7th Standard Maths Important Questions Chapter Important Questions

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Questions:

Generic 3

15 x 3 = 45

In ΔLMN, LM is extended to O. If ㄥL = 62⁰ and ㄥN = 31⁰, find ㄥNMO.

Find the value of
13²

What is the distance travelled by the tip of the seconds hand of a clock in 1 minute, if the length of the hand is 56 mm (use \(\pi =\frac { 22 }{ 7 } \)).

Praveen goes trekking with his friends. He has to record the distance in kilometres in his sports book. Can you help him?
The trekking record for four days are given below
3983 m

The floor of a building consists of 2000 tiles which are rhombus shaped and each of its diagonals are 40 cm and 25 cm. Find the total cost of polishing the floor, if the cost per m² = Rs.5.

In January the high temperature recorded was 90°F and the low temperature was -2°R Find the difference between the high and the low temperatures?

In the figure EF parallel to GH

If x and y vary inversely as each other and x = 10 when y = 6. Find y when x = 15.

If A = 2a²- 4b - 1; B = 5a²+ 3b - 8 and C = 2a²+ 9b + 3 then find the value of A - B + C.

10.

Each day, the workers drill down 22 feet further until they hit a pool of water. If the water is at 110 feet, on which day will they hit the pool of water?

11.

Construct a perpendicular bisector of the line segment AB = 6 cm.

12.

Raghavan wants to change the front elevation of his house using the tiles made up of tetromino shapes

1. How many tetrominoes are there in a tile ?
2. If the cost of a square tile is Rs. 52 then what will be the cost of the tiles that Raghavan buys for the front elevation?

13.

Anbu bought 2 notebooks for Rs 24. How much money will be needed to buy 9 such notebooks?

14.

If x = 3, y = 2 find the value of
(i) 4x + 7y
(ii) 3x + 2y − 5
(iii) x − y

15.

Find the area and perimeter of the parallelogram given in the figures
.

Multiple Choice Question

25 x 1 = 25

16.

What is the sum of the elements of nineth row in the Pascal’s Triangle?

(a)

128

(b)

254

(c)

256

(d)

126

17.

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

x	-2	-1	0	1	2	...
y	6	3	0	-3	-6	...

(a)

y = −2x

(b)

y = +2x

(c)

y = +3x

(d)

y = −3x

18.

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

19.

The unit digit of (32 x 65)⁰ is

(a)

(b)

(c)

(d)

20.

If two plane figures are congruent then they have

(a)

same size

(b)

same shape

(c)

same angle

(d)

same shape and same size

21.

How many zeros are there in 100¹⁰?

(a)

(b)

(c)

(d)

22.

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

23.

The formula to find the width of the circular path is

(a)

( L − l ) units

(b)

( B − b ) units

(c)

( R − r ) units

(d)

( r − R ) units

24.

The formula used to find the area of the circle is ________ sq.units.

(a)

4πr²

(b)

πr²

(c)

2πr²

(d)

πr² + 2r

25.

If the circumference of a circle is 82π , then the value of ‘r’ is

(a)

41 cm

(b)

82 cm

(c)

21 cm

(d)

20 cm

26.

0.009 is equal to

(a)

0.90

(b)

0.090

(c)

0.00900

(d)

0.900

27.

The place value of 3 in 85.073 is _______

(a)

tenths

(b)

hundredths

(c)

thousands

(d)

thousandths

28.

The measure of \(\angle\)BOC is

(a)

90°

(b)

180°

(c)

80°

(d)

100°

29.

The sum of all angles at a point is

(a)

360°

(b)

180°

(c)

90°

(d)

0°

30.

In the given figure the angles ∠1 and ∠2 are

(a)

opposite angles

(b)

Adjacent angles

(c)

Linear pair

(d)

Supplementary angles

31.

In a trapezium if the sum of the parallel sides is 10 m and the area is 140 sq. m, then the height is

(a)

7cm

(b)

40 cm

(c)

14 cm

(d)

28 cm

32.

(-100) - 0 + 100 = _____

(a)

200

(b)

(c)

100

(d)

-200

33.

