AP 10th Class Maths Model Paper Set 3 with Solutions

Regularly solving AP 10th Class Maths Model Papers Set 3 contributes to the development of problem-solving skills.

AP SSC Maths Model Paper Set 3 with Solutions

Time : 3.15 Hours
Max. Marks : 100

Instructions

1. IN THE DURATION OF 3 HRS. 15 MINUTES, 15 MINUTES OF TIME IS ALLOTTED TO READ THE QUESTION PAPER.
2. ALL ANSWERS SHALL BE WRITTEN IN THE ANSWER BOOKLET ONLY.
3. QUESTION PAPER CONSISTS OF 4 SECTIONS AND 33 QUESTIONS.
4. INTERNAL CHOICE IS AVAILABLE IN SECTION – IV ONLY.
5. ANSWERS SHALL BE WRITTEN NEATLY AND LEGIBLY.

Section – I
(12 × 1 = 12 M)

Note:

1. Answer all the questions in one word or phrase.
2. Each question carries 1 mark.

Question 1.
The smallest number by which 1/13 should be multiplied so that its decimal expansion terminates after two decimal places is ………. (Ch.No-1)
Solution:
13/100

Question 2.
The graph of y = p(x) is given in the following figure, for some polynomial p(x). Find the number of zeroes of p(x). (Ch.No-2)

Solution:
The number of zeroes of p(x) is 3 as the graph of y = p(x) intersects the x-axis at three points.

Question 3.
For what values of p does the pair of equations 4x + py + 8 = 0 and 2x + 2y + 2 = 0 has unique solutions ? (Ch.No-3)
Solution:
The given equations are
4x + py + 8 = 0 and 2x + 2y + 2 = 0
a₁ = 4, b₁ = p, c₁ = 8 and
a₂ = 2, b₂ = 2, c₂ = 2
For unique solution,
\(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\) ⇒ \(\frac{4}{2}\) ≠ \(\frac{\mathrm{p}}{2}\) ⇒ p ≠ 4
Hence, the value of p can be all real number other than 4.

Question 4.
Assertion : If the nth term of an AP be (2n² – 1), then the sum of its first n terms is n³.
Reason: If a, 1 and n are first term, last term and number of terms of an AP, respectively, then S_n = \(\frac{n}{2}\)(a + 1). (Ch.No-5)
Now, choose the correct answer from the following
a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
b) If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
c) If Assertion is correct but Reason is incorrect.
d) If the assertion is incorrect but Reason is correct.
Solution:
a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 5.
The given shapes are mathematically similar. Calculate the unknown side. (Ch.No-6)

Solution:
Given, both figures are similar.
So, the ratio of corresponding sides will be equal.
∴ \(\frac{3}{x}\) = \(\frac{5}{15}\) ⇒ x = \(\frac{3 \times 15}{5}\) = 9 cm

Question 6.
The measure of the angle of elevation of the top of the tower which is 75 \(\sqrt{3}\) m high from a point at a distance of 75 m from the foot of the tower in a horizontal plane is …….. (Ch.No-9)
Solution:
60°

Question 7.
In the following figure, OP = diameter of the circle. Then ∠APB is equal to (Ch.No-10)

(a) 30°
(b) 60°
(c) 45°
(d) none of these
Solution:
(c) 45°

Question 8.
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed, then the volume of remaining solid is (Ch.No-12)
a) 280 πcm³
b) 330 πcm³
c) 240 πcm³
d) 440 πcm³
Solution:
c) 240 πcm³

Question 9.
Statement (A): Probability theory is nothing but common sense reduced to calculation.
Statement (B) : There are certain experiments whose outcomes have equal chance of occurring which are called equally likely outcomes. (Ch.No-14)
A) Both A and B are true
B) A is true, B is false
C) A is false, B is true
D) Both A and B are false
Solution:
A) Both A and B are true

Question 10.
The zeroes of the polynomial x² – 3x – m (m + 3) are (Ch.No-2)
a) m, m + 3
b) -m, m + 3
c) m, -(m + 3)
d) -m, -(m + 3)
Solution:
b) -m, m + 3

