CBSE Class 10 Maths Standard Question Paper 2025 with Solutions

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CBSE Class 10 Maths Standard Board Question Paper 2025 with Solutions

Time : 3 hrs
Max. Marks : 80

General Instructions :

• This question paper has 5 Sections A,B,C,D and E .
• Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E has 3 Case Based integrated units of assessment ( 4 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 2 marks, 2 Qs of 3 marks and 2 Qs of 5 marks has been provided. An internal choice has been provided in the 2 marks question 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

Directions Q. No. 1 to 20 are multiple choice questions of 1 mark each.

Question 1.
√0.4 is a/an
(a) natural number
(b) integer
(c) rational number
(d) irrational number
Answer:
(d) We have, \(\sqrt{0.4}=\sqrt{\frac{4}{10}}=\frac{\sqrt{4}}{\sqrt{10}}=\frac{2}{\sqrt{10}}\)
Since, √10 is irrational, \(\frac{2}{\sqrt{10}}\) is also irrational.
Hence, √0.4 is an irrational number.

Question 2.
Which of the following cannot be the unit digit of 8ⁿ, where n is a natural number?
(a) 4
(b) 2
(c) 0
(d) 6
Answer:
(c) We have, 8ⁿ = (2 x 2 x 2)ⁿ = 2³ⁿ
We know that if a number ends with digit 0, then it should be divisible by both 2 and 5.
Here, prime factorisation of 8ⁿ does not contain 5. Therefore, unit digit of 8ⁿ cannot be zero.

Question 3.
Which of the following quadratic equations has real and equal roots ?
(a) (x+1)²=2 x+1
(b) x²+x=0
(c) x²-4=0
(d) x²+x+1=0
Answer:
(a) We have, (x + 1)² = 2x + 1
⇒ x² + 1+ 2x = 2x + 1 ⇒ x² =0 ………. (i)
On comparing Eq. (i) with ax² + bx + c = 0, we get <7 = 1, b = 0 and c = 0
∴ Discriminant, D = b² – Aac = 0-4 × 1 × 0 = 0
So, this equation has real and equal roots.

Question 4.
If the zeroes of the polynomial \(a x^2+b x+\frac{2 a}{b}\) are reciprocal of each other then the value of b is
(a) 2
(b) \(\frac{1}{2}\)
(c) -2
(d) \(-\frac{1}{2}\)
Answer:
(a) Given polynomial is ax² + bx + \(\frac{2 a}{b}\)
Let α and \(\frac{1}{\alpha}\) be the zeroes of the given polynomial.

Question 5.
The distance of the point A(-3,-4) from X-axis is
(a) 3
(b) 4
(c) 5
(d) 7
Answer:
(b) Given the point is A(-3,-4).
Let the perpendicular from A on the X-axis intersects at B.
Point B lies on the X-axis, so the coordinate of point B is (-3,0).
Distance of the given point from X-axis =A B
Using distance formula,

Question 6.
In the given figure, P Q || X Y|| B C, A P=2 cm, PX=1.5 cm and B X=4cm. If QY=0.75 cm then AQ+CY is equal to

(a) 6 cm
(b) 4.5 cm
(c) 3 cm
(d) 5.25 cm
Answer:

Question 7.
Given, ΔABC ∼ ΔPQR, ∠A=30° and ∠ Q=90°. The value of (∠R+∠B) is…………..
(a) 90°
(b) 120°
(c) 150°
(d) 180°
Answer:
(c) Given, ΔABC ~ ΔPQR, ∠A = 30° and ∠Q = 90° Here, ∠B = ∠Q= 90°
In AABC, ∠A + ∠B+ ∠C= 180°
⇒ 30° + 90° + ∠C = 180° ⇒ ∠C = 60°
Also, ∠C = ∠R = 60°
Now, ∠R+ ∠B = 60° + 90° = 150°

