Answer: (a)

Answer: (b)

Answer: (d)

Answer: (c)

Answer: (a)

Answer: (d)

Answer: (d)

Answer: (a)

Answer: (b)

Answer: (d)

Answer: (a)

Answer: (d)

Answer: (c)

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (a)

Answer: (d)

Important Questions

Question 1.
Write four such English letters which have no line of symmetry.
Solution:
The 4 English letters having no line of symmetry are: P, F, G and Z.

Question 2.
Write 5 such English letters which have horizontal line of symmetry.
Solution:
The required letters are:
E, B, C, D,K

Question 3.
Write 4 English letters which have vertical lines of symmetry.
Solution:
The required English letters are A, H, I, M

Question 4.
How many symmetrical lines do they have?

Solution:
(a) The figure is a square. It has 4 lines of symmetry.

(b) It is an isosceles triangle. It has one vertical line of symmetry.

(c) The given figure is a rectangle which has two lines of symmetry.

Question 5.
Find in which of the following, the dotted line is a line of symmetry.

Solution:
(a) The dotted line is a line of symmetry.
(b) The dotted line is a line of symmetry.
(c) The dotted line is not a line of symmetry.
(d) The dotted line is not a line of symmetry.
(e) The dotted line is a line of symmetry.
(f) The dotted line is a line of symmetry.

Symmetry Class 6 Extra Questions Short Answer Type

Question 6.
How many lines of symmetry does (a) a rectangle (b) a square (c) a parallelogram have? What about right angled triangle?
Solution:
(a) Rectangle has two lines of symmetry
(i) Horizontal
(ii) Vertical

(b) Square has four lines of symmetry.
(i) 1 horizontal
(ii) 1 vertical
(iii) 2 diagonally

(c) Parallelogram has no lines of symmetry

Right angled triangle has no line of symmetry.

Question 7.
Fill in the blanks:
(a) An angle is symmetrical about its …….. .
(b) A circle has …….. lines of symmetry.
(c) A rectangle …….. has lines of symmetry.
(d) A square has …….. lines of symmetry.
(e) A letter D has …….. lines of symmetry.
(f) A letter N has …….. lines of symmetry.
(g) A kite is symmetrical about its longer …….. .
(h) A scalene triangle has …….. lines of symmetry.
(i) A line of symmetry of an object is also called its …….. of symmetry.
(j) An equilateral triangle has …….. lines of symmetry.
(k) A regular pentagon has …….. lines of symmetry.
(l) A line of …….. divides a figure into two similar parts.
Solution:
(a) bisector
(b) infinite
(c) two
(d) four
(e) one
(f) no
(g) diagonal
(h) no
(i) axis
(j) three
(k) five
(l) symmetry

Question 8.
Show by figure the English alphabets which look the same in their reflected image.
Solution:
The following English alphabets look same after reflections.

Answer: (a)
Hint:
10 = 5 × 2

Answer: (a)
Hint:
80000 = 2 × 40000

Answer: (a)
Hint:
2030 = 23 = 2 : 3

Answer: (a)
Hint:
2525+25 = 2550 = 12 = 1 : 2

Answer: (b)
Hint:
150120 = 1512 = 54 = 5 : 4

Answer: (a)
Hint:
50,0003,00,000 = 16 = 1 : 6

Answer: (b)
Hint:
62 = 31 = 3 : 1

Answer: (a)
Hint:
1 m = 100 cm
40100 = 25 = 2 : 5

Answer: (a)
Hint:
48 = 12 = 1 : 2

Answer: (c)
Hint:
2 : 3 = 23 = 2×33×3 = 69 = 6 : 9

Answer: (a)
Hint:
10 : 5 = 105 = 10÷55÷5 = 21 = 2 : 1
20 ÷ 10 = 2010 = 20÷1010÷10 = 21 = 2 : 1
30 ÷ 15 = 3015 = 30÷1515÷15 = 21 = 2 : 1

Important Questions

Question 1.
Find the ratio of 75 cm to 1.5 m.
Solution:
The given numbers are not in the same units. So, converting them into same units.
1.5 m = 1.5 x 100 cm = 150 cm
[∵ 1 m = 100 cm]
∴ The required ratio is 75 cm : 150 cm.

