Kerala Syllabus Class 9 Maths Model Question Paper Set 1 English Medium

Teachers recommend solving Maths Model Question Paper for Class 9 Kerala Syllabus Set 1 to improve time management during exams.

Kerala Syllabus Std 9 Maths Model Question Paper Set 1 English Medium

Time: 2½ Hours
Max Score: 80 Marks

Instructions:

• There is a ‘cool off’ time 15 minutes in addition to the writing time. Use this time to get familiar with questions and plan your answers.
• Read the instructions carefully before answering the questions.
• Keep in mind, the score and time while answering the questions. Give explanations wherever necessary.
• No need to simplify irrationals like √2, √3, π etc. using approximations unless you are asked to do so.

Answer any 3 Questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
a) x + 4 = 10, x = ___________
b) Find a pair of natural numbers satisfying the equation x + y = 10
Answer:
a) x = 10 – 4 = 6
b) (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)

Question 2.
What is the side of a square of area 10 cm².
Answer:
(side)² = 10
side = \(\sqrt{10}\) cm

Question 3.
In the figure, BP = PC and area of the triangle ABP is 6cm².

a) What is the area of triangle APC?
b) What is the area of triangle ABC?
Answer:
a) area of triangle APC = area of triangle ABP = 6 cm²
b) area of triangle ABC = area of triangle APC + area of triangle ABP = 6 + 6= 12 cm²

Question 4.
In the figure, lines AP, BQ, CR, and DS are parallel. AB = BC = CD and PS = 12cm.

a) PQ = __________
b) QR = __________
Answer:
a) PQ = \(\frac{\mathrm{PS}}{4}\) = \(\frac{12}{4}\) = 3 cm
b) QR = \(\frac{\mathrm{PS}}{4}\) = \(\frac{12}{4}\) = 3 cm

Answer any 4 Questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
The price of a table and 5 chairs is Rs 7000. The price of a table and 2 chairs is Rs 4000.
a) What is the price of a chair?
b) What is the price of a table?
Answer:
Let t be the price of a table and c be the price of a chair. From the question we have,
t + 2c = 4000 ……… (1)
t + 5c = 7000 ……….. (2)
a) (2) – (1) → 3c = 3000
c = 1000
Thus, the price of a chair = 1000 rupees
b) (1) → t + 2 × 1000 = 4000
t = 4000 – 2000 = 2000
Thus, the price of a table = 2000 rupees

Question 6.
The sum of two numbers is 19, and their difference is,5. Find the numbers.
Answer:
Let m and n be the numbers. From the question we have,
m + n = 19 ………. (1)
m – n = 5 ……….. (2)
(1) + (2) → 2m = 24
m = 12
(1) → 12 + n = 19
n = 19 – 12 = 7
Thus, the numbers are 12 and 7.

Question 7.
In the figure, the area of the small square is 2 cm².

a) What is the area of the large square drawn on the diagonal of the small square?
b) What is the side of the larger square?
Answer:
a) Area of the large square = 2 × area of the small square
= 2 × 2 = 4 cm²

b) Area = (side)²
(side)² = 4
side = √4 = 2 cm.

Question 8.
Write the following numbers as the differences of two perfect squares.
a) 9
b) 13
Answer:
a) 9 = 25 – 16
b) 13 = 49 – 36

Question 9.
In the figure, P is the midpoint of BC, and Q is the midpoint of AP. If the area of triangle BQP = 10 cm²,

а) What is the area of triangle ABQ?
b) What is the area of triangle ABP?
c) Find the area of triangle ABC.
Answer:
a) Area of ΔABQ = Area of ΔBPQ = 10 cm²

b) Area of ΔABP = Area of ΔABQ + Area of ΔBPQ
= 10 + 10 = 20 cm²

c) Area of ΔABC = 2 × Area of ΔABP
= 2 × 20 = 40 cm²

Question 10.
Draw an equilateral triangle of perimeter 10 cm.
Answer:

Answer any 8 Questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
a) Write the decimal form of \(\frac{1}{9}\)
b) Write the sum 0.111 …… + 0.222 …… in decimal form.
c) Write the sum 0.333 …….+ 0.777…… in decimal form.
d) Write the product 0.333 …… × 0.666 …… in decimal form.
Answer:
a) \(\frac{1}{9}\) = 0.111 …..

b) 0.111… +0.222… = \(\frac{1}{9}\) + \(\frac{2}{9}\)
= \(\frac{3}{9}\)
= 0.333 …

c) 0.333….. + 0.777…. = \(\frac{3}{9}\) + \(\frac{7}{9}\)
= \(\frac{10}{9}\) = 1\(\frac{1}{9}\) = 1.111…..

d) 0.333… × 0.666… = \(\frac{3}{9}\) × \(\frac{6}{9}\) = \(\frac{1}{3}\) × \(\frac{2}{3}\) = \(\frac{2}{9}\) = 0.222…

Question 12.
The difference of the two smaller angles of a right triangle is 40°
a) What is the sum of the two smaller angles?
b) Calculate the smaller angles.
Answer:
Let x and y be the smaller angles.
a) x + y + 90° = 180° (Sum of all the angles in a triangle is 180°)
x + y = 90°

b) We have, x + y – 90°……… (1)
x – y = 40° ……….. (2)
(1) + (2) → 2x = 130° x = 65°
(1) → 65° + y = 90° y = 90° – 65° = 25°
Thus, the smaller angles are 65° and 25°.