Question 8.
Two coins are tossed simultaneously. The probability of getting atleast one head is
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{3}{4}\)
(d) 1
Answer:
(c) When two coins are tossed simultaneously, then possible outcomes are (H, H),(H, T),(T, H) and (T,T).
∴ Total number of outcomes = 4 Let E be the event of getting atleast one head i.e. one head or two head. Then, outcomes favourable of E are (H, T), (T, H) and (H, H)
i.e. number of outcomes favourable of E = 3
∴ P(E) =\(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of outcomes }} \)
=\(\frac{3}{4}\)

Question 9.
In the given figure, PA and PB are tangents to a circle with centre O such that ∠P=90°. If AB=3 √2cm then the diameter of the circle is

Answer:
(a) 3 √2cm
(b) 6 √2cm
(c) 3 cm
(d) 6 cm
Answer:
(d) Given, PA and PB are tangent to a circle with centre O. Such that ∠P = 90° and AB = 3√2 cm. Since, PA and PB are tangents from the same point P, they are equal in length i.e. PA = PB … (i)
In AAPB, AB² = PA² + PB²
[using Pythagoras theorem] ⇒ (3√2)² = PA² + PA²  [using Eq. (i)]
⇒ 2PA² =18 ⇒ PA² = 9 ⇒  PA = 3 cm
Since ∠P = 90° and OA ⊥ PA, OB ⊥ PB
∴ Quadrilateral OAPB is a square.
So, OA = OB= PA= PB = 3cm
Now, radius of the circle, OA = 3 cm and diameter of the circle
= 20A = 2 × 3 = 6 cm

Question 10.
For a circle with centre O and radius 5 cm, which of the following statements is true?
P : Distance between every pair of parallel tangents is 5 cm
Q : Distance between every pair of parallel tangents is 10 cm.
R : Distance between every pair of parallel tangents must be between 5 cm and 10 cm.
S : There does not exist a point outside the circle from where length of tangent is 5 cm.
(a) P
(b) Q
(c) R
(d) S
Answer:
(b) Given, a circle with centre Oand radius 5 cm. We know that the distance between two parallel tangents to a circle is equal to diameter of the circle.
Here, radius − 5 cm
∴ Diameter = 2×5 = 10 cm
Hence, distance between every pair of parallel tangents is 10 cm.

Question 11.
In the given figure, T S is a tangent to a circle with centre O. The value of 2 x° is

(a) 22.5
(b) 45
(c) 67.5
(d) 90
Answer:
(b) In the given figure, TS is a tangent.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OS ⊥ TS
So, ∠OST = 90°
Now, in ΔOST, ∠OST + ∠STO + ∠TOS= 180°
[angle sum property of a triangle] ⇒ 90° + r° + 3x° = 180°
⇒ 4x° = 90° ⇒ x° = 22.5°
Therefore, 2x° = 2 x 225° = 45°

Question 12.
If \(x\left(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\right)=y\left(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\right)\) then x: y is equal to
(a) 1: 1
(b) 1: 2
(c) 2: 1
(d) 4: 1
Answer:

Question 13.
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is 10√3m away from the base of the tree then angle of depression of the snake from the eye of the peacock is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) Let height of tree, A B=10m  and  B C=10√3 m

Again, let θ be the angle of depression.
Therefore, ∠XAC=∠ACB=θ

Hence, angle of depression is 30°

Question 14.
If a cone of greatest possible volume is hollowed out from a solid wooden cylinder then the ratio of the volume of remaining wood to the volume of cone hollowed out is
(a) 1: 1
(b) 1: 3
(c) 2: 1
(d) 3: 1
Answer:
(c) Let r and h be the radius and height of the cylinder, respectively.
∴ Volume of cylinder = nr²h
Since, the cone is hollowed out from the cylinder with greatest possible volume, therefore it has same radius and height as the cylinder.
∴ Volume of cone = \(\frac{1}{3}\) πr²h
Volume of remaining wood

Hence, the ratio of volume of remaining wood to the volume of cone hollowed out is 2: 1.

