RBSE 10th Maths Model Paper Set 3 with Answers in English 

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 3 with Answers in English provided here.

RBSE Class 10 Maths Model Paper Set 3 with Answers in English

Time: 2:45 Hours
Maximum Marks: 80

General Instructions:

• All the questions are compulsory.
• Write the answer of each question in the given answer book only.
• For questions having more than one part the answers to their parts are to be written together in continuity.
• Candidate must write first his/her Roll No. on the question paper compulsorily.
• Question numbers 17 to 23 have internal choices.
• The marks weightage of the questions are as follows :

Section	Number of Questions	Total Weightage	Marks for each question
Section A	1 (i to xii), 2(i to vi), 3(i to xii) = 30	30	1
Section B	4 to 16 =13	26	2
Section C	17 to 20 = 4	12	3
Section D	21 to 23 = 3	12	4

Section – A

Question 1.
Multiple Choice Questions :
(i) The total number of factors of prime numbers is : (1)
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

(ii) The quadratic polynomial, the sum of whose zeroes is – 5 and their product is 6, is: (1)
(a) x² + 5x + 6
(b) x² – 5x + 6
(c) x² – 5x – 6
(d) – x² + 5x + 6
Answer:
(a) x² + 5x + 6

(iii) The value of k for which the system of equations 2x + 4y = 5, 6x + ky – 9 has no solution, is : (1)
(a) 10
(b) 9
(c) 12
(d) 11.
Answer:
(c) 12

(iv) (x² + 1)² – x² = 0 has : (1)
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real roots.
Answer:
(c) no real roots

(v) The value of x for which 2x, (x + 10) and (3x + 2) are there consecutive terms of an AP are : (1)
(a) 6
(b) -6
(c) 18
(d) -18
Answer:
(a) 6

(vi) In the given figure A1 divi4es the line segment in the ratio: (1)

(a) 1 : 1
(b) 1 : 2
(c) 1 : 4
(d) 1 : 5
Answer:
(c) 1 : 4

(vii) The point P on x-axis equidistant from the A(- 1, 0) and B(5, 0) is : (1)
(a) (2, 0)
(b) (0, 2)
(c) (3, 0)
(d) (2, 2)
Answer:
(a) (2, 0)

(viii) If 4 tan 0 = 3, then \(\left[\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta+1}\right]\) is : (1)
(a) \(\frac{13}{5}\)
(b) \(\frac{5}{21}\)
(c) \(\frac{13}{21}\)
(d) \(\frac{12}{13}\)
Answer:
(c) \(\frac{13}{21}\)

(ix) sin 0 cos (90° – θ) + cos θ sin (90° – θ) is equal to : (1)
(a) \(\frac{\sqrt{3}}{2}\)
(b) 0
(c) 1
(d) 2
Answer:
(c) 1

(x) Mode of data 2, 3, 5, 2, 3, 6, 5, 2, 2, 5, 7, 4, 4, is : (1)
(a) 5
(b) 3
(c) 2
(d) 4
Answer:
(c) 2

(xi) For a given data with 120 obsevations, the ‘less than ogive and the ‘more than o give’ interesect at (42.5, 60). The median of the data is: (1)
(a) 60
(b) 42.5
(c) 120
(d) 21.25
Answer:
(b) 42.5

(xii) Which of the following cannot be probability of an event? (1)
(a) 5%
(b) 0.9
(c) 1.1
(d) 0.1
Answer:
(c) 1.1

Question 2.
Fill in the blanks :
(i) If the lines are ________ then the pair of liner equations in two variables has no solution. (1)
Answer:
Parallel

(ii) In a AP, the nth term is given by a_n = a + (n – 1) d. (1)
Answer:
term

(iii) Number of tangents to the circle from a point outside the circle are ________ (1)
Answer:
two

(iv) The distance of a point from the x-axis is called its ________ (1)
Answer:
ordinates

(v) If sin θ – cos θ = 0, then value of θ is ________ (1)
Answer:
45°

(vi) If mean of 5 observations x, x- + 2, x + 4, x + 6 and x + 8 is 11, then the Value of x is ________ (1)
Answer:
7

Question 3.
Very Short Answer Type Questions :
(i) After how many decimal places will the decimal expansion of \(\frac{23}{2^{4} \times 5^{3}}\) terminate? (1)
Answer:
We have

Hence, \(\frac{23}{2^{4} \times 5^{3}}\) will terminate after 4 decimal places.

