AP 10th Class Maths Model Paper Set 4 with Solutions

Regularly solving AP 10th Class Maths Model Papers Set 4 contributes to the development of problem-solving skills.

AP SSC Maths Model Paper Set 4 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

Note :

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

Question 1.
Find the value of x in the factor tree given below :

Solution:
The given factor tree when completed becomes.

Question 2.
Choose the correct matching.

(A) A-iii, B-i, C-ii
(B) A-ii, B-iii, C-i
(C) A-ii, B-i, C-iii
(D) A-iii, B-i, C-ii
Solution:
(B) A-ii, B-iii, C-i

Question 3.
Statement (A): Every solution of the equation is a point on the line representing it. (Ch.No-3) Statement (B): (1,1) is a solution of 2x+3y = 5 so (1,1) is a point on the line 2x + 3y = 5.
(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 4.
The AP whose sum of n terms is 2n² + n is given by (Ch.No-5)
a) 3,7,11,15, ………..
b) 3, 6, 9,12, ………
c) 3,8,13,18, ………
d) 3, 4, 5, 6, ………..
Answer:
a) 3,7,11,15, ………..

Question 5.
Assertion; In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², then ∠ACD = 90°.
Reason : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. (Ch.No-6)
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 Assertion is incorrect but Reason is correct.
Answer:
a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 6.
If ∆ ABC ~ ∆ PQR, perimeter of ∆ ABC = 32 cm, perimeter of ∆ PQR = 48 cm and PR = 6 cm, then the length of AC is equal to : (Ch.No-9)
a) 9 cm
b) 4 cm
c) 8 cm
d) 18 cm
Solution:
b) 4 cm

Question 7.
In the following figure, find AC if AB = 5 cm (Ch.No-10)

a) 2.5 cm
b) 3 cm
c) 2 cm
d) 1 cm
Solution:
c) 2 cm

Question 8.
The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10
cm, then its lateral surface area is ………. (Ch.No-12)
Answer:
260 πcm²

Question 9.
A child has a block in the shape of a cube with one letter written on each face as follows :

The cube is thrown once. What is the probability of getting A ?
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{4}\)
Solution:
(a) \(\frac{1}{3}\)

Question 10.
If the product of zeroes of the polynomial ax² – 6x – 6 is 4, find the value of a. (Ch.No-2)
Solution:
The given polynomial is
ax² – 6x – 6
Comparing it with Ax² + Bx + C, we get
A = a, B = -6, C = -B
According to the question,
Product of zeroes = 4
⇒ \(\frac{C}{A}\) = 4 ⇒ \(\frac{-6}{\mathrm{a}}\) = 4
⇒ a = –\(\frac{6}{4}\) = –\(\frac{3}{2}\)

Question 11.
If 7 tan θ = 4, then find the value of \(\frac{7 \sin \theta-3 \cos \theta}{7 \sin \theta+3 \cos \theta}\) (Ch.No-8)
Solution:
Hint: tan θ = \(\frac{4}{7}\), now divide numerator and denominator by cos θ, we get \(\frac{7 \tan \theta-3}{7 \tan \theta+3}\). Then, put value of tan θ and simplify = \(\frac{1}{7}\)

Question 12.
The ratio of the roots of the Quadratic equation x² + 12x + 35 = 0 is ……… (Ch.No-4)
Solution:
7 : 5

Section – II
(8 × 2 = 16 M)

Note:

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

Question 13.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it. (Ch.No-12)

Solution:
Radius of open side (r₁) = 10 cm
Radius of upper base (r₂) = 4 cm
Slant height (l) = 15 cm
Area of material used for making = Curved surface area of frustum of cone + area of closed side

Question 14.
If and β are the zeroes of the polynomial 2x² – 4x + 5, then find the value of α³ + β³. (Ch.No-2)
Solution:
If α and β are the zeroes of the polynomial 2x² – 4x + 5, then
α + β = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)
= \(\frac{-(-4)}{2}\) = 2 ……. (1)
and αβ = \(\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^2}\)
= \(\frac{5}{2}\) ……. (2)
Now, α³ + β³
= (α + β)³ – 3αβ(α + β)
= (2)³ – 3(\(\frac{5}{2}\))(2)
[From (1) and (2)]
= 8 – 15 = -7