Question 15.
If the mode of some observations is 10 and sum of mean and median is 25 then the mean and median, respectively are
(a) 12 and 13
(b) 13 and 12
(c) 10 and 15
(d) 15 and 10
Answer:
(b) Given, Mode = 10
and Mean + Median = 25
⇒ Mean = 25-Median   ……………… (i)
We know that
Mode = 3 Median – 2 Mean ⇒ 10 = 3 Median – 2 (25- Median) [from Eq. (i)]
⇒ 10 = 3 Medain -50+2 Median
⇒ 5 Median = 60 ⇒ Median = 12
∴ Now, Mean = 25 – Median = 25 – 12 = 13
Hence, the mean and median are 13 and 12, respectively.

Question 16.
If the maximum number of students has obtained 52 marks out of 80, then
(a) 52 is the mean of the data.
(b) 52 is the median of the data.
(c) 52 is the mode of the data.
(d) 52 is the range of the data.
Answer:
(c) Given, maximum number of students has obtained 52 marks out of 80.
We know that the mode is the value that appears most often in a set of data.
Since, the score 52 is obtained by the maximum number of students, therefore 52 is the mode of the data.

Question 17.
The system of equations 2 x+1=0 and 3y-5=0 has
(a) unique solution
(b) two solutions
(c) no solution
(d) infinite number of solutions
Answer:
(a) Given, pair of equation is
2x+1=0 and 3y-5=0
On comparing the given equations with standard form of pair of linear equations i.e.

∴ The system of given equations has unique solution.

Question 18.
In a right ΔABC, right-angled at A, if sin B=\(\frac{1}{4}\) then the value of sec B is
(a) 4
(b) \(\frac{\sqrt{15}}{4}\)
(c) \(\sqrt{15}\)
(d) \(\frac{4}{\sqrt{15}}\)
Answer:

Directions In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following
(a) Both Assertion and Reason are true and Reason is correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

Question 19.
Assertion (A) For any two prime numbers p and q, their HCF is 1 and LCM is p+q
Reason (R) For any two natural numbers, HCF × LCM = Product of numbers.
Answer:
(d) We know that HCF of two prime numbers P and Q is 1 and LCM of two prime numbers P and Q is P Q.
Hence, Assertion is false.
Clearly, Reason is true.

Question 20.
In an experiment of throwing a die,
Assertion (A) Event E₁: getting a number less than 3 and Event E₂: getting a number greater than 3 are complementary events.
Reason (R) If two events E and F are complementary events then P(E)+P(F)=1
Answer:
(d) When a die is thrown, there are six possible outcomes i.e. 1, 2, 3, 4, 5, 6
So, sample space, S={1,2,3,4,5,6}
Here, E₁= Event of getting a number less than 3 = {1,2}
and E₂= Event of getting a number greater than 3 ={4,5,6}
Now, E₁∪E₂={1,2,4,5,6} ≠S
∴ E₁ and E₂ are not complementary events.
Thus, Assertion is false.
Clearly, Reason is true.

Section B
Section B consists of 5 Questions of 2 marks each

Directions This section has 5 very short answer type questions of 2 marks each.

Question 21.
(a) Solve the following pair of equations algebraically 101 x+102 y=304 ; 102 x+101 y=305
Or
(b) In a pair of supplementary angles, the greater angle exceeds the smaller by 50°. Express the given situation as a system of linear equations in two variables and hence obtain the measure of each angle.
Answer:
(a) Given, pair of linear equation is
101 x+102 y=304 ………….. (i)
102 x+101 y=305 ………….. (ii)
On adding Eqs. (i) and (ii), we get
203 x+203 y=609
x+y=3  ………….. (iii)
[dividing both sides by 203]
On subtracting Eq. (i) from Eq. (ii), we get
x-y=1 ………….. (iv)
Now, adding Eqs. (iii) and (iv), we get
(say) 2 x=4
⇒ x=2
On substituting x=2 in Eq. (iii), we get
y=3-2=1
Hence, x=2 and y=1 is the required solution.