Question 13.
Find four fractions greater than √2 and less than √3.
Answer:
√2 = 1.414…
√3 = 1.732…
1.414… < 1.5 < 1.6 < 1.7 < 1.71 < 1.732…
√2< \(\frac{15}{10}\) < \(\frac{16}{10}\) < \(\frac{17}{10}\) < \(\frac{171}{100}\) < √3

Question 14.
The perpendicular sides of a right triangle are 1m and √2 m.
a) What is the length of its hypotenuse?
b) Calculate the perimeter of this triangle, correct to a centimetre.
Answer:
a) hypotenuse = \(\sqrt{(\sqrt{2})^2+1^2}\) = \(\sqrt{2+1}\) = √3 m
b) Perimeter = 1 + √2 + √3 = 1 + 1.41 + 1.73 = 4.14 m

Question 15.
Draw a square of perimeter 13 centimetres.
Answer:

Question 16.
In the picture, the midpoints of a triangle are joined to form a smaller triangle inside. The area of triangle PQR is 3 cm².

a) What is the area of the triangle APQ?
b) What is the area of the parallelogram BPQR?
c) What is the area of the triangle ABC?
Answer:
a) Area of ΔAPQ = Area of ΔPQR = 3 cm²
b) Area of parallelogram BPQR = 2 × area of ΔPQR = 2 × 3 = 6 cm²
c) Area of ΔABC = 4 × area of APQR = 4 × 3 = 12 cm²

Question 17.
The perimeter of a rectangle is 30 cm and one side is 3 cm longer than the other.
a) longer side + shorter side = ________
b) longer side – shorter side = ________
c) Calculate the length of the sides.
Answer:
a) Perimeter = 30
2 (longer side + shorter side) = 30
longer side + shorter side = 15 cm

b) longer side – shorter side = 3 cm

c) Let the shorter side = x
longer side = y
We have, y – x = 3 ……. (1)
y + x = 15 …….. (2)
(1) + (2) → 2y = 18 y = 9
(2) → 9 + x = 15 x = 15 – 9 = 6
Thus, the shorter side = 6 cm the longer side = 9 cm

Question 18.
A person invested 10,000 rupees in two banks with interest rates of 9% and 7%.
After one year he got 850 rupees as interest from both these together. How much did he invest in each bank?
Answer:
Let, the amount invested in the first bank = x
The amount invested in the second bank = y
From the question we have,
x + y = 10000 …………… (1)
\(\frac{9}{100}\) × x + \(\frac{7}{100}\) × y = 850
⇒ 9x + 7y = 85000 ………………. (2)
(1) × 9 → 9x + 9y = 90000 …………….. (3)
(3) – (2) → 2y = 5000
y = 2500
(1) → x + 2500 = 10000
x = 10000-2500 = 7500
Thus, the amount invested in the first bank = Rs. 7500
the amount invested in the second bank = Rs. 2500

Question 19.
The product of two numbers is 1271, and their sum is 72. What is the product of the numbers next to each?
Answer:
Let x and y be the numbers. It is given that,
x + y = 72
xy = 1271
the number next to x = x + 1
the number next to y = y + 1
(x + 1 )(y + 1) = xy + (x + y) + 1
= 1271 + 72 + 1 = 1344

Question 20.
The perimeter of a rectangle is 40 cm and its area is 91 cm2. Find the area of the rectangle with each side 2 cm shorter.
Answer:
Let, length = x
breadth = y
It is given that, 2(x + y) = 40
⇒ x + y = 20
Area = 91cm² ⇒ xy = 91
Length of the rectangle with each side 2 cm shorter = x – 2
Breadth of the rectangle with each side 2 cm shorter = y – 2
Area = (x – 2)(y – 2)
= xy – 2x – 2y + 4
= xy – 2(x + y) + 4
= 91 – 2 × 20 + 4
= 91 – 40 + 4
= 55 cm²

Question 21.
a) (x + \(\frac{1}{2}\))(y + \(\frac{1}{2}\)) = xy + x × \(\frac{1}{2}\) + \(\frac{1}{2}\) × y + \(\frac{1}{4}\) = xy + \(\frac{1}{2}\) (_ + _) + _
b) 10\(\frac{1}{2}\) × 4\(\frac{1}{2}\) =
c) 7 \(\frac{1}{4}\) × 5\(\frac{1}{4}\) =
Answer:
a) xy + \(\frac{1}{2}\) (x + y) + \(\frac{1}{4}\)

b) 10 × 4 + \(\frac{1}{2}\) (10 + 4) + \(\frac{1}{4}\) = 40 + 7 + \(\frac{1}{4}\)
= 47 + \(\frac{1}{4}\)

c) 7 × 5 + \(\frac{1}{4}\) (7 + 5) + \(\frac{1}{16}\) = 35 + 3 + \(\frac{1}{16}\)
= 38 + \(\frac{1}{16}\)

