Students must start practicing the questions from RBSE 10th Maths Model Papers Set 8 with Answers in English provided here.

## RBSE Class 10 Maths Model Paper Set 8 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) If p is a prime number and p divides k^{2}, then p divides: (1)

(a) 2k^{2}

(b) k

(c) 3k

(d) none of these

Answer:

(b) k

(ii) A cubic polynomial whose zeroes are – 2, -3 and – 1 is: (1)

(a) x^{3} + 11x^{2} + 6x + 1

(b) x^{3} + 6x^{2} + 11x + 6

(c) x^{3} + 11x^{2} + x + 6

(d) x^{3} + 6x^{2} + 6x + 11

Answer:

(b) x^{3} + 6x^{2} + 11x + 6

(iii) The point of intersection of the lines y = 3x and x = 3y is: (1)

(a) (3, 0)

(b) (0, 3)

(c) (3, 3)

(d) (0, 0)

Answer:

(d) (0, 0)

(iv) If the equation x^{2} + 9x + P = 0 has real roots, then: (1)

(a) p > \(\frac{81}{4}\)

(b) P < \(\frac{81}{4}\) (c) P ≥ \(\frac{-81}{4}\) (d) P ≥ \(\frac{9}{2}\)

Answer:

(a) p > \(\frac{81}{4}\)

(v) If the numbers 3P + 4, 7P + 1 and 12 P – 5 are in AP, then the value of P is: (1)

(a) 5

(b) 3

(c) 0

(d) 8

Answer:

(b) 3

(vi) To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distance points are marked on the ray AX such that the minimum number of these points is : (1)

(a) 8

(b) 10

(c) 11

(d) 8

Answer:

(d) 8

(vii) If the distance of P(x, y) from A(- 1, 5) and B(5, 1) are equal, then (1)

(a) y = 2x

(b) 5x = 2y

(c) 3x = 2y

(d) 2x = 3y

Answer:

(c) 3x = 2y

(viii) If tan A = \(\frac{3}{4}\), then sin^{2}A – cos^{2}A is equal to : (1)

(a) 1

(b) \(\frac{-7}{25}\)

(c) \(\frac{7}{25}\)

(d) \(\frac{3}{25}\)

Answer:

(b) \(\frac{-7}{25}\)

(ix) (sec θ + tan θ) (1 – sin θ) is equal to : (1)

(a) cos θ

(b) sin θ

(c) sec θ

(d) cosec θ

Answer:

(a) cos θ

(x) For a frequency distribution, mean, median and mode are connected by which of the following relations ? (1)

(a) Mode = 3 mean 2 median

(b) Mode = 2 median – 3 mean

(c) Mode = 3 median – 2 mean

(d) Mode = 3 median + 2 mean

Answer:

(c) Mode = 3 median – 2 mean

(xi) ‘More than’ give is : (1)

(a) An ascending curve

(b) A descending curve

(c) First ascending curve and than descending curve

(d) First descending curve and than ascending curve

Answer:

(b) A descending curve

(xii) A book of 100 pages is opened at random. The probability that a doublet page is found is : (1)

(a) \(\frac{9}{10}\)

(b) \(\frac{1}{5}\)

(c) \(\frac{9}{100}\)

(d) \(\frac{1}{10}\)

Answer:

(c) \(\frac{9}{100}\)

Question 2.

Fill in the blanks :

(i) If the two lines cross each other, then at least _________ point is common in both of them. (1)

Answer:

one

(ii) 9th term of the sequence defined by a_{n} = (- 1)^{n-1}. n^{3} is _________ (1)

Answer:

729

(iii) The sides of a triangle, (in cm) for which the construction of triangle is not possible are _________ (1)

Answer:

8, 6, 4

(iv) The distance between the points A(a sin α, a cos α) and B (a cos α, a sin α) = _________ (1)

Answer:

a√2

(v) tan^{2} θ + 1 is equal to _________ (1)

Answer:

sec^{2} θ

(vi) The area of triangle formed by the points (a, c + a), (a^{2} + c^{2}) and (-a, c – a) will be zero if these points are _________ (1)

Answer:

Collinear

Question 3.