∴ Required ratio = 1 : 2

Question 2.
Give two equivalent ratios of 3 : 5.
Solution:

Thus, 9 : 15 and 6 : 10 are the two equivalent ratios of 3 : 5.

Question 3.
Fill in the blank box.

Solution:

Question 4.
Check whether the given ratios are equivalent or not. 27, 621
Solution:

∴ They are equivalent ratios.

Question 5.
Divide 60 in the ratio of 2 : 3.
Solution:
Sum = 2 + 3 = 5
∴ First part = 25 x 60 =24 5
∴ Second part = 35 x 60 =36 5
Thus, the required two parts = 24 and 36.

Question 6.
Find the ratio of the following:
(a) 56 to 63.
(b) 55 to 120.
Solution:

Question 7.
Ramesh deposited ₹ 2050 in a bank and in the month of January he withdrew ₹ 410 from his account on the last date of the month. Find the ratio of
(a) Money withdrawn to the total money deposited.
(b) Money withdrawn to the remaining amount in the bank.
Solution:
Total money deposited = ₹ 2050
Amount of money withdrawn = ₹ 410
Amount of money left in the bank = ₹ 2050 – ₹ 410 = ₹ 1640
(a) Ratio of money withdrawn to the total money deposited

∴ Required ratio = 1 : 5
(b) Ratio of money withdrawn to the money left in the bank

∴ Required ratio = 1 : 4

Question 8.
There are 180 students in a class. Number of girls are 75. Find the ratio of the girls to the number of boys.
Solution:
Total number of students = 180
Number of girls = 75
Number of boys = 180 – 75 = 105
∴ Ratio of number of girls to the number of boys

Required ratio = 5 : 7

Ratio and Proportion Class 6 Extra Questions Short Answer Type

Question 9.
Green paint is made by mixing blue, yellow and white paints in the ratio 2 : 7 : 1. How much blue paint is needed to make 64 litres of green paint?
Solution:
Here, sum of ratios = 2 + 7 + 1 = 10
∴ Total quantity of green paint = 64 litres
Quantity of blue paint = 210 x 64 = 12.8 litres
Therefore, the required blue paint = 12.8 litres.

Question 10.
From the figure, find the ratio of
(a) The number of squares to the number of triangles.
(b) The number of circles to the number of rectangles.

Solution:
(a) Number of squares = 2
Number of triangles = 3 2
∴ Ratio = 23 or 2 : 3

(b) Number of circles = 3
Number of rectangles = 3
∴ Ratio = 33 or 1 : 1

Question 11.
In each of the following figures, find the ratio of the shaded region to the unshaded region.

Solution:
(a) Number of shaded parts = 4
Number of unshaded parts = 12

Required ratio = 1 : 3

(b) Number of shaded parts = 2
Number of unshaded parts = 4

Required ratio = 1 : 2

Question 12.
Are 20, 25, 12, 15 in proportion?
Solution:
We have 20, 25, 12, 15
Product of extremes = 20 x 15 = 300
Product of middles = 25 x 12 = 300
Since both the products are same.
∴ The four numbers 20, 25, 12, 15 are in proportion.

Question 13.
The first, second and fourth terms in a proportion are 32, 112, 217 respectively. Find the third term.
Solution:
Let the third term be x.
∴ 32, 112, x and 217 are in proportion.
∴ 32 : 112 :: x : 217

Thus, the third term = 62.

Question 14.
Find the value of x, if
(а) 8, x, x, 50 are in proportion.
(b) 36, 90, 90, x are in proportion.
Solution:
(a) Since 8, x, x, 50, are in proportion.
∴ x × x = 8 × 50
⇒ x² = 400
∴ x = 20

(b) Since 36, 90, 90, x are in proportion.
∴ 36 × x = 90 × 90
⇒ x = 90×9036 = 225
∴ x = 225

Question 15.
The cost of 10 tables is ₹ 7500. Find the number of tables that can be purchased with ₹ 9000.
Solution:
Number of tables purchased in ₹ 7500 = 10
Number of tables purchased in ₹ 1 = 107500
∴ Number of tables purchased in ₹ 9000
= 10×90007500 = 12

