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.

Answer: (b)

Answer: (a)

Answer: (d)

Answer: (a)

Answer: (a)

Answer: (a)

Answer: (b)

Answer: (c)

Answer: (c)

Answer: (c)

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (b)

Answer: (b)

Answer: (d)

Answer: (d)

Important Questions

Question 1.
Draw a rough sketch of:
(a) open curve
(b) closed curve
Solution:

Question 2.
Draw a rough sketch of closed curve made up of line segments.
Solution:
Required curve is ABCD closed with the line segments AB¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯, CD¯¯¯¯¯¯¯¯ and DA¯¯¯¯¯¯¯¯.

Question 3.
Draw two different angles having common point and a common arm.
Solution:
∠AOB and ∠COB are two different angles with common point O and common arm OB−→− .
IMG

Question 4.
Identify the points which are:
(i) in the interior
(ii) in the exterior
(iii) on the closed curve in the given figure.

Solution:
(i) Points P, Q and R are in the interior of the closed curve.
(ii) points S and T are in the exterior of the closed curve.
(iii) U and V are on the closed curve

Question 5.
Identify the following in the given figure:
(a) Sector
(b) Chord
(c) Diameter
(d) Segment.

Solution:
(a) OPR (shaded) is the sector of the circle.
(b) MN¯¯¯¯¯¯¯¯¯¯ is the chord.
(c) PQ¯¯¯¯¯¯¯¯ is the diameter.
(d) MXN (shaded) is the segment.

Question 6.
In the given figure, name all the possible triangles.

Solution:
Possible triangles are:
(i) ∆ABC
(ii) ∆ABD
(iii) ∆ABE
(iv) ∆ACD
(a) ∆ACE
(vi) ∆ADE

Question 7.
Name all the angles in the given figure.
Solution:
In the given figure, the names of all the angles are:

(i) ∠ABC
(ii) ∠BCD
(iii) ∠CDA
(iv) ∠DAB

Question 8.
In the given figure, name all the line segments:

Solution:
In the given figure, the name of the line segments are:
AB¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯, CD¯¯¯¯¯¯¯¯, DE¯¯¯¯¯¯¯¯, EA¯¯¯¯¯¯¯¯, DA¯¯¯¯¯¯¯¯, DB¯¯¯¯¯¯¯¯ and EC¯¯¯¯¯¯¯¯.

Basic Geometrical Ideas Class 6 Extra Questions Short Answer Type

Question 9.
Using the given figure, name the following:

(a) Line containing point M.
(b) Line passing through four points.
(c) Line passing through three points.
(d) Two pairs of intersecting lines.
Solution:
(a) MC←→− is the line containing the point M.
(b) AN←→ is the line passing through four points A, B, C and N.
(c) PQ←→ is the line passing through three points P, B and Q.
(d) Pairs for intersecting lines are
(i) AN←→ and PQ←→
(ii) AN←→ and MC←→−

Question 10.
On the given line, some points are given, write down the names of all segments.

Solution:
Segments are:
PQ¯¯¯¯¯¯¯¯ , PR¯¯¯¯¯¯¯¯ , PS¯¯¯¯¯¯¯ , PT¯¯¯¯¯¯¯ , QR¯¯¯¯¯¯¯¯ , QS¯¯¯¯¯¯¯ , QT¯¯¯¯¯¯¯¯ , RS¯¯¯¯¯¯¯ , RT¯¯¯¯¯¯¯ , ST¯¯¯¯¯¯¯ .

Question 11.
How many lines can pass through
(i) one given point?
(ii) two given points?
(iii) three non-collinear points
Solution:
(i) Infinite number of lines can be passed through one given point.
(ii) Only one line can pass through two given points.
(iii) Three lines can pass through three non- collinear points.

Question 12.
Look at the given figure and answer the following:

(a) Name the sides of the polygon ABCDEF.
(b) Name any two pairs of adjacent sides. .
(c) Name all the segments which intersect each other at one point.
(d) Name all the diagonals of the given polygon.
Solution:
(a) The sides of the polygon are:
AB¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯, CD¯¯¯¯¯¯¯¯, DE¯¯¯¯¯¯¯¯, EF¯¯¯¯¯¯¯¯ and FA¯¯¯¯¯¯¯¯.
(b) AB¯¯¯¯¯¯¯¯ and BC¯¯¯¯¯¯¯¯, BC¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯are the pairs of adjacent sides.
(c) AD¯¯¯¯¯¯¯¯, BE¯¯¯¯¯¯¯¯ and CF¯¯¯¯¯¯¯¯ intersect each other at O.
(d) Name of the diagonals are:
AD¯¯¯¯¯¯¯¯, BE¯¯¯¯¯¯¯¯ and CF¯¯¯¯¯¯¯¯.