(b) Let x and y be the measure of the greater angle and smaller angle, respectively.
We know that the sum of two supplementary angles is equal to 180°.
∴ x+y=180° ………….. (i)
According to the question,
x=y+50° ………….. (ii)
⇒ x-y=50° ………….. (iii)
On adding Eqs. (i) and (iii), we get
2 x=230°
⇒ x=115°
On substituting x=115° in Eq. (i), we get
y=180°-115°=65°
Hence, the measure of the greater angle and smaller angle are 115° and 65°.

Question 22.
(a) If a secθ+b tanθ=m and b secθ+a tan θ=n then, prove that a²+n²=b²+m²
Or
(b) Use the identity sin² A+cos² A=1 to prove that tan² A+1=sec² A
Hence, find the value of tan A, where sec A=\(\frac{5}{3}\), where A is an acute angle.
Answer:
(a) Given, a secθ+b tan θ =m
b sec θ +a tan θ =n
On squaring and subtracting Eqs. (i) and (ii), we get

Question 23.
Prove that abscissa of a point P which is equidistant from points with coordinates A(7,1) and B(3,5) is 2 more than its ordinate.
Answer:
Let the coordinate of point P be (x, y).
∵ Point P is equidistant from points with coordinates A(7,1) and B(3,5).
So, by distance formula

Question 24.
P is a point on the side B C of ΔABC such that ∠APC=∠BAC. Prove that AC²=B C.C P
Answer:
Given, ∠APC=∠BAC

Question 25.
The number of red balls in a bag is three more than the number of black balls. If the probability of drawng a red ball at random from the given bag is \(\frac{12}{23}\), find the total number of balls in the given bag.
Answer:
Let the number of red and black balls in a bag be x and y, respectively.
∴ Total number of balls =x+y
According to the question
x=y+3
→ x-y=3 ……………..(i)
If the probability of drawing a red ball is \(\frac{12}{23}\)

⇒12 y-11 y=33 ⇒ y=33
On substituting y=33 in Eq. (i), we get
x=3+33=36
Hence, the total number of balls in bag is 33+36 i.e. 69

SECTION C
Section C consists of 6 Questions of 3 marks each

Directions This section has 6 short answer type questions of 3 marks each.

Question 26.
(a) Prove that √5 is an irrational number.
Or
(b) Let p, q and r be three distinct prime numbers.
Check whether p⋅q.r+q is a composite number or not.
Further, give an example for 3 distinct primes p, q and r such that
(i) p . q . r+1 is a composite number.
(ii) p . q . r+1 is a prime number.
Answer:
Let us assume on the contrary that √5 is a rational number. Then, there exist positive integers a and b such that \(\sqrt{5}=\frac{a}{b}\) where a and b are coprime.
Now, \(\sqrt{5}=\frac{a}{b}\)
⇒ 5= \(\frac{a^2}{b^2}=5 b^2=a^2\)
⇒ 5 divides a²
⇒ 5 divides a.
So, 5 is a factor of a.  …………….(i)
⇒ a = 5c for some integer c.
⇒ a² = 25c² = 5b² = 25c²    [∵ a² = 5b²]
⇒b² = 5c² ⇒ 5 divides b²
⇒ 5 divides b.
So, 5 is a factor of b.  …………….(ii)
From Eqs. (i) and (ii), we observe that a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprime. This means that our assumption is not correct.
Hence, √5 is an irrational number.
Hence Proved

(b) We know that a prime number is a number that has only two factors.
Let the values of p, q and r are 2,3 and 5, respectively.
∴ p.q.r+q=2 × 3 × 5+3=33,
which is a composite number.
[since, a composite number is a natural number or a positive integer which has more than two factors]
(i) Let the values of p, q and r be 3,5 and 7, respectively. Then,
p × q × r+1  =3 × 5 × 7+1
= 105+1=106
which is a composite number.

(ii) We have, p.q.r+1=3 × 5 × 7+1=106,
which is not a prime number.