Answer any 6 Questions from 22 to 29. Each question carries 5 scores. (6 × 5 = 30)

Question 22.
The sum of numbers 1 more than two natural numbers is 37. The difference of numbers 1 less than these natural numbers is 5. If we take two natural members as x and y and x > y,
a) What are the numbers 1 more than these?
b) What are the numbers 1 less than these?
c) Find the numbers.
Answer:
a) (x + 1) and (y + 1)

b) (x – 1)and (y – 1)

c) It is given that,
(x + 1) + (y + 1) = 37
x + y + 2 = 37
x + y = 35 ………………… (1)
(x – 1) – (y – 1) = 5
x – y = 5 ……………….. (2)
(1) + (2) → 2x = 40
x = 20
(1) → 20 + y = 35
y = 35 – 20 = 15
Thus, the numbers are 20 and 15.

Question 23.
a, b, c, d are four consecutive natural numbers.
a) b × c – a × d =
b) If b = 11, what are the values of a, c, and d?
c) Write four natural numbers such that the difference between the products of the two numbers in the middle and the two numbers at the ends is two.
Answer:
Let a = x
b = x + 1
c = x + 2
d = x + 3

a) b × c = (x + 1 )(x + 2) = x² + 3x + 2
a × d = x(x + 3) = x² + 3x
b × c – a × d = x² + 3x + 2 – (x² + 3x) = 2

b) Given that b = 11, that is x + 1 = 11.
⇒ x = 10
Thus, a = 10
c = 10 + 2 = 12
d = 10 + 3 = 13

c) 1, 2, 3, 4

Question 24.
The price of 6 pens and 4 pencils is 110 rupees. The price of 4 pens and 6 pencils is 90 rupees.
a) What is the price of a pen?
b) What is the price of a pencil?
Answer:
Let, the price of a pen = x
the price of a pencil = y
From the question we have,
6x + 4y = 110
⇒ 3x + 2y = 55 ……………. (1)
4x + 6y = 90 ……………. (2)
a) (1) × 3 → 9x + 6y = 165 ……………. (3)
(3) – (2) → 5x = 75
x = 15
Thus, the price of a pen = 15 rupees

b)
(1) × 4 → 12x + 8y = 220 …………… (4)
(2) × 3 → 12x + 18y = 270 …………….. (5)
(5) – (4) → 10y = 50
y = 5
Thus, the price of a pencil = 5 rupees

Question 25.
In the figure, squares are drawn on the sides of the right triangle ABC.
The area of the square ABPQ is 36 cm2, area of the square BCSR is 64 cm².

a) What is the area of the square ACNM?
b) What is the length of a side of the square ACNM?
c) What is the area of the square drawn on the diagonal CM of the square ACNM?
Answer:
a) Using Pythagoras theorem, area of square ACNM = 36 + 64 = 100 cm²
b) Length of a side of the square ACNM = \(\sqrt{100}\) = 10 cm
c) Area = 2 × area of the square ACNM = 2 × 100 = 200 cm²

Question 26.
The sides of a rectangle are (√3 + 1) cm and (√3 – 1) cm.

a) Calculate the perimeter of this rectangle.
b) Calculate the area of this rectangle.
Answer:
a) Perimeter = 2((√3 + 1) + (√3 – 1)) = 2 × 2√3 = 4√3 cm

b) Area = (√3 + 1) × (√3 – 1)
= 3 – √3 + √3 – 1
= 2 cm²

Question 27.
In the figure, P is the midpoint of BC, Q the midpoint of AC and R is the midpoint of AB.

AP, BQ, CR are called __________
b) What is the ratio of areas of triangles ABP and APC?
c) The point of intersection of the medians is called _________ of the triangle.
d) If the area of the triangle BGP is 10 cm², what is the area of the triangle ABC?
Answer:
a) medians
b) 1 : 1
c) centroid
d) Area of the ΔABC = 6 × area of the ΔBGP
= 6 × 10 = 60 cm² (∵ Three medians of a triangle divides it into 6 equal parts)

Question 28.
The sum of the digits of a two-digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
Let 10x + y be the two-digit number. Then
x + y = 11 …………….. (1)
If the digits is interchanged, then the number will be 10y + x
Thus,
10y + x = (10x + y) + 27
9y – 9x = 27
y – x = 3 …………….. (2)
(1) + (2) → 2y = 14
⇒ y = 7
∴ x = 4
Thus, the two-digit number is 47.

Question 29.
Look at the pattern given below.
2² – 1² = 3
3² – 2² = 5
4² – 3² = 7
……………..
……………..
a) Write the next line.
b) Write 25 as the difference of two perfect squares.
c) Draw a square of area 13 cm².
Answer:
a) 5² – 4² = 9
b) 25 = 13² – 12²
c) We know that 7² – 6² = 13