Question 11.
Find the value of cos 60° cos 30° + sin 60° sin 30°.
Solution:

Question 12.
Choose the correct matching :

A) A – i, B – ii, C – iii
B) A – ii, B – iii, C – i
C) A – i, B – ii, C – iii
D) A – iii, B – ii, C – i
Solution:
A) A – i, B – ii, C – iii

Section – II
(8 × 2 = 16M)

Note:

1. Answer all the questions.
2. Each question carries 2 marks.

Question 13.
Two cones with same base radius 8 cm and height 15 cm are joined together along their bases.
Find the surface area of the shape so formed. (Ch.No-12)
Solution:
Hint : If two cones with same base and height are joined together along their bases, then the shape so formed looks likes as figure shown.

Given that radius of cone, r = 8 cm and height of cone h = 15 cm
So, Surface area of the shape so formed = Curved area of first cone + Curved surface area of second cone
= 2 (Curved surface area of cone)
[Since , both cones are identical]
= 855 cm² (approx.)

Question 14.
Find a quadratic polynomial, the sum and product of whose zeroes are 0 and \(\frac{4}{5}\) respectively. Hence find the zeroes. (Ch.No-2)
Solution:
The required quadratic polynomial is K[x² – (Sum of the zeroes) x + Product of the zeroes],
where K(≠ 0) is real.
= K[x² – (0)x +\(\frac{4}{5}\)]
= \(\frac{K}{5}\)(5x² + 4)
Zeroes are given by
\(\frac{K}{5}\) (5x² + 4) = 0
⇒ 5x² + 4 = 0
which gives no real values of x.
So, this polynomial does not have any real zero.

Question 15.
Find the roots of the quadratic equation x² – 3x + 2 = 0. (Ch.No-4)
Solution:
The given quadratic equation is
x² – x – 2x + 2 = 0
Splitting the middle term – 3x,
x² – x – 2x + 2 = 0
⇒ x(x – 1) – 2(x – 1) = 0
⇒ (x – 1)(x – 2) = 0
⇒ x – 1 = 0 or x – 2 = 0
⇒ x = 1 or x = 2
⇒ x = 1, 2
Hence, 1 and 2 are the roots of the quadratic equation x² – 3x + 2 = 0.

Question 16.
In the given figure, AD = 2 cm, BD = 3 cm, AE = 3.5 cm and AC = 7 cm. Is DE parallel to BC ?
(Ch.No-6)

Solution:
\(\frac{A D}{B D}\) = \(\frac{2}{3}\)
∴ CE = AC – AE = 7 – 3.5 = 3.5 cm
So, \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = \(\frac{3.5}{3.5}\) = 1 ; \(\frac{\mathrm{AD}}{\mathrm{BD}}\) ≠ \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
We know that ifa line does not divides any two sides of a triangle in the same ratio, then the line is not parallel to the third side.
Hence, DE is not parallel to BC.

Question 17.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3). (Ch.No-7)
Solution:

Question 18.
If sin θ = \(\sqrt{3}\) cos θ, find the value of \(\frac{\tan \theta-1}{\tan \theta+1}\).
Solution:
sin θ = \(\sqrt{3}\) cos θ

Question 19.
A peacock is sitting on the top of a tree. It observes a serpent on the ground making an angle of depression of 30°. The peacock catches the serpent in 12s with the speed of 300 m/min. What is the height of the tree ?(CBSE 2015) (Ch.No-9)
Solution:
Hint: Let C be the position of peacock and A be the position of serpent.