Question 27.
Find the zeroes of the polynomial p(x)=3 x²-4 x-4. Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of p(x).
Answer:
Given the polynomial p(x) = 3x² – 4x – 4
= 3x² – (6 – 2)x – 4 [by middle term splitting] = 3x² -6x + 2x -4 = 3x(x – 2) + 2(x-2)
= (3x + 2)(x – 2)
For zeroes, put p(x) = 0
∴ (3x + 2)(x-2) = 0
⇒ 3x + 2 = 0 or x-2 = 0
⇒ x = \(\frac{-2}{3}\) or i=2
So, the zeroes of the polynomial are \(\frac{-2}{3}\) and 2.
As per giyen condition, the zeroes of the new

Question 28.
Check whether the following pair of equations is consistent or not. If consistent, solve graphically.
x+3 y=6
3 y-2 x=-12
Answer:
The given equations can be rewritten as x + 3y – 6 = 0 and -2x + 3y + 12 = 0.
On comparing with standard form of pair of linear equations,
we get a₁= 1 b₁ = 3, c₁= – 6 and a₂ = – 2, b₂ = 3, c₂= 12

Hence, the pair of linear equation is consistent.
Now, table for linear equation x+3 y=6 is

Plot the points A and B and join them to the line AB.
Table for linear equation -2 x+3 y=-12 is

Two lines A B and C D intersect at the point B(6,0).
Hence, x=6 and y=0 is the required solution of the pair of linear equations.

Question 29.
If the points A(6,1), B(p, 2), C(9,4) and D(7, q) are the vertices of a parallelogram ABCD then find the values of p and q. Hence, check whether A B C D is a rectangle or not.
Answer:
Given, ABCD is a parallelogram in which the coordinates of the vertices are A(6,1), B(p, 2), C(9,4) and D(7, q).
∵ ABCD is a parallelogram, the diagonals bisect each other.
∴ The mid-point of the diagonals of the parallelogram will coincide.
∴ Coordinate of mid-point of AC = Coordinate of mid-point of BD.


Hence, the length of all sides of parallelogram are equal.
∴ ABCD is a square as well as rectangle.

Question 30.
(a) Prove that \(\frac{\cos \theta-2 \cos ^3 \theta}{\sin \theta-2 \sin ^3 \theta}+\cot \theta\)=0.
Or
(b) Given that sinθ a+ cosθ = x, prove that sin⁴ θ + cos⁴ θ= \(\frac{2-\left(x^2-1\right)^2}{2}\).
Answer:

Question 31.
In the given figure, T P and T Q are tangents drawn to a circle with centre O. If ∠OPQ=15° and ∠PTQ=θ then find the value of sin 2θ.

Answer:
Given, ∠OPQ = 15° and ∠PTQ = 0 TP= TQ
[∵ length of tangents drawn from an external point are equal]
and ∠OPT = 90°
[tangent to a circle is perpendicular to radius through the point of contact].
∠QPT = ∠OPT – ∠OPQ = 90° – 15° = 75°
In ΔPTQ,
By angle sum property,
PQT + ∠QPT + ∠PTQ = 180
∠PQT + ∠QPT + θ = 180°
Since, ∠PQT = ∠QPT = 75°
[angle opposite to equal sides are also equal i.e. TP=TQ]
75° = + 75° + θ = 180°
⇒ 150° + θ = 180°
⇒ θ = 180° – 150° = 30°
The value of 2θ = 2 x 30° = 60°

Section D
Section D consists of 4 Questions of 5 marks each

Directions This section has 4 long answer questions of 5 marks each.

Question 32.
(a) There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter.