Very Short Answer Type Questions :

(i) HCF and LCM of two integer are 12 and 336 respectively. If one integer is 48, then find another integer. (1)

Answer:

LCM = 336, HCF = 12 and one integer = 48

We know that

Another integer = \(\frac{\mathrm{LCM} \times \mathrm{HCF}}{\mathrm{I}^{\text {st }} \text { integer }}\)

= \(\frac{336 \times 12}{48}\)

= 7 × 12 = 84

(ii) Divide the polynomial 2x^{2} + 3x + 1 by polynomial x + 2. (1)

Answer:

Hence, quotient = 2x – 1

and remainder = 3

(iii) Find that polynomial whose zeros are – 5 and 4. (1)

Answer:

Let α = – 5 and β = 4

Then α + β = – 5 + 4 = -1

and αβ = – 5 x 4 = – 20

Quadratic polynomial x^{2} – (α + β) x + αβ = 0

x^{2} – (- 1)x + (-20) = 0

x^{2} + x – 20 = 0

(iv) Find the value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 has unique solution. (1)

Answer:

Given equation are :

x + 2y – 3 = 0

And 5x + ky + 7= 0

The condition for unique solution is :

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

⇒ \(\frac{1}{5} \neq \frac{2}{k}\)

⇒ k ≠ 0

(v) For what values of k, the roots of the equation x^{2} + 4x + k = 0 are real ? (1)

Answer:

The given equation is : x^{2} + Ax + k = 0

∵ The given equation has real roots

D ≥ 0

⇒ b^{2} – 4ac ≥ 0

⇒ (4)^{2} – 4 × 1 × k ≥ 0

⇒ 16 – 4k ≥ 0

⇒ – 4k ≥ – 16

⇒ – k ≥ – 4

⇒ – k ≥ – 4

(vi) Find the root of the quadratic equation x^{2} – 0.09 = 0. (1)

Answer:

x^{2} – 0.09 = 0

⇒ x^{2} – (0.3)^{2} = 0

⇒ (x + 0.3) (x – 0.3) = 0

⇒ x – 0.3 – 0.3

and x – 0.3 = 0

⇒ x = – 0.3

and x = 0.3

⇒ x = ± 0.3

(vii) Construct a triangle with sides 4 cm, 5 cm and 7 cm and then another triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the first triangle. (1)

Answer:

(viii) Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). (1)

Answer:

Let the coordinates of point A be (x, y).

∵ O is the mid point of AB

2 = \(\frac{x+1}{2}\) and -3 = \(\frac{y+4}{2}\)

⇒ 4 = x + 1

and -6 = y + 4

⇒ x = 4 – 1 = 3

⇒ y = -6 – 4 = – 10

Hence, coordinates of point A are (3, -10)

(ix) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. (1)

Answer:

sin 67° + cos 75°

= sin (90° – 23°) + cos (90 – 15°)

= cos 23° + sin 15°

(x) Evaluate: \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\) (1)

Answer:

(xi) If x_{i}s are the mid points of the class intervals of grouped data, f_{i}s are the corresponding frequencies and x̄ is the mean, then find Σ(f_{i}x_{i} – x̄). (1)

Answer:

We know that the mean of the data is given by

x̄ = \(\frac{\sum f_{i} x_{i}}{x}\) where n = Σf_{i}

Σf_{i}x_{i} – x̄ = Σf_{i}x_{i} – Σx̄

= nx̄ – nx̄

= 0 [∵ x_{i} = nx̄]

(xii) A child has a die whose six faces shows the letters as given below. (1)

The die is thrown once. What is the probability of getting (i) A ? (ii) D ?

Answer:

In throwing the die any one of the six faces may come upward.

Total number of possible outcomes = 6

(i) Since there are two faces with letter A

∴ Number of favourable outcomes = 2

∴ P(getting A) = \(\frac{2}{6}=\frac{1}{3}\)

(ii) There is one face with letter D

Number of favourable outcome = 1

P(gettings D) = \(\frac{1}{6}\)

Section – B

Question 4.

Show that √3 is an irrational number. (2)

Answer:

Let us assume, that 3√2 is a rational. It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0

∴ 3√2 = \(\frac{a}{b}\) [HCF of a and b is 1 and b ≠ 0],

⇒ √2 = \(\frac{a}{3b}\) ……..(i)

⇒ \(\frac{a}{3b}\) = rational number [∵ a, b are positive integers]

So, from (i) √2 is a rational number.

But this contradicts the fact √2 is irrational number. So, our assumption

that 3√2 is rational number, is wrong. Hence, 3√2 is an irrational number.

Question 5.

Find the zeroes of a quadratic polynomial x^{2} – 2x – 15. (2)

Answer:

Let p(x) = x^{2} – 2x – 15

= x^{2} – (5 – 3) x – 15

= x^{2} – 5x + 3x – 15

= x(x – 5) + 3(x – 5) = (x – 5) (x + 3)

To find zeroes, p(x) = 0

(x – 5) (x + 3) = 0 ⇒ x = -5, -3

Thus, required zeroes are 5 and -3.

Question 6.