Question 16.
39 packets of 12 pens each costs ₹ 374.40. Find the cost of 52 packets of 10 pens each.
Solution:
Number of pens in 1 packet = 12
Number of pens in 39 packets = 12 x 39 = 468
Number of pens in 1 packet = 10
Number of pens in 52 packets = 10 x 52 = 520
Now cost of 468 pen = ₹ 374.40

Answer: (a)

Answer: (b)

Answer: (c)

Answer: (b)

Answer: (a)

Answer: (b)

Answer: (c)

Answer: (a)

Answer: (d)

Answer: (c)

Answer: (a)

Answer: (c)

Answer: (c)

Answer: (c)

Answer: (d)

Answer: (d)

Important Questions

Question 1.
Six less than a number equals to two. What is the number?
Solution:
Let the number be ‘x’.
According to condition, we have x – 6 = 2
By inspections, we have 8 – 6 = 2
∴ x = 8
Thus, the required number is 8.

Question 2.
Write an algebraic expression for each of the following:
(a) 3 subtracted from a number y.
(b) 5 is added to three times a number x.
Solution:
(a) The required expression is y – 3
(b) The required expression is 5 + 3x

Question 3.
Write an algebraic expression for the following expressions:
(a) The sum of a number x and 4 is doubled.
(b) One fourth of a number x is added to one third of the same number.
Solution:
(a) The required expression is 2 x (x + 4)
(b) The required expression is 14x + 13x

Question 4.
Think of a number x. Multiply it by 3 and add 5 to the product and subtract y subsequently. Find the resulting number.
Solution:
Required number is (3x + 5)
Now we have to subtract y from the result i.e., 3x + 5 – y

Question 5.
Here is a pattern of houses with matchsticks:

Write the general rule for this pattern.
Solution:
One house is made of 6 matchsticks i.e. 6 x 1
Two houses are made of 12 matchsticks i.e. 6 x 2
Three houses are made of 18 matchsticks i.e. 6 x 3
∴ Rule is 6n where n represents the number of houses.

Question 6.
If the side of an equilateral triangle is x, find its perimeter.
Solution:
We know that the three sides of an equilateral triangle are equal.
∴ x + x + x = 3x.
Thus, the required perimeter = 3x units.

Question 7.
If x = 3, find the value of the following:
(i) x + 5
(ii) 2x – 3
(iii) x – 7
(iv) x3 – 1
Solution:
Given that x = 3
(i)x + 5 = 3 + 5 = 8
(ii) 2x – 3 = 2 x 3 – 3 = 6 – 3 = 3
(iii) x – 7 = 3- 7 = -4
(iv) x3 – 1 = 33 -1 = 1 – 1 = 0

Question 8.
If x = 2, y = 3 and 2 = 5, find the value of;
(a) 2x + y + z
(b) 4x -y + z
(c) x – y + z
Solution:
(a) Given that: x – 2, y = 3 and z = 5
∴ 2x + y + 2 = 2 x 2 + 3 + 5
= 4 + 3 + 5 = 12
(b) 4x – y + z = 4 x 2 – 3 + 5
= 8 – 3 + 5 = 5 + 5 = 10
(c) x – y + z = 2 – 3 + 5 = -1 + 5 = 4

Question 9.
State which of the following are equations with a variable?
(a) 12 = x – 5
(b) 2x > 7
(c) x2 = 5
(d) 5 + 7 = 3 + 9
(e) 7 = (11 x 5) – (12 x 4)
Solution:
(a) 12 = x – 5 is an equation with a variable x.
(b) 2x > 7 is not an equation because it does not have ‘=’ sign.
(c) x2 = 5 is an equation with a variable x.
(d) 5 + 7 = 3 + 9 is not an equation because it has no variable.
(e) 7 = (11 x 5) – (12 x 4) is not an equation because it has no variable.

Question 10.
Think of a number, add 2 to it and then multiply the sum by 6, the result is 42.
Solution:
Let the number be x.
∴ Sum of x and 2 = x + 2
Now by multiplying the sum by 6, we get
6 × (x + 2) = 42
⇒ 6 × x + 6 × 2 = 42
⇒ 6x + 12 = 42
By inspection, we get
6 × 5 + 12 = 42
⇒ 30 + 12 = 42
∴ 42 = 42
So, the required number = 5

Algebra Class 6 Extra Questions Short Answer Type

Question 11.
The side of a regular hexagon is s cm. Find its perimeter.
Solution:
Each side of a regular hexagon = s
∴ its perimeter = s + s + s + s + s + s = 6s cm

Question 12.
If a = 3, b = 12 and c = 14, find the value of

Solution:

Question 13.
Complete the table and find the solution of the equation 19 – x = 13

Solution:
By inspection, we have

Thus, the required solution is 6.