An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B, respectively.
Or
(b) Find the smallest value of p for which the quadratic equation x²-2(p+1) x+p²=0 has real roots. Hence, find the roots of the equation so obtained.
Answer:
(a) Given, the diameter of the circular park = 65 m According to the question,
Distance of P from A = Distance of P from B + 35
PA = PB + 35
We know that angle in a semi-circle is a right angle.
∠APB = 90°
In AAPB, by Pythagoras theorem AB² = PA² + PB²
⇒ (65)² = PA² + PB²  [∵AB = 65 m]
⇒ 4225 = (PB + 35)² + PB² [from Eq. (i)]
⇒ 4225 = PB² + (35)² + 7QPB + PB²
⇒ 4225 = 2PB² + 1225 + 70PB
⇒ 2PB² + 70PB = 4225 – 1225
⇒ 2 PB² + 70 PB = 3000
⇒ PB² + 35 PB = 1500
⇒ PB² + 35 PB – 1500 = 0
⇒ PB² + 60 PB – 25 PB -1500 = 0
[by middle term splitting]
⇒ PB(P =B+60)-25(P =B+60)=0
⇒ (PB+60)(PB-25)=0
⇒ P B+60=0 or P B-25=0
⇒ PB = -60 or P B=25
Since, distance cannot be negative.
∴ PB=25 m and
PA = 25+35=60m
Hence, distance of point P from A and B are 60 m and 25 m
Or
(b) Given the quadratic equation is
x²-2(p+1) x+p²=0
On comparing with a x²+b x+c=0, we get
a=1, b=-2(p+1) and  c=p²
Now, discriminant (D)=b²-4 a c
= [-2(p+1)]²-4 × 1 × p²
= 4(p+1)²-4 p²
= 4 p²+4+8 p-4 p²
= 4+8 p
Since, the roots are real.
So, the discriminant must be greater than or equal to zero.
∴ 4+8 p ≥ 0
8p ≥ – 4 ⇒ p ≥ – \(\frac{1}{2}\)

So, the smallest value of p for which the equation has real roots is p=- \(\frac{1}{2}\).

Question 33.
(a) If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side. State and prove the converse of the above statement.
Or
(b) In the given figure, ΔCAB is a right triangle, right angled at A and AD⊥BC. Prove that ΔADB ~ ΔCDA. Further, if BC = 10 cm and CD=2 cm, find the length of AD.

Answer:
(a) Given ΔABC in which a line D E parallel to B C intersects AB at D and AC at E, i.e.
DE || BC
To prove \(\frac{AD}{DB}=\frac{AE}{EC}\)
Construction Join BE and CD.
Draw EF ⊥A B and DG⊥AC

Since,  ΔBDE and ΔDEC stand on the same base D E and between same parallel lines DE and BC.
∴(ΔBDE)=ar(ΔDEC) …….(iii)
From Eqs. (i), (ii) and (iii), we get
\(\frac{A D}{D B}=\frac{AE}{EC}\) …….(iv)
Hence Proved
Or
(b) Given, ΔCAB is a right angled triangle at A and AD⊥BC
In ΔADB,
∠ADB+∠B A D+∠DBA  =180°
⇒ 90°+∠BAD+∠DBA =180°
[since, AD ⊥BC ⇒ ∠ADB=90°]
⇒ ∠DBA=90°-∠BAD
Now, ∠CAB=90° [given]
⇒∠CAD+∠BAD  =90°
⇒ ∠CAD =90°-∠B A D
In ΔADB and ΔCDA,
∠ADB=∠CD A=90° [AD⊥BC]
∠DBA=∠CAD  [from Eqs. (i) and (ii)] }
∴ Δ ADB sim Δ CDA [by AA similarity criterion]
Now, \(\frac{A D}{CD} =\frac{BD}{AD} \)
[corresponding sides of similar triangle are proportional]
⇒ \(\frac{AD}{2}=\frac{BC-C D}{AD}\)
⇒ AD²=2 ×(10-2)=2 × 8=16 ⇒ AD=4cm
Hence, the length of AD is 4 cm

Question 34.
From one face of a solid cube ot side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use π=\(\frac{22}{7}, \sqrt{5}=22\)
Answer:
Given, the side of the cube = 14 cm
∴ The volume of the cube = (side)³= (14)³
= 2744 cm³
The largest possible cone is carved out in one face of cube.
∴ The radius (r) of cone is half the side of the cube
i.e. r = \(\frac{14}{2}\) = 7cm
and height (h) of the cone is equal to the side of the cube i.e. h = 14 cm
∴ The volume of the cone = \(\frac{1}{3}\) x πr²h

So, the surface area of the remaining solid = Surface area of cube
+ Curved surface area of cone
– Surface area of base of the cone
= 1176-154+338.8=1360.8 cm²

Question 35.
Following distribution shows the marks of 230 students in a particular subject. If the median marks are 46 then find the values of x and y.

