Kerala Syllabus Class 9 Maths Model Question Paper Set 5 English Medium

Teachers recommend solving Maths Model Question Paper for Class 9 Kerala Syllabus Set 5 to improve time management during exams.

Kerala Syllabus Std 9 Maths Model Question Paper Set 5 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.
The point of intersection of all three medians of a triangle is called ……………….. of the triangle.
Answer:
Centroid.

Question 2.
What is the side of a square of area 5 square metres?
Answer:
√5 m.

Question 3.
If we take two positive numbers as x, y and x > y;
a) write their sum.
b) write their difference.
Answer:
a) x + y
b) x – y

Question 4.
Write two examples of an irrational number.
Answer:
√2, √3

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

Question 5.
The sum of two numbers is 42 and their difference is 24. Find the numbers.
Answer:
Let x and y be two numbers
Then x + y = 42 …………… (i)
x – y = 24 ……………… (ii)
By adding (i) and (ii)
We get 2 x = 66
∴ x = 33
Substitute the value of x in (i)
33 + y = 42
y = 42 – 33 = 9
The numbers are 33 and 9.

Question 6.
In the figure lines AP, BQ, CR and DS are parallel and equally spaced.

If QR = 2cm, what is the length of PS?
Answer:
In the figure all the four parallel lines are cut by the line PS and it is given that QR = 2 cm and it is given that the parallel lines are placed equally.
∴ AB = BC = CD
So, PQ = QR = RS = 2 cm
If three or more parallel lines cut a line into equal parts, they will cut any line into equal parts.
∴ PS = PQ + QR + RS = 2 + 2 + 2 = 6 cm

Question 7.
In the picture, the side of the smallest square is 1 centimetre.

a) Calculate the area of the largest square.
b) What is the length of the side of the largest square?
Answer:
a) From the figure it is given that the side of the smallest square is 1 cm
So, AC² = 1² + 1² = 2
AC = √2 cm
EC² = √2² + √2² = 2 + 2 = 4 cm
EC = 2 cm
EH² = (2)² +(2)² = 4 + 4 = 8 cm
EH = 2√2 cm
Area of largest squared = (side)² = EH²
= (2√2)² = 8 cm²

b) Side of the largest square = EH = 2√2 cm.

Question 8.
The price of a table and ‘chair is 5000 rupees. The price of the table and three chairs is 7000 rupees.
a) What’s the price of a chair?
b) What’s the price of a table?
Answer:
x + y = 5000 …………. (i)
x + 3y = 7000 ………….. (ii)
By subtracting (i) and (ii)
we get, 2y = 2000
y = 1000
By substituting the value of y in (i) we get,
1000 + x = 5000
∴ x = 5000 – 1000
= 4000
a) Price of a chair = Rs. 1000
b) Price of a table = Rs. 4000

Question 9.
The sides of a rectangle are √2 + 1 centimetres and √2 – 1 centimetres.
a) What is the perimeter of this rectangle?
b) What is the area of this rectangle?
Answer:

a) Perimeter of the rectangle = 2(l + b)
= 2 ( (√2 + 1 )+ (√2 – 1))
= 2(2√2)
= 4√2 cm

b) Area of the rectangle = length × breadth
= (√2 + 1)(√2 – 1)
= √2² – 1²
= 2 – 1
= 1 cm²

Question 10.
Draw an equilateral triangle of perimeter 13 centimetres.
Answer:

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

Question 11.
Five years ago, Milan was thrice as old as Miya. Ten years later, Milan will be twice as old as Miya. How old are Milan and Miya?
Answer:
Let, age of Milan = x
age of Miya = y
Five years ago,
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -10 …………… (1)
10 years later,
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10 …………….. (2)
(2) – (1) → y = 20
(2) → x – 40 = 10
x = 50
Milan’s age = 50
Miya’s age = 20

Question 12.
Draw a rectangle of perimeter 13 centimetres and sides in the ratio 3 : 4.
Answer:

Question 13.
a) Write the decimal form of \(\frac{1}{3}\)
b) Write the decimal form of \(\frac{1}{9}\)
c) Write the sum 0.111…+ 0.222… in decimal form.
d) Write the product 0.333… × 0.666… in decimal form.
Answer:
a) \(\frac{1}{3}\) = \(\frac{3}{9}\) = 0.333…..
b) \(\frac{1}{9}\) = 0.111……..
c) 0.111…… + 0.222… = 0.333…
d) 0.333…+ 0.666… = \(\frac{3}{9}\) × \(\frac{6}{9}\) = \(\frac{1}{3}\) × \(\frac{2}{3}\) = \(\frac{2}{9}\) = 0.222…

Question 14.
The perimeter of a rectangle is 26 centimetres and its length is 3 centimetres more than the breadth.
a) length + breadth = ………………….
b) length – breadth = ………………….
c) Calculate the lengths of the sides.
Answer:
a) Length + breadth = \(\frac{26}{2}\) = 13 centimetres.

b) According to the question, length = breadth + 3
so, breadth + 3 + breadth = 13 (from (a))
2 breadth = 13 – 3 = 10
Breadth = \(\frac{10}{3}\) = 5 centimetres.
Length = 13 – 5 = 8 centimetres.
Now, length – breadth = 8 – 5 = 3 centimetres.

c) length = 8 centimetres,
breadth = 5 centimetres.