Question 20.
In the given figure, if ∠ACB = 50°, then find ∠ATO. (Ch.No-10)

Solution:
Hint: ∠OAT = 90°
[∵ angle between radius and tangent]
Now, ∠BOA = 100° [angle subtended by an arc at centre is twice the angle subtended at remaining part of circle]
⇒ ∠ATO = 180° – (∠TOA) + (∠OAT)
[∵ angles property of a triangle]
= 40°

Section – III
(8 × 4 = 32 M)

Note:

1. Answer all the questions.
2. Each question carries 4 marks.

Question 21.
Two different dice are thrown together. Find the probability that the numbers obtained have
i) even sum, and (Ch.No-14)
ii) even product
Solution:
Number of all possible outcomes = 6 × 6 = 36

i) Let E₁ be the event that the numbers obtained have even sum.
Then, outcomes favourable to E₁ are
(1, 1), (1, 3), (1, 5), (2, 2), (2, 4),
(2, 6), (3, 1), (3, 3), (3, 5), (4, 2)
(4, 4), (4, 6) (5, 1), (5, 3), (5, 5)
(6, 2), (6, 4) and (6, 6)
∴ Number of outcomes favourable to E₂ is 18.
∴ P(E₁) \(=\frac{\text { Number of outcomes favourable to } \mathrm{E}_1}{\text { No. of all possible outcomes }}\)
= \(\frac{18}{36}\) = \(\frac{1}{2}\)

ii) Let E₂ be the event that the numbers obtained have even product.
Then outcomes favourable to E₂ are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3)
(2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3)
(6, 4), (6, 5) and (6, 6).
∴ Number of outcomes favourable to E₂ = 27
∴ P(E₂) \(=\frac{\text { No. of outcomes favourable to } \mathrm{E}_2}{\text { No. of all possible outcomes }}\)
= \(\frac{27}{36}\) = \(\frac{3}{4}\)

Question 22.
The given distribution shows the number of runs scored by some top batsmen of the world in
one-day international cricket matches. (Ch.No-13)

Find the mode of the data.
Solution:
Here the maximum frequency from the given table is f₁ = 18 and hence the maximum number of batsmen have their runs scored in the corresponding interval 4000 – 5000, so the modal class is 4000 – 5000.
Therefore, l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9

Hence, the mode of the data is 4608.7 runs (approx.)

Question 23.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. (Ch.No-12)
Solution:
Given: radius of metallic sphere = 4.2 cm.

∴ Volume = \(\frac{4}{3}\)π(4.2)³ ………. (i)
∴ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder = πr²h = π(6)² × h …… (ii)
According to question,
Volume of the cylinder = Volume of the sphere
310.46 cm³ = \(\frac{22 \times 36}{7}\) h
310.46 cm³ = 113.142 cm³h
h = \(\frac{310.464}{113.142}\) cm³
h = 2.74 cm.
Height of cylinder = 2.74 cm.

Question 24.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 13. Find his present age. (Ch.No-4)
Solution:
Let the present age of Rehman be x years
3 years ago Rehman’s age was (x – 3) years
5 years from now Rehman’s age will be = (x + 5) years
According to question,
\(\frac{1}{x-3}+\frac{1}{x+5}\) = \(\frac{1}{3}\)
⇒ \(\frac{x+5+x-3}{(x-3)(x+5)}\) = \(\frac{1}{3}\)
⇒ \(\frac{2 x+2}{x^2+2 x-15}\) = \(\frac{1}{3}\)
⇒ 6x + 6 = x² + 2x – 15
⇒ x² – 4x – 21 = 0
⇒ x² – 7x – 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x + 3) (x – 7) = 0
⇒ x + 3 = 0 or x – 7 = 0
⇒ x = 7 or -3 (-3 is rejected)

Question 25.
In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. (Ch.No-8)
Solution:
Let QR = x and PR = y
Then, PR + QR = 25 [given]
⇒ y + x = 25
⇒ y = 25 – x ……..(i)
In right angled ∆PQR,

Question 26.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP. (Ch.No-5)
Solution:
Let a be the first term and d be the common difference of given AP.
Given, a₄ + a₈ = 24
∴ (a + 3d) + (a + 7d) = 24
[∵ a_n = a + (n – 1)d] …… (i)
⇒ a + 5d = 12
[dividing both sides by 2]
and a₆ + a₁₀ = 44
⇒ (a + 5d) + (a + 9d) = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22 ……. (ii)
[dividing both sides by 2]
On subtracting Eq. (i) from Eq. (ii), we get
2d = 10 ⇒ d = 5
On putting d = 5 in Eq. (i), we get
a + 25 = 12 ⇒ a = -13
Hence, the first three terms are
a, (a + d) and (a + 2d)
i.e., -13, (-13 + 5) and (-13 + 2 × 5)
i.e., -13, -8 and -3.