Question 15.
If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.
(Ch.No-4)
Solution:
Hint: Given quadratic equation is
x² – 2kx – 6 = 0 ……… (i)
Since, x = 3 is one of the root of the given quadratic equation. Then, it satisfies the given equation.
So, put x = 3 in Eq. (i), we get
(3)² – 2k(3) – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ 6k = 3
⇒ k = \(\frac{1}{2}\)

Question 16.
Hint : Give two different examples of pair of (Ch.No-6)
i) similar figures
ii) non-similar figures.
Solution:
i)

• Pair of the equilateral triangles are similar figures.
• Pair of the squares are similar figures.

ii)

• A triangle and a quadrilateral form a pair of non-similar figures.
• A square and a circle form a pair of non-similar figures.

Question 17.
Find the area of the triangle whose vertices are : (Ch.No-7)
i) (2, 3), (-1, 0), (2,-4)
ii) (-5, -1), (3, -5), (5, 2)
Solution:
i) Vertices of the triangle :

ii) Vertices of the triangle :

Question 18.
If \(\) tan θ = 3 sin θ, then find the value of sin² θ – cos² θ. (Ch.No-8)
Solution:

Question 19.
If two towers of heights x m and y m subtend angles of 30° and 60° respectively at the centre of a line joining their feet, then find the ratio of x : y. (Ch.No-9)
Solution:
Hint: Let AB be the tower of height x m, and CD be the tower of height y m.

Let E be the mid-point of the line AC.
Then, ∠AEB = 30° and ∠CED = 60°.
Also, AE = EC = a m (let)
In right angled ∆BAE, tan 30° = \(\frac{P}{B}\) = \(\frac{\mathrm{AB}}{\mathrm{AE}}\) = \(\frac{x}{a}\) and in right angled ∆DCE,
tan 60° = \(\frac{\mathrm{DC}}{\mathrm{CE}}\) = \(\begin{aligned}
& \frac{y}{a} \\
&
\end{aligned}\) = 1 : 3

Question 20.
PA is a tangent to the circle with centre O. If BC = 3 cm, AC = 4 cm and ∆ACB ~ ∆PAO, then find OA and \(\frac{\mathrm{OP}}{\mathrm{AP}}\) (Ch.No-10)

Solution:

Section – III
(8 × 4 = 32 M)

Note:

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

Question 21.
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. (Ch.No-14)
Solution:
Number of all possible outcomes
= 4 × 4 = 16

Let E be the event that product of x and y is less than 16.
Then, outcomes favourable to E are (1, 1), (2, 1), (3, 1), (4, 1), (1, 4), (2, 4), (3, 4) and (1, 9).
∴ Number of outcomes favourable to E is 8.
∴ P(E) = \(\frac{\text { Number of outcomes favourable to } \mathrm{E}}{\text { No. of all possible outcomes }}\)
= \(\frac{8}{16}\) = \(\frac{1}{2}\)

Question 22.
To find out the concentration of SO₂ in the air (in parts power million i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: (Ch.No-13)

Find the mean concentration of SO₂ in the air.
Answer:
Table for the given data is

We have, a = 0.10, h = 0.04, N = 30 and Σf_iu_i = -1
By step deviation method,
Mean
\((\overline{\mathrm{x}})\) = a + \(\left(\frac{\Sigma f_i u_i}{N}\right)\) × h = 0.10 + \(\left(\frac{-1}{30}\right)\) × 0.04
= 0.10 – \(\frac{0.04}{30}\) = 0.10 – 0.001 = 0.099
Hence, the mean concentration of SO₂ in air is 0.099 ppm.

Question 23.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. (Ch.No-12)
Solution:
Given, solid is a combination of a cone and a hemisphere,

Also, radius of the cone, r = Radius of the hemisphere = 1 cm.
Height of the cone, h = 1 cm
∴ Required volume of the solid = Volume of the cone + Volume of the hemisphere

Question 24.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in ₹) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, then find the number of articles produced and the cost of each article. (Ch.No-4)
Solution:
Let the number of pottery articles produced on a particular day be x.
Then, cost of production of each article = ₹x(2x + 3)
[by given condition]
So, the total cost of production Number of pottery articles x Cost of production of each article = ₹ x(2x + 3)
According to the question,
x(2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
[∵ 15 × (-12) = -180 and 15 – 12 = 3]
⇒ x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
⇒ 2x + 15 = 0 or x – 6 = 0
⇒ x = –\(\frac{15}{2}\) or x = 6
But x cannot be negative, as number of pottery articles should be positive.
∴ x = 6
Hence, the number of articles produced = 6
and the cost of each article = 2 × 6 + 3
= ₹ 15