Question 15.
The product of two numbers is 300 and their sum is 35.
a) What is the product of the numbers just after each?
b) What is the product of the numbers just before each?
Answer:
Let the numbers be x and y.
Given that xy = 300 and x + y = 35
(x + 1 )(y + 1) = xy + x + y +1 = 300 + 35 + 1 = 336
(x – 1)(y – 1) = xy – (x + y) + 1 = 300 – 35 + 1 = 266

Question 16.
The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. Find the number.
Answer:
Let the number be 10x + y.
Given that x + y = 9 ……………. (1)
According to the question,
10x + y + 27 = 10y + x
9x – 9y = -27
x – y = \(\frac{-27}{9}\) = -3 ……………. (2)
adding (1) and (2), we get
2x = 6
x = \(\frac{6}{2}\) = 3
Put x = 3 in (1), we get
3 + y = 9
y = 9 – 3 = 6
Therefore, the numbers are 3 and 6.

Question 17.
The perimeter of a rectangle is 40 centimetres and its area is 99 square centimetres. Find the area of the rectangle with each side 2 centimetres shorter.
Answer:
Let l be the length and b be the breath of the rectangle.
Perimeter of the rectangle = 2 (l + b) = 40 centimetres.
(l + b) = 20 centimetres.
Area of the rectangle = l × b = 99 square centimetres. ,
Area of the rectangle with each side 2 centimetres shorter is = (l – 2 )(b – 2)
= lb – 2(l + b) + 4
= 99 – 2 (20) + 4
= 99 – 40 + 4
= 63 square centimetres.

Question 18.
In the picture, the midpoints of a triangle ABC are joined to form a smaller triangle PQR inside:

a) What part of the area of triangle ABC is triangle PQR?
b) How many times the perimeter of triangle PQR is the perimeter of triangle ABC?
Answer:
a) For Δ ABC
Let AB = x,BC = y and AC = c.
Hence AP = PB = \(\frac{x}{2}\)
BR = RC = \(\frac{y}{2}\)
AQ = QC = \(\frac{z}{2}\)
Since Δ PQR are at the midpoint of Δ ABC. Therefore,
PQ = \(\frac{y}{2}\)
QR = \(\frac{x}{2}\)
PR = \(\frac{z}{2}\)
Consider Δ PBR and Δ PQR where,
PB = QR = \(\frac{x}{2}\)
BR = PQ = \(\frac{y}{2}\)
PR = PR
Thus Δ PBR ~ Δ PQR
Therefore, area of Δ PQR = \(\frac{1}{4}\) (area of Δ ABC)
Hence area of triangle PQR is \(\frac{1}{4}\)^th part of the area of triangle ABC.

b) Perimeter of Δ PQR = \(\frac{y}{2}\) + \(\frac{x}{2}\) + =\(\frac{z}{2}\) = \(\frac{1}{2}\)(x + y + z)
Perimeter of Δ ABC = x + y + z
Therefore,
Perimeter of Δ PQR = \(\frac{1}{2}\) (Perimeter of Δ ABC)

Question 19.
Find the length of the longest stick that can fit in a cubical vessel (without bending) of side 10 centimetres.
Answer:
Given, length of the side of the cubical vessel = 10 cm

The stick can be fit in the cubical vessel without bending is through its diagonal. Thus,
For Δ EFG
EG² = EF² + FG²
= 10² + 10² = 200
EG = \(\sqrt{200}\) = 10√2
For Δ AEG ,
AG² = EG² + AE²
= (10√2)² + 10²
= 200 + 100 = 300
AG = \(\sqrt{300}\) = 10√3
Therefore, length of the longest stick = 10√3 cm

Question 20.
In the figure, P is the midpoint of BC and Q is the midpoint of AC.

If the area of triangle APQ is 5 square centimetres,
a) What is the area of triangle QPC? ,
b) What is the area of triangle ABC?
Answer:
a) Given, area of ΔAPQ = 5 square centimeter
We know a line from the vertex of a triangle divides the length of the opposite side and the area of the triangle in the same ratio. Thus, if the line if from the vertex to the midpoint of the opposite side then the area of triangle is same.
Here, Q is the midpoint AC
Therefore, area of ΔQPC = 5 square centimetre

b) Area of ΔAPC = 10 square centimeter
Since P is the midpoint of BC from the vertex A
Therefore, area of ΔABC = 2 × 10 = 20 square centimeter.

