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

Answer: (c)

Answer: (c)

Answer: (c)

Answer: (c)

Answer: (d)

Answer: (b)

Answer: (b)

Answer: (d)

Answer: (c)

Answer: (b)

Answer: (a)

Answer: (a)

Answer: (d)

Answer: (a)

Answer: (c)

Important Questions

Integers Class 6 Extra Questions Very Short Answer Type

Question 1.
Represent the following on number line:
(a) -5
(b) 4
Solution:
(a) – 5

(b) 4

Question 2.
Identify the negative integers from the given numbers.
– 5, 3, 0, 5, – 6, 7, 3, 4, – 4, – 7
Solution:
Negative integers are – 5, – 6, – 4 and – 7 .

Question 3.
What is the additive identity of – 20?
Solution:
Additive identity of – 20 is 20.

Question 4.
What is negative of 0?
Solution:
The negative of 0 is 0 it self.

Question 5.
What is the absolute value of |- 6| ?
Solution:
Absolute value of |- 6| is 6.

Question 6.
What is the absolute value of |0| ?
Solution:
Absolute value of |0| is 0.

Question 7.
What is the negative of – 13?
Solution:
Negative of (- 13) is – (- 13) = 13.

Question 8.
What is the successor of – 7?
Solution:
The successor of -7 = -7 + 1 = -6

Question 9.
What is the predecessor of – 5?
Solution:
The predecessor of -5 is – 5 – 1 = -6.

Question 10.
Write the opposite of 50 km towards north?
Solution:
Opposite of 50 km towards north is 50 km towards south.

Integers Class 6 Extra Questions Short Answer Type

Question 11.
Write the following integers in their increasing order.
– 3, 0, – 6, 5, – 4, 6, 3, – 8
Solution:
The required increasing order is:
– 8, – 6, – 4, – 3, 0, 3, 5, 6

Question 12.
Comparing the following pairs of number use > or < .

Solution:
(a) 0 > – 6
(b) – 10 < – 2
(c) – 100 < 100
(d) 2 > – 2

Question 13.
Write all the integers between the following pair of integers:
(а) 0 and – 4
(B) – 5 and 5
(c) – 8 and – 13
(d) 3 and 6
Solution:
(a) Integers between 0 and – 4 are: -3, -2, -1
(B) Integers between – 5 and 5 are: -4, -3, -2, -1, 0, 1, 2, 3, 4.
(c) Integers between – 8 and – 13 are: -12, -11, -10, -9
(d) Integers between 3 and 6 are: 4 and 5

Question 14.
Find the solution of the following additions using number line:
(a)(- 3) + 5
(B) (- 5) + (-2)
Solution:
(a)(- 3) + 5

∴ (-3) + 5 = 2

(B) (- 5) + (-2)

∴ (-5) + (-2) = (-7)

Question 15.
Find the sum of the following integers:
(a) (- 8) + (+ 5) + (- 3) + (- 2)
(B) (- 7) + (- 9) + (+ 4) + (+ 3)
Solution:
(a) (- 8) + (+ 5) + (- 3) + (- 2)
= (- 8) + (+ 5) – (3 + 2)
= (- 8) + (+ 5) – (5)
= (- 8) + 0 = – 8 [ ∵ (+ a) + (- a) = 0]

(b) (- 7) + (- 9) + (+ 4) + (+ 3)
= (- 7) + (- 9) + (4 + 3)
= (- 7) + (- 9) + (+ 7)
= (- 7) + (+ 7) + (- 9)
= 0 + (- 9) = – 9 [ ∵ (- a) + (+ a) = 0]

Integers Class 6 Extra Questions Long Answer Type

Question 16.
Ramesh thinks of an integer. He subtracts 12 from it and gets the result as – 6. What was the integer he thought of?
Solution:
The given sum can be written as under.
(___) – (12) = – 6
The required integer is 12 – 6 = 6.

Question 17.
Fill in the blanks:
(a) To subtract (- 8) from 13, we add ……… to 13.
(b) To subtract 5 from (- 12), we add ……… to (-13)
(c) The negative of a negative integer is a ……… integer.
(d) An integer when added to its opposite gives ……… as the sum.
(e) -4 + …….. = 1
(f) 4 – (- 3) = ………
(g) ……… + (- 79) = 19
Solution:
(a) 8
(b) – 4
(c) positive
(d) 0
(e) 5
(f) 7
(g) 98

Question 18.
Determines:
(a) | 5 | – |- 3 |
(b) | 5 – 6|+ | -1 |
(c) – 7 + | – 3 |
(d) | 5 | + | – 12 |
Solution:
(a) | 5 | – | – 3 | = 5 – 3 = 2 [∵ |a| = a and |- a| = a]
(b) |5 – 6 | + |- 1| = |- 1| + |-1| = 1 + 1 = 2
(c) – 7 + | – 3 | = -7 + 3 = -4
(d) | 5 | + | – 12 | = 5 + 12 = 17

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

Question 19.
If * is an operation such that for two integers p and q, p * q = p + q – 2, then find:
(a) 6 * 2
(b) (- 2) * (- 3)
(c) (- 2) * (4)
(d) (+ 3) * (- 1)
Solution:
(a) Given that: p * q = p + q – 2
⇒ 6 * 2 = 6 + 2 – 2 = 6 + 0 = 6
Thus, 6 * 2 = 6.

(b) Given that: p * q = p + q – 2
⇒ (- 2) * (- 3) = (- 2) + (- 3) – 2
= -5 – 2 = -7.
Thus, (- 2) * (- 3) = – 7.