Marks	Number of Students
10-20	12
20-30	30
30-40	X
40-50	65
50-60	y
60-70	25
70-80	18

Answer:
Given, median =46 and n=∑ f=230
∴ Sum of frequencies =230
∵ 12+30+x+65+y+25+18=230
⇒ 150+x+y=230
⇒ x+y=230-150=80
⇒ x+y=80 ……………(i)
Now, the cumulative frequencey table for given distribution is

Here, n=230
⇒ \(\frac{n}{2}\)=115
Given, median =46, which belongs to the class 40-50.
So, the median class is 40-50.
Then, l=40, h=10, f=65 and cf=42+x

Hence, the values of x and y are 34 and 46, respectively.

Section E
Section E consists of 3 Questions of 4 marks each

Directions This section has 3 case study based questions of 4 marks each.

Question 36.
Anurag purchased a farmhouse which is in the form of a semi-circle of diameter 70m. He divides it into three parts by taking a point P on the semi-circle in such a way that ∠PAB=30° as shown in the following figure, where O is the centre of semi-circle.

In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges. Based on given information, answer the following questions.
(i) What is the measure of ∠POA?
(ii) Find the length of wire needed to fence entire piece of land.
(iii) (a) Find the area of region in which saplings of mango tree are planted.
Or
(b) Find the length of wire needed to fence the region III.
Answer:
Given, ∠PAB = 30°
The angle subtended by a diameter on the semi-circle is 90°
∴  ∠APB = 90°
Now, join Pto O.
Here, OP = OA [radii of same semi-circle]
∴ ∠OPA = ∠OAP = 30°
[angles opposite to equal sides are equal]

In APOA,
∠POA + ∠OPA + ∠OAP = 180°
[by angle sum property] ⇒ ∠POA + 30°+ 30° = 180°
⇒ ∠POA + 60° = 180°
⇒ ∠POA = 180° – 60° = 120°
So, the measure of ∠POA is 120°.

(ii) The length of wire needed to fence entire piece of land = The length of the arc APB + Diameter

Also, PB = 35 m
So, ΔPOB is an equilateral triangle.
∴ Area of an equilateral,
ΔOPB=\(\frac{\sqrt{3}}{4}\) (side)²
\(\frac{\sqrt{3}}{4}\) 35 x 35 = 530.42 m²
So, the area of region I
= Area of sector OPBQ – Area of AOPB
[∵ each angle of an equilateral triangle is 60°]
= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}\) 35× 35 – 530.42
= 641.67 – 530.42 = 111.25 m²
Hence, the area of region in which saplings of mango tree are planted is 111.25 m²
Or
(iii) (b) Now, the length of arc PB= \(\frac{\theta}{360^{\circ}}\) × 2πr
= \(\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 35\) =36.66
And the perimeter of semi-circle = \(\frac{1}{2}\) × 2πr
=\(\frac{1}{2} \times 2 \times \frac{22}{7} \times\) 35 = 110
So, length of arc AP= Perimeter of semi-circle – Length of arc  PB
= 110-36.66
= 73.34
Hence, the length of wire needed to fence the Region III = PA+Length of arc AP
=35√3 + 73.34
=60.62+73.34 [∵ √3=1.732]
=133.96 m

Question 37.
In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below

The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.
Based on given information, answer the following questions, using concept of Arithmetic Progression.
(i) What is the length of the 6th lane?
(ii) How long is the 8th lane than that of 4th lane?
(iii) (a) While practicing for a race, a student took one round each in first six lanes. Find the total distance covered by the student.
Or
(b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student.
Answer:
Given, the length of innermost lane i.e. first lane = 400m
And each subsequent lane is 7.6 m longer than the preceding lane.
So, the list of numbers is 400, 407.6, 415.2, …
and its form an AP, where a = 400 and d = 7.6
(i) 6th lane = a₆
= a + (6-1)d  [∵ a_n = a + (n-1)d]
= 400 + 5 x 7.6
= 400+38 = 438
So, the length of the 6th lane is 438 m.

