Class 6 : Chapter 6 : Exercise 6.1

1 . Write Opposites of the following :

a) Increase in weight
        Decrease in weight

b) 30km North
        30km South

c) 80m east
        80m West

d) Loss of Rs 700
         Profit of Rs 700

e) 100m above sea level
         100m below sea level


2. Represent the following numbers as integers with appropriate signs.

a) An aeroplane is flying at a height two thousand meter above the ground level.

  Solution

        Given, above the ground level  which implies +
   so, the given expression can be represent as +2000m

b) A submarine is moving at a depth , eight hundred meter below the sea level.

   Solution

        Given, below the sea level  which implies -
   so, the given expression can be represent as -800

c) A deposit of Rs 200. 

  Solution

        Given,  deposit which means saving so its +
    so, the given expression can be represent as + Rs 200

d) Withdrawal of Rs 700.

Solution

         Given,  withdrawal which means takingout so its -
    so, the given expression can be represent as -Rs 700

3 . Represent the following numbers on a number line.

a) +5
b) -10
c) +8
d) -1
e) -6

Solution

 

Exponents and the Laws of Exponents

Exponents

The Exponent of a number says how many times to use that number in a multiplication.
That is,
Exponents are the mathematical shorthand that tells us to multiply the same number by
itself for a specified number of times.

Example

The simpler way to write 5 * 5 * 5 * 5 * 5 * 5  as 5^6 or 56 .

Base : The number being multiplied by itself.
in our example, 5 the base .

Exponent : The number of times we are multiplying the base .
in our example , 6 is the exponent. Exponent are also called as power or index .  

Laws of Exponents 

1 . x ^ 1 = x  [ identity exponent ]

Example : 

• 5 ^ 1 = 5
• 8 ^ 1 = 8
• 0 ^ 1 = 0
• (-3) ^ 1 = -3 

2 . x ^ -n =( 1 / x)  ^ n [ Negative exponent ]

Note : 

x ^ -1 = 1 / x
( x / y ) ^ -n = ( y / x ) ^ n

Example

(5/3)^- 2 = (3/5)^2

                  = (3/5) * (3/5)
                  = (3*3) /( 5*5)
                  = 9 / 25 

3 . x ^ a * x ^ b = x ^ ( a + b ) [ product of powers ]

Example

3^2 * 3^3 = 3^(2+3)
  9   *  27    = 3^5
     243        = 243

4 . x ^ m / x ^ n = x ^ ( m - n ) [ Quotient of powers ]

Example

3 ^ 5 / 3 ^ 2 =  3 ^ ( 5 - 2 )
 243 /    9      =  3 ^ 3                    or         3*3*3*3*3 / 3*3 =  3 ^ 3
      27             =  27                        or                   3 * 3 * 3       =  27
                                                                                    27             =  27 

  5 . ( x ^ m ) ^ n = x ^ ( m * n )  [ power of power ]

Example

( 5 ^ 3 ) ^ 2  =  5 ^ ( 3 * 2 )
      125  ^ 2   =  5 ^ ( 6 )
       15,625    =  15,625

6 . ( x * y ) ^ m = (x ^ m) * (y ^ m) [ power of a product ]

Example

( 5 * 4 ) ^ 2  =  (5 ^ 2) * (4 ^ 2)
    20     ^ 2   =       25    *    16
         400      =           400

7 . ( x / y ) ^ m = x ^ m / y ^ m [ power of a quotient ]

Example

      (5/3)^2     =   (5^2) / (3^2)
(5/3) * (5/3)  =     25 / 9
    25 / 9          =    25 / 9

8 . x ^ 0 = 1   [ zero exponent ]

Note:
3/3 = 1
that is, 3^1/3^1 = 1  ( by law 4 )
                3^(1-1) = 1
                3^0 = 1

 
Example

•     5^0    =  1
• (1/5)^0 =  1

9 . x^(m/n) =  (x ^m)^1/n =             [Fractional exponent]

Example

5^(3/2) = (5^3) ^ (1/2)
               = 125 ^ (1/2)

           

 

Class 10 : Exercise 3.3

1 . Solve the following pair of linear equations by the substitution method

i) x + y = 14
   x - y = 4

Solution

Lets name the given equations are

   x + y = 14    -----------------> (1)

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

we have to solve the given equations using substitution method

Step 1:

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 'x'

x + y = 14    -----------------> (1)

to move y variable to another side we have to  subtract y on both sides

 x + y = 14

     - y    -y    

 x       = 14-y 

now lets name the above equation as (3)

so, x = 14 - y   -------------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 

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

(14 -y ) - y = 4

14 -y -y = 4

14 - 2y = 4 

-14       = -14    ( subtract 14  on both sides )

 0  - 2y  = -10

      

-2y = -10

divide both sides by -2

 -2y = -10

 -2       -2

y = 5

Step 3:

Substitute y variable in any one of the given equations

lets take equation (1) and substitute y = 5

x + y = 14    -----------------> (1)

x + 5 =14   ( to move the constant 5 to another side we have to subtract 5 on both sides)

   - 5 = -5  

x + 0 =  9 

hence

 x = 9 and y = 5

ii) s - t = 3

   (s/3) + (t/2) = 6

Solution 

Lets name the given equations are

               s - t = 3 -----------> (1)

   (s/3) + (t/2) = 6 -----------> (2)

Step 1

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 's'

        s - t = 3  ------------> (1)

to move t variable to another side we have to add t on both sides

  s - t = 3

    +t   +t    

 s       = 3+t 

now lets name the above equation as (3)

s = 3 + t ----------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 

(s/3) + (t/2) = 6 -----------> (2)

(3+t)/3 + (t/2) =6

3/3 + t/3 + t/2 = 6

1 + t/3 + t/2 = 6  (to move 1 to right side,we have to subtract 1 on both sides) 

so, t/3 + t/2 = 6-1

     t/3 + t/2 = 5   ( Lcm of 3 and 2 = 6)

that is, 

     ( 2t + 3t ) / 6 = 5 

        5t / 6 = 5    ( multiply by 6 on both sides )

         5t = 5*6

         5t = 30  ( to find t we have to divide both sides by 5 )

          5      5

          t = 6

Step 3:

Substitute t variable in any one of the given equations

lets take equation (1) and substitute t = 6

  s - t = 3 -----------> (1)

  s - 6 = 3

    + 6   +6

  s      = 9  

Hence

 s = 9 and t =6


iii)  3x - y = 3
      9x - 3y = 9 

Solution

Lets name the given equations as

      3x - y = 3    ------------> (1)
      9x - 3y = 9  ------------> (2)

Step 1

Consider any one of the equation and isolate any one of the variable

here, lets take the first equation and isolate the variable 'x'

 3x - y = 3    ------------> (1)
    + y = +y   (to move y to right side we just added y on both sides)

 3x    = 3 + y
 to make 3x as 3 we have to divide both sides by 3
that is,
 3x  = (3 + y) 
  3          3

x = (3+y) / 3 -----------> (3)

Step 2:

Now substitute equation (3) in another given equation

here, lets substitute equation 3 in 2

that is, 
     9x - 3y = 9  ------------> (2)

    3 9(3 + y)  - 3y = 9

          3
        3(3+y) -3y = 9
          9 + 3y -3y = 9
          9  + 0 = 9
              9  =  9 ( which is always true )

Hence, the given pair of equations has infinite possible number of solutions.

Note :

When we try to solve both equations if we get any number = same number
then that equation has infinite number of solutions
 
When we try to solve both equations if we get any number != same number
then that equation has NO solutions 
(!= meaning 'not equal')
 

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