHS of (i) is \( \sqrt{3} \) which is irrational.
This is not possible. This contradiction has arisen due to our wrong assumption that 5 + 2\( \sqrt{3} \)is rationaL So, 5+2\( \sqrt{3} \) is irrational.

Question 27.
If the zeroes of the polynomial x² + px + q are double in value to the zeroes of the polynomial 2x² – 5x – 3, then find the values of p and q. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Find the sum of zeroes of the polynomial (0.5m) Find the product of zeroes of the polynomial (0.5m) Find the sum of zeroes of the polynomial (0.5m) Calculate p (0.5m) Find the product of zeroes of the polynomial (0.5m) Calculate q (0.5m)

Answer:
Let a and p be the zeros of the polynomial 2x² – 5x – 3
Then α + β = \(\frac{5}{2}\)
And αβ = \(– \frac{3}{2}\)
Let 2α and 2β be the zeroes x² + px + q
Then 2α + 2β = -p
2(α + β) = -p
2 × \(\frac{5}{2}\) = -p
So P = -5
And 2α × 2β = q
4αβ = q
So q = 4 × \(– \frac{3}{2}\)
= -6

Question 28.
A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. Find the length of the journey.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the actual speed of the train and actual time taken as x km/h and y hr (0.5m) Form the pair of linear equations from the given situation (1m) Solve the two equations simultaneously to get the values of x and y (1m) Calculate the length of journey (0.5m)

OR
Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate off per chocolate, and got a total of ₹400. If he had sold the first lot at the rate of per chocolate, and the second lot at the rate of 4 for 5 chocolates, his total collection would have been ₹460. Find the total number of chocolates he had.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the number of chocolates in both lots in form of x and y (0.5m) Form the pair of linear equations from the given situation (1m) Solve the two equations simultaneously to get the values of x and y (1m) Calculate the total number of chocolates Anuj had (0.5m)

Answer:
Let the actual speed of the train be x km/hr and let the actual time taken be y hours Distance covered is xy km If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x + 6) km/hr, time of journey is (y – 4) hours.
∴ Distance covered = (x + 6) (y – 4)
⇒ xy = (x + 6)(y – 4)
⇒ -4x + 6y – 24 = 0
⇒ -2x + 3y – 12 = 0 ……(i)
Similarly xy = (x – 6)(y + 6)
⇒ 6x – 6y – 36 = 0
⇒ x – y – 6 = 0
Solving (i) and (ii) we get x = 30 and y = 24
Putting the values of x and y in equation (i), we obtain
Distance = (30 × 24) km = 720 km.
Hence, the length of the journey is 720 km.

Let the number of chocolates in lot A be x
And let the number of chocolates in lot B be y total number of chocolates = x + y
Price of 1 chocolate = = ₹ \(\frac{2}{3}\), so for x chocolates = \(\frac{2}{3}\) x and price of y chocolates at the rate of ₹ 1 per chocolate = y.
∴ by the given condition \(\frac{2}{3}\)x + y = 400
⇒ 2x + 3y = 1200
Similarly
x + \(\frac{4}{3}\) y = 460
⇒ 5x + 4y = 2300 …(ii)
Solving (i) and (ii) we get
x =300 and y = 200
∴ x + y = 300 + 200 = 500
So, Anuj had 500 chocolates.

Question 29.
Prove the following that (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Use trigonometric identities to start simplifying LHS (1m) Eventually reach LHS = RHS stage (2m)

Answer:

Question 30.
Prove that a parallelogram circumscribing a circle is a rhombus (3)

Total Marks

Breakdown
(As per CBSE Marking Scheme)

3m
(SA-II)

Form equations using properies of tangents with diagram (1m)

Use parallelogram property with above equations to achieve ‘equal sides’ (1.5m)

Mention the rhombus property and conclude (0.5m)

OR
In the figure XY and XT’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C interesting XY at A and X’Y’ at B, what is the measure of ∠AOB.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Do necessary construction in the figure (0.5m) Apply the properties of tangent (0.5m) Use SSS congruency criterion to prove ∆OPA ≅ ∆OCA (lm) Calculate the angle betweem the tangents (0.5m) Use angle sum property, find ∆AOB (0.5m)

Answer:
Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively.