(ii) 4th lane = a₄
= a + (4-1)d      [∵ a_n = a + (n-1)d]
= 400 + 3 x 7.6
= 400+22.8 = 422.8
So, the length of the 4th lane is 422.8 m.
8th lane = a_s = a + (8 – 1)d
= 400+7 x 7.6
= 400+ 53.2 = 453.2
So, the length of the 8th lane is 453.2 m.
Hence, 8th lane is 453.2 m – 4228 m = 30.4 m longer than 4th lane.

(iii) (a) Total distance covered by the student
= The sum of length of first 6 lanes
= S₆ =|[2 x400 + (6-1)7.6]
[∵ sum of first n terms of a_n AP is s_n = \(\frac{n}{2}\)[2a + (n-1)d]
= 3 [800 + 38] = 3 x 838 = 2514
So, the distance covered by the student is 2514 m.
0r
(b) Total distance covered by the student
= The sum of length of first 8 lanes – Sum of length of first 3 lanes
= S₈ – S₃ = \(\frac{8}{2}\) [2 x 400+(8-1)7.6] – \(\frac{3}{2}\) [2 x 400 + (3-1)7.6]
= 4[800 + 53.2] –\(\frac{3}{2}\) [800 + 15.2]
= 4 x 853.2 – \(\frac{3}{2}\) x 815.2
= 3412.8 – 1222.8 = 2190 m
So, the distance covered by the student is 2190 m.

Question 38.
The Statue of Unity situated in Gujarat is the world’s largest Statue which stands over a 58 m high base. As part of the project, a student constructed an inclinometer and wishes to find the height of Statue of Unity using it.
He noted following observations from two places
Situation I The angle of elevation of the top of statue from Place A which is 80√3m away from the base of the statue is found to be 60°.
Situation II The angle of elevation of the top of Statue from a Place B which is 40 m above the ground is found to be 30° and entire height of the Statue including the base is found to be 240m.

Based on given information, answer the following questions.
(i) Represent the Situation I with the help of a diagram.
(ii) Represent the Situation II with the help of a diagram.
(iii) (a) Calculate the height of statue excluding the base and also find the height including the base with the help of Situation I.
Or (b) Find the horizontal distance of Point B (Situation II) from the statue and the value of tanα, where α is the angle of elevation of top of base of the statue from Point B.
Answer:
Given, the height of base of the statue is 58 m.
(i) Let PQ be the height of base of the statue and PR be the height of statue including base.

(ii) Let MO be the height of base of the statue and SM be the height of statue including base.

(iii) (a) From part (i), in Δ APR,
tan 60° = \(\frac{PR}{AP} \)
√3  =\(\frac{RQ+P Q}{80 \sqrt{3}} \)
⇒ √3 × 80 √3  =RQ+58 ⇒ 240=RQ+58
⇒  RQ =240-58=182m
Now, PR=182+58=240m
So, the height of statue excluding base is 182 m, and the height of statue including base is 240 m.

Or
(iii) (b) From part (ii), we have, MS=240m and  M=58 m}
∴  OS=MS – OM=240-58=182 m
Also, BL=40m
∴ ON = O M-N M=58-40 [∵ NM=BL]
= 18m
Now, SN=O S+O N=182+18=200m

In ΔBNS,
tan 30°  = \(\frac{SN}{BN}\) ⇒ \(\frac{1}{\sqrt{3}}=\frac{200}{B N}\)
⇒ BN =200 √3m
So, the horizontal distance of point B from the statue is 200 √3m.
Now, in ΔBNO
tan α = \(\frac{O N}{B N}=\frac{18}{200 \sqrt{3}} \)
⇒ tan α = \(\frac{3 \sqrt{3}}{100}\)