(ii) If α and β are zeroes of a polynomial x² – 4√3x + 3, then find the value of α + β – αβ. (1)
Answer:
Let p(x) = x² – 4√3x + 3
Since, α and β are zeroes of p(x), then α + β = sum of zeroes = \(\frac{-(-4 \sqrt{3})}{1}\) = 4√3

And α × β = product zeroes = \(\frac{3}{1}\) = 3
Now, α + β – αβ = 4√3 – 3.

(iii) If two zeroes of the polynomial f(x) = x³ – 4x² – 3x + 12 are √3 and -√3 , then find its third zero. (1)
Answer:
(iii) Let α = √3 and β = – √3 the third zeroes be γ, then
α + β+ γ = –\(\left(\frac{-4}{1}\right)\)
⇒ √3 – √3 + γ = 4
⇒ γ = 4
Third, third root is 4.

(iv) Use elimination method to find all possible solutions of the following pair of linear equations 2x + 3y = 8 and 4x + 6y = 7 : (1)
Answer:
The given equations are :
2x + 3y = 8 …(1)
4x + 6y = 7 …(2)
Multiplying the equation (1) by 2 then subtract equation (2), we get

Which is a false statement.
Therefore, the pair of equations has no solution.

(v) Write the discriminant of the quadratic equation (x + 5)² = 2(5x – 3) (1)
Answer:
The given equaiton is
(x + 5)² = 2(5x – 3)
⇒ x² + 10x + 25 = 10x – 6
⇒ x² + 10x – 10x + 25 + 6 = 0
⇒ x² + 31 = 0
Here, a = 1, b = 0 and c = 31

Discriminant = b² – 4ac
= 0² – 4 × 1 × 31 = – 124

(vi) Find the roots of the quadratic equation 3x² – 2√6x + 2 = 0. (1)
Answer:
The given equation is :
3x² – 2√6x + 2 = 0
⇒ (√3x)² + (√2)² – 2 × √3 × √2x = 0
⇒ (√3x – √2)² = 0
⇒ x = \(\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}\)
Hence, x = \(\sqrt{\frac{2}{3}}\) and x = \(\sqrt{\frac{2}{3}}\) are the roots of the given quadratic equation.

(vii) Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. (1)
Answer:

(viii) Find the area of the triangle formed by the points A (5, 2), B (4, 7) and C (7, – 4). (1)
Answer:
Here, x₁ = 5, y₁ = 2, x₂ = 4, y₂ = 7, x₃ = 7, y₃ = – 4.
∴ Area of ∆ABC
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[5(7 + 4) + 4(- 4 – 2) + 7(2 – 7)]
= \(\frac{1}{2}\)[5 × 11 + 4 × (- 6) + 7 × (- 5)]
= \(\frac{1}{2}\)(55 – 24 – 35) = \(\frac{1}{2}\)(55 – 59)
= \(\frac{1}{2}\) × (- 4) = – 2 = 2 sq. units

(ix) If cosec A = \(\frac{17}{8}\), then calculate tan A. (1)
Answer:
∵ cosec A = \(\frac{17}{8}\)
∴ sin A = \(\frac{8}{17}\)

By using phythagoras Theorem, we have

AC² = AB² + BC²
⇒ (17k)² = AB² + (8k)²
⇒ (17k)² – (8k)² = AB²
⇒ (17k + 8k) (17k – 8k) = AB²
⇒ 25k x 9k = AB²
⇒ 225k² = AB²
⇒ AB = \(\sqrt{225 k^{2}}\) = 15k
∴ tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{8 k}{15 k}=\frac{8}{15}\)