(-5) - (-18) = ____

(a)

(b)

-13

(c)

(d)

-23

34.

4 typists are employed to complete a work in 12 days. If two more typists are added, they will finish the same work in _______ days.

(a)

(b)

(c)

(d)

35.

When we subtract ‘a’ from ‘-a’, we get _______

(a)

(b)

(c)

-2a

(d)

-a

36.

The addition of 3mn, −5mn, 8mn and −4mn is

(a)

(b)

-mn

(c)

2mn

(d)

3mn

37.

Suppose 3 kg. of sugar is used to prepare sweets for 50 members, then ____ kg. of sugar is required for 150 members.

(a)

(b)

(c)

(d)

38.

The height of the rhombus whose area 96 sq. m and side 24 m is

(a)

8 m

(b)

10 m

(c)

2 m

(d)

4 m

39.

Choose the pair of like terms

(a)

7p, 7x

(b)

7r, 7x

(c)

−4x, 4

(d)

−4x, 7x

40.

The temperature at 12 noon at a certain place was 18° above zero. If it decreases at the rate of 3° per hour at what time it would be 12° below zero?

(a)

12 mid night

(b)

12 noon

(c)

10 am

(d)

10 pm

Generic 5

10 x 5 = 50

41.

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.

42.

Find the unit digit of the large numbers:
64¹⁰

43.

Find the unit digit of the following exponential numbers:
81¹⁰⁰

44.

Find the area of a hula loop whose diameter is 28 cm (use \(\pi =\frac { 22 }{ 7 } \)).

45.

Express the following as fractions
A capsule contains 0.85 mg of medicine.

46.

Ani is scuba diving. She descends 5 feet below sea level. She descends the same distance 4 more times. What is Anis final elevation?

47.

In the adjoining figure, the lines \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) intersect at 'O'. If ∠COD = 50°, find the measures of the other three angles.

48.

Sum a weaves 25 baskets in 35 days. In how many days will she weave 110 baskets?

49.

Find the perimeter of a square whose side is y - 2 units.

50.

The floor of an office building consists of 200 rhombus shaped tiles and each of its length of the diagonals are 40 cm and 25 cm. Find the total cost of polishing the floor at 45 per sq m.

Generic 2

30 x 2 = 60

51.

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

52.

Tabulate and find the relationship between the variables (x and y) for the following patterns.

53.

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

54.

In the given figure find the values of x and y.

55.

If a = 3 and b = 2, then find the value of the following.
a^a − b^b

56.

State whether the two triangles are congruent or not. Justify your answer.

57.

Express each of the following numbers using exponential form.
729

58.

A path 5 m wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth of the field. If the area of the path is 500 m² then find the length and breadth of the field.

59.

The figure shown is the aerial view of the pathway. Find the area of the pathway.

60.

Represent the decimal numbers 0.07 and 0.7 on a number line.

61.

There are 26 boys and 24 girls in a class. Express the fractions of boys and girls as decimal numbers.

62.

Write each of the following as decimal numbers.
900 + 50 + 6 +\(\frac { 3 }{ 100 } \)

63.

In the following figure, PQRS is a parallelogram find x and y.

64.

When Malar woke up her temperature was 10rF. Two hours later it was 30 lower, what was her temperature then?

65.

Can two right angles form a linear pair?

66.

Find the Shortest distance to reach the play ground from school

Also find out all possible routes

67.

A call taxi charges Rs.130 for 100 km. How much would one travel for Rs. 390?

68.

Write any three expressions each having 4 terms:

69.

From a water tank 100 litres of water is used every day. After 10 days there is 2000 litres of water in the tank. How much water was there in the tank before 10 days?

70.

Find the shortest route to Vivekanandar Memorial Hall from the Mandapam using the given map.

71.

During summer, the level of the water in a pond decreases by 2 inches every week due to evaporation. What is the change in the level of the water over a period of 6 weeks?

72.

The taxi charges in a city comprise of a fixed charge of Rs.100 for 5 kms and Rs.16 per km for every additional km. If the amount paid at the end of the trip was Rs.740 find the distance travelled.

73.

Using the given tetrominoes with numbers, complete the 4 x 4 magic square

74.