(c) Given that: p * q = p + q – 2
⇒ (- 2) * (4) = (- 2) + (4) – 2 = 2 – 2 = 0.
Thus, (- 2) * (4) = 0.

(d) Given that p*q – p + q – 2
⇒ (+ 3) * (- 1) = (+ 3) + (- 1) – 2 = 2 – 2 = 0
Thus, (+ 3) * (- 1) = 0.

Question 20.
Complete the table:

Solution:
The required table can be completed as under:

Answer: (b)

Answer: (b)

Answer: (a)

Answer: (c)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (c)

Answer: (c)

Answer: (b)

Answer: (a)

Answer: (a)

Answer: (b)

Answer: (c)

Answer: (b)

Answer: (a)

Answer: (c)

Answer: (d)

Answer: (b)

Answer: (c)

Important Questions

Question 1.
Which of the following line-segments is longer?

Solution:
By using divider, CD¯¯¯¯¯¯¯¯ seems to be longer than AB¯¯¯¯¯¯¯¯ .

Question 2.
How many line segments are used in making a triangle?

Solution:
Three line segments are used to make a triangle.

Question 3.
What is the measure of straight angle?
Solution:
The measure of straight angle is 180°.

Question 4.
What is complete angle?
Solution:
The angle for one revolution is called a complete angle.

Question 5.
Find the number of right angle turned through by the hour hand of a clock when it goes from 3 to 6.

Solution:
When the hour hand of a clock goes from 3 to 6, it turns through a right angle.

Question 6.
Draw the rough sketch of the following:
(a) Acute angle
(b) Obtuse angle
(c) Reflex angle
Solution:

(a) Acute angle

(b) Obtuse angle

(c) Reflex angle

Question 7.
Identify the types of angle from the given figures:


Solution:
(a) Obtuse angle
(b) Acute angle
(c) Right angle
(d) Straight angle
(e) Reflex angle

Question 8.
What are the degree measures of the following angles?
(a) Right angle
(b) A complete angle
(c) Straight angle
Solution:
(a) Degree measure of a right angle is 90°.
(b) Degree measure of a complete angle is 360°.
(c) Degree measure of a straight angle is 180°.

Question 9.
What are the types of the given triangles on the basis of angles?

Solution:
(a) Acute angled triangle.
(b) Obtuse angled triangle.
(c) Right angled triangle.

Question 10.
What are the types of the following triangles on the basis of sides?

Solution:
(a) Scalene triangle.
(b) Equilateral triangle.
(c) Isosceles triangle.

Basic Geometrical Ideas Class 6 Extra Questions Short Answer Type

Question 11.
In the given figure, name the following angles as acute, obtuse, right, straight or reflex.
(a) ∠QOY
(b) ∠YOP
(c) ∠ROX
(d) ∠QOX
(e) ∠POQ

Solution:
(a) ∠QOY = acute angle.
(b) ∠YOP = obtuse angle.
(c) ∠ROX = right angle.
(d) ∠QOX = reflex angle.
(e) ∠POQ = straight angle.

Question 12.
In the given figure, find the measure of the angles marked with a, b, c, d, e and f.
Solution:
∠a = 180° -129° = 51°
∠b = 180° – (51° + 92°)
= 180° – 143° = 37°
∠c = 180° – 88° = 92°
∠d = 180° – 152° = 28°
∠e = 180° – 143° = 37°
∠f= 180° – (∠e + ∠d)
= 180° – (37° + 28°)
= 180°- 65°= 115°
∠g = 180° – ∠f = 180° – 115° = 65°

Question 13.
Classify the given triangles whose sides are indicated on them.

Solution:
(a) All sides are different. So, it is a scalene triangle.
(a) Lengths of two sides of the triangle are same. So, it is an isosceles triangle.
(b) All sides are unequal and one angle is right angle. So it is scalene right angled triangle.
(c) Two sides of this triangle are equal. So, it is an isosceles triangle.

Question 14.
Complete each of the following, so as to make a true statement:
(a) A ……….. is a rectangle with a pair of adjacent sides equal.
(b) A parallelogram with a pair of adjacent sides equal is called a ………. .
(c) A quadrilateral having exactly one pair of parallel sides is called a …….. .
(d) A quadrilateral having both pairs of opposite sides parallel, is called a ……… .
(e) A parallelogram whose each angle is a right angle is called a ………… .
Solution:
(a) Square
(b) Rhombus
(c) Trapezium
(d) Parallelogram
(e) Rectangle.

Basic Geometrical Ideas Class 6 Extra Questions Long Answer Type

Question 15.
Verify the ‘Euler’s formula’ V + F = E + 2 for the given figures.
(a) A triangular prism having 5 faces, 9 edges and 6 vertices.
(b) A rectangular prism with 6 faces, 12 edges and 8 vertices.
(c) A pentagonal prism with 7 faces, 15 edges and 10 vertices.
(d) A tetrahedron -with 4 faces, 6 edges and 4 vertices.
Solution:
(a) Here, F = 5, E = 9 and V = 6
∴ V + F = E + 2
⇒ 6 + 5 = 9 + 2
⇒ 11 = 11
Hence, verified.

(b) Here, F = 6, E = 12 and V = 8
∴ V + F = E + 2
⇒ 8 + 6 = 12 + 2
⇒ 14 = 14
Hence, verified.

(c) Here, F = 7, E = 15 and V = 10
∴ V + F = E + 2
⇒ 10 + 7 = 15 + 2
⇒ 17 = 17
Hence, verified.

(d) Here, F = 4, E = 6 and V = 4
∴ V + F = E + 2
⇒ 4 + 4 = 6 + 2
⇒ 8 = 8
Hence, verified.