Question 14.
If x = –12, y = 14 and z = 0, find the value of the given expressions
(a) 8z + 2x -y
(b) z – y + 3x
Solution:

Question 15.
Fill in the blanks:
(a) 5 added to -5 = ……….
(b) If x = 3, then 3x – 5 = ……….
(c) If x = 1 and y = 2, then 2x + 3y = ……….
(d) If 10x – 6 = 14, then x = ……….
(e) 4 less than a number x = ……….
Solution:
(a) 0
(b) 4
(c) 8
(d) 2
(e) x – 4

Algebra Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 16.
A starts his car from Delhi at 6.00 am to Amritsar. The uniform speed of his car is x km/h. At 12.00 noon, he finds that he is still 50 km away from Amritsar. Find the distance between Delhi and Amritsar.
Solution:
Time taken by A to reach Amritsar = 12.00 noon – 6.00 am = 6 hour.
The uniform speed of the car = x km/ hr
∴ Total distance covered by A = Time x speed = 6x km.
∴ Distance between Delhi and Amritsar = (6x + 50) km.

Question 17.
Anshika’s Score in Science is 15 more than the two-third of her score in Sanskrit. If she scores x marks in Sanskrit, find her score in Science.
Solution:
Anshika’s score in Sanskrit = x
∴ Her marks in Science = 23x +15
∴ Thus, Anshika’s score in Science = 23x + 15

Answer: (c)

Answer: (a)

Answer: (b)

Answer: (a)

Answer: (a)

Answer: (d)

Answer: (d)

Answer: (b)
Hint:
Distance covered = 2(20 + 10) = 60 m

Answer: (b)
Hint:
Perimeter = 2(4 + 3) = 14m

Answer: (a)
Hint:
Perimeter = 4 × 2 = 8m

Answer: (a)
Hint:
Perimeter = 2(25 + 20) = 90 cm

Answer: (d)
Hint:
Perimeter = 4(1 + 2 + 2) = 20cm

Answer: (a)
Hint:
Perimeter = 4 + 3 + 2 + 2 + 3 + 6 = 20 cm

Answer: (d)
Hint:
Length of the lace = 2(2 + 1) = 6 m

Answer: (b)
Hint:
Perimeter = 2(9 + 1) = 20 cm

Answer: (a)
Hint:
Total distance covered = 10 × 2 × (40 + 30) = 1400 m

Answer: (a)
Hint:
Cost= 10 × 2 × (10 + 5) = ₹ 300

Important Questions

Question 1.
The perimeter of a square is 64 cm. Find the length of each side.
Solution:
Perimeter of the square = 64 cm

Question 2.
Length and breadth of a rectangular table-top are 36 cm and 24 cm respectively. Find its perimeter.
Solution:
Length of the rectangular table-top = 36 cm
and its breadth = 24 cm.
∴ Perimeter of the table-top = 2 [length + breadth]
= 2 [36 cm + 24 cm]
= 2 x 60 cm = 120 cm.

Question 3.
Which of the following figure has greater perimeter?

Solution:
Fig. (i) Perimeter of the square = 4 x side
= 4 x 4 cm = 16 cm
Fig. (ii) Perimeter of the rectangle
= 2 [length + breadth]
= 2[8 cm + 3 cm]
= 2 x 11 cm = 22 cm
Since 22 cm > 16 cm
∴ Rectangle has greater perimeter than the square.

Question 4.
How much distance will you have to travel in going around each of the following figures?