A man has to build a rhombus shaped swimming pool. One of the diagonal is 13 m and the other is twice the first one. Then find the area of the swimming pool and also fi nd the cost of cementing the floor at the rate of Rs. 15 per sq.cm.

75.

10 farmers can plough a field in 21 days. Find the number of days reduced if 14 farmers ploughed the same field?

76.

Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

77.

If the three angles at a point are in the ratio 1 : 4 : 7, find the value of each angle?

78.

Find the angle \(\angle\)GEH from the given figure.

79.

The shadow of a pole with the height of 8 m is 6m. If the shadow of another pole measured at the same time is 30m, find the height of the pole?

80.

Identify the like terms among the following : 7x, 5y, −8x, 12y, 6z, z, −12x, −9y, 11z.

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Answers Key:

Generic 3

15 x 3 = 45

Let, ㄥNMO = y
Exterior angle = sum of two interior opposite angles
y = 62⁰ + 31⁰
= 93⁰

13² = 13 x13 = 169

Here the distance travelled by the tip of the seconds hand of a clock in 1 minute is the circumference of the circle and the length of the seconds hand is the radius of the circle. So, r = 56 mm
Circumference of the circle, C = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 56
= 2 x 22 x 8
= 352 mm
Therefore, distance travelled by the tip of the seconds hand of a clock in 1 minute is 352 mm.

3983 m = \(\frac { 3983 }{ 1000 } \)km = 3.983 km
So, Praveen's trekking record is 0.004 km, 0.028 km, 0.537 km, 3.983 km.
Since, 1m = \(\frac { 1 }{ 1000 } \)km = 0.001 km

Area of each tile = \(\frac { 1 }{ 2 } \) x d₁ X d₂sq. units

∴ Area of 2000 tiles = 500 x 2000 = 10,00,000 cm² = 100 m²
Cost of polishing 1 m² = Rs. 5
∴ Cost of polishing 100 m² = 5 x 100 = Rs. 500

The high temperature recorded = 90°F
The low temperature recorded = -2°F
Difference = 90°F - (-2°F)
= 90°F + (Additive inverse of -2°F)
= 90°F + (+2°F) = 92°F

LEAB = 60° and LACD = 105°
Determine
(i) LCAF and
(ii) LBAC

(i) Since EF II GH and AC is a transversal
⇒ LCAF + LACH = 180
⇒ LCAF + 105° = 180°
=75°
(ii) ∴ EF II GH and AC is transversal.
∴ LEAC = LACH
⇒ LBAC = 105°
⇒ LBAC + LBAB = 105°
⇒ LBAC + 60° = 105°
⇒ LBAC = 105°- 60°
= 45°
∴ LCAF = 75° and LBAC = 45°

Since x and y vary inversely as each other
xy = constant
10 x 6 = 15 xy
60 = 15y

y = 4

Given:
A = 2a²- 4b-1; B = 5a²+ 3b - 8 and C = 2a²+ 9b + 3
A + B + C = (2a²- 4b - 1) - (5a²+ 3b- 8)+(2a²- 9b + 3)
= 2a²- 4b - 1 + (5a²- 3b + 8) + 2a - 9b + 3
= 2a²- 4b - 15a ²- 3b + 8 + 2a²- 9b + 3
= 2a²- 5a²+ 2a²- 4b - 3b - 9b -1 + 8 + 3
= (2 - 5 + 2)a²- 4b - 3b - 9b - 1 + 8 + 3
= -a²- 16b + 10

10.

Depth drilled in one day = –22 feet
Depth of water = –110 feet
Number of days required = –110 ÷ –22 = 5
Hence the workers wil reach resource in 5 days.

11.

Step 1: Draw a line. Mark two points A and B on it so that AB = 6 cm.

Step 2: Using compass with A as center and radius more than half of the length of AB, draw two arcs of same length, one above AB and one below AB.

Step 3: With the same radius and B as center draw two arcs to cut the arcs drawn in step 2. Mark the points of intersection of the arcs as C and D.

Step 4: Join C and D. CD will intersect AB. Mark the point of intersection as O CD is the required perpendicular bisector of AB.
Measure ∠AOC. Measure the length of AO and OB. What do you observe?