Solve the pair of equations 7x – 15y = 2 and x + 2y = 3 by substitution method. (2)

Answer:

The given equations are

7x – 15y = 2 …(1)

x + 2y = 3 …(2)

From equation (2), we get

x = 3 – 2y

Substituting the value of x in equation (1), we get

7(3 – 2y) – 15y = 2

⇒ 21 – 14y – 15y = 2

⇒ – 29y = 2 – 21

⇒ – 29y = – 19

⇒ y = \(\frac{-19}{-29}\)

⇒ y = \(\frac{19}{29}\)

Putting the value of y in equation (2), we get

x + 2 × \(\frac{19}{29}\) = 3 ⇒ x + \(\frac{38}{29}\) = 3

Question 7.

The diagonal of a rectangular field is 25 metres more than the shorter side. If longer side is 23 metres more than the shorter side, find the sides of the field. (2)

Answer:

Let the shorter side be x m.

Then diagonal = (x + 25)m and longer side = (x + 23 )m.

We known that each angle of a rectangle is 90°.

In a right ∆ABC, we have (x + 25)^{2} = (x + 23)^{2} + x^{2}

x^{2} + 50x + 625 = x^{2} + 46x + 529 + x^{2}

⇒ 2x^{2} + 46x + 529 – x^{2} – 50x – 625 = 0

⇒ x^{2} – 4x — 96 = 0

⇒ x^{2} – (12 – 8)x – 96= 0

⇒ x^{2} – 12x + 8x – 96 = 6

⇒ (x^{2} – 12x) + (8x – 96) = 0

⇒ x(x – 12) + 8(x – 12) = 0

⇒ (x – 12) (x + 8) = 0

⇒ x – 12 = 0 and x + 8 = 0

⇒ x = 12 and x= – 8 (Reject)

Hence, longer side = 12 + 23 i.e., 35 m and shorter side = 12 m.

Question 8.

Find the 10th term of the AP : 2, 7, 12, … (2)

Answer:

The given sequence of AP is 2, 7, 12, …

Here, a = 2, d = a_{2} – a_{1} = 7 – 2 = 5

We know that nth term of AP is

a_{n} = a + (n – 1 )d

⇒ a_{n} = 2 + (10 – 1) × 5

⇒ a_{10} = 2 + 9 × 5

⇒ a_{10} = 2 + 45

⇒ a_{10} = 47

Hence, 10th term of given AP = 47

Question 9.

Find the sum of all the natural numbers divisible by 5 between 2 and 101. (2)

Answer:

The numbers divisible by 5 between 2 and 101 are 5, 10, 15, ……………. , 100

∵ a_{2} – a_{1} = 10 – 5 = 5,

a_{3} – a_{2} = 45 – 10 = 5

The sequence form an AP Here, a = 5, d = 5, a_{n} = l = 100 we know that

⇒ 100 =5 + (n – 1) × 5

⇒ 95 = 5n – 5

⇒ 100 = 5n

⇒ n = \(\frac{100}{5}\) = 20

Now,

S_{n} = \(\frac{n}{2}\)[a + l]

S_{20} = \(\frac{20}{2}\)[5 + 100]

= 10 × 105 = 1050

Hence, S_{20} = 1050.

Question 10.

Explain the meaning of scale factor with the help of two examples. (2)

Answer:

Scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle e.g.

- Scale factor \(\frac{2}{3}\) means the sides of the constructed triangle is \(\frac{2}{3}\)of the sides of given triangle.
- Scale factor \(\frac{5}{2}\) means the sides of the constructed triangle is \(\frac{5}{4}\) of sides of given triangle.

Question 11.

Draw a circle of radius 3 cm with centre O and take a point P outside the circle such that OP = 8 cm. From P, draw two tangents to the circle. (2)

Answer:

Steps of Construction :

- Draw a circle with O as the centre and radius 3 cm
- Mark a point P outside the circle such that OP = 8 cm.
- Join OP and draw prependicular bisector of PO meeting PO at M.

- Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R.
- Join PQ and PR then PQ and PR are required tangents.

Question 12.

Show that tan^{4}θ + tan^{2}θ = sec^{4}θ – sec^{2}θ (2)

Answer:

We have

L.H.S. = tan^{4}θ + tan^{2}θ

= tan^{2}θ [tan^{2}θ + 1]

= (sec^{2}θ – 1) (sec^{2}θ) [∵ 1 + tan^{2}θ = sec^{2}θ]

= sec^{4}θ – sec^{2}θ

= R.H.S.

Question 13.