Question 25.
If cot θ = \(\frac{7}{8}\), evaluate
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
ii) cot² θ
Solution:
Given, cot θ = \(\frac{7}{8}\) ⇒ \(\frac{B}{P}\) = \(\frac{7}{8}\)
Let, B = 7k and P = 8k
where, k is any positive integer.
Draw a right angled ∆PQR, right angled at Q.
In right angled ∆WQR, H² = B² + P²
⇒ H² = (7k)² + (8k)²
[by using Pythagoras theorem]
⇒ H² = 49k² + 64k² = 113k²
⇒ H = k \(\sqrt{113}\)
[taking positive square root since, side cannot be negative]

Question 26.
Determine the AP whose 3^rd term is 16 and the 7^th term exceeds the 5^th term by 12. (Ch.No-5)
Solution:
Let a be the first term and d be the common difference of given AP.
Given that the third term of the AP is
a₃ = 16
⇒ a + 2d = 16 [∵ a_n = a + (n – 1)d] …….(i)
Also, it is given that
7^th term of an AP = 12 + 5^th term of an AP i.e., a₇ = 12 + a₅ ⇒ a₇ – a₅ = 12
⇒ (a + 6d) – (a + 4d) = 12
⇒ 2d = 12 ⇒ d = 6
On putting d = 6 in Eq. (i), we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
We know that general form of an AP is a, a + d, a + 2d, a + 3d,…
Then, the required AP is
4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6,….
i.e., 4, 10, 16, 22,….

Question 27.
In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°. (Ch.No-10)

Solution:
Given XY and X’Y’ are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X’Y’ at B.
To prove: ∠AOB = 90°
Proof : We know that, tangents drawn from an external point to a circle are equal in length.
∴ AP = AC [∵ A is an external point]……(i)
Thus, in ∆APO and ∆ACO,
AP = AC [from Eq. (i)]
AO = AO [common sides]
OP = OC [radii of circle]
∆APO ≅ ∆ACO [by SSS congruence rule]
Then, ∠OAP = ∠OAC [by CPCT] …… (ii)
⇒ ∠PAC = 2∠CA0 …….(iii)
Similarly, we can prove that
∠CBO = ∠OBQ
⇒ ∠CBQ = 2 ∠CBO ….. (iv)
Since, XY || X’Y’ [given]
∴ ∠PAC + ∠QBC = 180°
[∵ sum of interior angles on the same side of transversal is 180°]
⇒ 2∠CAO + 2∠CBO = 180°
[from Eqs. (iii) and (iv)]
⇒ ∠CAO + ∠CBO = 90°
Now, in ∆AOB, ∠CAO + ∠CBO + ∠AOB = 180°
[by angle sum property of triangle]
⇒ ∠CAO + ∠CBO = 180° – ∠AOB …….(vi)
From Eqs. (y) and (vi), we get
180° – ∠AOB = 90° ⇒ ∠AOB = 90°
Hence proved.

Question 28.
Find the zeroes of the polynomial x² + 7x – 8 and verify the relationship between the zeroes and the coefficients. (Ch.No 2)
Solution:
Let p(x) = x² + 7x – 8
Zeroes of p(x) are given by
p(x) = 0
⇒ x² + 7x – 8 = 0
⇒ x² + 8x – x – 8 = 0
⇒ x(x + 8) -1(x + 8) = 0
⇒ (x + 8) (x – 1) = 0
⇒ x + 8 = 0 or x – 1 = 0
⇒ x = -8 or x = 1
⇒ x = -8, 1
Hence the zeroes of p(x) are -8 and 1.
Comparing p(x) = x² + 7x – 8 with ax² + bx + c, we get a = 1, b = 7, c = -8
Now,
Sum of zeroes = (-8) + 1 = -7 = \(²\)
Product of zeroes = (-8) (1) = -8 = \(\frac{c}{a}\)
Hence the relationship between the zeroes and the coefficients is verified.

