Kerala Syllabus Class 9 Maths Model Question Paper Set 3 English Medium

Teachers recommend solving Maths Model Question Paper for Class 9 Kerala Syllabus Set 3 to improve time management during exams.

Kerala Syllabus Std 9 Maths Model Question Paper Set 3 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.
If the sum of two numbers is 30 and the difference is 4, then what are the numbers?
Answer:
Let x and y be the two numbers.
Then x + y = 30 ………….. (1)
x – y = 4 ……………. (2)
Adding (1) and (2) we get, 2x = 34
Substituting the value of x in (1) x = 17
(1) → 17 + y = 30
∴ y = 13
Thus, the numbers are 17 and 13.

Question 2.
In an equilateral triangle one side is (√3 + 1) cm. Find its perimeter.
Answer:
Perimeter = 3 × side
= 3(√3 + 1)
= 3√3 + 3
≈ 3 × 1.732 + 3
= 5.196 + 3
= 8.196 cm

Question 3.
Find y = x⁴ + x³ + x² + x + 2 if x = -1
Answer:
y = (-1)⁴ + (-1)³ + (-1)² + (-1) + 2
= 1 + (-1) + 1 + (-1) + 2
= 1 – 1 + 1 -1 + 2
= 2

Question 4.
The product of two numbers is 1400 and their sum is 81. What is the product of the numbers next to each?
Answer:
Let one number be x and the other be y. It is given that,
xy = 1400
x + y = 81
We can take consecutive counting numbers next to the given numbers as x + 1 and y + 1 respectively.
(x + 1)(y + 1) = xy + (x + y) + 1
= 1400 + 81 + 1
= 1482

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

Question 5.
Compute y = x² + 9x – 5 by taking x as the given number.
a) x = 1
b) x = – 3
c) x = 0
Answer:
a) x = 1
y = x² + 9x – 5 = 1² + (9 × 1) – 5 = 1 + 9 – 5 = 10 – 5 = 5

b) x = – 3
y = x² + 9x – 5 = (- 3)² + 9(- 3) – 5 = 9 – 27 – 5 = -23

c) x = 0
y = x² + 9x – 5 = 0² + 9 × 0 – 5 = -5

Question 6.
Draw the triangle with angles the same as those of the triangle shown below, and sides scalepl by 1\(\frac{1}{2}\)

Answer:
6 × \(\frac{3}{2}\) = 9 cm
7 × \(\frac{3}{2}\) = 10.5 cm

Question 7.
In the figure, D, E, and F are midpoints of sides of the triangle ABC.
a) If BC = 8 cm, find DF.
b) If the perimeter of triangle ABC is 20 cm, find the perimeter of triangle DEF.
c) If the area of triangle ABC ¡s 16 cm², find the area of triangle DEF.

Answer:
a) DF = \(\frac{B C}{2}\) = \(\frac{8}{2}\) = 4 cm
b) Perimeter of triangle ABC = 20 cm
∴ Perimeter of triangle DEF = \(\frac{20}{2}\) = 10 cm.

c) Area of triangle ABC =16 cm²
∴ Area of triangle DEF = \(\frac{16}{4}\) = 4 cm²

Question 8.
The product of two odd numbers is 899, and the sum is 60. What is the product of the two odd numbers that come just after each of these two odd numbers?
Answer:
Let the odd numbers be x and y, then,
xy = 899 and x + y = 60
(x + 2 )(y + 2) = xy + 2(x + y) + 2²
= 899 + 2 × 60 + 4
= 1023

Question 9.
In the figure ∠B = 90°, AB = BC.

a) AB = BC = 1 cm. Then, find the length of AC.
b) Find the perimeter of triangle ABC
c) Find the area of triangle ABC
d) If a square with side AC is drawn, then what is its Area?
Answer:
a) AC = \(\sqrt{1^2+1^2}\) = √2
b) Perimeter = 1 + 1 + √2 = 2 + √2 ≈ 2 + 1.414 ≈ 3.414 cm
c) Area = \(\frac{1}{2}\) × 1 × 1= \(\frac{1}{2}\) cm²
d) Area = 4 × \(\frac{1}{2}\) cm²