Question 13.
Fill in the blanks.
(a) ……… is the largest chord of a circle.
(b) ……… divides the circle into two equal semi circles.
(c) Rectangle is a ……… curve.
(d) Triangle has ……… angles and three ……… .
(e) Only one line can be drawn through ……… points.
(f) ……… number of lines can pass through a given point.
(g) A closed figure made up of entirely of line segments is called a ……… .
(h) A curve which does not cross itself is called a ……… curve.
(i) The length of boundary of a circle is called its ……… .
Solution:
(a) Diameter
(b) Diameter
(c) Closed
(d) three, sides
(e) two
(f) Infinite
(g) polygon
(h) simple
(i) circumference

Basic Geometrical Ideas Class 6 Extra Questions Long Answer Type

Question 14.
Draw the medians of a AABC and answer the following:
(а) Name the three medians.
(b) Do the medians intersect each other at the same point?
(c) What is that point called?
(d) Can this point be outside the triangle?

Solution:
(a) Names of the medians are AD¯¯¯¯¯¯¯¯, BE¯¯¯¯¯¯¯¯ and CF¯¯¯¯¯¯¯¯ .
(b) Yes, the medians intersect each other at the same point G.
(c) The point of intersection of the medians of a triangle is called ‘Centroid’.
(d) No, this point cannot be out of the triangle.

Basic Geometrical Ideas Class 6 Extra Questions Multiple Choice Type

Question 15.
Least number of line segments required to make a polygon is
(a) 1
(b) 2
(c) 3
(c) 4
Solution:
Correct option is (c).

Question 16.
How many lines can be drawn through given two points?
(a) Only one
(b) 2
(c) 4
(d) Countless
Solution:
Correct option is (a).

Question 17.
How many vertices are there in a triangle?
(a) 1
(b) 2
(c) 3
(c) 4
Solution:
Correct option is (c).

Question 18.
Say ‘true’ or ‘False’.
(а) Two diameters of a circle will necessarily intersect.
(b) The centre of a circle always lies in the interior.
(c) The diameter is half of the radius of a circle.
(d) Longer chord is nearer to the centre of the circle.
Solution:
(a) True
(b) True
(c) False
(d) True

Basic Geometrical Ideas Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 19.
Draw an equilateral ∆ABC of any size. Draw AD as its median and an altitude AM.
(i) Does AD coincide with AM?
(ii) Name the point on the median which divides it in the ratio 1:2.
(iii) What is the measure of ∠ADC and ∠ADB?
(iv) Are D and M the same points?
Solution:
(i) Yes, AD coincides with AM.

(ii) The point on the median which divides it in the ratio 1 : 2 is called centroid of the triangle.
(iii) ∠ADC = ∠ADB = 90°
(iv) Yes, D and M are the same points.

Question 20.
In the given figure, l, m and n are three parallel lines, x andy intersect these lines.
(i) Name the points lying on the line x.
(ii) Name the points lying on the liney.
(iii) Name the points inside the quadrilateral ABED.
(iv) Name the points outside the quadrilaterals ABED and BCFE.
(v) Name the lines passing through three points.
Solution:
(i) A, B and C lie on the line x.
(ii) D, E and F lie on the liney.
(iii) Q is the point inside ▢ABED

(iv) Points R and S are outside the quadrilaterals ABED and BCFE.
(v) Lines x andy pass through the three points A, B, C and D, E, F respectively.

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (c)

Answer: (c)

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (d)

Answer: (b)

Answer: (d)

Answer: (d)

Answer: (b)

Answer: (d)

Answer: (a)

Answer: (c)

Answer: (a)

Importanrt Questions

Question 1.
What is the sum of any two
(a) even numbers
(b) odd numbers?
Solution:
(a) The sum of any two even numbers is even.
Example: 4 (even) + 6 (even) = 10 (even)
(b) The sum of any two odd numbers is even.
Example: 5 (odd) + 7 (odd) = 12 (even)

Question 2.
Which of the following numbers is divisible by 3?
(a) 1212
(b) 625
Solution:
(a) Given number = 1212
Sum of the digits = 1 + 2 + 1 + 2 = 6, which is divisible by 3.
Hence, 1212 is also divisible by 3.
(b) Given number = 625
Sum of the digits = 6 + 2 + 5 = 13, which is not divisible by 3.
Hence, 625 is not divisible by 3.

Question 3.
If the LCM and HCF of any two numbers are 15 and 4 respectively, find the product of the numbers.
Solution:
We know that the product of the number = LCM x HCF = 15 x 4 = 60
Hence, the product of the given numbers = 60.

Question 4.
Find the HCF of 5 and 7.
Solution:
Given numbers are 5 and 7. We observe that 5 and 7 are co-prime numbers.
Hence, the HCF is 1.

Question 5.
Write first 3 multiples of 25.
Solution:
We have 25 x 1 = 25; 25 x 2 = 50; 25 x 3 = 75
Hence, the required multiples are 25, 50 and 75.

