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Questions:

Generic 3

7 x 3 = 21

Simplify and find the degree of the expression
(4m² + 3n)− (3m + 9n² )− (3m² − 6n² ) + (5m− n)

In the following figures, polygons are formed by increasing the number of sides using matchsticks as given below.

Find the number of sticks required to form the next three shapes by tabulation and generalisation.

Can the following angles form a triangle?
(i) 80⁰, 70⁰, 50⁰
(ii) 56⁰, 64⁰, 60⁰

Express the following numbers in exponential form with the given base:
243, base 3.

Find the perimeter of the given shape (Figure) (Take \(\pi =\frac { 22 }{ 7 } \)).

Two automatic ice cream machines A and B designed to fill cups with 100 ml of ice cream were tested. Two cups of ice creams, one from machine A and one from machine B are weighed and found to be 99.56 ml from machine A and 99.65 ml from machine B. Can you say which machine gives more quantity of ice creams?

The height of a person is 165 cm. Express this height in metre.

Multiple Choice Question

10 x 1 = 10

If p(x) and q(x) are two expressions of degree 3, then the degree of p(x) + q(x) is

(a)

(b)

(c)

(d)

Undefined

Choose the correct relationship between x and y from the given table.

x	-2	-1	0	1	2	...
y	4	5	6	7	8	...

(a)

y = x + 4

(b)

y = x + 5

(c)

y = x + 6

(d)

y = x + 7

10.

Identify the correct relationship between x and y from the given table.

x	1	2	3	4	...
y	4	8	12	16	...

(a)

y = 4x

(b)

y = x + 4

(c)

y = 4

(d)

y = 4 × 4

11.

The unit digit of the numeric expression 10⁷¹ + 10⁷² + 10⁷³ is

(a)

(b)

(c)

(d)

12.

In the given figure, AB is parallel to CD. Then the value of b is

(a)

112⁰

(b)

68⁰

(c)

102⁰

(d)

62⁰

13.

The angles of a triangle are in the ratio 2: 3: 4. Then the angles are

(a)

20, 30, 40

(b)

40, 60, 80

(c)

80, 20, 80

(d)

10, 15, 20

14.

The formula used to find the area of the rectangular path is

(a)

π(R²− r²) sq. units

(b)

(L × B) − (l × b) sq. units

(c)

LB sq. units

(d)

lb sq. units

15.

In the formula, C = 2πr, ‘r’ refers to

(a)

circumference

(b)

area

(c)

rotation

(d)

radius

16.

The simplest form of 0.35 is

(a)

\(\frac { 35 }{ 1000 } \)

(b)

\(\frac { 35 }{ 10 } \)

(c)

\(\frac { 7 }{ 20 } \)

(d)

\(\frac { 7 }{ 100 } \)

17.

3 + \(\frac { 4 }{ 100 } +\frac { 9 }{ 1000 } \) = ?

(a)

30.49

(b)

3049

(c)

3.0049

(d)

3.049

Generic 2

13 x 2 = 26

18.

Verify whether the following hexagonal shapes form a part of the Pascal’s Triangle.

19.

Verify whether the following hexagonal shapes form a part of the Pascal’s Triangle.

20.

Find the trianglular numbers from the Pascal’s Triangle and colour them.

21.

Find the degree of the following terms.
-7ab

22.

Using the given figure, prove that the triangles are congruent. Can you conclude that AC is parallel to DE.

23.

II. Construct a triangle ABC with given conditions.
(i) AB = 7 cm, AC = 6.5 cm and ∠A = 120°.
(ii) BC = 8 cm, AC = 6 cm and ∠C = 40°.
(iii) An isosceles obtuse triangle with equal sides 5 cm

24.

Simplify and express each of the following in exponential form:
4⁵ x 4² x 4⁴

25.

A goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.

26.

The cost of fencing a circular race course at the rate of Rs.8 per metre is Rs.2112. Find the diameter of the race course.

27.

A picture is painted on a ceiling of a marriage hall whose length and breadth are 18 m and 7 m respectively. There is a border of 10 cm along each of its sides. Find the area of the border.