Solution:
Distance travelled in going around Fig. (i)
= 12 cm + 3 cm + 12 cm + 3 cm = 30 cm
Distance travelled in going around Fig. (ii)
= 6 cm + 4 cm + 4 cm + 4 cm = 18 cm

Question 5.
Find the perimeter of a square whose side is 15 cm.
Solution:
Side of the square = 15 cm
∴ Perimeter of the square = 15 cm x 4 = 60 cm

Question 6.
Find the cost of fencing a rectangular park 300 m long and 200 m wide at the rate of ₹4 per metre.
Solution:
Length of the park = 300 m
Breadth = 200 m
∴ Perimeter of the park = 2 [length + breadth]
= 2 [300 m + 200 m]
= 2 x 500 m = 1000 m.
Cost of fencing the rectangular park = 1000 x 4 = ₹4000

Question 7.
Find the area of a square field whose each side is 150 m.
Solution:
Side of the square field = 150 m
∴ Area of the square field = Side x Side
= 150 m x 150 m
= 22500 sq m.

Question 8.
Length and breadth of a rectangular paper are 22 cm and 10 cm respectively. Find the area of the paper.
Solution:
Length of the rectangular paper = 22 cm
Breadth = 10 cm
∴ Area of the rectangular paper = length x breadth
= 22 cm x 10 cm
= 220 sq cm.

Mensuration Class 6 Extra Questions Short Answer Type

Question 9.
Find the length of a rectangle given that its perimeter is 880 m and breadth is 88 m.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
∴ 2 [length + breadth] = 880
length + breadth = 880 ÷ 2 = 440
∵ Breadth = 88 m
∴ Length = 440 m – 88 m = 352 m
Hence, the required length = 352 m.

Question 10.
How many trees can be planted at a distance of 6 metres each around a rectangular plot whose length is 120 m and breadth is 90 m?
Solution:
Length of the rectangular plot = 120 m
Breadth = 90 m
∴ Perimeter of the rectangular plot
= 2 [length + breadth]
= 2 [120 m + 90 m]
= 2 x 210 m = 420 m
Now distance between two trees = 6 m
∴ Number of trees around the rectangular plot = 420 m ÷ 6 m = 70

Question 11.
A rectangular park is 30 metres long and 20 metres broad. A steel wire fence is put up all around it. Find the cost of putting the fence at the rate of ₹15 per metre.
Solution:
Length of the rectangular park = 30 m
Breadth = 20 m
∴ Perimeter of the rectangular park = 2(length + breadth)
= 2 [30 + 20] = 2 x 50 m = 100 m
∴ Cost of fencing all around the park = ₹15 x 100 = ₹1500

Question 12.
Find the area of the figures A, B, C and D drawn on a squared paper in the following figure by counting squares.

Solution:
(A) Counting the squares, we have 8 squares
∴ Area = 8 sq units
(B) Counting the squares, we have 4 squares
∴ Area = 4 sq units
(C) Counting the squares, we have 5 squares
∴ Area = 5 sq units
(D) Counting the squares, we have 7 squares
∴ Area = 7 sq units

Mensuration Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 13.
A rectangle and a square have the same perimeter 100 cm. Find the side of the square. If the rectangle has a breadth 2 cm less than that of the square. Find the breadth, length and area of the rectangle.
Solution:
Perimeter of the square = 100 cm
Perimeter 100

= 25 cm.
∴ Breadth of the rectangle = 25 cm – 2 cm = 23 cm
Now perimeter of the rectangle = 100 cm
∴ 2 [length + breadth] = 100
length + breadth = 100 ÷ 2 = 50 cm
But breadth = 23 cm
∴ Length = 50 cm – 23 cm = 27 cm
Now, Area of the rectangle
= length x breadth = 27 cm x 23 cm
= 621 sq cm.

Question 14.
Fencing the compound of a house costs ₹5452. If the rate is ₹94 per metre, find the perimeter of the compound. If the breadth is 10 m, find its length.
Solution:
Cost of fencing the compound = ₹5452
and the rate of fencing = ₹94 per metre
∴ Perimeter of the compound = 5452 ÷ 94 = 58 metres
Now breadth of the compound = 10 m.
2 [length + breadth] = 58 m
∴ length + breadth = 58 + 2 m = 29 m
∴ Length of the compound = 29 m – 10 m = 19 m.