Section – IV

Note:

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

Question 29.
a) Prove that (\(\sqrt{\mathrm{p}}\) + \(\sqrt{\mathrm{q}}\)) is irrational, where p and q are primes. (Ch.No-1)
(OR)
b) In the given figure, altitudes AD and CE of ∆ABC intersect each other at the point P. (Ch.No-6)

Show that
i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆ADB
iv) ∆PDC ~ ∆BEC
[Use AAA similarity criterion to prove the result.
Solution:
Given, AD and CE are altitudes which intersect each other at the point P.

i) In ∆AEP and ∆CDP,
∠AEP = ∠CDP [each 90°
and ∠APE = ∠CPD
[vertically opposite angles]
∴ ∆AEP ~ ∆CDP
[by AA similarity criterion]

ii) If ∆ABD and ∆CBE,
∠ADB = ∠CEB [each 90°
and ∠ABD = ∠CRE [common angles]
∴ ∆ABD ~ ∆CRE
[by AA similarity criterion]

iii) In ∆AEP and ∆ADB,
∠AEP = ∠ADB [each 90°]
and ∠PAF = ∠BAD [common angles]
∴ ∆AEP ~ ∆ADB
[by AA similarity criterion]

iv) In ∆PDC and ∆BEC,
∠PDC = ∠BEC [each 90°]
and ∠PCD = ∠BCE [common angles]
∴ ∆PCD ~ ∆BEC
[by AA similarity criterion]

Question 30.
a) Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts. (Ch.No-7)
(OR)
b) From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm². (Ch.No-12)
Solution:
Let P, Q and R be the points on line segment AB such that
AP = PQ = QR = RB
Let AP = PQ = QR = RB = k

(OR)

Given, diameter of cylinder = Diameter of conical cavity
= 1.4 cm

= CSA of conical cavity + CSA of cylinder + Area of the base of the cylinder
= πrl + 2πrh + πr²
= πr(l + 2h + r)
= 22 × 0.7 × (2.5 + 2 × 24 + 0.7)
= 22 × 0.1 × (2.5 + 4.8 + 0.7)
= 2.2 × 8
= 17.6 ≈ 18 cm²

Question 31.
a) A lot of 20 books contains 8 books on Mathematics, 5 books on English and the rest are novels. A book is selected at random. Find the probability that (Ch.No-14)
i) it is a novel.
ii) it is not a Mathematics book.
iii) it is a novel or a Mathematics book.
iv) it is an English book.
(OR)
b) A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and form the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. (Ch.No-9)
Solution:
a) Total number of books in the lot 20
∴ Number of all possible outcomes = 20
Out of these,
Number of books on mathematics = 8
Number of books on English 5
Number of books that are novels
= 20 – (8 + 5) = 20 – 13 = 7

i) Let E₁ be the event that it is a novel.
Then, number of outcomes favourable to E₁ = 7
∴ P(E₁)
No. of outcomes favourable to E₁ = 7
\(=\frac{\text { No. of outcomes favourable to } E_1}{\text { No. of all possible outcomes }}\) = \(\frac{7}{20}\)

ii) Let E₂ be the event that it is not a mathematics book. Then, number of out comes favourable to E₂
= 5 + 7 = 12
∴ P(E₂)
\(=\frac{\text { No. of outcomes favourable to } E_2}{\text { No. of all possible outcomes }}\) = \(\frac{12}{20}\) = \(\frac{3}{5}\)

iii) Let E₃ be the event that it is a novel or a mathematics book.
Then, number of outcomes favourable to E₃
= 7 + 8 = 15
∴ P(E₃)
\(=\frac{\text { No. of outcomes favourable to } E_3}{\text { No. of all possible outcomes }}\) = \(\frac{15}{20}\) = \(\frac{3}{4}\)

iv) Let E₄ be the event that it is an English book.
Then, number of outcomes favourable to E₄ = 5
∴ P(E₄)
\(\frac{\text { No. of outcomes favourable to } E_4}{\text { No. of all possible outcomes }}\) = \(\frac{5}{20}\) = \(\frac{1}{4}\).