28.

Write the following decimal numbers in words.
0.7

29.

Represent the following decimal numbers on the number line.
2.1

30.

Compare the following decimal numbers and find out the smaller number.
0.99, 1.9

Fill in the blanks

5 x 1 = 5

31.

Degree of the constant term is _________.

32.

The degree of the term a³b²c⁴d² is ________.

33.

When the unit digit of the base and its expanded form of that number is 9, then the exponent must be ________ power.

34.

When base is 12 and exponent is 17, its exponential form is___________

35.

The exponential form 14⁹ should be read as_________

Generic 5

2 x 5 = 10

36.

Tabulate the 3^rd slanting row of the Pascal’s Triangle by taking the position of the numbers in the slanting row as x and the corresponding values as y.

x	1	2	3	4	5	6	...
y	1	3	6	10	15	21	...

Verify whether the relationship, y = \(\frac { x(x+1) }{ 2 } \) between x and y for the given values is true.

37.

Find the unit digit of the following exponential numbers:
25²³

Match the following

5 x 1 = 5

38.

39.

Match the given patterns of shapes with the appropriate number pattern and its generalization.

40.

729⁷²⁵

41.

121¹¹

42.

20¹⁰

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Answers Key:

Generic 3

7 x 3 = 21

(4m² + 3n) − (3m+ 9n² ) − (3m² − 6n² ) + (5m−n)
= 4m² + 3n − 3m − 9n² − 3m² + 6n² + 5m − n
= (4m² − 3m² ) + (3n − n) + (−3m+ 5m) + (−9n² + 6n²)
= m² + 2n + 2m − 3n²
Hence, the degree of the expression is 2.

In the above pattern of polygons, in the first shape (x = 1), we get a closed shape called a triangle. Similarly the second shape (x = 2 ) gives a four sided polygon and the third shape (x = 3) is a five sided polygon and continuing in the same way two more shapes are formed. If the number of match sticks required to form each of the shapes is taken as y, then the values of x and y are tabulated as given below.

x	1	2	3	4	5	...
y	3	4	5	6	7	...

Observe the table and express y in terms of x as below:
when x = 1, y = 3 = 1 + 2
when x = 2 , y = 4 = 2 + 2
when x = 3, y = 5 = 3 + 2
when x = 4 , y = 6 = 4 + 2
when x = 5, y = 7 = 5 + 2
Hence, each of the values of y which we get from the table is 2 more than x. That is
y = x + 2 .
Therefore,
6^th shape (x = 6) will have y = 8 = 6 + 2 (8 match sticks)
7^th shape (x = 7) will have y = 9 = 7 + 2 (9 match sticks)
8^th shape (x = 8) will have y = 10 = 8 + 2 (10 match sticks)
We clearly see that the next three shapes will require 8, 9 and 10 matchsticks.

(i) Given angles 80⁰, 70⁰, 50⁰
Sum of the angles = 80⁰+ 70⁰+ 50⁰ = 200⁰ ≠ 180⁰
The given angles cannot form a triangle.
(ii) Given angles 56⁰, 64⁰, 60⁰
Sum of the angles = 56⁰+ 64⁰+ 60⁰ = 180⁰
The given angles can form a triangle.

243 = 3 x 3 x 3 x 3 x 3 = 3⁵

In this shape, we have to calculate the circumference of semicircle on each side of the rectangle. The outer boundary of this figure is made up of semicircles of two different sizes. Diameters of each of the semicircles are 7 cm and 14 cm. We know that the circumference of the circle = πd units.
Circumference of the semicircular part = \(\frac { 1 }{ 2 } \)πd units
Hence, the circumference of the semicircle having diameter 7 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 7 = 11 cm
Circumference of the pair of semicircular parts (II and IV) = 2 x 11 = 22 cm
Similarly, circumference of the semicircle having diameter 14 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 14 = 22 cm
Circumference of the pair of semicircular parts (I and III) = 2 x 22 = 44 cm.
Perimeter of the given shape = 22 + 44 = 66 cm.