Question 21.
Draw a rectangle of perimeter 30 centimeters and sides in the ratio 5 : 3.
Answer:

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

Question 22.
A vertical pole of length 6 m casts a shadow 4 m long on the ground arid at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Answer:

Here both the triangles are similar. So,
\(\frac{A C}{P R}\) = \(\frac{A B}{P Q}\)
\(\frac{6}{P R}\) = \(\frac{4}{28}\)
PR = 6 × 7 = 42 m

Question 23.
The sum of the digits of a two-digit number is 12. The number got by interchanging the digits is 36 more than the original number. What is the number?
Answer:
Let 10x + y be the two-digit number then x + y = 12 ……………. (1)
If the number is interchanged, then the digit will be 10y + x
Thus,
10y + x = (10x + y) + 36
9y – 9x = 36
y – x = 4 ……………… (2)
(1) + (2) → 2y = 16
⇒ y = 8
∴ x = 4
Thus, the two-digit number is 48.

Question 24.
a) What is the hypotenuse of a right triangle with perpendicular sides √2 cm and √3 cm? How much larger than the hypotenuse is the sum of the perpendicular sides?
b) Sides of a triangle are √3 + √2, √3 – √2 and 2. Find the approximate perimeter of the triangle.
c) Side of a square is √2 + 1. What is the area of the square?
d) Area of a square is 2 m². What is the length of its diagonal?
Answer:
a) Hypotenuse = \(\sqrt{(\sqrt{2})^2+(\sqrt{3})^2}\) = \(\sqrt{2+3}\) = √5
Hypotenuse – sum of the perpendicular sides = √5 – (√2 + √3) = √5 – √2 – √3
b) Perimeter of the triangle = √3 + √2 + √3 – √2 + 2 = 2√3 + 2 = 2 × 1.732 + 2 = 5.464
c) Area = (side)² = (√2 + 1)² = 2 + 2√2 + 1 = (3 + 2√2) m²
d) If area is 2 then the side length is √2 m.
Length of the diagonal = side × √2 = √2 × √2 = 2m.

Question 25.
\(\frac{1}{p}\) + \(\frac{1}{q}\) = \(\frac{7}{12}\)
\(\frac{1}{p}\) – \(\frac{1}{q}\) = \(\frac{1}{12}\)
a) If \(\frac{1}{p}\) = x and \(\frac{1}{q}\) = y then writes the equations.
b) Find x and y.
c) What is p and q?
Answer:
a) x + y = \(\frac{7}{12}\) ……………… (1)
x – y = \(\frac{1}{12}\) …………… (2)

b) Adding (1) and (2) we get,
(1) + (2) gives 2x = \(\frac{8}{12}\)
x = \(\frac{4}{12}\) = \(\frac{1}{3}\)
Substituting this in (1) we get, y = \(\frac{7}{12}\) – \(\frac{4}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

c) \(\frac{1}{p}\) = \(\frac{1}{3}\). So, p = 3.
\(\frac{1}{q}\) = \(\frac{1}{4}\) So, q = 4.

Question 26.
Divide a 7 cm long line into three equal pieces.
Answer:

Question 27.
We can make a trapezium by cutting a square and an equilateral triangle with sides twice that of the square, and rearranging the pieces as below:

Answer:
Consider one part of the square. Its measures are:

Consider one part of the triangle. Its measures are:

Now, the measures of the trapezium are;

Perimeter of the trapezium = 2√2 + 2 + 2√3 + 2 + 2√2 + 2√3
= 4 + 4√2 + 4√3
= 4(1 + √2 + √3)
≈ 4(1 + 1.414 + 1.732)
≈ 4 × 4.146
≈ 16.584 cm

Area of the trapezium = \(\frac{1}{2}\) × sum of the lengths of the parallel sides × height
= \(\frac{1}{2}\) × (2 + 2√3 + 2 + 2√3) × 2
= 4 + 4√3
= 4(1 + √3)
≈ 4(1 + 1.732)
≈ 4 × 2.732
≈ 10.928 cm²

Question 28.
Draw a triangle and mark a point inside it. Join this point to the vertices and extend each of them by half its original length. Join the end points of these lines to form another triangle:

Prove that the sides of the larger triangle are one and a half times the sides of the original triangle.
Answer:

Consider ΔBDC and ΔQDR where,
BD : QD = 2x : 3x = 2 : 3
CD : RD = 2y : 3y = 2 : 3
Since the 2 sides are scaled by the same factor and one angle is common, then the two triangles are similar.
BC : QR = 2 : 3
\(\frac{B C}{Q R}\) = \(\frac{2}{3}\)
QR = \(\frac{3}{2}\)BC = 1\(\frac{1}{2}\)BC
Similarly, PR = \(\frac{3}{2}\)AC = 1\(\frac{1}{2}\)AC
PQ = \(\frac{3}{2}\)AB = 1\(\frac{1}{2}\)AB

Question 29.
Calculate y when x = -2 and x = –\(\frac{1}{2}\) in the equation y = \(\frac{1}{x-1}\) + \(\frac{1}{x+1}\)
Answer: