Reducing the Fractions to their Lowest Terms.

REDUCING A FRACTION:

Reducing a fraction means making the fraction as simple as possible.

Reducing a fraction is called as simplifying the fraction. We say the fraction in its Lowest terms.       

We need to divide the numerator and denominator by a common factor.

We can keep reducing a fraction till the numerator and the denominator in the reduced fraction have no common factor other than 1.

 

Simplify (or) Reduce the Fractions (or) the fractions to their Lowest Terms:

1) 48/108 =

Solution:

(i) We need to find out the common factor of both Numerator and Denominator. The Common factor is 2.

(ii) Divide the numerator and denominator with that common factor:2

Numerator: N = 48 ÷ 2 = 24.

Denominator:D = 108 ÷ 2 = 54.

(iii) We need to simplify the fraction with their common factor, till we get the common factor as 1.

Numerator: N    = 48 ÷ 2  =  24 ÷ 2  = 12 ÷ 3  = 4.

Denominator: D = 108 ÷ 2 =  54 ÷ 2  = 27 ÷ 3  = 9.

The Simplified or Reduced fraction is: 4/9, for which the common factor is 1.   

     

2) 168/144 =

Solution:

(i) The common factor of the given fraction numerator, N: 168 and denominator, D: 144 is 4.

(ii) Divide the numerator and denominator with the common factor in each step.

(iii) Reduced the fraction till the common factor of the fraction is 1.

Numerator: N    = 168 ÷ 4 = 42 ÷ 3 = 14 ÷ 2 = 7.

Denominator: D = 144 ÷ 4 = 36 ÷ 3 = 12 ÷ 2 = 6.

The Common factor of the Reduced fraction: 7/6  is 1.   

Thus, the simplified or Reduced fraction of the given fraction is:

168/144 =  42/36  =  14/12 =  7/6.

 

3) 20/24 =

Solution:

(i) The common factor of the given fraction is: 4.

(ii) Dividing the given fraction with their common factor at each step till the common factor of the reduced fraction is 1.

Numerator: N    = 20 ÷ 4 = 5.

Denominator: D = 24 ÷ 4 = 6. 

Thus, the Reduced fraction = 20/24 = 5/6.

                             (OR)

If we choose the common factor for the given fraction as 2, then

Numerator: N    = 20 ÷ 2 = 10 ÷ 2 = 5.

Denominator: D = 24 ÷ 2 = 12 ÷ 2 = 6.

The Reduced fraction = 20/24 = 10/12 = 5/6.

 

Note: We can choose different common factor in reducing the fraction, the Reduced fraction is same.

 

4) 32/60 =

Solution:

Let the common factor of the given fraction =  4.

Numerator: N    = 32 ÷ 4 = 8.

Denominator: D = 60 ÷ 4 = 15.   

Thus, the Reduced Fraction is : 32/60 = 8/15.

                             (OR)

Let the common factor of the given fraction = 2.

Numerator: N     = 32 ÷ 2  =  16 ÷ 2  =  8.

Denominator: D = 60 ÷ 2  =  30 ÷ 2  = 15.

Thus, the Reduced fraction in its lowest terms:

32/60 = 16/30 = 8/15.

 

 5) 27/81=

Solution:

If the common factor of the given fraction is: 9.

Numerator: N    =  27 ÷ 9  =  3 ÷ 3  =  1.

Denominator: D =  81 ÷ 9  =  9 ÷ 3  =  3.

Thus, the given fraction in its Lowest terms = 27/81 = 3/9 = 1/3.

                          (OR)

If the common factor of the given fraction is: 3.

Numerator: N    =  27 ÷ 3  =  9 ÷ 3    =  3 ÷ 3  =  1.

Denominator: D =  81 ÷ 3  =  27 ÷ 3  =  9 ÷ 3  =  3.

Thus, the given fraction in its Lowest terms =

27/81  =   9/27  =  3/9  =  1/3.   

Equivalent Fractions.

EQUIVALENT FRACTIONS:

The fractions, that has same value but has a different numerator and denominator.

In the above figure, the fractional parts are different but has the same value.  

FORMING EQUIVALENT FRACTIONS:

The Equivalent fraction of the given fraction can be formed either by multiplying or dividing the given fractions.

a) Forming Equivalent Fractions by using Multiplication:

1) 2/3

Solution:

The numerator and denominator of the given fraction: 2/3, doesn’t have common factor.

 To convert into equivalent fraction we need to multiply the given fraction with the same number i.e; 2.

Numerator    = 2 * 2 = 4 * 2  = 8.

Denominator = 3 * 2 = 6 * 2 = 12.

2) 3/7 =

 

3) 2/9 =

 

 

b) To Simplify the Fraction using Division:

The given fraction is simplified by dividing the numerator and denominator with the same common factor.

 

1) Simplify the fraction: 12/30.

Solution:

The numerator and denominator of the given fraction has common factor.

Numerator = 12 ÷ 2    =  6 ÷ 3  = 2.

Denominator = 30 ÷ 2 = 15 ÷ 3 = 5.

 

2) 8/12 =

 

The numerator and denominator of the given fraction has common factor.

Numerator    =  8 ÷ 2  =  4 ÷ 2  =  2.

Denominator = 12 ÷ 2 =  6 ÷ 2  =  3.

Fractions.

FRACTIONS:

Fractions represent equal parts of a whole or a collection.

 Fraction of a whole: When we divide a whole into equal parts, each part is a fraction.

Fraction notation: (p/q)

A fraction has two parts: Numerator and Denominator.

The number on the top of the line is called the numerator. It tells how many equal parts of the whole or collection are taken. 

 The number below the line is called the denominator.  It shows the total divisible number of equal parts the whole into or the total number of equal parts which are there in a collection. 

Examples:

 

Figure(i): It represents the whole part.

Figure(ii): The number of parts are 2. The shaded part is 1.The Fraction of parts shaded = 1/2.

Figure(iii): The number of parts are 3.The shaded parts are 2.  The Fraction of parts shaded = 2/3.

Figure(iv): The numbers of parts are 4.The shaded parts are 3.  The Fraction of parts Shaded = 3/4.

Figure(v): The number of  diamonds = 6. The number of diamonds coloured= 3. The Fraction of diamonds coloured = 3/6.

Figure(vi): The number of stars = 7. The number of stars coloured = 5. The Fraction of stars coloured = 5/7.

Therefore,

Fraction = Number of parts coloured or shaded / Total number of parts.

Finding Greatest number which is exactly divided and When divided leaves remainders.

Finding Greatest number using H.C.F in below cases:

(i) Finding Greatest number which is divisible exactly or leaves no remainder.

(ii) Finding Greatest number when on divided by numbers a, b, c  leaves Remainder’K’ in each case.

(iii) Finding Greatest number when on divided leaves Remainders x, y, z respectively.

 

 

Case(i) Steps to find greatest number, when on divided leaves no remainder:

We need to find the H.C.F of the given numbers: a, b, c which gives the greatest number when on divided by the numbers leaves no remainder.

Required Greatest number = H.C.F( a, b, c ).

 

Find the greatest number which is divisible exactly:

1) 1365, 1560, 1755.

Solution:

Given: Numbers = 1365, 1560, 1755.

We need to find the H.C.F of the numbers: H.C.F( 1365, 1560, 1755 ) = 195.

Greatest number = H.C.F( 1365, 1560, 1755 ) = 195.

 

2) 96, 528, 792

Solution:

Given: Numbers = 96, 528, 792.

We need to find the H.C.F(96, 528, 792 ) = 24.

Greatest number = H.C.F(96, 528, 792 ) = 24.

 

 

Case(ii) Steps to find Greatest number, when on divided by the numbers    a, b, c  leaves Remainder x, y, z respectively.

1. Find the Difference between the numbers and the remainders: (a – x) ,    (b – y) , (c – z ).

2. Find the H.C.F of the difference of the numbers and the remainders, which gives the required Greatest number when on divided leaves the remainders x, y, z respectively.

Greatest number = H.C.F ( a-x , b-y, c-z ).

 

1) Find the greatest number that will divide 728 ,900 leaving the remainders 8 ,4 respectively?

Solution:

Given: Numbers: a, b = 728, 900; Remainders: x, y = 8, 4.

1. Difference between the numbers and the remainders: a – x, b – y .

a – x = 728 – 8 = 720 ;  b – y = 900 – 4 = 896.

2. Find out H.C.F( 720, 896 ) = 16.

Greatest number = H.C.F( 720, 896 ) = 16.

 

2) Find the greatest number when on divided by 38, 45, 52 leaves remainders 2, 3, 4 respectively?

Solution:

Given: Numbers: a, b, c = 38, 45, 52; Remainders: x, y, z = 2, 3, 4.  

1. Difference between the numbers and the remainders: a – x , b – y , c – z ;

a – x = 38 – 2 = 36 ; b – y = 45 – 3 = 42 ; c – z = 52 – 4 = 48.

2. Find out the H.C.F( 36, 42, 48 ) = 6.

Greatest number = H.C.F( 36, 42, 48 ) = 6.

 

 

Case(iii) Steps to find the Greatest number, when on divided by the numbers a, b, c leaves same remainder ‘ K’ in each case.

1. Find the difference between the numbers: a, b, c and the remainder ’K’ :

a – K , b – k , c – k.

2. Find out the H.C.F of the difference of the numbers: a, b, c and the remainder ‘K’

i.e; H.C.F( a – K , b – K , c – K ).

Greatest Number = H.C.F( a – K , b – K , c – K ), when on divided by the numbers: a, b, c leaves the remainder ‘K’ in each case.

 

Find the greatest number when on divided by the numbers 260, 720, 1410 so as to leave the remainder 7 in each case?

Solution:

Given: Numbers: a, b, c = 260, 720, 1410; Remainder ‘K’ = 7.

1. Difference between the numbers: a – K , b – K , c – k

    a – K = 260 – 7 = 253 ; b – K = 720 – 7 = 713 ; c – K = 1410 – 7 = 1403.

2. Finding H.C.F( a-k , b-k, c-k ) = H.C.F ( 253, 713, 1403 ) = 23.

Factors of 253  = 11 * 23,

Factors of 713  = 31 * 23,

Factors of 1403 = 61 * 23.   

H.C.F ( 253, 713, 1403 ) = 23.

Greatest number = H.C.F ( 253, 713, 1403 ) = 23.

Finding Greatest and Smallest n - digit number when divided leaves Remainder'K' .

C) Greatest n – digit number and Smallest n – digit number, when divided by the numbers a, b, c leaves remainder x, y, z such that a – x = b – y = c – z = K . using L.C.M method.

(i) Steps to find Greatest n – digit number when divided by the given numbers leaves remainder ’K’ respectively.

1.We need to find the L.C.M of the given numbers: L.C.M ( a, b, c ) = L.

2. Divide the greatest n – digit number with the L.C.M ‘L’:                                          (Greatest n-digit number ÷ L.C.M),which leaves Remainder ‘R’.

3. Subtract the Remainder‘R’ from the greatest n-digit number and Add Remainder 'K' to it which gives the Greatest n – digit number when divided by the numbers a, b, c, leaves remainders x, y, z  respectively.   

Required Greatest n – digit number = (Greatest n- digit number – R) + K.   

                           

 Find the greatest number of six digits which on being divided by 6, 7, 8, 9 and 10 leaves 4, 5, 6, 7 and 8 as remainder respectively.

Solution:

Given: Numbers : a, b, c, d, e = 6, 7, 8, 9 ,10 ; 

Remainders :x, y, z, u, v = 4, 5, 6, 7, 8.  Greatest 6 digit number = 9,99,999.

Check : a-x = b-y = c-z = d-u = e-v = K

              6-4 = 7-5 = 8-6 = 9-7 = 10- 8 = 2 = K.   

1. We need to find the L.C.M of the given numbers:

L.C.M ( 6, 7, 8, 9, 10 ) = 2520 = L.

 

2. Divide the greatest 6 digit number = 999999 with the L.C.M = 2520.                          ( 999999 ÷ 2520 ), which leaves Remainder ‘R’ = 2079.

3. Subtract the Remainders ‘R’ from the Greatest n-digit number and Add Remainder'K' to it which gives required greatest n-digit number.

Required Greatest 6 digit number = (Greatest 6 digit number – R) + K.

  = 999999 – 2079 + 2 = 9,97,922.

 

(ii) Steps to find the Smallest n –digit number when on divided by the numbers a, b, c leaves remainders x, y, z; such that a – x = b – y = c – z = ‘K’.

1. We need to find the L.C.M of the numbers; L.C.M(a, b, c) = L.

2. Divide the Smallest n – digit number with L.C.M :                                                    (Smallest n–digit number ÷ L.C.M), which leaves Remainder ‘R’.   

3. Subtract the Remainder ‘R’ from L.C.M ‘L’  : L – R .

4. Add the Remainder ‘K’ and ( L – R ) to the Smallest n – digit number which gives the required Smallest n – digit number , when on divided by the numbers a, b, c leaves remainders x, y, z respectively.

Required Smallest n–digit number = Smallest n-digit number + (L–R) + K.

 

 

Find the least number of 4 digits when on divided by the numbers 12, 16, 18 and 20 leaves the remainder 21 in each case?

