In a school one day the maths teacher told the students of class IX about the number systems. She drew a number line and told them that the number line represents various types of numbers on it.
Rational numbers can be represented on the number
line. A number is called a rational number if it can be written in the form of
p / q , where p and q are integers and q ≠0.
Based on the above information, answer the
following questions.
Question | Answer |
1 | b |
2 | d |
3 | c |
4 | b |
5 | c |
On the basis of above information, answer the following questions.
a) x10
b) x10 + x9 + 3
c) x10 + 4
d) x + 10
Q5) The value of the polynomial x2 - 3x + 2 at x = -1 is
Question | Answer |
1 | d |
2 | b |
3 | a |
4 | c |
5 | b |
a) (3, 3), (- 2, 3), (- 2, - 2), (4, - 2)
b) (4, 3), (- 2, 3), (- 2, - 2), (4, - 2)
c) (4, 3), (- 2, - 2), (- 2, - 2), (4, - 2)
d) (4, 3), (- 2, 3), (- 2, - 4), (4, - 2)
a) 2 units b) 3 units c) 6 units d) 8 units
a) 2 b) - 4 c) -2 d) 4
a) Square b) Rectangle c) Parallelogram d) Rhombus
Q5) The area of the shape formed by joining the positions of four friends.
a) 30 b) 36 c) 25 d) 32
Question | Answer |
1 | b |
2 | c |
3 | d |
4 | b |
5 | a |
Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in the figure.
Question 1
What are the coordinates of A and B respectively?
a) A(3, 5); B(7, 8)
b) A(5, 3); B(8, 7)
c) A(3, 5); B(7, 9)
d) A(5, 3); B(9, 7)
Answer c
Question 2
What are the coordinates of C and D respectively?
a) C(11, 5); D(7, 1)
b) C(5, 11); D(1, 7)
c) C(5, 11); D(7, 1)
d) C(5, 11); D(-1, 7)
Answer a
Question 3
What is the distance between B and D?
a) 5 units b) 14 units c) 8 units d) 10 units
Answer c
Question 4
What is the distance between A and C?
a) 5 units b) 14 units c) 8 units d) 10 units
Answer c
Question 5
What are the coordinates of the point of intersection of AC and BD?
a) (7, 5) b) (5, 7) c) (7, 7) d) (5, 5)
Answer a
|
|
Answer Key |
|
Q1 |
(0, 0) |
|
Q2 |
(4, 6) |
|
Q3 |
(6, 5) |
|
Q4 |
(16, 0) |
|
Q5 |
(-12, 6) |
Prime Minister's National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and man-made disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc. Two friends Sita and Gita, together contributed Rs. 200 towards Prime Minister's Relief Fund. Answer the following
Q1) Which out of the
following is not the linear equation in two variables ?
(a) 2x = 3
(b) x2 + x = 1
(c) 4 = 5x – 4y
(d) x – √2y = 3
Q2) How to represent the above situation in linear equations in two variables ?
(a) 2x + y = 200
(b) 200x = y
(c) x + y = 200
(d) 200 + x = y
Q3) If Sita contributed Rs. 76, then how much was contributed by Gita ?
a) Rs. 120 b) Rs. 124 c) Rs. 123 d) Rs. 125
Q4) If both contributed equally, then how much is contributed by each?
(a) Rs. 50, Rs. 150
(b) Rs. 50, Rs. 50
(c) Rs. 100, Rs. 100
(d) Rs. 120, Rs. 120
Q5) Which is the standard form of linear equations x = – 5 ?
(a) x + 5 = 0
(b) 1.x – 5 = 0
(c) 1.x + 0.y + 5 = 0
(d) 1.x + 0.y = 5
Question | Answer |
1 | b |
2 | c |
3 | b |
4 | c |
5 | c |
Two friends Pankaj and Rohit went to a market. Pankaj bought 3 notebooks and 2 pens for Rs. 80. Rohit also bought the same types of notebooks and pens as Pankaj. He paid 110 for 4 notebooks and 3 pens.
From the above
information answer the following questions
Q1) Form the pair of
linear equations in two variables from this situation by taking
cost of one notebook as
Rs. x and cost of one pen as Rs. y.
(a) 3x + 2y = 80 and 4x +
3y = 110
(b) 2x + 3y = 80 and 3x +
4y = 110
(c) x + y = 80 and x + y
= 110
(d) 3x + 2y = 110 and 4x
+ 3y = 80
Solution
Here, the cost of one
notebook be Rs. x and that of pen be Rs. y.
According to the
statement, we have
3x + 2y = 80 and
4x + 3y = 110
Q2) Which is the solution satisfying both the equations formed in (i)?
(a) x = 10, y = 20
(b) x = 20, y = 10
(c) x = 15, y = 15
(d) none of these
Solution
3x + 2y = 3(20) + 2(10) =
60 + 20 = 80
4x + 3y = 4(20) + 3(10) =
80 + 30 = 110
Ans: (b) x = 20, y = 10
Q3) Find the cost of one pen?