Question 27.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. (Ch.No-10)
Solution:
Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

To prove : ∠QOR = 180° – ∠QPR
or ∠QOR + ∠QPR = 180°
Proof: In ∆OQP and ∆ORP,
PQ = PR [∵ tangents drawn from an external point are equal in length]
OQ = OR [radii of circle]
OP = OP [common sides]
∴ ∆OQP ≅ ∆ORP [by SSS congruence rule]
Then, ∠QPO = ∠RPO [by CPCT]
and ∠POQ = ∠POR [by CPCT]
⇒ ∠QPR = 2∠OPQ
and ∠QOR = 2∠POQ ……. (i)
Now, in right angled ∆OQP, ∠QPO + ∠QOP = 90°
⇒ ∠QOP = 90° – ∠QPO
⇒ 2∠QOP = 180° – 2∠QPO
[multiplying both sides by 2]
⇒ ∠QOR = 180° – ∠QPR [from Eq.(i)]
⇒ ∠QOR + ∠QPR = 180°
Hence proved.

Question 28.
Find the zeroes of the quadratic polynomial 2x² – (2\(\sqrt{3}\) – 1)x – \(\sqrt{3}\) and verify the relationship between the zeroes and the coefficients. (Ch.No-2)
Solution:
Let p(x) = 2x² – (2\(\sqrt{3}\) – 1)x – \(\sqrt{3}\)
Zeroes of p(x) are given by p(x) = 0
⇒ 2x² – (2\(\sqrt{3}\) – 1)x – \(\sqrt{3}\) = 0
⇒ 2x² – 2\(\sqrt{3}\)x + x – \(\sqrt{3}\) = 0
⇒ 2x(x – \(\sqrt{3}\)) + 1(x – \(\sqrt{3}\)) = 0
⇒ (x – \(\sqrt{3}\))(2x + 1) = 0
⇒ x – \(\sqrt{3}\) = 0 (or) 2x + 1 = 0
⇒ x = \(\sqrt{3}\) (or) x = –\(\frac{1}{2}\)
⇒ x = \(\sqrt{3}\), –\(\frac{1}{2}\)
Hence the zeroes of p(x) are \(\sqrt{3}\) and –\(\frac{1}{2}\)
Comparing p(x) = 2x² – (2\(\sqrt{3}\) – 1)x – \(\sqrt{3}\) with ax² + bx + c, we get
a = 2, b = -(2\(\sqrt{3}\) – 1), c = –\(\sqrt{3}\)
Now,
Sum of zeroes
= (\(\sqrt{3}\)) + (-\(\frac{1}{2}\)) = \(\frac{2 \sqrt{3}-1}{2}\) = \(-\frac{b}{a}\)
product of zeroes
= (\(\sqrt{3}\))(-\(\frac{1}{2}\)) = –\(\frac{\sqrt{3}}{2}\) = \(\frac{c}{a}\)
Hence the relationship between the zeroes and the coefficients is verified.