(x) Find the value: \(\frac{2 \cos 67^{\circ}}{\sin 23^{\circ}}-\frac{\tan 40^{\circ}}{\cot 50^{\circ}}\) – cos 0°
Answer:
We have

(xi) Find the mode of the following distribution : (1)

Answer:
The class interval 60 – 80 has maximum frequency. So, it is the modal class.
∴ l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20

Hence, mode = 65.625

(xii) A dice is thrown once, find the probability of getting prime numbers. (1)
Answer:
A dice is thrown once, the possible numbers are : 1, 2, 3, 4, 5, 6
The numbers of all possible outcomes = 6
The prime number are 2, 3, 5.
Let E₂ be event of a getting of a prime number
Number of outcomes to favourable to event = 3
∴ P(E₂) = \(\frac{3}{6}=\frac{1}{2}\)

Section – B

Question 4.
If n is an odd integer, then show that n² – 1 is divisible by 8. (2)
Answer:
We know that any odd positive integer n can be written in form 4q + 1 or 4q + 3
So, according to the question

Case I : When n = 4q + 1
Then, n² – 1 =(4q + 1)² – 1
= 16q² + 89 + 1 – 1
= 8q(2q + 1) …(1)
Which is divisible by 8.

Case II : When n = 4q + 3
Then, n² – 1 = (4q + 3)² – 1
= 16q² + 24q + 9 – 1
=16q² + 24q + 8
=8(2q² + 3q + 1) …(2)
Which is divisible by 8. Therefore, from equations (1) and (2), it is clear that, if n is an odd positive integer, n² – 1 is divisible by 8.

Question 5.
Verify g(x) = x³ – 3x + 1 is a factor of P(x) = x⁵ – 4x³ + x² + 3x + 1 or not. (2)
Answer:
If g(x) = x³ – 3x + 1 is a factor of p(x) = x⁵ – 4x³ + x² + 3x + 1, then p(x) is divisible by g(x) completely i.e., obtained remainder is zero.

Now we divide p(x) by g(x) as follows

Since, the obtained remainder is not zero. So, g(x) is not a factor of p(x).

Question 6.
Find c if the systems of equations ex + 3y + (3 – c) = 0; 12x + cy – c = 0 has infinitely many solutions. (2)
Answer:
Here, a₁ = c, b₁ = 3, c₁ = 3 – c, a₂ = 12, b₂ = c and c₂ = – c.
Condition for infinitely many solution
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

⇒ \(\frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}\)
⇒ Taking \(\frac{c}{12}=\frac{3}{c}\)
⇒ c² = 12 × 3
c² = 36
c = \(\sqrt{36}\)
⇒ c = + 6, -6
Hence, c = ± 6

Question 7.
If ad ≠ bc, then prove that the equation (a² + b²)x² + 2 (ac + bd)x + (c² + d²) = 0 has no real roots. (2)
Answer:
The given equation is :
(a² + b²)x² + 2(ac + bd)x + (c² + d²) = 0

Here, A = (a² + b²), B = 2(ac + bd) and C = (c² + d²)
⇒ B² – 4AC = 0
⇒ 12(ac + bd)]² – 4 × (a² + b²) × (c² + d²) = 0
⇒ 4(a²c² + b²d² + 2abcd) – 4(a²c² + a²d² + b²c² + b²d²) = 0
⇒ 4a²c² + 4b²d² + 8abcd – 4a²c² – 4a²c² – 4b²c² – 4b²d² = 0
⇒ – 4a²d² – 4b²c² + 8abcd = 0
⇒ – 4 (a²d² + b²c² – 2abcd) = 0
⇒ – 4(ad – bc)² = 0
Since ad ≠ bc = 0 Therefore D < 0
Hence, the equation has no real roots.