12.

1.

Therefore, there are nine tetrominoes in a tile.
2. Given, the cost of a tile is Rs. 52
There are six tiles in the front elevation
Therefore, the total cost = 6 x 52 = Rs. 312

13.

Using unitary method we can solve this as follows:
The cost of 2 notebooks = Rs. 24
The cost of 1 notebook = \(\frac { 24 }{ 2 } \) = Rs.12
Therefore, the cost of 9 notebooks = 9 x Rs.12
= Rs.108
Hence, Anbu has to pay Rs.108 for 9 notebooks.

14.

i) 4x + 7y = 4 (3) + 7 (2) = 12 + 14 = 26
ii) 3x + 2y – 5 = 3 (3) + 2 (2) – 5 = 9 + 4 – 5 = 8
iii) x – y = 3 – 2 = 1

15.

(i) From the Fig.2.4
Base of a parallelogram (b) = 15 cm, Height of a parallelogram (h) = 4 cm
Area of a parallelogram = b x h sq.units. Therefore, Area = 15 × 4 = 60 sq. cm.
Thus, area of the parallelogram is 60 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (15 + 5 +15 + 5) = 40 cm.
(ii) From the Fig.2.5
Base of a parallelogram (b) = 9 m, Height of a parallelogram (h) = 16 cm.
Area of a parallelogram = b × h sq.units Therefore, Area = 9 × 16 = 144 sq. cm
Thus, area of the parallelogram is 144 sq. cm.
Perimeter of the parallelogram = sum of the length of the four sides.
= (18 + 9 +18 + 9) = 54 cm.

Multiple Choice Question

25 x 1 = 25

16.

17.

(d)
y = −3x

18.

(a)
y = 4x

19.

(d)
1

20.

(d)
same shape and same size

21.

(d)
20

22.

(b)
40, 60, 80

23.

24.

(b)
πr²

25.

(a)
41 cm

26.

27.

(d)
thousandths

28.

29.

(a)
360°

30.

31.

(d)
28 cm

32.

(b)
0

33.

34.

(b)
8

35.

36.

37.

(a)
9

38.

(d)
4 m

39.

(d)
−4x, 7x

40.

(d)
10 pm

Generic 5

10 x 5 = 50

41.

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.

42.

64¹⁰
Unit digit of base 64 is 4 and power is 10 (even power). Therefore, unit digit of 64¹⁰ is 6.

43.

81¹⁰⁰
Unit digit of base 81 is 1 and power is 100
Thus, the unit digit of 81¹⁰⁰ is 1.

44.

Given the diameter (d) = 28 cm
Radius (r) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of a circle = πr² sq.units
= \(\frac { 22 }{ 7 } \) x 14 x 14
So, the area of the circle = 616 cm².

45.

0.85 = 0 + \(\frac { 8 }{ 10 } \)+\(\frac { 5 }{ 100 } \)
= \(\frac { 85 }{ 100 } \)=\(\frac { 17 }{ 20 } \)
A capsule holds \(\frac { 17 }{ 20 } \) mg of medicine

46.

Ani descends 5 feet below sea level once she descends 4 more times
\(\therefore\) She descends (5 x 4) + 5 feet in total = 20 + 5 = 25 feet below sea level

47.

∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° - 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles
∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC = 130°
∠BOD = 130°

48.

To weave 25 baskets time taken = 35days
\(\therefore\) To weave 1 basket time taken = \(\cfrac { 35 }{ 25 } days\)

49.

Perimeter = (y - 2) + (y - 2) + (y - 2) + (y -2)
= y - 2 + y - 2 + y - 2 + y - 2 = 4y - 8
Perimeter of the square = 4y- 8units.

50.

Given, the length of the diagonals of a rhombus shaped tile are 40 cm and 25 cm
The area of one tile = \(\cfrac { 1 }{ 2 } \times \left( { d }_{ 1 }\times { d }_{ 2 } \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 40\times 25\)
= 500 sq.cm
Therefore, the area of 200 such tiles = 200 x 500
= 100000 sq. cm
= \(\cfrac { 100000 }{ 10000 } \)(1sq.m = 1000 sq.cm)
= 10 sq.m
Therefore, the cost of polishing 200 such tiles at the rate of Rs 45 per sq. cm
= 10 x 45 = Rs. 450.

