Class 10 : Exercise 3.1

1. Aftab tells his daughter , "seven years ago, i was seven times as old as you were. Also three years from , I shall be three times as old as you will be ." Represent this situation algebraically and graphically . 

Solution

Let us assume, the current age of Aftab = x
                         the current age of his daughter = y

 Seven years ago ,
                      the age of Aftab = x-7
                      the age of his daughter = y-7
Given,
   seven years ago, i was seven times as old as you were
that is,
       x-7 = 7(y-7)
       x - 7 = 7y -49
 now lets keep x and y on one side
that is, lets move y to the left side
to move 7y to the  left side, we have to subtract 7y on both sides
so the above equation becomes
       x-7y - 7 = -49
now lets move the constant to the right side
so, to move -7 to right side, we have to add 7 on both sides
which becomes,
       x - 7y = -49 + 7
       x - 7y = -42 
lets name the above equation as (1)

       x - 7y = -42 -----------------> ( 1 )

After 3 years ,  the age of Aftab = x + 3
                         the age of his daughter = y + 3
Also given , three years from , I shall be three times as old as you will be
that is ,    x + 3 = 3 ( y + 3 )
                x + 3 = 3y + 9
lets keep variables on left side and constant on right side
so to move 3y to left side, we have to subtract 3y on both sides 
                x - 3y + 3 = 9
                x - 3y = 9 - 3
                x - 3y = 6 
lets name the second equation as ( 2 )

                x - 3y = 6 ----------------> ( 2 )

Therefore the algebraic equations are

            x - 7y = -42 -----------------> ( 1 )
            x - 3y = 6    ---------------->   ( 2 )

Now , lets represent the above equations graphically

Lets take the first equation
        x - 7y = -42
lets do the table
lets assume x values as -7 , 0 , 7 and find its corresponding y values by substituting x values in the above equation

      x    -7     0     7 

      y      5     6     7        

Lets take the second equation
          x - 3y = 6
lets do the table
lets assume x values as -3 , 0 , 3 and find its corresponding y values by substituting x values in the above equation

     x    -3     0     3 

     y    -3     -2     -1       






2 . The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later she buys another bat and 3 more balls of the same kind for Rs 1300 . Represent this situation algebraically and graphically . 

Solution

let us assume,
       Cost of a bat = x
       Cost of a ball = y
Given,
       cricket team buys 3 bats and 6 balls for Rs 3900
which means
       3x + 6y = 3900   -------------->  (1)
also given,
       she buys another bat and 3 more balls of the same kind for Rs 1300
that is
       x + 3y = 1300  -----------------> (2)

Therefore the algebraic equations are 

            3x + 6y = 3900   -------------->  (1)
            x + 3y = 1300  -----------------> (2)

Now lets represent the above equations graphically

Lets take our first equation

           3x + 6y = 3900   -------------->  (1)


lets do the table
lets assume x values as 100 , 0 , -100 and find its corresponding y values by substituting x values in the above equation

x     100     0     -100

y     600    650   700

Lets take the second equation

           x + 3y = 1300  -----------------> (2)

lets do the table
lets assume x values as 400 , 100 , -200 and find its corresponding y values by substituting x values in the above equation

x     400    100    -200

y     300    400     500 

Lets do the graph
 

3 . The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160 . After a month , the cost of 4 kg of apples and 2 kg of grapes are Rs 300 . Represent this situation algebraically and graphically.  

Solution

lets say,
   Cost of one kg of apples = x
   Cost of one kg of grapes =  y

Given,
The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160 
that can be written as 

2x + 1y = 160  -------------------> (1)

also given,
the cost of 4 kg of apples and 2 kg of grapes are Rs 300
which is

4x + 2y = 300 -------------------> (2)

Therefore the algebraic equations are 

 
2x + 1y = 160  -------------------> (1)
4x + 2y = 300 -------------------> (2)

Now lets represent the above equations graphically

Lets take our first equation

         2x + 1y = 160 -------------->  (1)

lets do the table
lets assume x values as 50 , 60 , 70 and find its corresponding y values by substituting x values in the above equation

x     50     60    70

y     60     40    20 

Lets take the second equation

          4x + 2y = 300  -----------------> (2)

lets do the table
lets assume x values as 400 , 100 , -200 and find its corresponding y values by substituting x values in the above equation

x    30    40    50

y    90    70    50

Lets do the graph

Distributive Laws

It says that multiplying a number by a group of numbers added together is same as doing each multiplication separately. 