Question 8.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc …, the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty if he has delayed the work by 30 days? (2)
Answer:
We have,
Penalty for the first day = ₹ 200
Penalty for the second day = ₹ 250
Penalty for the third day = ₹ 300
So, the sequence of penalty is: 200, 250, 300,…………..
Here, a = 200

The penalty for each succeeding day is ₹ 50 more than for each the preceding day.
So, d= 250 – 200 = 50 and n = 30
Now, S₃₀ = \(\frac{n}{2}\) [2 × 200 + (30 – 1) × 50] [∵ S_n = \(\frac{n}{2}\){2a + (n – 1)d}]
⇒ S₃₀= 15 [400 + 29 × 50]
⇒ S₃₀ = 15 [400 + 1450]
⇒ S₃₀ = 15 × 1850
⇒ S₃₀ = 27750
Hence, the penalty paid for 30 days is ₹ 27750.

Question 9.
Find the sum of the odd numbers between 0 and 50. (2)
Answer:
The odd numbers between 0 and 50 are 1, 3, 5, 7,…, 49.
Here we see that a = 1, d = a₂ – a₁
= 3 – 1 = 2, l = a_n = 49
⇒ a + (n – 1 )d = l
⇒ 1 + (n – 1) × 2 = 49
⇒ 1 + 2n – 2 = 49
⇒ 2n – 1 = 49
⇒ n = 25
Now, S_n = \(\frac{n}{2}\)[a + l]
⇒ S₂₅ = \(\frac{25}{2}\)[1 + 49]
⇒ S₂₅ = 25 × 25 = 625
Hence, the sum of odd numbers between 0 and 50 = 625.

Question 10.
Construct an equilateral ∆ABC with each sides 5 cm. Then construct another triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of ∆ABC.
Answer:

Steps of Construction

• Draw a line segment BC = 5 cm
• With B as centre and radius 5 cm draw an arc.
• With C as centre and radius 5 cm draw another arc to intersect the previous arc at A.
• Join AB and AC to get equilateral AABC.
• Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
• Along BX mark 3 points B₁( B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃
• Join B₃C
• From B₂ draw B₂C’ ∥ B₃C meeting BC at C’
• From C’ draw C’A’ ∥ CA meeting BA at A’. Then A’ BC’ is required triangle, each of whose side is \(\frac{2}{3}\) of corresponding sides of ∆ABC.

Question 11.
Construct a pair of tangents to a circle of radius 4 cm from an external point at a distance 6 cm from the centre of the circle. (2)
Answer:
Steps of Construction :

• Draw a circle with O as the centre and radius 4 cm.
• Mark a point P outside the circle such that OP = 6 cm.
• Join OP and draw perpendicular bisector of PO meeting PO at M.

• Draw a circle with M as the centre and radius equal to PM = MO intersecting the given circle at points Q and R.
• Join PQ and PR.

Then PQ and PR are the required tangents.

Question 12.
If sin A + sin² A = 1, then find the value of the expression cos²A + cos⁴A. (2)
Answer:
We have, sin A + sin²A = 1
⇒ sin A = 1 – sin²A
⇒ sin A = cos²A
squaring both sides, we get
sin²A = cos⁴A
⇒ 1 – cos² A = cos⁴ A
⇒ 1 = cos² A + cos⁴ A
⇒ cos² A + cos⁴ A = 1

Question 13.
If A, B and C are interior angles of a triangle ABC, then show that sin\(\frac{B+C}{2}\) = cos\(\frac{A}{2}\). (2)
Answer:
We know that the sum of angles of a triangle is 180°.
∴ A + B + C = 180°
⇒ B + C = 180° – A
⇒ \(\frac{B+C}{2}=\frac{180^{\circ}-A}{2}\)
⇒ \(\frac{B+C}{2}\) = 90° – \(\frac{A}{2}\)
⇒ Sin\(\left(\frac{B+C}{2}\right)\) = Sin(90 – \(\frac{A}{2}\))
⇒ sin \(\left(\frac{B+C}{2}\right)\) = cos \(\frac{A}{2}\) [∵ sin (90° – θ) = cos θ]

Question 14.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. (2)

Which method did you use for finding the mean, and why ? (2)
Answer:

From the table, we have Σf_i = 20, Σf_ix_i = 162
∴ x̄ = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\) = x̄ = \(\frac{162}{20}\) = 8.1
Hence, mean number of plants per house = 81 plants.
We used direct method to find the mean because numerical values of f_is and x_i‘s were small.