Question 6.
What are the possible factors of (a) 12 (b) 18?
Solution:
(a) Possible factors of 12 are:
12 = 1 x 12; 12 = 2 x 6; 12 = 3 x 4
Hence, the factors of 12 are 1, 2, 3, 4, 6 and 12.

(b) Possible factors of 18 are:
18 = 1 x 18; 18 = 2 x 9; 18 = 3 x 6
Hence, the factors of 18 are 1, 2, 3, 6, 9 and 18.

Question 7.
Write first three multiples of 11.
Solution:
First three multiples of 11 are:
11 x 1 = 11; 11 x 2 = 22; 11 x 3 = 33
Hence, the required multiples are: 11,22 and 33.

Question 8.
Write pairs of twin prime numbers less than 20.
Solution:
Pairs of twin prime numbers are: (3, 5), (5, 7), (11, 13), (17, 19).

Question 9.
Write the number which is even as well as prime.
Solution:
2 is the only even number which is prime number also.

Question 10.
What is the fundamental theorem of arithmetic?
Solution:
Every number greater than 1 has exactly one prime factorisation.

Playing With Numbers Class 6 Extra Questions Short Answer Type

Question 11.
Simplify: 32 + 96 ÷ (7 + 9)
Solution:
Given that: 32 + 96 ÷ (7 + 9)
= 32 + 96 ÷ 16 (Using BODMAS)
= 32 + 6 = 38

Question 12.
Simplify: 18 + {1 + (5 – 3) x 5}
Solution:
Given that: 18 + {1 + (5 – 3) x 5} (Using BODMAS)
= 18 + {1 + 2 x 5} = 18 + {1 + 10}
= 18 + 11 = 29.

Question 13.
Without actual division, show that 11 is a factor of 1,10,011.
Solution:
Here 1,10,011 = 1,10,000 + 11
= 11 x 10,000 + 11 x 1
= 11 x (10,000 + 1)
= 11 x 10,001
It is clear that 11 is a factor of 11 x 10,001.
Hence, 11 is a factor of 1,10,011.

Question 14.
The sum of two numbers is 25 and their product is 144. Find the numbers.
Solution:
The product of two numbers is 144.
∴ The possible factors are 1 x 144, 2 x 72, 3 x 48, 4 x 36, 6 x 24, 8 x 18, 9 x 16, 12 x 12
Here, we observe that out of these factors, we take 9 and 16.
Product = 9 x 16 = 144 and sum = 9 + 16 = 25
Hence, the required numbers are 9 and 16.

Question 15.
Is 80136 divisible by 11?
Solution:
Sum of the digits at odd places = 6 + 1 + 8 = 15
Sum of the digits at even places = 3 + 0 = 3
Difference of the two sums = 15 – 3 = 12,
which is neither 0 nor the multiple of 11.
Hence, 80136 is not divisible by 11.

Question 16.
The HCF and LCM of two numbers are 6 and 120 respectively. If one of the numbers is 24, find the other number.
Solution:
Given that: HCF = 6
LCM = 120
Let the two numbers be a and b, where a = 24, b = ?
We know that: a x b = HCF x LCM
⇒ 24 x b = 6 x 120
⇒ b = 6x12024
⇒ b = 30
Hence, the other number is 30.

Question 17.
Find the LCM of 12 and 30.
Solution:
Given numbers are 12 and 30
12 = 2 x 2 x 3;
30 = 2 x 3 x 5
∴ LCM = 2 x 2 x 3 x 5 = 60
Hence, the LCM of 12 and 30 = 60.

Question 18.
Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Solution:
Given numbers are 18, 24 and 32, we have

Thus, LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
The smallest 4-digit number = 1000
Now, we write multiples of 288, till we get a 4-digit number.
288 x 1 = 288, 288 x 2 = 576,
288 x 3 = 864, 288 x 4 = 1152
Hence, 1152 is the required number.

Question 19.
Find the greatest number which divides 82 and 132 leaving 1 and 6, respectively as remainders.
Solution:
Given numbers are 82 and 132 and the remainders are 1 and 6 respectively.
We have, 82 – 1 = 81 and 132 – 6 = 126
So, we need to find the HCF of 81 and 126

Common factor is 3 (occurring twice).
∴ HCF = 3 x 3 = 9
Hence, the required number is 9.

Question 20.
Find the greatest number that will divide 455, 582 and 710 leaving remainders 14, 15 and 17 respectively.
Solution:
Given numbers are 455, 582 and 710 and the respective remainders are 14, 15 and 17.
We have 455 – 14 = 441, 582 – 15 = 567 and 710 – 17 = 693.
Now let us find their HCF.

Common factors are 3 and 7.
∴ HCF = 3 x 7 = 21
Hence, the required number is 21.

