Dividing a Quantity into Three Parts in a Given Ratio

We will discuss here how to solve different types of word problems on dividing a quantity into three parts in a given ratio.


1. Divide $ 5405 among three children in the ratio 1\(\frac{1}{2}\) : 2 : 1\(\frac{1}{5}\).

Solution:

Given ratio = 1\(\frac{1}{2}\) : 2 : 1\(\frac{1}{5}\)

                = \(\frac{3}{2}\) : 2 : \(\frac{6}{5}\)

Now multiply each term by the L.C.M. of the denominators

                 = \(\frac{3}{2}\) × 10 : 2 × 10 : \(\frac{6}{5}\) × 10, [Since, L.C.M. of 2 and 5 = 10]

                 = 15 : 20 : 12

So, the amount received by three children are 15x, 20x and 12x.

15x + 20x + 12x = 5405

              ⟹ 47x = 5405

                 ⟹ x = \(\frac{5405}{47}\)

      Therefore, x = 115

Now,

15x = 15 × 115 = $ 1725

20x = 20 × 115 = $ 2300

12x = 12 × 115 = $ 1380

Therefore, amount received by three children are $ 1725, $ 2300 and $ 1380.



2. A certain sum of money is divided into three parts in the ratio 2 : 5 : 7. If the third part is $224, find the total amount, the first part and second part.

Solution:

Let the amounts be 2x, 5x and 7x

According to the problem,

        7x = 224

      ⟹ x = \(\frac{224}{7}\)

Hence, x = 32

Therefore, 2x = 2 × 32 = 64 and 5x = 5 × 32 =160.

So, the first amount = $ 64 and the second amount = $ 160

Hence, total amount = First amount + Second amount + Third amount

                             = $ 64 + $ 160 + $ 224

                             = $ 448



3.  A bag contains $ 60 of which some are 50 cent coins, some are $ 1 coins and the rest are $ 2 coins. The ratio of the number of respective coins is 8 : 6 : 5. Find the total number of coins in the bag.

Solution:

Let the number of coins be a, b and c respectively.

Then, a : b : c is equal to 8 : 6 : 5

Therefore,  a = 8x, b = 6x, c = 5x 

Therefore, the total sum = 8x × 50 cent + 6x × $ 1 + 5x × $ 2

                                  = $ (8x × \(\frac{1}{2}\) + 6x ×  1 + 5x ×  2)

                                  = $ (4x + 6x + 10x)

                                  = $ 20x

Therefore, according to the problem,

$ 20x = $ 60

  ⟹ x = \(\frac{$ 60}{$ 20}\)

  ⟹ x = 3

Now, the number of 50 cent coins = 8x = 8 × 3 = 24

The number of $ 1 coins = 6x = 6 × 3 = 18

The number of $ 2 coins = 5x = 5 × 3 = 15

Therefore, the total number of coins =  24 + 18 + 15 = 57.



4. A bag contains $ 2, $ 5 and 50 cent coins in the ratio 8 : 7 : 9. The total amount is $ 555. Find the number of each denomination.

Solution:

Let the number of each denomination be 8x , 7x and 9x respectively.

The amount of $ 2 coins = 8x × 200 cents = 1600x cents

The amount of $ 5 coins = 7x × 500 cents = 3500x cents

The amount of 50 cent coins = 9x × 50 cents = 450x cents

The total amount given = 555 × 100 cents = 55500 cents

Therefore, 1600x + 3500x + 450x = 55500

                                  ⟹ 5550x = 55500

                                         ⟹ x = \(\frac{55500}{5550}\)

                                         ⟹ x = 10

Therefore, the number of $ 2 coins = 8 × 10 = 80

The number of $ 5 coins = 7 × 10 = 70

The number of 50 cent coins = 9 × 10 = 90

 










10th Grade Math

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