(a) Rs. 20 (b) Rs. 10 (c) Rs. 5 (d) Rs. 15
Ans: (b) Rs. 10
Q4) Which is the
solution satisfying both the equations formed in (i)?
(a) x = 10, y = 20
(b) x = 20, y = 10
(c) x = 15, y = 15
(d) none of these
Solution
3x + 2y = 3(20) + 2(10) =
60 + 20 = 80
4x + 3y = 4(20) + 3(10) =
80 + 30 = 110
Ans: (b) x = 20, y = 10
Q5) Find the total cost if they will purchase the same type of 15 notebooks and 12 pens.
(a) Rs. 400 (b) Rs. 350 (c) Rs. 450 (d) Rs. 420
Solution
Total cost = Rs. 15 x 20
+ Rs. 12 x 10
= 300 + 120 = Rs. 420
Mohan has a piece of land in the shape of a given figure. He divides the land into two parts to produce different crops.
On the basis of the
above information, answer the following questions:
Q1) The length of AC is
a) 40m b) 50m c) 30m d) 80 m
Q2) The area of â–³ABC is
a) 400 m2
Q3) The area of â–³ACD is
a) 1000 m2 b) 500
m2 c) 1500
m2 d) 1000 m2
Q4) The cost of ploughing the land at the rate of ₹ 20 / m2 is (Take
a) ₹ 12000 b) ₹ 34640 c) ₹ 46640 d) None of these
Q5) The cost of fencing all around the land with barbed wire @ of ₹ 30 / m is
a) ₹ 6000 b) ₹ 5000 c) ₹ 6500 d) ₹ 6600
Question | Answer |
1 | b |
2 | c |
3 | a |
4 | c |
5 | d |
Case Study-Based Question: Cone
Riya is planning to make a birthday party hat in the shape of a cone for her younger sister’s birthday. She uses a thick sheet of paper with a radius of 7 cm for the base of the cone and a slant height of 25 cm. To decorate the hat, she plans to cover it with coloured paper, excluding the circular base. She also needs to calculate the volume of the cone to figure out how much space it will occupy.
Read the case study carefully and answer the following questions:
Q1. Calculate the curved surface area (CSA) of the cone.
Q2. If the cost of coloured paper is ₹2 per square cm, what is the total cost to cover the curved surface of the cone?
Q3. Find the volume of the cone.
Q4. Riya decides to add a circular base to the hat with the same radius. What will be the total surface area of the cone including the base?
Q5. If Riya makes 10 such hats, calculate the total cost of coloured paper required to decorate all the hats (only the curved surface).
Answer Key
Given Data: Radius (r) = 7 cm, Slant height (l) = 25 cm
Cost of coloured paper = ₹2 per square cm, π =3.14
Answer1
Curved surface area = πrl = 3.14 × 7 × 25 = 549.5 cm2.
Answer2
Cost to cover curved surface area = CSA × Cost per square cm = 549.5 × 2 = 1099
Answer 3:
Volume of the cone :
First calculate the height by using Pythagoras theorem we
get h = 24 cm
Volume = 1/3 π r2h = 1/3 × 3.14 × 7 x 7 × 24 = 1232.16 cm3.
Total surface area including base area TSA = CSA + Area of Base
= πrl + πr2 = 549.5 + 153.86 = 703.36 cm2.
Answer 5 Cost of 10 hats (CSA only) = 10 × CSA × cost per cm2.
Mathematics teacher of a school took her 9th standard students to show Red fort. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Red fort to students. Then the teacher said in this monument one can find combination of solid figures. There are 2 pillars which are cylindrical in shape. Also 2 domes at the corners which are hemispherical.7 smaller domes at the centre. Flag hoisting ceremony on Independence Day takes place near these domes.
i) How much cloth material will be required to cover 2 big domes each of radius 2.5 m Take π = 22/7
Ans: 78.57 m2
ii) Write the formula to find the volume of a cylindrical pillar.
Ans: πr2h
iii) Find the lateral surface area of two pillars if height of the pillar is 7m and radius of the base is 1.4m
Ans: 123.2 m2
iv) How much is the volume of a hemisphere if the radius of the base is 3.5m ?
Ans: 89.83 m3
v) What is the ratio of sum of volumes of two hemisphere of radius 1 cm each to the volume of a sphere of radius 2 cm ?
Ans: 1:8
Height Intervals (in cms) | No. of students |
131-140 | 1 |
141-150 | 7 |
151-160 | 5 |
161-170 | 9 |
171-180 | 9 |
181-190 | 10 |
Total | 41 |
Q1) Class size of third class interval is
a) 8 b) 9 c) 9.5 d) 10
Q2) Upper limit of 5th class interval is
a) 180 b)170.5 c) 180.5 d)179.5
Q3) Class mark of 6th class interval is
a) 184.5 b)185 c)185.5 d)186
Q4) How many students have their height more than 160 cm
a) 19 b) 18 c) 27 d) 28
Q5) How many students have their height less than or equal to 180 cm ?
a) 31 b) 19 c)29 d) 22
Question | Answer |
1 | d |
2 | c |
3 | c |
4 | d |
5 | a |
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