Question 10.
In a box, there are 17 coins, including ten rupees coins and five rupees coins. If the total value of these is 105 rupees, how many coins each?
Answer:
Let the number of ten rupee coins = x
Number of five rupee coins = y
x + y = 17 ………….. (1)
10x + 5y = 105 …………… (2)
(1) × 10 → 10x + 10y = 170 ……………. (3)
(3) – (2) → 5y = 65
y = 13
(1) → x + 13 = 17
x = 17 – 13 = 4
Thus, the number of ten rupee coins = 4
number of five rupee coins = 13

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

Question 11.
The larger side of a rectangle is 7 cm longer than the smaller side. The diagonal is 13 cm. Find
the length of the sides.
Answer:
Let x be the larger side and y be the smaller side. It is given that x – y = 7 …………… (1)
Diagonal length = 13 ⇒ x² + y² = 13²
We know that, (x – y)² = x² + y² – 2xy
⇒ 7² = 13² – 2xy
xy = \(\frac{13^2-7^2}{2}\) = \(\frac{120}{2}\) = 60
We have, (x +y)² = x² + y² + 2xy
= 169 + 120
= 289
x + y = \(\sqrt{289}\) = 17 …………….. (2)
(1) + (2) → 2x = 24
x = 12 cm
(1) → 2y = 10
y = 5 cm
Thus, 12 cm and 5 cm are the side lengths.

Question 12.
Draw a square of area 10 cm²
Answer:
We know that, x = So, we can write,
10 = (\(\frac{10+1}{2}\))² – (\(\frac{10-1}{2}\))²
10 = (5.5)² – (4.5)²
Now, draw as per the figure.

Here, the square of one side of the square is \(\sqrt{10}\) cm. Therefore, its area = 10 cm².

Question 13.
Find the value of z from the equation z = x – y with the given values.
a) x = 7, y = 2
b) x = -3, y = -6
c) x = -8, y = 3
d) x = -4, y = 9
Answer:
a) x = 7, y = 2
z = x – y = 7 – 2 = 5

b) x = -3 y = -6
z = x – y = -3 – (-6) = – 3 + 6 = 3

c) x = -8 y = 3
z = x – y = – 8 – 3 = -11

d) x = -4 y = 9
z = x – y = – 4 – 9 = -13

Question 14.
x = \(\sqrt{0.5}\), y = \(\sqrt{32}\), z = \(\sqrt{128}\)
a) Find xy, yz and xz.
b) Find xy + yz + xz
c) Prove that y = 8x
Answer:

b) xy + yz + xz = 4 + 64 + 8
= 76

c) 8x = 8 × \(\sqrt{0.5}\)
= \(\sqrt{32}\)
= y

Question 15.
What is the remainder when the product of two numbers is divided by 5, given that one number leaves a remainder of 1 and the other leaves a remainder of 2 when each is divided by 5?
Answer:
Let 5m + 1 be a number that leaves a remainder of 1 when divided by 5 and 5n + 2 be a number
that leaves a remainder of 2 when divided by 5.
Then the product become = (5m + 1)(5n + 2)
⇒ (5m + 1)(5n + 2) = 25mn + 10m + 5n + 2
= 5(5mn + 2m + n) + 2
It is clear that, remainder is 2 when the product is divided by 5.

Question 16.
ABC is a triangle right angled at B. P, Q, R the mid points of the sides of Δ ABC. PR = 3 cm, PQ = 4 cm.

a) What is the length of QR?
b) What are the sides of triangle ABC?
c) Suggest a suitable name for PQBR.
Answer:
a) ΔPQR is a right triangle.
∴ QR = \(\sqrt{P Q^2+P R^2}\)
= \(\sqrt{4^2+3^2}\) = \(\sqrt{25}\) = 5 cm

b) Sides of triangle ABC,
AB = 2 × PQ = 2 × 4 = 8 cm
BC = 2 × PR = 2 × 3 = 6cm
AC = 2 × QR = 2 × 5 = 10 cm

c) Rectangle

Question 17.
In the figure ∠B = ∠D = 90 AB = 15 cm, AD = 5 cm

a) If ∠DAE = 40°, find ∠AED, ∠BAC.
b) What is the measure of ∠C?
c) \(\frac{B C}{D E}\) = …… …… …….. [3, 4, 5]
Answer:
a) ∠AED = 90 – 40 = 50°
∠BAC = 40°

b) ∠C = 50°

c) ΔADE and ΔABC are similar
\(\frac{A D}{A B}\) = \(\frac{A B}{A D}\) = \(\frac{15}{5}\) = \(\frac{3}{1}\)
So,
\(\frac{B C}{D E}\) is also \(\frac{3}{1}\) = 3

Question 18.
A boy of height 90 cm walking away from the base of a lamp post at the speed of 1.2 m/s. If the
lamp is 3.6 m above the ground then find the length of his shadow after 4 second.
Answer:

Speed = 1.2 m/s
After 4 seconds, he covers 4 × 1.2 meters= 4.8 m
\(\frac{A B}{B E}\) = \(\frac{C D}{D E}\)
\(\frac{3.6}{(4.8+x)}\) = \(\frac{3.6}{x}\)
3.6x = 4.8 × 0.9 + 0.9x
2.7x = 4.32
x = 1.6 m

Question 19.
Construct a square of perimeter 10.5 cm.
Answer:
Draw a line of length 10.5cm. Divide this line into four equal parts. Draw the square as given below:

Question 20.
50 × 40 = 2000 then find
a) 51 × 41 = ……………….
b) 52 × 42 = ……………….
c) 49 × 39 = ……………….
d) 48 × 38 = ……………….
Answer:
a) 51 × 41 = 50 × 40 + (50 + 40) + 1² = 2000 + 90 + 1 = 2091
b) 52 × 42 = 50 × 40 + 2(50 + 40) + 2² = 2000 + 180 + 4 = 2184
c) 49 × 39 = 50 × 40 – (50 + 40) + 1² = 2000 – 90 + 1 = 1911
d) 48 × 38 = 50 × 40 – 2(50 + 40) + 2² = 2000 – 180 + 4 = 1824

Question 21.
A total of 21 vehicles including bikes and autorickshaw ^vere parked in a cinema theatre. If the total number of wheels is 49, what is the number of autorickshaws and bike each?
Answer:
Let, the number of bikes = x
Number of autorickshaws = y
x + y = 21 ……………. (1)
Since bikes have 2 wheel and autorickshaw have 3 wheels then,
2x + 3y = 49 ……………… (2)
(1) × 3 → 3x + 3y = 63 ……………. (3)
(3) – (2) → x = 14
Therefore, y = 21 – 14 = 7
Thus, number of bikes =14
Number, of autorickshaws = 7

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

Question 22.
On a base AB, 7cm in length, draw a triangle CAB. Mark X on AB such that AX = 4.2 cm. Through X draw a line XY parallel to BC to meet AC at Y.
(a) Draw a rough diagram
(b) Calculate
(i) \(\frac{A X}{B X}\) and \(\frac{A Y}{C Y}\)
(ii) \(\frac{A B}{A X}\) and \(\frac{A C}{A Y}\)
(iii) \(\frac{A B}{B X}\) and \(\frac{A C}{C Y}\)
Answer:

Question 23.
Construct a right triangle having perimeter 15 cm and sides are in the ratio 3 : 4 : 5.
Answer:
Draw a line of length 15 cm, divide it in the ratio 3 : 4 : 5. Construct the triangle.

Question 24.
If the sides of a rectangle are decreased by one metre, its area would be 240 square metres; if increased by one metre, it would be 306 square metres.
i) What is the area of the rectangle?
ii) What is its perimeter?
iii) What are the lengths of its sides?
Answer:
Let length be x and y be breath, now we can write
(x – 1 )(y – 1) = 306 ⇒ xy – (x + y) + 1 = 240 …………… (1)
(x + 1 )(y + 1) = 240 ⇒ xy + (x + y) + 1 = 306 ………….. (2)
(1) + (2) → 2xy + 2 = 546
⇒ xy = \(\frac{546 – 2}{2}\) = 272
(2) – (1) → 2(x + y) = 66
⇒ x + y = \(\frac{66}{2}\) = 33 …………… (3)
We know that,
(x – y)² = (x + y)² – 4xy
= 33² – (4 × 272) = 1
⇒ x – y = 1 …………… (4)
(3) + (4) → 2x = 34
⇒ x = 17
(3) – (4) → 2y = 32
⇒ y = 16
i) Area of rectangle -xy = 272 square metres
ii) Perimeter = 2(x + y) = 66 metre
iii) Side lengths are; 17 m and 16 m

Question 25.
Write in ascending o(der.
(i) 3√5 and 4√3
(ii) 2√5, 5√2 and 3√7
Answer:

Question 26.
In the figure ABCD is a rectangle BC = 24 cm, DP = 10 cm, CD = 15 cm. Find AQ and CQ.

Answer:
CD = 15 cm and DP = 10 cm ⇒ PC = 5 cm.
Let x be the length of CQ.
BQ = x + 24
Triangles ΔABQ and PCQ are similar.
\(\frac{P C}{A B}\) = \(\frac{C Q}{B Q}\)
Since ABCD is a rectangle
AB = CD = 15 cm
\(\frac{5}{15}\) = \(\frac{x}{x+24}\)
Solving we get, x = 12 cm
AQ² = AB² + BQ² = 15² + 36² = 1521
AQ = 39 cm

Question 27.
If x = 4 y = -3 and z = 8 find the value of
a) (x + y) + z
b) x + (y + z)
c) xyz
d) (x + y)z
e) xy + xz
Answer:
a) (x + y) + z = (4 – 3) + 8
= (4 – 3) + 8
= 1 + 8
= 9

b) x + (y + z) = 4 + (-3 + 8)
= 4 + 5
= 9

c) yz = 4 × (-3) × 8
= -12 × 8
= -96

d) (x + y)z = (4 – 3) × 8
= 1 × 8
= 8

e) xy + xz = (4 × (-3)) + (4 × 8)
= – 12 + 32
= 20

Question 28.
Check whether the equations are identities. Write the patterns got from each, on taking x = 1, 2, 3 and x = -1, -2, -3
a) – x + (x + 3) = 3
b) (x + 2) – (x + 3) = -1
c) – x – (x + 1) + 2x + 1 = 0
Answer:
If x = 1
a) – x + (x + 3) = – 1 + (1 + 3)
= – 1 + 4
= 3

b) (x + 2) – (x + 3) = (1 + 2) – (1 + 3)
= 3 – 4
= -1

c) – x – (x + 1) + 2x + 1 = -1 – (1 + 1) + 2 + 1
= – 1 – 2 + 2 + 1
= 0

Similarly, if we take x = 2, 3 we can see all these identities are true always. Now, if x = -1
a) – x + (x + 3) = – (- 1) + (- 1 + 3)
= 1 + 2
= 3

b) (x + 2) – (x + 3) = (- 1 + 2) – (- 1 + 3)
= 1 – 2
= -1

c) – x – (x + 1) + 2x + 1 = – (- 1) – (- 1 + 1) – 2 + 1
= 1 – 0 – 1
= 0
Similarly, if we take x = -2, -3 we can see all these identities are true always.

Question 29.
Manju has some four-wheel toys and three-wheel toys. Total number of wheels is 29 and total number of toys is 8
a) Write the suitable equations by taking number of toys of different types as x and y.
b) Find the number of three-wheel toys and four-wheel toys by solving the equations.
Answer:
a) x + y = 8 …………….(1)
4x + 3y = 29 ………………. (2)
b) (1) × 4 → 4x + 4y = 32 ……………. (3)
(3) – (2) → y = 3
(1) → x + 3 = 8
x = 8 – 3 = 5