Section – IV
(5 × 8 = 40 M)

Note:

1. Answer all the questions.
2. Each question carries 8 marks.
3. There is an internal choice for each question.

Question 29.
a) Let a, b, c and p be the rational numbers such that p is not a perfect cube if a + bp^1/3 + cp^2/3 = 0, then prove that a = b = c. (Ch.No-1)
(OR)
b) CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides
AB and FE of ∆ABC and ∆EFG, respectively. If ∆ABC ~ ∆FEG, show that (Ch.No-6)
i) \(\frac{\mathrm{CD}}{\mathrm{GH}}\) = \(\frac{\mathrm{AC}}{\mathrm{FG}}\)
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Solution:
a) Hint : We have, a + bp^1/3 + cp^2/3 = 0 …… (i)
Multiplying both sides by p^1/3, we get
ap^1/3 + bp^2/3 + cp = 0 …… (ii)
Multiplying Eq. (i) by b and Eq. (ii) by c and subtracting, we get
(ab + b²p^1/3 + bcp^2/3) – (acp^1/3 + bcp^2/3 + c²p) = 0
⇒ (b² – ac)p^1/3 + ab – c²p = 0
⇒ b² – ac = 0 and ab – c²p = 0
[∵ p^1/3 is irrational]
⇒ b² = ac and ab = c²p
⇒ a² (ac) = c⁴p² [putting b² = ac in a²b² = c⁴p²]
⇒ a³c – c⁴p² = 0 ⇒ (a³ – c³p²)c = 0
⇒ a³ = p²c³ or c = 0 ⇒ p² = \(\frac{a^3}{c^3}\)
⇒ (p²)^1/3 = \(\left(\frac{a^3}{c^3}\right)^{1 / 3}\) = \(\frac{a}{c}\) ⇒ (p^1/3)² = \(\frac{a}{c}\)
This is not possible as p^1/3 is irrational and \(\frac{a}{c}\) is rational.
∴ a³ – p² c³ ≠ 0, hence c = 0.
Putting c = 0 in b² – ac = 0, we get b = 0
b = 0, c = 0, then a = 0
Hence, a = b = c = 0.

OR

b) Draw ∆ABC and ∆FEG and then draw bisectors CD and GH of ∠ACB and ∠EGF such that D lies on AB and H lies on FE.

i) Given, ∆ABC ~ ∆FEG
So, all corresponding angles are equal.
∴ ∠CAB = ∠GFE
or ∠CAD = ∠GFH ……. (i)
and ∠ACB = ∠FGE
⇒ \(\frac{1}{2}\)∠ACB = \(\frac{1}{2}\)∠FGE
[dividing both sides by 2]
⇒ ∠ACD = ∠FGH …….. (ii)
From Eqs. (i) and (ii),
∆ACD ~ ∆FGH
[by AA similarity citerion]
∴ \(\frac{\mathrm{CD}}{\mathrm{GH}}\) = \(\frac{\mathrm{AC}}{\mathrm{FG}}\)
[since, corresponding sides of two similar triangles are proportional]

ii) In ∆DCB and ∆HGE, ∠DBC = ∠HEG …… (iii)

From Eqs. (iii) and (iv),
∆DCB ~ ∆HGF
[by AA similarity criterion]

iii) In ∆DCA and ∆HGF
∠DAC = ∠HFG ……. (v)

Question 30.
a) Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3). (Ch.No-7)
(OR)
b) Rachel, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm, and its length is 12 cm. (Ch.No-12)
If each cone has a height of 2 cm, then find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model are nearly the same.)
Solution:
a) We know that equation of circle is
x² + y² + 2gx + 2fy + c = 0
where (-g, -f) are the centre of circle.
(6, -6) passes through the circle
36 + 36 + 12 g + 2f(-6) + r = 0
72 + 12g – 12f + c = 0 ……(1)
(3, -7) passes through the circle
(3)² + (-7)² + 2g(3) + 2f(-7) + c = 0
9 + 49 + 6g – 14f + c = 0
58 + 6g – 14f + c = 0 …….(2)
(3, 3) passes through the circle
(3)² + (3)² + 2g(3) + 2 f(3) + c = 0
18 + 6g + 6f + c = 0 ……. (3)
From (2) and (3)’

(OR)

b) Given, model is a combination of a cylinder and two cones. Clearly, volume of the air will be equal to the sum of the volumes of two cones and one cylinder.