We know that the tangents drawn to a circle from an exterior point are equal in length
∴ AP = AS ……..(i)
BP = BQ …(ii)
CR = CQ …(iii)
DR = DS …(iv)
Adding (i), (ii), (iii) and (iv) we get
AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC …(v)
Since AB = DC and AD = BC (opposite sides of parallelogram ABCD)
putting in (v) we get, 2AB = 2AD
or AB = AD.
∴ AB = BC = DC = AD
Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus
OR

Join OC
In ∆OPA and ∆OCA
OP = OC (radii of same circle)
PA = CA
(length of two tangents from an external point)
AO = AO (Common)
Therefore, ∆OPA ≅ ∆OCA
(By SSS congruency criterion)
Hence, ∠1 = ∠2 (CPCT)
Similarly
∠3 = ∠4
∠PAB + ∠QBA = 180°
(co interior angles are supplementary as XY || X’Y)
2∠2 + 2∠4 = 180°
∠2 + ∠4 = 90° …(1)
∠2 + ∠4 + ∠AOB = 180° (Angle sum property)
Using (i), we get, ∠AOB = 90°

Question 31.
Two coins are tossed simultaneously. What is the probability of getting
(A) At least one head?
(B) At most one tail?
(C) A head and a tail? (3)
Answer:
(A) P (At least one head) = \(\frac{3}{4}\)
(C) P(A head and a tail) = \(\frac{3}{4}\)
(B) P(At most one tail) =\(\frac{2}{4}\) = \(\frac{1}{2}\)

SECTION – D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Identify co-relation in time taken by both pipes (0.5m) Form a quadratic equation in x (2m) Solve it to find the values of x (1m) Pick the positive/realistic value of x (0.5m) Calculate the time taken by larger pipe and smaller pipe (1m)

OR
In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr from its usual speed and the time of the flight increased by 30 min. Find the scheduled duration of the flight. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Assuming the usual speed of plane and the reduced speed of the plane as x km/hr and (x-200) km/hr hours (0.5m) Form a quadratic equation in x (2m) Solve it to find the values of x (1m) Pick the positive/realistic value of x (0.5m) Calculate scheduled duration of the flight (1m)

Answer:
Let the time taken by larger pipe alone to fill the tank = x hours
Therefore, the time taken by the smaller pipe = x + 10 hours
Water filled by larger pipe running for 4 hours = \(\frac{4}{x}\) litres
Water filled by smaller pipe running for 9 hours = \(\frac{9}{x+10}\) litres
We know that, \(\frac{4}{x}\) + \(\frac{9}{x+10}\) = \(\frac{1}{2}\)

Which on simplification gives:
x² – 16x – 80 = 0
x² – 20x + 4x – 80 =0
x(x – 20) + 4(x – 20) = 0
(x + 4)(x – 20) = 0
x = -4, 20
x cannot be negative.
Thus, x = 20
x + 10 = 30
Larger pipe would alone fill the tank in 20 hours and smaller pipe would fill the tank alone in 30 hours.

Let the usual speed of plane be x km/hr and the reduced speed of the plane be (x – 200) km/hr
Distance = 600 km [Given]
According to the question,
(time taken at reduced speed) – (Schedule time) = 30 minutes = 0.5 hours
\(\frac{600}{x-200}\) + \(\frac{600}{x}\) = \(\frac{1}{2}\)

Which on simplification gives:
x² – 200x – 240000 = 0
x² – 600x + 400x – 240000 = 0
x(x – 600) + 400(x – 600) = 0
(x – 600)(x + 400) = 0
x= 600 or x= -400
But speed cannot be negative.
∴ The usual speed is 600 km/hr and the scheduled duration of the flight is \(\frac{600}{600}\) = 1 hour

