WBBSE Solutions For Class 8 Maths Chapter 24 A Few Interesting Problems

A Few Interesting Problems

Question 1. Interesting problems with matchsticks.

1. I constructed an equilateral triangle with three matchsticks.
Solution:

2. My brother constructed 6 equilateral triangles with 12 matchsticks.
Solution:

Replacing only 4 sticks from the 12 sticks I will construct 3 equilateral triangles where their areas are not equal.

First triangle is A, Second triangle is (A + B) and 3rd triangle is (A + B + C). All three are equilateral triangles whose measurements are not equal.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. Megha made an arrangement with 26 matchsticks as shown in the figure beside.
Solution:

Replacing only 14 matchsticks in this arrangement I constnucted 3 squares where their areas are not same.

A, B and C are three squares whose areas are not same.

Question 3. Rokeya had 20 matchsticks. She made a square type house with 4 matchsticks as the figure beside and she also made a square type fence around the garden with the remaining 16 sticks.
Solution:

Given

Rokeya had 20 matchsticks. She made a square type house with 4 matchsticks as the figure beside and she also made a square type fence around the garden with the remaining 16 sticks.

Let’s divide the garden into five equal shapes and sizes with 10 other matchsticks.

I inserted Rokeya’s arrangement with 10 new sticks 1, 2, 3, 4, 5, 6, 7,8, 9 and 10; and divided this garden into five equal parts A, B, C, D, and E.

Question 4. Let’s put numbers from 1 to 19 in the circles of the following wheel in such a way that the sum of the numbers in 3 circles along a straight line segment is 30.
Solution:

Question 5. 

1. 

Solution:

= \(\frac{3}{2}-1=\frac{1}{2}\)

= \(\frac{8}{3}-2=\frac{2}{3}\)

= \(\frac{19}{3}-3=\frac{10}{3}\)

2. 

Solution:

(4+9) × 2 = 26

(9 + 16) × 2 = 50

(16 + 4) × 2 = 40

Question 6. 

1.

Solution:

\((2)^2+(4)^2=20\),

\((1)^2+(5)^2=26\),

\((3)^2+(9)^2=90\)

\((1)^3+(2)^2=9\),

\((4)^3+(5)^3=189\),

\((2)^3+(3)^3=35\)

Question 7.

1. 

Solution:

1. 7×3 + 8 = 29

2. 4×3 + 7 = 19

3. 5×? + 6 = 31

∴ ? = (31 – 6)÷5 = 5

2.

Solution:

1. 4×2 – 1 =7

2. 5×3-3 = 12

3. 6×7 – ? = 39

∴ ? = 42 – 39 = 3

Solution:

1. (10-9)+(15-12)

Solution:

(10-9)+(15-12) = 1+3

(10-9)+(15-12) = 4

(10-9)+(15-12) = 4

2. (16-20)+(28-12)

Solution:

(16-20)+(28-12)  =-4+16

(16-20)+(28-12)  =12

(16-20)+(28-12) =12

3. (15-16)+(23-11)

Solution:

(15-16)+(23-11) =-1+12

(15-16)+(23-11) =11

(15-16)+(23-11) =11

Let’s see the rules of the game and find out the correct numbers.

Question 1. If the sign ‘÷’ stands for the sign ‘×’ and the sign ‘+’ stands for the sign ‘÷’ and the sign ‘#’ stands for the sign ‘+’, then let’s write which of the following numbers will be the value of 2÷5+5#100.

1. 100
2. 102
3. 108
4. 105

Solution:

2 × 5 ÷ 5 + 100

= 2 × 1 + 100

= 102

Question 2. If 7 * 1= 64 and 3*9=144, then let’s write which one of the following numbers will be the value of 5*6 :

1. 2
2. 45
3. 101
4. 121

Solution:

7*1 = 64

= \((7+1)^2\) = 64 :

= \((3+9)^2\) = 144

∴  5 * 6

= \((5+6)^2\)

= 121

Question: 3. If 84 ‘+’ 72 = 45 and 73 ‘×’ 41=43, then find the value of 94 ‘×’ 72 from the following:

1. 55
2. 59
3. 56
4. 66

1. 84 = 8-4 = 4

Solution:

72 = 7-2 = 5

∴ Number = 45

2. 73 = 7-3= 4

Solution:

41=4-1=3

∴ Number = 43

3. 94 = 9-4 = 5

Solution:

72 = 7-2 = 5

∴ Number = 55

Question 4. If the sign ‘÷’ and the sign ‘+’ and the numbers ‘6’ and ‘3’ interchange their positions, then find which one of the following relations is true-

1. 3+6-2=5

Solution:

=6÷3 + 2

= 2 + 2

= 4

2. 6÷3+2=8

Solution:

= 3 + 6 ÷ 2

= 3 + 3

= 6

3. 3+6÷5=7

Solution:

6÷3 + 5

=2+5

=7

4. 3÷6+1 =6

Solution:

= 6 + 3÷1

= 6 + 3

= 9

∴ 3. Relation is true

Question 5. If the sign ‘+’ and the sign ‘-‘ and the numbers ‘4’ and ‘8’ interchange their positions, then find which of the following relations is true

1. 4+8-12=16

Solution:

8-4+12

=20-4

= 16

2. 4-8+12=6

Solution:

8 + 4-12

= 12-12

= 0

3. 8+4-12=24

Solution:

4 – 8 + 12

= -4+12

= 8.

4. 8-4+12=8

Solution:

4 + 8-12

= 12-12

= 0

∴ 1. Relation is true

Question 6. Let’s find out few more interesting numbers –

Solution:

Let’s see why 142857 is a (Revolving Number) –

142857×1 = 142857

142857×2 = 285714

142857×3 = 428571

142857×4 = 571428

142857×5= 714285

142857×6= 857142

Question 7. Let’s find 24 using a one digit number thrice: 3³-3 = 24 Let’s find 24 using another one digit number thrice except 3.

Solution:

8 + 8 + 8 = 24

8. Let’s find 30 using a one digit number thrice: 3³+3 = 30. Let’s find 30 using another one digit number thrice except 3.

Solution:

6 x 6 – 6 = 30

5 x 5 + 5 = 30

9. Imran arranged 8 pieces of papers in two columns writing the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9.

Solution:

Let’s try to get the same sum from the columns replacing only 2 pieces of paper.

Question 10. Maria’s father kept an envelope with some money on the table to 10.buy books and went to office. Maria returned home and saw that 98 is written on the envelope. So she went to the book shop and bought a book priced Rs. 92. However, at the time of paying the money she saw that there were less than 92 rupees in the envelope. How did this happen? Let’s think and write.
Solution:

Given

Maria’s father kept an envelope with some money on the table to 10.buy books and went to office.

Maria returned home and saw that 98 is written on the envelope. So she went to the book shop and bought a book priced Rs. 92.

However, at the time of paying the money she saw that there were less than 92 rupees in the envelope.

Maria’s father had written 86 on the envelope but Maria had seen the envelope in opposite direction.