1. a * ( b + c ) = a * b + a * c

Examples : 

1. 7 * (50 - 2 )
   = 7 * 48
   = 336
lets rearrange
( 7 * 50 ) - (7*2)        [ here, a * ( b - c ) = a * b - a * c]
=350 -14
= 336 

2. 625 * (-35) + ( -625 ) * 65 
 that is ,  [ 625 * (-35) ] +  [ ( -625 ) * 65 ]
              = (-21,875 ) + ( -40,625 )
              = - 62,500
lets rearrange
[ 625 * (-35) ] +  [ ( -625 ) * 65 ]
here, 625 is common in both , so lets take 625 out
that is, 625 * ( -35 - 65 ) 
          = 625 * -100
          = - 62,500

Associative Laws

Associative Laws

It means that it doesn't matter how we group the numbers.

1. ( a + b ) + c = a + ( b + c )
2. ( a * b ) * c = a * ( b * c ) 

Examples :

1. ( 3 + 2 ) + 5
   = 5 + 5
   = 10
lets regroup it,
 3 + ( 2 + 5 )
= 3 + 7
= 10

2. ( 3 * 2 ) * 5
   = 6 * 5
   = 30 
lets regroup it,
3 * ( 2 * 5 )
= 3 * 10
 = 30 

Commutative Law

Commutative Law

The Commutative laws which means we can swap numbers over and still get the same answer when we add or multiply.

1. a + b = b + a
2. a * b = b * a

Examples : 

1. 3 + 4 = 7
now, lets swap
  4 + 3 = 7
even if we swap we are getting same answer 7.
 
2. 3 * 4 = 12
    4 * 3 = 12
 
3. 8 * 53 * (-125)
   = 424 * (-125)
   = -53,000
lets swap
   53 * (8*(-125))
 = 53 * (-1000)
 = -53,000
even if we swap we are getting same answer -53,000

NOTE :
 - *- = + 
 + * + = +
 + * - = -
 - *+ = -

4. 15 * (-25) * (-4) * (-10)
   = [15 * (-25) * (-4) ] * (-10)
   = 1500 * (-10)
   = -15,000 
lets swap
15 * (-25) * (-4) * (-10)
   =  (-25) * [15 * (-4)  * (-10)]
   =  -25 * 600
   = -15,000

Factors and Multiples

Factors

        Factors can be defined as the numbers which we can multiply together to get another number.

Example

2 * 3 = 6 , here 2 and 3 are the factors 6

2  *4 = 8 , 2 and 4 are the factors of 8.

Note : A number can have many factors

Example

1 * 12 = 12

2 * 6 = 12

3 * 4 = 12

 so the factors of 12 are 1 , 2 , 3 , 4 , 6 and 12.

Multiples

        A multiple is the result of multiplying a number by an integer.

Example

Multiples of 2

2 * 1 = 2

2 * 2 = 4

2 * 3 = 6 ,...

so the multiples of 2 are 2 , 4 , 6 , 8 , ...

Note : Factors and multiples are different but they both involve multiplication.

Class 6 : Excersice 10.3

10. By splitting the following figures into rectangles, find their areas ( The measures are given in cms )

a.

Solution: 

First we have to broken the given figure into rectangles as shown above (in blue ).

now lets find each rectangles area

Area of 1st Rectangle :
length = 4cm , width = 2cm
Area = 4 * 2 = 8 cm square

Area of 2nd rectangle :
length = 6cm , width = 1cm
Area = 6 * 1 = 6 cm square

Area of 3rd rectangle :
length = 3cm , width = 2cm
Area = 3 * 2 = 6 cm square

Area of 4th rectangle :
length = 4cm , width = 2cm
Area = 4 * 2 = 8 cm square

Area of the given figure = area of 1st rectangle + area of 2nd rectangle
                                         + area of 3rd rectangle + area of 4th rectangle  
                                       = 8 + 6 + 6 + 8 cm square
                                       = 28 cm square