Answer: (a)

Answer: (d)

Answer: (a)
Hint:
62, 62

Answer: (c)
Hint:
95 – 25 = 70

Answer: (c)
Hint:
95, 78, 75

Answer: (d)
Hint:
55, 36, 42, 25

Answer: (c)
Hint:
62,62,73

Answer: (c)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (c)
Hint:
12 – 3 = 9

Answer: (a)
Hint:
3 : 6 = 1 : 2

Answer: (d)
Hint:
6 : 12 = 1 : 2

Important Questions

Question 1.
A die was thrown 35 times and the following numbers were obtained:
5, 1, 4, 2, 3, 2, 6, 6, 1, 4, 2, 5, 4, 5, 3, 6, 1, 5
2, 6, 2, 5, 4, 1, 3, 2, 1, 4, 1, 6, 2, 6, 3, 3, 3
Prepare a frequency table for the data.
Solution:
From the given data, we have the following table.

Question 2.
The result of a Mathematics test is as follows:
80, 90, 70, 80, 80, 60, 80, 70, 90, 65, 100, 60, 70, 60, 70, 85, 65, 70, 70, 85, 90, 60, 65, 80, 60
Make a frequency table for the above data and answer the following questions:
(a) What is the maximum marks obtained?
(b) How many students score less than 75 marks?
(c) How many students scored 80 marks or above?
(d) How many students appeared in the test?
Solution:
From the above information, we have the following table.

(a) Maximum marks obtained by a student = 100
(b) 5 + 3 + 6 = 14 students obtained marks less than 75.
(c) 5 + 2 + 3 + 1 = 11 students scored marks 80 or above 80.
(d) Total 25 students were appeared in the test.

Question 3.
Mr. Rajan made a pictograph given below to show the number of cars washed at a car washing station during three days of a week.

From the pictograph, find that:
(a) How many cars were washed on
(i) Friday
(ii) Saturday
(iii) Sunday?
(b) On which day the maximum number of cars were washed at the station?
(c) On which day the minimum number of cars were washed at the station?

(d) How many more cars were washed on Saturday than on Friday?
Solution:
(a) (i) On Friday – 4 x 5 = 20 cars
(ii) On Saturday – 9 x 5 = 45 cars
(iii) On Sunday – 7 x 5 = 35 cars.
(b) On Saturday, the maximum number of cars,
i.e., 9 x 5 = 45 were washed at the stations.
(c) On Friday, the minimum number of cars,
i.e., 4 x 5 = 20 were washed on the station.
(d) 45 – 20 = 25 more cars were washed on Saturday than on Friday.

Question 4.
Read the pictograph given below and answer the following questions:
Persons employed in one year

(a) What is the number of persons employed in government service?
(b) How many more person were employed in government service than in private service?
(c) In which service, were the maximum number of persons employed?
Solution:
(a) Number of persons employed in government service = 10 x 3000 = 30,000
(b) 10 x 3000 – 6 x 3,000 = 30,000 – 18,000 = 12,000 persons were employed more in government service than in private service.
(c) In government service, the maximum number of persons were employed.

Question 5.
In March 2012, children for six colonies of Meerut were given pulse polio Drops. The colony wise number of children were as follows:

Represent the data by pictograph.
Solution:
Pictograph:

Question 6.
The given bar graph represents the frequency of a, e, i, o, and u in a piece of English writing.

(a) Which letter occurred the maximum number of times?
(b) Which letter occurred 40 times?
(c) Which letter occurred less than 30 times?
(d) Write down the five letters in the decreasing order of frequencies.
Solution:
(a) a letter occurred the maximum number of times.
(b) i letter occurred 40 times.
(c) u letter occurred less than 30 times.
(d) a, e, o, i, u is the decreasing order of their frequencies.

Question 7.
The marks obtained by six students in Mathematics are given below. Represent the. data by a bar graph. Use a scale of 0.5 cm for each name on the horizontal axis and 0.5 cm for 10 marks on the vertical axis.

Solution:
The required bar graph is given as below:

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (b)

Answer: (b)

Answer: (b)

Answer: (c)

Answer: (a)

Answer: (a)

Answer: (b)

Important Questions

Question 1.
Write the following in decimals.
5 hundreds 3 tens 8 ones 4 tenths
Solution:
5 Hundreds + 3 Tens + 8 Ones + 4 Tenths
= 5 x 100 + 3 x 10 + 8 x 1 + 4 x 110
= 500 + 30 + 8 + 410
= 538 + 410 = 538.4

Question 2.
Write 14.3 in place value table.
Solution:

Question 3.
Write the following in decimals.