Question 15.
The following data gives the information on the observed life times (in hours) of 200 electrical components. (2)

Determine the modal life times of the components.
Answer:
The class interval 80 – 100 ha maximum frequency. So, it is the modal class.
∴ l = 80, f₁ = 65, f₀ = 38, f₂ = 24, h = 20

= 80 + 27 × 20 = 80 + 7.94 = 87.94
Hence, modal lifetimes of component = 87.94 hours.

Question 16.
A piggy bank contains hundred 50 P coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 P coin ? (ii) will not be a ₹ 5 coin ? (2)
Answer:
Total number of coins = 100 + 50 + 20 + 10 = 180
Total number of possible outcomes = 180
(i) Number of 50 paise coins = 100
Number of favourable outcomes = 100
∴ P(50 paise coins) = \(\frac{100}{180}=\frac{5}{9}\)

(ii) Number of coins other than
₹ 5 coins = 100 + 50 + 20 = 170
Number of favourable outcomes = 170
P (will not be a ₹ 5 coins) = \(\frac{170}{180}=\frac{17}{18}\)

Section – C

Question 17.
If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its n terms. (3)
Or
Find the sum of all three digit numbers which one divisible by 11 or multiples of 11. (3)
Answer:
Let first terms be a and common difference be d
S₄ = 40 (given)
⇒ \(\frac{4}{2}\)[2a+ (4 – 1) x d] = 40
⇒ 2 [2a + 3d] = 40
⇒ 2a + 3d = 20 ….(1)
And S₁₄ = 280
⇒ [2a + (14 – 1)d] = 280
⇒ 2a + 13d = \(\frac{280}{7}\)
⇒ 2a + 13d = 40 ………..(2)

Substracting the eq. (2) from eq. (1),
we get

⇒ d = \(\frac{-20}{-10}\) =2
Substituting the value of d in the eq.(1), we get
2a + 3 × 2 =20
⇒ 2a + 6 = 20
⇒ 2a = 20 – 6 = 14
⇒ a = \(\frac{14}{2}\) = 7
its sum of nth term (S_n)
= \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{n}{2}\)[2 × 7 + (n – 1) × 2]
= \(\frac{n}{2}\)[14 + 2n – 2]
= \(\frac{n}{2}\)[12 + 2n]
= n(6 + n) = n² + 6n
Hence, sum of its nth = n² + 6n

Question 18.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. (3)
Or
Find the value of k, the points A (8, 1), B (3, – 4) and C (2, k) are collinear. (3)
Answer:
The given points are P(2, – 3) and Q(10, y)
and PQ = 10 units
∴ PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
⇒ 10 = \(\sqrt{(8)^{2}+y^{2}+9+6 y}\)
⇒ 10 = \(\sqrt{64+y^{2}+9+6 y}\)
⇒ 10 = \(\sqrt{73+y^{2}+6 y}\)
⇒ 100 = 73 + y2 + 6y
⇒ y² + 6y + 73 – 100 = 0
⇒ y² + 6y – 27 = 0
⇒ y² + (9- 3)y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9) (y – 3)= 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = – 9 or y = 3.
Hence, y = – 9 or 3.