If tan(A + B) = 1 and tan(A – B) = \(\frac{1}{\sqrt{3}}\); 0° < (A + B) ≤ 90°, A > B, find A and B. (2)

Answer:

We have,

tan (A + B) = 1

⇒ tan (A + B) = tan 45°

⇒ A + B = 45° … (1)

And tan (A – B) = \(\frac{1}{\sqrt{3}}\)

⇒ tan (A – B) = tan 30°

⇒ A – B = 30° … (2)

Adding equations (1) and (2), we get A + B = 45°

A – B = 30°

2A = 75°

⇒ A = \(\frac{75}{2}\) = 37.5°

Substituting the value of A in equation (1), we get

37.5° + B = 45°

⇒ B = 45° – 37.5° = 7.5°

Hence, A = 37.5° and B = 7.5°.

Question 14.

Find the mode of the following frequency distribution. (2)

Answer:

The class interval 30 – 40 has maximum frequency. So, it is the modal class.

∴ l = 30, f_{1} = 16, f_{0} = 10, f_{2} = 12, h = 10

Question 15.

If the median of the following frequency distribution is 32.5. Find the values of f_{1} and f_{2}.

Answer:

Let us prepare the cumulative frequency distribution table as given below.

Here, n = 40

⇒ f_{1} + f_{2} + 31 = 40

⇒ f_{1} + f_{2} = 9 ……(i)

Median is 32.5, which lies in the class interval 30-40, So the median class is 30 – 40.

l = 30, f = 12, c.f. = f_{1} + 14, h = 10

⇒ 2.5 × 12 = 60 – 10f_{1}

⇒ 10f_{1} = 60 – 30

⇒ f_{1} = \(\frac{30}{10}\) = 3

Putting the value of f_{1} in equation (i), we get

3 + f_{2} = 9 ⇒ f_{2} = 6

Hence, f_{1} = 3, f_{2} = 6

Question 16.

A bag contains cards with numbers written on it form 1 – 80. A card is pulled out at random. Find the probability that the card shows a perfect square. (2)

Answer:

The number of cards written on it

form 1 – 80 = 80

The numbers of all possible outcomes = 80

The number of cards shown a perfect square = 1, 4, 9, 16, 25, 36, 49, 64 i.e. 8 cards

Let E be event of shows a perfect square

Number of outcomes to favourable E = 8

∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of all possible outcomes }}\)

= \(\frac{8}{80}=\frac{1}{8}\)

Section – C

Question 17.

Find the sum of first 24 terms of the list of numbers whose rath term is given by a_{n} = 3 + 2n (3)

Or

If S_{n} denotes, the sum of the first n terms of an AP. Prove that S_{12} = 3(S_{8} – S_{4}) (3)

Answer:

We have,

a_{2} = 3 + 2n

a_{1} = 3+ 2 × 1 = 5

a_{2} = 3 + 2 × 2 = 7

a_{3} = 3 + 2 × 3 = 9 …… and so on.

List of numbers become 5, 7, 9, …

a_{2} – a_{1} = 7 – 5 = 2

a_{3} – a_{2} = 9 – 7 = 2

∵ a_{2} – a_{1} = a_{3} – a_{2}

The sequence forms an AP.

Here, a = 5

d = a_{2} – a_{1} = 7 – 5 = 2

n = 24

We know that S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

⇒ S_{24} = \(\frac{24}{2}\) [2 × 5 + (24 – 1) × 2]

⇒ S_{24}= 12 [10 + 23 × 2]

⇒ S_{24} = 12 [10 + 46]

⇒ S_{24} = 12 × 56 = 672

Hence, the sum of first 24 terms of list of numbers = 672.

Question 18.

Find the area of triangle ABC with A(1, -4) and the mid points of sides through A being (2, -1) and (0, -1). (3)

Or

Find the area of that triangle whose vertices are (-3, -2), (5, -2) and (5, 4). Also prove that it is a right angled triangle. (3)

Answer:

Let the coordinates of vertices B be (x_{2}, y_{2}) and C be (x_{3}, y_{3})

Let E (2, – 1) and F (0, – 1) are mid points of AB and AC respectively.

