Problems on Partnership, Mixture and Alligations

Worked out problems on partnership, mixture and alligations are explained here with the step by step explanations. Follow the examples to get a fair idea on how to solve these word problems.

Word problems on partnership, mixture and alligations:

1. In what ratio must a grocer-man mix two varieties of rice @ $2.02/kg and $2.44/kg so as to make a mixture worth $2.26/kg?

Solution:

Quantity of cheaper/Quantity of dearer

= (2.44 – 2.26)/(2.26 – 2.02)

= 0.18/0.24

= 18/24

= 3/4

The above relationship can also be represented as under:

Problems on Partnership Math

Therefore the required ratio = 3 : 4.


2. Three friends X, Y and Z started a business by investing amount in the ratio of 5 : 7 : 6 respectively. After a period of six months Z withdrew half of the amount invested by him. If the amount invested by X is $4000 and the total profit earned at the end of one year is $3300, what is Z’s share in profit?

Solution:

Let the amount invested be 5x, 7x, 6x

5x = 4000

x = 4000/5

Therefore, x = 800

If Y invested 7x then 7 × 800 = 5600

If Z invested 6x then 6 × 800 = 4800

In one year the amount invested by the three friends;

4000 × 12 : 5600 × 12 : 4800 × 6 + 2400 × 6

48000 : 67200 : 28800 + 14400

48000 : 67200 : 43200

Now we will reduce to its lowest terms, divide each by 4800;

10 : 14 : 9

We know, the total profit earned at the end of one year is $3300.

Know we will calculate the Z’s share in profit

= 9/(10 + 14 + 9) × 3300

= 9/33 × 3300

= 9 × 100

= 900

Therefore, Z’s share in profit = $900.


3. A grocer buys two varieties of pulses at $1.80 and $1.20 per kg respectively. In what ratio should these be mixed, so that by selling the mixture at $1.75 per kg., when 25% may be gained by the grocer?

Solution:

Cost price of the mixture of pulses

= (S.P × 100)/(100 + gain %)

= (1.75 × 100)/(100 + 25%)

= 175/125

= $1.40 per kg

Word problems on partnership, mixture

Therefore the required ratio = 0.4 : 0.2 = 2 : 1.

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