Question 33.
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Draw a figure with necessary construction (1m) Use AA similarity criterion to prove ∆ADE ~ ∆ABC (1m) Apply basic proportionality theorem to prove AD/ DB = AE/ EC (1.5m) Apply basic proportionality theorem to prove DE/EA= CF/FB (1.5m)

Answer:
For the Theorem :
Given, To prove, Construction and figure
Proof

Let ABCD be a trapezium DC || AB and EF is a line parallel to AB and hence to DC.
To prove : \(\frac{DE}{EA}\) = \(\frac{CF}{FB}\)
Construction : Join AC, meeting EF in G.
Proof: In ∆ABC, we have
GF || AB
\(\frac{CG}{GA}\) = \(\frac{CF}{FB}\) [By BPT] …(i)
In ∆ADC, we have
EG || DC (EF || AB & AB || DC)
\(\frac{DE}{EA}\) = \(\frac{CG}{GA}\) [By BPT] …(ii)
From (i) & (ii), we get,
\(\frac{DE}{EA}\) = \(\frac{CF}{FB}\)

Detailed Answer:
We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

We need to prove that \(\frac{AD}{AE}\) = \(\frac{DB}{EC}\)
Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.
Now, area of ∆ADE (\(\frac{1}{2}\) base × height) = \(\frac{1}{2}\) AD × EN. area of AADE is denoted as ar(ADE).
So, ar(ADE) = \(\frac{1}{2}\) AD × EN

Note that ABDE and ADEC are on the same base DE and between the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) …(iii)
Therefore, form (i), (ii) and (iii), we have:
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
Let ABCD be a trapezium with AB||DC
AD and BC are non-parallel sides.
Let, EF be a line parallel to AB, DC.
Join AC such that ∆ACD and ∆ACB are triangles and AC meets EF at G.

Apply Thales theorem on ∆ACD and ∆ACB Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side to intersect the other sides distinct points divides two sides in same ratio.
In ∆ACD
Here, EG||DC. According to Basic Proportionality Theorem
\(\frac{AE}{ED}\) = \(\frac{AG}{GC}\) ……….(i)

In ∆ACB
Here, GF||AB. According to Basic Proportionality Theorem
\(\frac{EC}{BF}\) = \(\frac{AG}{GC}\)
⇒ \(\frac{BF}{FC}\) = \(\frac{AG}{GC}\)
Equating equation (i) and equation (ii)
\(\frac{AE}{ED}\) = \(\frac{BF}{FC}\)
Hence proved that a line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

Question 34.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively decided to provide place and the canvas for 1500 tents and share the whole expenditure equally. The lower part of each tent is cylindrical with base radius 2.8 m and height 3.5 m and the upper part is conical with the same base radius, but of height 2.1 m. If the canvas used to make the tents costs ?120 per m², find the amount shared by each school to set up the tents. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Write values and slant height of conical part formula (1m) Calculate height (1m) Deduct Area of Canvas formula and find value (2m) Use unit cost to calculate final value (1m)

OR
There are two identical solid cubical boxes of side 7cm. From the top face of the first cube a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. and
(A) the ratio of the total surface area of the two new solid formed
(B) volume of each new solid formed. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Calculate the surface area new solid of first (1m) Calculate the surface area of second new solid (1m) Find the ratio of the total surface area of two new solids (1m) (B) Calculate the volume for first new solid (1m) Calculate the volume for second new solid (1m)

Answer:
Radius of the base of cylinder (r) = 2.8 m = Radius of the base of the cone (r)
Height of the cylinder (h) = 3.5 m
Height of the cone (H) = 2.1 m.
Slant height of conical part (l) = \( \sqrt{r^{2} + H^{2}} \)
= \( \sqrt{2.8^{2} + 2.1^{2}} \)
= \( \sqrt{7.84 + 4.41} \)
= \( \sqrt{12.25} \) = 3.5 m

