SUMMER EXERCISE (F1-F2)

1. Let x be the number of boys in the class

So, x/2 is the number of girls in the class

Total number of students = 39

So, x + x/2 = 39

3x/2 = 39

x = 26

So, there are 26 boys in the class.


2. Two adjacent sides of the square are equal

So, x = 2x - 3

x - 2x = -3

-x = -3

x = 3

Therefore, the perimeter of the square

= 4 X 3

= 12 cm


3. Let y be the cost of one of the more expensive cat

So, y - 3 is the cost of one of the less expensive one

3y + 4(y - 3) = 100 - 7

7y - 12 = 93

7y = 93 + 12

7y = 105

y = 15

So, the more expensive ceramic cat costs $15, and the other kind costs 15 - 3 = $12

1.IN a badminton class ,the number of boys is 2 times of the number of girls.If the total number of students in the badminton class is 39 , how many boys are there in the class?

Lex x be the number of boy in badminton class
The number of girl in badminton class be 0.5x
Since x + 0.5x = 39
1.5x = 39
x = 26
Conclusion:
The number of boy in badminton class is 26
The number of girl in badminton class is 26 / 2 or 13

2.If the two adjacent sides of a square measure x cm and (2x-3)cm
respectively, find the perimeter of the aquare.

The perimeter = (Length + width) * 2
Let x be the length
Let 2x-3 be the width
Perimeter = (x + 2x - 3) * 2
= (3x - 3) * 2
= 6(x-1)

3. ann had $100.She used part of her money to buy 7 ceramic cats, in which 3 of them are more expensive than the rest by $3.If she had $7 left , how much do the two type of ceramic cat cost?
Let x be the cost of more expensive ceramic cats
The cost less expensive ceramic cats will be x - 3
Since the cost of total 7 ceramic cats is $93 or $100 - $7
Therefore 3x + 4 * (x - 3) = 93
3x + 4x - 12 = 93
7x = 105
x = 15
Conclusion
The cost of more expensive ceramic cats is $15
the cost of less expensive ceramic cats is $12 or ($15 - $3)