Solution:

Question 4.
Write the following as decimals.
(a) Three tens and eight-tenths
(b) Fifteen point seven
Solution:
(a) Three tens and eight-tenths = 3 x 10 + 8 x = 30 + 0.8 = 30.8
(b) Fifteen point seven = 15.7

Question 5.
Write the following as decimals.

Solution:
(a) 200 + 50 + 6 + 210 = 256 + 0.2 = 256.2
(b) 150 + 30 + 810 = 180 + 0.8 = 180.8

Question 6.
Write in decimals:

Solution:

Question 7.
Write as fractions in lowest form.
(a) 0.05
(b) 20.25
Solution:

Question 8.
Write the following in decimals.
Three hundred six and seven hundredths.
Solution:
Three hundred six and seven hundredths

Question 9.
Which is greater 0.4 or 0.5?
Solution:
0.4 = 4 and 0.5 = ±
Here 510 > 410
∴ o.5 > 0.4

Question 10.
Express the following as rupees using decimals:
(a) 7 paise
(b) 625 paise
Solution:

Decimals Class 6 Extra Questions Short Answer Type

Question 11.
Represent 1.3,3.8 and 4.1 on the number line.
Solution:

A represents 1.3
B represents 3.8
and C represents 4.1

Question 12.
Read the numbers from the place value table and write them in decimals;

Solution:
(a) 4 Thousands + 0 Hundreds + 2 Tens + 3 Ones + 6 Tenths + 0 Hundredths
= 4 x 1000 + 0 x 100 + 2 x 10 + 3 x 1 + 6 x 110 + 0 x 1100
10 100
= 4000 + 0 + 20 + 3 + 0.6 + 0 = 4023.6

(b) 2 Thousands + 8 Hundreds + 8 Tens + 0 Ones + 3 Tenths + 4 Hundredths
= 2 x 1000 + 8 x 100 + 8 x 10 + 0 x 1 + 3 x 110 + 4 x 1100
= 2000 + 800 + 80 + 0.3 + 0,04
= 2880 + 0.34 = 2880.34

(c) 6 Thousands + 4 Hundreds + 2 Tens + 8 Ones + 4 Tenths + 3 Hundredths
= 6 x 1000 + 4 x 100 + 2 x 10 + 8 x 1 + 410 + 3100
= 6000 + 400 + 20 + 8 + 0.4 + 0.03
= 6428 + 0.43 = 6428.43

Question 13.
Write each of the following as decimals:

Solution:

Question 14.
Convert the following unlike decimals into like decimals:

Solution:
LCM of 10, 50, 5 and 20 = 100

Question 15.
Write the following decimals in their expanded form:
(a) 27.65
(b) 102.05
(c) 36.36
(d) 0.507
Solution:
(a) 27.65

(b) 102.05

(c) 36.36

(d) 0.507

Question 16.
Write each of the decimals as a mixed fraction.
(a) 95.8
(b) 15.78
(c) 0.015
(d) 19.91
Solution:

Question 17.
Express each of the following in terms of litres (L) using decimals:
(a) 625 mL
(b) 760 mL
(c) 11 L 125 mL
(d) 7 L 350 mL
Solution:
(a) We know that 1000 mL = 1 L
∴ 625mL = 6251000L = 0.625 L
Thus, 625 mL = 0.625 L

(b) We know that 1000 mL = 1 L
∴ 760 mL = 7601000 L
= 0.760 L = 0.76 L
Thus, 760 mL = 0.76 L

(c) We know that 1000 mL = 1 L
∴ 11L 125L = 11 L + 1251000 L
= (11 +0.125) L = 11.125 L
Thus, 11 L 125 mL = 11.125 L

(d) We know that 1000 mL = 1 L
∴ 7 L 350 mL = 7 L + 3501000 L
= (7 + 0.350) L = 7.350 L = 7.35 L
Thus, 7L 350 mL = 7.35 L

Question 18.
Write 3.03, 2.75 and 2.5 in ascending order.
Solution:
The given decimals are unlike.
∴ Their corresponding like decimals are 3.03, 2.75 and 2.50.
Now neglecting the decimals, we have 303, 275 and 250.
Since, 303 > 275 > 250,
we have 3.03 > 2.75 > 2.50
∴ Ascending order is 2.50 < 2.75 < 3.03

Question 19.
Find the value of the following:
(а) 15 – 9.363
(B) 5.28 – 1.4 + 3.116
Solution:

Hence, 5.28 – 1.4 + 3.116 = 6.996

Decimals Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 20.
Mr. Ranjan purchased 15.500 kg rice, 25.750 kg flour and 3.250 kg sugar. Find the total weight of his purchases. Is it 50 kg or less? If less, how much less?
Solution:
Weight of rice = 15.500 kg
Weight of flour = 25.750 kg
Weight of sugar = 3.250 kg
Total weight of this purchases = 15.500 kg + 25.750 kg + 3.250 kg
= 44.500 kg
We see that the total weight of his purchases is less than 50 kg.