Area of canvas used to make tent = CSA of cylinder + CSA of cone
2 × π × 2.8 × 3.5 + π × 2.8 × 3.5
= 61.6 + 30.8 = 92.4 m²
Cost of 1500 tents at 120 per sq.m = 1500 × 120 × 92.4 = 16,632,000

Share of each school to set up the tents \(\frac{16632000}{50}\)
= ₹ 332,640

(A) SA for first new solid (S₁):
6 × 7 × 7 + 2π × 3.52 – π × 3.5²
= 294 + 77 – 38.5 = 332.5 cm²

SA for second new solid (S₂):
6 × 7 × 7 + 2 π × 3.5² – π × 3.5²
= 294 + 77 – 38.5 = 332.5 cm²
So S₁ : S₂ = 1 : 1

(B) Volume for first new solid (V₁)
= 7 × 7 × 7 – \(\frac{2}{3}\) π × 3.5³
= 343 – \(\frac{539}{6}\)
= \(\frac{1519}{6}\) cm³

Volume for second new solid (V₂)
= 7 × 7 × 7 + \(\frac{2}{3}\) π × 3.5³
= 343 + \(\frac{539}{6}\) = \(\frac{2597}{6}\) cm³

Question 35.
The median of the following data is 525. Rnd the values ofx and y, if the total frequency is 100 (5)

Class interval	Frequency
0-100	2
100-200	5
200-300	x
300-400	12
400-500	17
500-600	20
600-700	y
700-800	9
800-900	7
900-1000	4

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Make frequency distribution table with all given data (2m) Using the sum of frequency, make an equation with variables x and y (0.5m) Apply the formula of median to form another equation with variables x and y (1.5m) Solve the two equations to find the values of x and y (1m)

Median = 525, so Median Class = 500 – 600

Total Marks

Breakdown
(As per CBSE Marking Scheme)

5m
(LA)

(A) Apply the distance formula (0.5m)
(B) Apply the mid-point formula (0.5m)
Find the coordinates of Z (0.5m)
(C) Equate the distances of the point (x,y) from two points (0.5m)
Apply the distance formula (0.5m)
Get the coordinates of Z on y-axis (1m)

Find the length of GH and AF (1m)
Calculate the area of trapezium AFGH (1m)

76 + x + y = 100
x + y = 24
Median = \(l+\frac{\frac{n}{2}-c f}{f} \times h\)
Since, l = 500, h = 100, f = 20, cf = 36 + x and n = 100
Therefore, putting the value in the Median formula, we get; 525 = \(\frac{\frac{50-(36+x)}{20}} \times 100\)
so x = 9, y = 24 – x [from eq. (i)]
y = 24 – 9 = 15
Therefore, the value ofx = 9 and y = 15.

SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
A tiling or tessellation of a flat surface is the covering of a plane using one or more geometric shapes, called tiles, with no overlaps and no gaps. Historically, tessellations were used in ancient Rome and in Islamic art. You may find tessellation patterns on floors, walls, paintings etc. Shown below is a tiled floor in the archaeological Museum of Seville, made using squares, triangles and hexagons.

A craftsman thought of making a floor pattern after being inspired by the above design. To ensure accuracy in his work, he made the pattern on the Cartesian plane. He used regular octagons, squares and triangles for his floor tessellation pattern

Use the above figure to answer the questions that follow:
(A) What is the length of the line segment joining points B and F? (1)
(B) The centre ‘Z’of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z? (1)
(C) What are the coordinates of the point on y axis equidistant from A and G?

What is the area of Trapezium AFGH?