Compare 99.56 and 99.65
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 5 < 6.
Therefore, 99.56 < 99.65

Given, height of the person = 165 cm.
Hence, the height of the person = \(\frac { 165 }{ 100 } \) = 1.65 m
Since, 1 cm = \(\frac { 1 }{ 100 } \) m = 0.01 m

Multiple Choice Question

10 x 1 = 10

10.

(a)
y = 4x

11.

(a)
0

12.

(b)
68⁰

13.

(b)
40, 60, 80

14.

(b)
(L × B) − (l × b) sq. units

15.

(d)
radius

16.

17.

(d)
3.049

Generic 2

13 x 2 = 26

18.

8 × 45 × 84 = 28 × 9 × 120

19.

1 × 13 × 66 = 11 × 1 × 78

20.

21.

Degree 2

22.

BC = BD
BE = AB
<CBA = <DBE
By SAS criterion \(\triangle C D B \cong \angle E B D\)
The two sides AC is parallel to DE.

23.

(i)

Steps:
1. Draw a line segmentAB = 7 cm.
2. At A, draw a ray AX making an angle of 120⁰ with AB.
3. With A as centre, draw an arc of radius 6.5 em to cut the ray AX at C.
4. Join BC.
(ii)

Steps:
1. Draw a line segment \(\overline { BC } =8cm\)
2. At e draw a ray ex making an angle of radius.
3. With e as centre draw an arc of radius 6 cm.
4. Join AB.
(iii)

Steps:
1. Draw a line segment AB = 5 cm.
2. At B, draw a ray BX making an angle of 110° with AB.
3. With B as centre, draw an arc of radius 5 em to cut the ray BX at C.
4. Join AC.

24.

4⁵⁺²x 4⁴ = 4⁷ x 4⁴= 4⁷⁺⁴
= 4¹¹

25.

R =.52 m, r = 45 m
Area of the grass land that the goat cannot graze \(=\pi\left(R^{2}-r^{2}\right) \)
\(=\pi\left(52^{2}-45^{2}\right) \)
\(=\pi(2704-2025) \)
\(=\frac{22}{7}(679) \)
= 22 x 97 = 2134 m².

26.

Cost of fencing a circular race course at the rate of Rs 8 per metre = Rs 2112
\(\beta \times 2 \pi r =2112 \)
\(8 \times 2 \times \frac{22}{7} \times r =2112 \)
\(r=2112 \times \frac{1}{8} \times \frac{1}{2} \times \frac{7}{22}\)
r = 42 m
Diameter of the race course = 2 x 42 = 84 m

27.

Length of marriage hall (L) = 18 m
Breadth of marriage hall (B) = 7 m
Area of the border = Area of outer rectangle - Area of inner rectangle
= L x B - l x b
= 18 x 7 - 17.8 x 6.8.
= 126 - 121.04 = 4.96 m²

28.

seven tenths

29.

30.

Comparing the whole number parts of the two decimal numbers.
We get 0 < 1
0.99 < 1.9
Smaller number is 0.99.

Fill in the blanks

5 x 1 = 5

31.

()
0

32.

()
11

33.

()
odd

34.

()
12¹⁷

35.

()
14 power 9

Generic 5

2 x 5 = 10

36.

Observe the table carefully. To verify the relationship between x and y, let us substitute the values of x and get the values of y.
If x = 1, y = \(\frac { 1(1+1) }{ 2 } =\frac { 2 }{ 2 } =1\)
If x = 2 , y = \(\frac { 2(2+1) }{ 2 } =\frac { 6 }{ 2 } =3\)
If x = 3, y = \(\frac { 3(3+1) }{ 2 } =\frac { 12 }{ 2 } =6\)
If x = 4 , y = \(\frac { 4(4+1) }{ 2 } =\frac { 20 }{ 2 } =10\)
If x = 5, y = \(\frac { 5(5+1) }{ 2 } =\frac { 30 }{ 2 } =15\)
Hence, y = \(\frac { x(x+1) }{ 2 } \) is verified.