Playing With Numbers Class 6 Extra Questions Long Answer Type

Question 21.
Simplify the following:
40 + [20 – {28 ÷ 7 – 3 + (30 – 5 of 4)}]
Solution:
Using BODMAS Rule, we have
40 + [20 – {28 ÷ 7 – 3 + (30 – 5 of 4)}]
= 40 + [20 – {28 ÷ 7 – 3 + (30 – 20)}]
= 40 + [20 – {28 ÷ 7 – 3 + 10}]
= 40 + [20 – [4 – 3 + 10}]
= 40 + [20-11] = 40 + 9 = 49.

Question 22.
Three sets of English, Hindi, and Urdu books are to be stacked in such a way that the books are stored subjectwise and the height of each stack is the same. The numbers of English, Hindi and Urdu books are 336, 192 and 144 respectively. Assuring that the books have that same thickness, determine the number of stacks of English, Hindi and Urdu books.
Solution:
To arrange the books in the required way,
we have to find the greatest number that divides 336, 192 and 144 exactly.
So, HCF of 336, 192 and 144 is

Common factors are 2 x 2 x 2 x 2 x 3 = 48
∴ HCF = 48, i.e. each stack contains 48 books.
∴ Number of stacks of English books = 336 + 48 = 7
Number of stacks of Hindi books = 192 4- 48 = 4
Number of stacks of Urdu books = 144 4- 48 = 3

Question 23.
Which of the following statements are true?
(а) 1371 is divisible by 3
(b) 1155 is not divisible by 9
(c) 1478 is not divisible by 4
(d) 2470 is divisible by 5
(e) If a number is divisible by 9, it is also divisible by 3.
(f) If a number is divisible by 3, it is also divisible by 9.
(g) The sum of any two odd numbers is even.
(h) If a number is divisible by 8, it must be divisible by 6.
(i) If a number is divisible by 3 and 6, it is divisible by 18.
(j) 1758 is not divisible by 8.
Solution:
(a) Yes, 1371 is divisible by 3. So it is true statement.
(b) Yes, 1155 is not divisible by 9. So it is true statement.
(c) Yes, 1478 is not divisible by 4. So it is true statement.
(d) Yes, 2470 is divisible by 5. So it is true statement.
(e) Yes, it is true statement. if No, it is not true statement.
(g) Yes, it is true statement.
(h) No, it is not true statement.
(i) No, it is not true statement.
(j) Yes, it is true statement.

Playing With Numbers Class 6 Extra Questions Multiple Choice Type

Question 24.
Which of the following numbers is divisible by 11?
(a)112111
(b) 928389
(c) 12011
(d) 11111

Question 25.
Match column I with column II.
Column I                                                                               Column II
(a) A number divisible by 11                                                     (i) 2
(b) HCF of two consecutive odd numbers                            (ii) 4
(c) The difference between twin prime number                 (iii) product of the number
(d) Number of factors of a prime number                            (iv) 60
(e) Lowest composite number                                                 (v)2
(f) LCM of 12 and 5                                                                    (vi) 4587594
(g) The smallest prime number                                              (vii) 1
(h) Product of HCF and LCM is equal to                              (viii) 2
Solution:
(a) → (vi)
(b) → (vii)
(c) → (viii)
(d) → (i)
(e) → (ii)
(f) → (iv)
(g) → (v)
(h) → (iii)

Playing With Numbers Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 26.
Ill cows, 185 sheep and 296 goats are to be taken across a river. There is only one boat and the boatsman says; he will take the same number and same kind of animals in each trip. Find the largest number of animals in each trip and the number of trips he will have to make.
Solution:
We have
Number of cows = 111
Number of sheep = 185
Number of goats = 296
According to the condition of the boatsman, we need to find HCF of 111, 185 and 296
111 = 3 x 37;
185 = 5 x 37;
296 = 2 x 2 x 2 x 37
∴ the HCF = 37
So, the number of animals of same kind = 37.
Number of trips
= 11137 + 18537 + 29637
= 3 + 5 + 8 = 16
Hence, the number of animals in each trip = 37
and the number of trips = 16.

Question 27.
In a seminar, the number of participants in Mathematics, Physics and Chemistry are 60, 96 and 144 respectively. Find the number of rooms required if in each room, the same number of ‘ participants are to be seated and all of them are to be in the same subject.
Solution:
The number of participants in each room must be the HCF of 60, 96 and 144.
∴ 60 = 2 x 2 x 3 x 5
96 = 2 x 2 x 2 x 2 x 2 x 3
144 = 2 x 2 x 2 x 2 x 3 x 3
HCF = 2 x 2 x 3 = 12
Number of rooms required
= 6012 + 9612 + 14412
= 5 + 8 + 12 = 25
Hence, number of participants = 12
in each room and number of rooms required = 25.