Question 19.
Prove the following identify, where the angles involved are acute angles for which the expression is denned \(\frac{1+\cot ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\left(\frac{1-\cot \mathrm{A}}{1-\tan \mathrm{A}}\right)^{2}\) (3)
Or
If sin θ + cosec θ = 2, then find the value of sinⁿθ + cosecⁿθ. (3)
Answer:

Question 20.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency (f). (3)

Or
A survey conducted on 26 households in a locality by a group of students resulted in the following frequencey table for the number of family members in a household : (3)

Answer:
Here, h = 2

From the table, we have a = 18, h = 2, Σf_i = 44 + f, Σf_iu_i = -20 + f and mean (x) = ₹ 18
x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ 18 + \(\left(\frac{-20+f}{44+f}\right)\) × 2
⇒ 0 = \(\frac{-40+2 f}{44+f}\) ⇒ 0 = – 40 + 2f
⇒ 2f= 40 ⇒ f= \(\frac{40}{2}\) = 20
Hence, the missing frequency (f)
= 20.

Section – D

Question 21.
Romila went to a stationery shop and purchased 2 pencils and 3 erasers for ₹ 9. Her friend, Sonali saw the new variety of pencils and erasers with Romila and she also bought 4 pencils and 6 erasers of the same kind for ₹ 18. Represent this situation algebraically and graphically. (4)
Or
Solve graphically the system of equations 3x – 2y – 12 = 0 and 3x + 3y + 3 = 0. Find the coordinate of the vertices of the triangle formed by these two lines and the y-axis. (4)
Answer:
Let the cost of 1 pencil be ? x and cost of 1 eraser be ? y.
According to question,
2x + 3y = 9 …(1)
and 4x + 6y = 18 …(2)
The algebraic representation of this situation is 2x + 3y = 9, 4x + 6y = 18.

For representation of these equations graphically, we draw the graph of these equations as follows :
2x + 3y = 9
⇒ 3y = 9 – 2x
⇒ y = \(\frac{9-2 x}{3}\)
We put the different values of x in this equation then we get different values ofy and we prepare the table of x, y for the equation 2x + 3y = 9.

Table-1

and 4x + 6y = 18
⇒ 6y = 18 – 4x
⇒ y = \(\frac{18-4 x}{6}\)
Now, we put the different values of x in this equation then we get different values of y and we prepare the table of x, y for the equation 4x + 6y = 18.

Table-2

Now, we plot the values of x and y from Table-1 and Table-2 on the graph paper and we draw the graphs those passes through these values.

Question 22.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. (4)
Or
Construct ∆ABC in which BC = 5 cm, ∠CAB = 120° and ∠ABC 30°. Then construct another triangle where side are 4/5 times the corresponding sides of ∆ABC. (4)
Answer:
Steps of Construction :

• Construct ∆ABC such that BC = 8 cm, AB = 6 cm and ∠ABC = 90°.
• Draw BD ⊥ AC.
• ∠CDB = 90° So, BC is the diameter of circle passing through B, C, D
• Bisect line BC at O.
• O as the centre, CO = OB as radius draw a circle which passes through B, C, D.

• Join AO which intersects the circle at M.
• M as centre and MA = OM as radius draw another circle which intersect the previous circle at P and B.
• Join PA.
  Then, AP and AB are the required tangents.

Justification :
∠ABC = 90° [By Construction]
AB ⊥ OB.
OB is radius and AB is tangent. Similarly, AP is tangent.

Question 23.
If the median of the distribution given below is 28-5, find the values of x and y. (4)

Or
The marks distribution of 60 students in science examination are given below in the table. Find the median of this data. Also compare and interpret the mode and the mean. (4)

Answer:
For calculating the median, we prepare the cumulative frequency distribution table as given below :

n = 60
⇒ 45 + x + y = 60
⇒ x + y = 60 – 45
⇒ x + y = 15 …(1)
Median is 28-5, which lies in the class 20 – 30.
So, it is the median class.
∴ l = 20, f = 20, cf = 5 + x and h = 10.

Putting the value of x in equation (1), we get y = 7
⇒ y= 15 – 8 = 7
Hence, x = 8, y = 7