Generic 2

30 x 2 = 60

51.

8 × 45 × 84 = 28 × 9 × 120

52.

steps (x)	1	2	3	4	5
shapes (y)	1	3	5	7	9

The relationship between x and y is y = 2x - 1

53.

54.

By Exterior angle property
x + 28^o = 62^o
x = 62^o - 28^o = 34^o
Now in ABC
x + y + 28^o = 180^o
34^o + y + 28^o = 180^o
62^o + y = 180^o
y = 180^o - 62^o = 118^o
x = 34^o, y = 118^o

55.

3²- 2² = 3 x 3 x 3 -2 x 2
27-4 = 23

56.

congruent triangles by SAS

57.

58.

Let the length of the field = l
Breadth of the field = b
Width of the path = 5 m
Area of the path = 500 m²
Area of 2 horizontal paths + Area of 2 vertical paths = 500 m²
2 x 5 (l - 5) + 2 x 5 (b - 5) = 500
10(l - 5) + 10(b - 5) = 500
10[l - 5 + b - 5] = 500
l + b -10 = 50
l + b = 50 + 10
l + b = 60 ......(1)
Also given length = 3 x breadth
l = 3b
3b + b = 60
\(4 b=60 ; \ b=\frac{60}{4}=15 \mathrm{~m}\)
l + 15 = 60 , l = 60 - 15 = 45 m

59.

Area of pathway = Area of outer rectangle - Area of inner rectangle
= LB - lb
= 80 x 50 - 70 x 40
= 4000 - 2800 = 1200 m²

60.

\(0.07=0+0 \times \frac{1}{10}+7 \times \frac{1}{100}=\frac{7}{100}\)

61.

Total number of students = 26 + 24 = 50
Fraction of boys \(=\frac{26}{50}=\frac{26 \times 2}{50 \times 2} \)
\(=\frac{52}{100}=0.52 \)
Fraction of Girls \(=\frac{24}{50}=\frac{24 \times 2}{50 \times 2} \)
\(=\frac{48}{100}=0.48 \)

62.

900 + 50 + 6 + \(\frac { 3 }{ 100 } \) = \(956+0\times \cfrac { 1 }{ 10 } +3\times \cfrac { 1 }{ 100 } \)
= 956.03
( Since the tenth place value is not there, it is taken as '0')

63.

We know that in a parallelogram opposite sides are equal.
∴ 3x = 18
\(x=\frac { 18 }{ 3 } \)
x = 16
and
3y - 1 = 26
3y = 26 + 1 = 27
\(y=\frac { 27 }{ 3 } \)
y = 9

64.

Initially Malar's temperature = 102°F
After two hours it lowered 3° \(\Rightarrow\) -3°F
\(\therefore\) Here present temperature = 102° + (-3°) = 99°F

65.

Yes, because the sum of two right angles is 180° and form a linear pair.

66.

Route1: School \(\rightarrow \) Swimming pool \(\rightarrow \) canteen \(\rightarrow \) playground
Distance 9 km + 2 km + 10 km = 21 km.
Route 2 : School \(\rightarrow \) canteen \(\rightarrow \) playground
Distance : 10 km + 10 km = 20 km.
Route 3 : School \(\rightarrow \) garden \(\rightarrow \) playground
Distance: 2 km + 20 km = 22 km.
Route 4 : School \(\rightarrow \) garden ~ teashop \(\rightarrow \) playground
Distance: 2 km + 17 km + 2 km = 21 km.
Route 5 : School \(\rightarrow \) Park \(\rightarrow \) teashop \(\rightarrow \) playground
Distance: 12 km + 12 km + 2 km = 24 km.
\(\therefore \) Route 2 is the shortest distance.

67.

For Rs 130 distance travelled = 100 km

300 can be travelled with Rs 390

68.

