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.
The first triangle is A, the 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 arranged 26 matchsticks as shown in the figure beside.
Solution:
“WBBSE Class 8 Maths Chapter 24 solutions, A Few Interesting Problems”
Replacing only 14 matchsticks in this arrangement I constructed 3 squares where their areas are not same.
A, B, and C are three squares whose areas are not the 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.
“Class 8 WBBSE Maths Chapter 24 solutions, Interesting Problems study material”
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
“WBBSE Class 8 Maths Chapter 24, A Few Interesting Problems solved examples”
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.
- 100
- 102
- 108
- 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 :
- 2
- 45
- 101
- 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:
- 55
- 59
- 56
- 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
“WBBSE Class 8 Interesting Problems solutions, Maths Chapter 24”
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 a 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 for 10.buy books and go to the office. Maria returned home and saw that 98 was written on the envelope. So she went to the bookshop and bought a book priced at 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 for 10.buy books and go to the office.
Maria returned home and saw that 98 is written on the envelope. So she went to the bookshop and bought a book priced at Rs. 92.
“Class 8 WBBSE Maths Chapter 24, A Few Interesting Problems easy explanation”
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 othe pposite direction.