Thus, the total weight 44.500 kg is 5.500 kg less than 50 kg.

Question 21.
Ten years old Rahul can carry a maximum weight of 15 kg. If he bought 4 kg 900 g of apples, 2 kg 600 g of grapes and 5 kg 300 g of mangoes. Can he carry the total weight that he bought. If yes, then how much more weight he can carry with him?
Solution:
Weight of apples = 4 kg 900 g
Weight of grapes = 2 kg 600 g
Weight of mangoes = 5 kg 300 g
∴ Total weight of his purchases = 4 kg 900 g + 2 kg 600 g + 5 kg 300 g
= 4.900 kg + 2.600 kg + 5.300 kg
= 12.800 kg
But Rahul can carry a maximum weight of 15 kg.
Thus more weight that he can carry with him = 15 kg – 12.800 kg

= 2.200 kg

Answer: (a)

Answer: (b)

Answer: (c)

Answer: (c)

Answer: (d)

Answer: (a)

Answer: (b)

Answer: (b)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (d)

Answer: (c)

Answer: (b)

Answer: (a)

Important Questions

Question 1.
Represent the following fractions on number line.

Solution:

Question 2.
Write the fractions showing the shaded portions:

Solution:
(a) Shaded portion represents 14
Shaded portion represents 26

Question 3.
Colour the part according to the fraction given:

Solution:

Question 4.
Identify the proper and improper fractions:

Solution:
Proper fractions are: 56, 12 and 34
Improper fractions are: 72, 115 and 65

Question 5.
What fraction of these circles have ‘x’ in them?

Solution:
Fraction of the circles with ‘x’ in the given figure = 58.

Question 6.
Write all the natural numbers from 1 to 15. What fraction of them are prime numbers?
Solution:
Natural numbers from 1 to 15 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15 Prime numbers from 1 to 15 are 2, 3, 5, 7, 11, 13, i.e., 6 prime numbers.
∴ Fraction of prime numbers = 615

Question 7.
Identify the like fractions from the following:

Solution:
23 and 13 have the same denominator.
∴ 23 and 13 are the like fractions.

Question 8.
Identify the unlike fractions from the following:

Solution:
25 , 27 and 16 have different denominators.
∴ 25 , 27 and 16 are unlike fractions.

Question 9.
Convert the following improper fractions into mixed fraction.

Solution:

Question 10.
Convert the following mixed fractions into improper fractions:

Solution:

Fractions Class 6 Extra Questions Short Answer Type

Question 11.
Write the following fractions in ascending order:

Solution:
Here, the numerators of all the fractions are same.

Question 12.
Write any
(a) three proper and three improper fractions with denominator 7.
(b) two proper and two improper fractions with numerator 9.
Solution:
(a) Proper fractions with denominator 7 are: 27 , 37 and 57
Improper fractions with denominator 7 are: 97 , 117 and 136

(b) Proper fractions with numerator 9 are:
911 and 917
Improper fractions with numerator 9 are:
92 and 95

Question 13.
Compare the following fractions:

Solution:

Question 14.

Solution:
LCM of 12, 16 and 24 is 48

Question 15.
Find the sum of 123 and 325.
Solution:

Question 16.
Subtract 234 from 418.
Solution:

Question 17.
Insert > or < to make each of the following true.

Solution:

Fractions Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 18.
Find the difference between the greatest and the smallest fractions.

Solution:

Question 19.
Simran painted 23 of the wall space in her room. Her brother Rahul helped and painted 15 of the wall space. How much did they paint together? What part of the whole space is left unpainted?
Solution:
Space of the wall painted by Simran = 23
Space of the wall painted by Rahul = 15