∴ 2 = \(\frac{1+x_{2}}{2}\) and -1 = \(\frac{-4+y_{2}}{2}\)

⇒ 4 = 1 + x_{2} and -2 = -4 + y_{2}

⇒ x_{2} = 3 and y_{2} = 2

∴ (x_{2}, y_{2}) = (3, 2)

Again 0 = \(\frac{1+x_{3}}{2}\) and -1 = \(\frac{-4+y_{3}}{2}\)

⇒ 0 = 1 + x_{3} and -2 = -4 + y_{3}

⇒ x_{3} = -1 and y_{3} = 2

⇒ (x_{3}, y_{3}) = (- 1, 2)

Here, x_{1} = 1, y_{1} = -4, x_{2} = 3,

y_{2} = 2, x_{3} = – 1, y_{3} = 2

Area of ABC = \(\frac{1}{2}\)[x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= \(\frac{1}{2}\)[1(2 – 2) + 3(2 + 4) + (-1)(-4 – 2)]

= \(\frac{1}{2}\)[1 × 0 + 3 × 6 + (-1) × (-6)]

= \(\frac{1}{2}\)[0 + 18 + 6] = \(\frac{1}{2}\) × 24 = 12

Hence, area of ABC = 12 sq.units

Question 19.

Prove that: \(\frac{\tan ^{2} \mathrm{~A}}{\tan ^{2} \mathrm{~A}-1}+\frac{{cosec}^{2} \mathrm{~A}}{\sec ^{2} \mathrm{~A}-{cosec}^{2} \mathrm{~A}}=\frac{1}{1-2 \cos ^{2} \mathrm{~A}}\) (3)

OR

Prove that: \(\frac{\sin \theta}{\cos \theta+{cosec} \theta}=2+\frac{\sin \theta}{\cot \theta-{cosec} \theta}\) (3)

Answer:

Question 20.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components. (3)

Or

A class teacher has the following absent record of 40 students of a class for the whole term. Find the mean number of days a student was absent. (3)

Answer:

The class 60-80 has maximum frequency. So, it is the modal class.

∴ l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38 and h = 20

Hence, the modal lifetimes of the electrical components = 65.625 years

Section – D

Question 21.

The cost of 7 erasers and 5 pencils is ₹ 58 and the cost of 5 erasers and 6 pencils is ₹ 56. Formulate this problem algebrically and solve it graphically. (4)

Or

Solve the following by graphical method : (4)

9x – 10y = 14 and 5x – 3y = 11

Answer:

Let the cost of 1 erasers be ₹ x and cost of 1 pencil be ₹ y

∴ Cost of 7 erasers is 7x and cost of 5 pencils is

According to question

7x + 5y = 58

Cost of 5 erasers is 5x and cost of 6 pencils is 6y

According to question

5x + 6y = 56 …(ii)

The algebraic representation of this situtation is

7x + 5y = 58

And 5x + 6y = 56

For representation of these equations graphically, we draw the graphs of these equations as follows :

7x + 5y = 58 ⇒ y = \(\frac{58-7 x}{5}\)

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 7x + 5y = 58.

Table 1

And 5x + 6y = 56 ⇒ y = \(\frac{56-5 x}{6}\)

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 this equation 5x + 6y -• 56.

Table 2

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

Observe that we get two straight lines which intersect each other at point (4, 6)

Hence, cost of 1 eraser is ₹ 4 and cost of 1 pencil is ₹ 6.

Question 22.

Draw a circle of radius 5 cm. From a point 13 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Also verify the measurement by actual calculation. (4)

Or

Construct a triangle with sides 3 cm, 5 cm and 8 cm and then another triangle whose 8 sides are \(\frac{8}{5}\) of the corresponding sides of the first triangle. (4)

Answer:

Steps of Construction :

- Draw a circle with centre O of radius 5 cm.
- Mark a point A, 13 cm away from the centre.
- Join AO and bisect it at M.
- Draw a circle with M as the centre and radius equal to AM intersects the given circle at points P and Q.
- Join AP and AQ.

Then AP and AQ are the required tangents lengths of AP and AQ are 12 cm.

Justification:

Join OP

∴ ∠APO = 90° [Angle is a semicircle]

AP ⊥ OP

Therefore, AP is a tangent to the given circle. Similarly AQ is a tangent to the given circle.

In right triangle APO,

we have AO^{2} = PO^{2} + AP^{2}

13^{2} = 5^{2} + AP^{2}

169 – 25 = AP^{2}

AP = \(\sqrt{144}\)

AP = 12 cm

Question 23.

The median of the following data is 525. Find the values of x and y, if the total,frequency is 100. (4)

Or

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : (4)

Answer:

Let us prepare the cumulative frequency distribution table as given below :

Here, n = 100 (given)

⇒ 76 + x + y = 100

⇒ x + y = 100 – 76

⇒ x + y = 24 …(1)

Median is 525, which lie in the class interval 500 – 600 So, the median class is 500 – 600.

∴ l = 500, cf = 36 + x, f = 20, and h = 100

Putting the value of x in equation (1), we get

9 + y = 24

⇒ y = 24 – 9 = 15

Hence, x = 9 and y = 15.

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