(i) 2x³- 3x²+ 3xy + 8
(ii) 7x³+ 9y²- 2xy²
(iii) 9x²- 2x + 3xy - 1

69.

Water used for one day = 100 litres.
Water used for 10 days = 100 x 10 = 1000 litres.
Water in tank after 10 days = 2000 litres
\(\therefore\)Water in tank before 10 days = 2000 + 1000 = 3000 litres

70.

Possible routes from Mandapam to Vivekandar Memorial are
route 1:
(a) . Mandapam \(\longrightarrow \) Pullivasal Island \(\longrightarrow \) Krusadai Island \(\longrightarrow \) Vivekanandar Memorial Hall.
Distance = 6 Km + 2 Km + 1.5 Km = 9.5 Km
route 2:
(b) Mandapam\(\longrightarrow \) Krusadai Island\(\longrightarrow \) Vivekanandar Memorial Hall.
Distance = 7 Km + 1.5 Km = 8.5 Km
8.5 km < 9.5 km
\(\therefore \) Shortest route: Mandapam \(\longrightarrow \) Krusadai Island\(\longrightarrow \) Vivekanandar Memorial

71.

The level of the water in a pond decreases in a week = -2 inches.
Level of water decreases in 6 weeks = 6 x -2 = -12 inches
The change in the level of the water over a period of 6 week is -12 inches

72.

Total amount paid at the end of the trip = Rs. 740
Amount paid for 5 km = Rs. 100(-)
= Rs. 640
Let the distance travelled after completing 5 km = y
Charge for one additional km = Rs. 16
16 y = 640
\(\frac{16 y}{16}=\frac{640}{16}\) [ Divide by 16 on both sides]
y = 40
The total distance travelled = 5 km + 40 km
= 45 km

73.

74.

One diagonal d₁ = 13 cm
Other diagonal d₂ = twice the first one
= 2 x d₁
2 x 13 = 26 cm
Area of swimming pool
\(=\frac{1}{2} \mathrm{~d}_{1} \times \mathrm{d}_{2} \text { sq.units } \)
\(=\frac{1}{2} \times 13 \times 26=169 \mathrm{~cm}^{2} \)
Cost of cementing the floor per sq.cm = Rs. 15
Cost of cementing the floor for 169 sq.cm
= Rs. 15 x 169 = Rs. 2535

75.

Let x be the total number of days required if 14 farmers plough

Number of days	21	x
Number of farmers	10	14

This is inverse proportion,
\(x_{1} y_{1}=x_{2} y_{2}\)
10 x 21 = 14 x x
\(x=\frac{10 \times 21}{14} \Rightarrow x=15\)
Total number of reduced days
= 21 - 15 = 6 days

76.

Parallel sides, a = 24 cm
b = 20 cm
Distance between them (Height) h = 15 cm
Area of trapezium \(=\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) \text { sq.units } \)
\(=\frac{1}{2} \times 15(24+20) \)
\(=\frac{1}{2} \times 15 \times 44 \)
Area of the trapezium = 330 cm²

77.

Let the three angles be x, 4x and 7x .
The sum of three angles at a points is supplementary
∴ x + 4x + 7x = 360°
12x = 360°
x^o = \(\frac{{360}^o}{12}\)
x = 30°
If x= 30^othen
4x = 4 x 30° = 120°
7x = 7 x 30° = 210°.
∴ The three angles are 30°, 120° and 210°.

78.

From the figure
∠FEH = ∠FEG + ∠GEH
120° = 34° + ∠GEH
120° - 34° = ∠GEH
∠GEH = 86°

79.

Height of the pole (m)	8	x
Height of the shadows (m)	6	30

Since the given data are in direct proportion,
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}} \)
\(\frac{8}{6}=\frac{x}{30} \)
6x = 8 x 30
\(\therefore x=\frac{8 \times 30}{6}\)
x = 40 m
If the height of the shadow is 30 m height, then the height of the pole is 40 m.

80.

(i) 7x, -8x, -12x are like terms containing same algebraic variable 'x'.
(ii) 5y, 12 -9y are like terms containing same algebraic variable 'y'.
(ili) 6z, z and 11 z are like terms containing same algebraic variables 'z'.