Question 31.
a) A die is thrown once. Find the probability of getting (Ch.No-14)
(i) a prime number;
(ii) a number lying between and ;
(iii) an odd number.
(OR)
b) Two poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. (Ch.No-9)
Solution:
a) Number of all possible outcomes (1, 2, 3, 4, 5, 6) = 6

i) Let E₁ be the event of getting a prime number.
(A natural number p > 1 is said to be prime if only divisors of p are 1 and p).
Then, the outcomes favourable to E₁ are
Therefore, the number of outcomes favourable to E₁ is 3.
So, P(E₁) =

ii) Let E₂ be the event of getting a number lying between and .
Then, the outcomes favourable to E₂ are Therefore, the number of outcomes favourable to E₂ is 3.
So, P(E₂) =

iii) Let E₃ be the event getting an odd number.
Then, the outcomes favourable to E₃ are Therefore, the number of outcomes favourable to E₃ is 3.
So, P(E₃) =

Remark : for part (ii):

(i) “Between” 2 and 6
⇒ Excluding 2 and 6, i.e., 3, 4, 5
(ii) “From” 2 to 6
Including 2 and 6, i.e., 2, 3, 4, 5, 6.

(OR)

b)

Let AB = 80 m be the width of the road.
On both sides of the road,
poles AE = BD = h m are standing. Let C be any point on AB such that from point C, angles of elevation are
∠BCD = 60°, and
∠ACE = 30°.
Let BC = x m.
Then, AC = AB – BC = (80 – x)m
In right angled ∆CAE, tan 30° = \(\frac{P}{B}\) = \(\frac{\mathrm{AE}}{\mathrm{AC}}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{80-x}\) [∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]
⇒ 80 – x = h \(\sqrt{3}\) ⇒ h \(\sqrt{3}\) + x = 80 ….. (i)
and in right angled ∆ CBD,
tan 60° = \(\frac{\mathrm{BD}}{\mathrm{BC}}\) ⇒ \(\sqrt{3}\) = \(\frac{h}{x}\) [∵ tan 60° = \(\sqrt{3}\)]
⇒ h = \(\sqrt{3}\)x ……….. (ii)
On putting h = \(\sqrt{3}\)x in Eq. (i), we get
\(\sqrt{3}\)x (\(\sqrt{3}\)) + x = 80 ⇒ 3x + x = 80
⇒ 4x = 80
⇒ x = 20 m
On putting x = 20 m in Eq. (ii), we get
h = 20 \(\sqrt{3}\) m
Now, AC = 80 – x = 80 – 20 = 60 m
Hence, height of the poles is 20 \(\sqrt{3}\) m and the distances of the point C from the poles
are 60 m and 20 m.

Question 32.
a) The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the
data obtained is represented in the following table: (Ch.No-13)

Find the median length of the leaves.
(OR)
b) Find the 20^th term from the last term of the AP : 3, 8,13, …….., 253. (Ch.No-5)
Solution:
a) The data is in the form of discontinuous classes and hence needs to be converted to continuous classes for finding the median, since the formula for median assumes continuous classes.
For doing this let us subtract 0.5 from lower limit of each class (Rule: LS) and add 0.5 to upper limit of each class-interval (Rule: UA)
Hence the discontinuous classes change to continuous classes
117.5 – 126.5, 126.5 – 135.5, …… 171.5 – 180.5
∴ The given distribution after converting discontinuous classs to continuous classes is

Now, N = Σf_i = 40
So, \(\frac{\mathrm{N}}{2}\) = \(\frac{40}{2}\) = 20
Cumulative frequency just > \(\frac{\mathrm{N}}{2}\) = 20 is 29.
The corresponding class interval to this cf 29 is 144.5 – 153.5
So, 144.5 – 153.5 is the median class.
Therefore, l = 144.5, h = 9, f = 12, cf = 17
∴ Median = l + \(\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\) × h = 144.5 + \(\left(\frac{20-17}{12}\right)\) × 9
= 144.5 + \(\frac{1}{4}\) × 9
= 144.5 + 2.25
= 146.75 mm.
Hence, the median length of the leaves is 146.75 mm.