(OR)

Let BC = h m be the height of the pedestal and CD 1.6 m be the length of the statue, which is standing on the pedestal. Again, let point A be a fixed point on the ground such that the angles of elevation of the top of the statue and bottom of the statue (i.e., top of the pedestal) are
∠DAB = 60° and ∠CAB = 45°.
Also, let AB = x m.
In right angled ∆ ABD, tan 60° = \(\frac{P}{B}\) = \(\frac{\mathrm{BD}}{\mathrm{AB}}\)

Question 32.
a) A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

b) Find the number of terms in each of the following APs:
i) 7, 13, 19, ………., 205
ii) 18, 15\(\frac{1}{2}\), 13, ………, -47
Solution:
a) To calculate the median age, we need to find the class-intervals and their corresponding frequencies. It is shown below:

Now, N = Σf_fi = 100
So, \(\frac{\mathrm{N}}{2}\) = \(\frac{100}{2}\) = 50
Cumulative frequency just > \(\frac{\mathrm{N}}{2}\) = 50 is 70. The corresponding class-interval to this cf 70 is 35 – 40.
So, 35 – 40 is the median class.
Therefore, l = 35, h = 5, f = 33, cf = 45
∴ Median = l + \(\left(\frac{\frac{N}{2}-c f}{f}\right)\) × h = 35 + \(\left(\frac{50-45}{33}\right)\) × 5
= 35 + \(\frac{25}{33}\) = 35 + 0.76 = 35.76 years (approx.)
Hence, the median age is 35.76 years (approx.)

(OR)

b)
i) 7, 13, 19,…….., 205
Here, a = 7
d = 13 – 7 = 6
Let the number of terms be n.
∴ a_n = 205 → last term
⇒ a + (n – 1) d = 205
⇒ 7 + (n – 1) 6 = 205
⇒ 6(n – 1) = 205 – 7
⇒ 6(n – 1) = 198
⇒ n – 1 = \(\frac{198}{6}\)
⇒ n – 1 = 33
⇒ n = 33 + 1
⇒ n = 34
Hence, the number of terms of the given AP is 34.

ii) 18, 15\(\frac{1}{2}\), 13, ……., -47
Here, a = 18
d = 15\(\frac{1}{2}\) – 18 = \(\frac{31}{2}\) – 18 = –\(\frac{5}{2}\)
Let the number of terms be n.
Then, a_n = -47 → last term
⇒ a + (n – 1)d = 47
⇒ 18 + (n – 1)(-\(\frac{5}{2}\)) = -47
⇒ –\(\frac{5}{2}\)(n – 1) = -47 – 18
⇒ –\(\frac{5}{2}\)(n – 1) = -65
Cross multiplying
-5(n – 1) = -65 × 2 = -130
⇒ n – 1 = \(\frac{-130}{-5}\)
⇒ n – 1 = 26
⇒ n = 26 + 1
⇒ n = 27
Hence, the number of terms of the given AP is 27.

Question 33.
a) Solve the following pair of linear equations by the substitution and cross – multiplication methods : 8x + 5y = 9, 3x + 2y = 4
(OR)
b) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges, whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
ii) A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator. Find the fraction.
Solution:
a) We have : 8x + 5y = 9 …….(i).
3x + 2y = 4 ….. (ii)
Substitution method :
Multiplying equation (i) by 2 and equation (ii) by 5 and then subtracting the results,
we get:
16x – 15x = 18 – 20
⇒ x = – 2
Substituting x = -2 in equation (i), we get:
8(-2) + 5y = 9
⇒ -16 + 5y = 9
⇒ 5y = 25
⇒ y = 5
Thus, the required solution is x = -2 and y = 5.
Cross multiplication method:

(OR)

b) i) Let fixed monthly hostel charges = and charges per day = ₹ y
A. T. Q.
As per condition of student A
x + 20y = 1000
As per condition of student B
x + 26y = 1180
By cross multiplication method

∴ Fixed monthly hostel charges = ₹ 400
and charges per day = ₹ 30

ii) Let numerator x and denominator = y
∴ Fraction = \(\frac{x}{y}\)
A.T.Q.
1st condition :
\(\frac{x-1}{y}\) = \(\frac{1}{3}\) ⇒ 3x – 3 = y
⇒ 3x – y = 3 ……. (i)
2nd condition :
\(\frac{x}{y+8}\) = \(\frac{1}{4}\) ⇒ 4x = y + 8 ……… (ii)
⇒ 4x – y = 8
Solving (i) and (ii) for x and y
By cross multiplication method