Solution:

Given : Numbers = 12, 16, 18, 20 ; Remainder ‘K’ = 21.

Smallest 4 digit number = 1000.

1. We need to find the L.C.M of the numbers:

L.C.M ( 12, 16, 18, 20 ) = 720 = L.

2. Divide the Smallest 4 digit number = 1000 with L.C.M ’L’ = 720;                            ( 1000 ÷ 720 ) , which leaves Remainder ‘R’ = 280.

3. Subtract Remainder ‘R’ = 280 from L.C.M ‘L‘ = 720 :

 L – R = 720 – 280 = 440.

4. Add the Remainder ‘K’ and ( L – R ) to the Smallest 4 digit number = 1000  , gives the required Smallest 4 digit number, when divided by the given numbers leaves remainder 21 reapectively.

Required Smallest 4 digit number = Smallest 6 digit number + (L- R) + K.

  = 1000 + 440 + 21 = 1461.   

Finding the Smallest n- digit number which is divisible exactly.

B) ii) Smallest n – digit number when divided leaves no remainder using L.C.M:

Steps to find Smallest n – digit number, which is exactly divisible by the given numbers(leaves no remainder):

1. Find out the L.C.M of the given numbers, ‘L’ .

2. Divide the smallest n– digit number with the L.C.M,

(Smallest n-digit ÷ L.C.M )  which leaves Remainder,’R’.

3. Find out the difference between L.C.M,’L’ and Remainder’R’: ( L – R ); which is to be added to the Smallest n-digit number gives the required Smallest n-digit number.

Required Smallest n-digit number = Smallest n-digit number + (L – R).

 

1) Find the least no. of 5 digits which is exactly divisible by 4, 12, 15 and 18 is?

Solution:

Given : Numbers = 4, 12, 15, 18 ; Least 5 digit number = 10000.

1. Find out the L.C.M,L of the given numbers;L.C.M( 4, 12, 15, 18 ) =  180.

2. Divide the Smallest 5 digit number,10000 with the L.C.M,L : 180 ;                           ( 10000 ÷  180 ) which leaves Remainder R = 100.

3. Subtract Remainder R =100  from L.C.M ‘L’= 180.

i.e;  L – R = 180 – 100 = 80 , which is to be added to the Smallest 5 digit number 10000 gives the required Smallest 5 digit number.

Required Smallest 5 digit number = Smallest 5 digit number + ( L – R ).

 = 10000 + 80 = 10,080, which is exactly divisible by the given numbers.                     

 

 

2) Find the smallest 3-digit number, such that they are exactly divisible by 3, 4 and 5?

Solution:

Given : Numbers = 3, 4, 5 ; Smallest 3 digit number = 100.
1.We need to find the L.C.M of the given numbers; L.C.M(3, 4, 5) = 60.

As the given numbers are Consecutive numbers, the L.C.M of the numbers is the product of the given consecutive numbers.

Therefore, L.C.M( 3, 4, 5 ) = 3 * 4 * 5 = 60 = L.

2. Divide the Smallest 3 digit number: 100, with the L.C.M,L = 60 :

 ( 100 ÷  60 ) , which leaves Remainder R = 40.

3. Find the difference between L.C.M and Remainder: L – R = 60 – 40 =20 , which is to be added to the Smallest 3 digit number, gives the Required Smallest 3 digit number.

Required Smallest 3 digit number = Smallest 3 digit number + ( L – R ).

 = 100 + 20 = 120, which is divisible exactly by the given numbers.

 

3) Find the least number of 6 digits which is exactly divisible by 30, 36, 48, 72 is?

Solution:

Given: Numbers = 30, 36, 48, 72; Smallest 6 digit number = 100000.

 

1. We need to find out L.C.M of the given numbers:

L.C.M ( 30, 36, 48, 72 ) = 720 = L.

 

 

2. Divide the Smallest 6 digit number =100000 , with the L.C.M ‘L’ = 720.

( 100000 ÷  720 ) , which leaves Remainder ‘R’ = 640.

 3.Subtract the Remainder ‘R’ = 640 from the L.C.M ’L’ = 720 .

L – R = 720 – 640 = 80 , which is to be added to the Smallest 6 digit number 100000 to get the required Smallest 6 digit number which is divisible eactly by the given numbers.

 Required Smallest 6 digit number = Smallest 6 digit number + ( L – R ).

  = 100000 + 80 = 1,00,080, which is divisible exactly by the given numbers.