(OR)

b) The given AP is 3, 8, 13, …….., 253
Here, a = 3
d = 8 – 3 = 5
l = 253
Let the number of terms of the AP be n.
Then, nth term = l = 253
⇒ 3 + (n – 1) 5 = 253
| ∵ a_n = a + (n – 1) d
⇒ (n – 1) 5 = 253 – 3
⇒ (n – 1)5 = 250
⇒ n – 1 = \(\frac{250}{5^{\prime}}\)
⇒ n – 1 = 50
⇒ n = 50 + 1
⇒ n = 51
So, there are 51 terms in the given AP.
Now, m = 20^th term from the last term
= (n – m + 1)^th
= (51 – 20 + 1)^th
term from the beginning
(Remark : See 5.3 (2))
= 32^th term from the beginning
= 3 + (32 – 1)5 |∵ a_n = a + (n – 1)d
= 3 + 155 = 158
Hence, the 20^th term from the last term of the given AP is 158.
Alternative Solution : Let us write the given AP in the reverse order. Then the AP becomes
253, 253 – 5 = 248, 243, …….. 3
Here, a = 253
d = a₂ – a₂ = 248 – 253 = -5
Therefore, required term
= 20^th term of this new AP
= 253 + (20 – 1) (- 5)
∵ a_n = a + (n – 1) d
= 253 + 19(-5) = 253 – 95 = 158
Hence, the 20^th term from the last term of the given AP is 158.

Question 33.
a) Solve the following pairs of equations by reducing them to a pair of linear equations:

i) \(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}\) = \(\frac{13}{6}\)
ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1
(OR)
b) The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju. (Ch.No-3)
Solution:
a)
i) We have:
\(\frac{1}{2 x}+\frac{1}{3 y}\) = 2 ……. (1)
\(\frac{1}{3 x}+\frac{1}{2 y}\) = \(\frac{13}{6}\) ……. (2)
Substituting \(\frac{1}{\mathrm{x}}\) = u and \(\frac{1}{\mathrm{y}}\) = v in equations (1) and (2), we have :
\(\frac{u}{2}+\frac{v}{3}\) = 2 …… (3)
and \(\frac{u}{3}+\frac{v}{2}\) = \(\frac{13}{6}\) ……… (4)
Multiplying both sides of both equations by 6, we get:
3u + 2v = 12 ……. (5)
2u + 3v = 13 ……. (6)
Now multiplying equation (5) by 3 and (6) by 2, and then subtracting the results, we get:

ii) We have : \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2 …… (1)
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1 …… (2)
Substituting \(\frac{1}{\sqrt{\mathrm{x}}}\) = u and \(\frac{1}{\sqrt{\mathrm{y}}}\) = v in equations (1) and (2), we get:
2u + 3v = 2 …….. (3)
4u – 9v = -1 ……… (4)
Multiplying equation (3) by 3 and adding the result to equation (4), we get :

b) Let the ages of Ani and Biju be x years y years respectively.
If Ani is older than Biju
x – y = 3
If Biju is older than Ani
y – x = 3
-x + y = 3 [Given}
Dharm’s age = 2x years and Cathy’s age = \(\frac{y}{2}\) years
Clearly, Dharam is older than Cathy.
∴ 2x – \(\frac{y}{2}\) = 30
⇒ 4x – y = 60
Thus, we have the following two systems of linear equations:
x – y = 3 …….(i)
4x – y = 60 ……(ii)
And x – y = -3 ……. (iii)
4x – y = 60 ……(iv)
subtracting equation (i) from equation (ii) we get: 3x – 57 x = 19
Putting x = 19 in equation (i),
we get 19 – y = 3 ⇒ y = 16
Again subtracting equation (iv) from equation (iii), we get 3x = 63
⇒ x = 21
Putting x = 21 in equation (iii) we get
21 – y = -3
⇒ y = 24
Hence, Ani’s age is either 19 years or 21 years and Biju’s age is either 16 years or 24 years.