Total Marks

Breakdown
(As per CBSE Marking Scheme)

5m
(LA)

(A) Find ‘a’ and ‘d’ (0.5m)
Calculate nth term (0.5m)
(B) Apply the formula for sum of terms and substitute the values (1m)
Reject negative value and pick answer (1m)

Write the formula for sum of n terms of A.P. (1m)
For n=10, find the sum (1m)
(C) Find the middle row (0.5m)
Calculate the nth term (0.5m)

Answer:
(A)
B(1,2), F(-2,9)
BF² = (-2 – 1)² + (9 – 2)²
= (-3)² + (7)²
= 9 + 49
= 58

So, BF = \( \sqrt{58} \) units

(B) W(-6, 2), X(-4, 0), 0(5, 9), P(3, 11)
Clearly WXOP is a rectangle

Point of intersection of diagonals of a rectangle is the mid point of the diagonals. So the required point is mid point of WO or XP
= \(\left(\frac{-6+5}{2}, \frac{2+9}{2}\right)\)
= \(\left(\frac{-1}{2}, \frac{11}{2}\right)\)

(C) A(-2, 2), G(-4, 7)
Let the point on y-axis be Z(0, y)
AZ² = GZ²
(0 + 2)² + (y – 2)² = (0 + 4)² + (y – 7)²
(2)² + y² + 4 – 4y = (4)² + y² + 49 – 14y
8 – 4y = 65 – 14y
10y = 57
So, y = 5.7
i.e. the required point is (0, 5.7)

OR

A(-2, 2), F(-2, 9), G(-4, 7), H(-4, 4)
Clearly GH = 7 – 4 = 3units
AF = 9 – 2 = 7 units
So, height of the trapezium AFGH = 2 units
So, area of AFGH = \(\frac{1}{2}\)(AF + GH) × height
= \(\frac{1}{2}\)(7+3) × 2
= 10 sq. units

Question 37.
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

(A) If the first circular row has 30 seats, how many seats will be there in the 10th row? (1)
(B) For 1500 seats in the auditorium, how many rows need to be there?
OR
If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10th row? (2)
(C) If there were 17 rows in the auditorium, how many seats will be there in the middle row? (1)

Total Marks

Breakdown
(As per CBSE Marking Scheme)

(CBQ)

(A) Find ‘a’ and’d’ (0.5m)
Calculate nth term (0.5m)
(B) Apply the formula for sum of n
terms and substitute the values (lm)
Reject negative value and pick answer (lm)

Write the formula for sum of n terms of A.P. (lm)
For n=10, find the sum (lm)
(C) Find the middle row (0.5m)
Calculate the nth term (0.5m)

Answer:
(A) Since each row is increasing by 10 seats, so it is an AP with first term a = 30, and common difference d = 10.
So number of seats in 10th row
= a₁₀= a + 9d
= 30+ 9 × 10 = 120

(B) S_n = \(\frac{n}{2}\)[2a+(n-1)d]
1500 = \(\frac{n}{2}\)[2×30+(n-1)10]
3000 = 50n + 10n²
n² + 5n – 300 = 0
n² + 20n – 15n – 300 = 0
(n + 20) (n – 15) = 0
Rejecting the negative value, n = 15

No. of seats already put up to the 10th row = S₁₀
S₁₀ = \(\frac{10}{2}\)[2×30 + (10-1)10]
= 5(60 + 90) = 750
So, the number of seats still required to be put are 1500 – 750 = 750

Question 38.
We all have seen the airplanes flying in the sky but might have not thought of how they actually reach the correct destination. Air Traffic Control (ATC) is a service provided by ground-based air traffic controllers who direct aircraft on the ground and through a given section of controlled airspace, and can provide advisory services to aircraft in non- controlled airspace. Actually, all this air traffic is managed and regulated by using various concepts based on coordinate geometry and trigonometry.

At a given instance, ATC finds that the angle of elevation of an airplane from a point on the ground is 60°. After a flight of 30 seconds, it is observed that the angle of elevation changes to 30°. The height of the
plane remains constantly as 3000√2 m. Use the above information to answer the questions that follow:
(A) Draw a neat labelled figure to show the above situation diagrammatically. (1)
(B) What is the distance travelled by the plane in 30 seconds?
OR
Keeping the height constant, during the above flight, it was observed that after 15( √3 -1) seconds, the angle of elevation changed to 45°. How much is the distance travelled in that duration. (2)
(C) What is the speed of the plane in km/hr. (1 )