Question

three bottles of chemical weigh in the ratio 6:7:8.If the ratio of the weights must be changed to 8:7:6,then the chemicals at the first two must be increased by most fractions of themselves​

Answer 1

Given:

three bottles of chemical weigh in the ratio 6:7:8. If the ratio of the weights must be changed to 8:7:6,

To Find:

then the chemicals at the first two must be increased by most fractions of themselves​

Solution:

The given ratio is 6:7:8, first multiply the ratio by 10, which gives us

60:70:80

Now we need to change it to 8:7:6, keeping 80 as constant let change the ratio to 8:7:6. Now 80/6=40/3

And multiplying it with the ratio we have

$\frac{40}{3}*8:\frac{40}{3}*7:\frac{40}{3}*6$

So it is converted into the ratio of 8:7:6

Now the 1st fraction by which it changed will be,

$Fraction change=\frac{\frac{320}{3}-60 }{60} \\=\frac{7}{9}$

Now the 2nd fraction by which it changed will be,

$Fractionchange=\frac{\frac{280}{3}-70 }{70}\\=\frac{1}{3}$

Hence, the chemicals increased by a 7/9 fraction for the 1st ratio and for the 2nd they changed by 1/3 of themselves.

Answer 2

First chemical  is to  be increased by 7/9  of itself and second chemical  is to  be increased by 1/3 if 6:7:8 ratio to be changed to 8:7:6

Given:

• Three bottles of chemical weigh in the ratio 6:7:8.
• Ratio of the weights must be changed to 8:7:6

To Find:

• First two chemicals must be increased by  fractions of themselves

Solution:

Given Ratio

6 : 7 : 8

Multiply by 3

18 : 21 : 24

New Required ratio  

8 : 7 : 6

Multiply by 4

32 : 28 : 24

Old Ratio  18 : 21 : 24

New Ratio 32 : 28 : 24

No change in 3rd Chemical Hence,

First chemical 18 has to be made 32

and Second chemical  21 has to be made 28

First chemical  is to  be increased by ( 32 - 18)/18

= 14/18

= 7/9

Second chemical  is to  be increased by ( 28 - 21)/21

= 7/21

= 1/3

First chemical  is to  be increased by 7/9  of itself and second chemical  is to  be increased by 1/3 if 6:7:8 ratio to be changed to 8:7:6

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