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kipiarov [429]
3 years ago
11

Ella grew 3 over 4 of an inch last year. Ben grew 2 over 3 of an inch last year. How much more did Ella grow than Ben last year?

Mathematics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

Ella grew \frac{1}{12} of an inch more

Step-by-step explanation:

For this problem, you want to subtract Ben's growth from Ella's.

This expression would be \frac{3}{4}-\frac{2}{3}.

To work these out, they must have the same denominator (which should be the lowest common multiple or LCM).

In this case, the LCM is 12, so you want to multiply each side so the denominator is 12. This means the first fraction should be multiplied by \frac{3}{3} and the second by \frac{4}{4}.

This makes the expression \frac{9}{12}-\frac{8}{12}, which equals \frac{1}{12}.

**This content involves adding and subtracting fractions, which you may want to revise. I'm always happy to help!

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A chemist whishes to prepare 100 liters of 45% purity of sulphuric acid .He has two kinds of acid solutions in stock ,one is 55%
tester [92]

Answer:

the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

Step-by-step explanation:

From the given information,

Let x be the litres of 55% pure solution

Let y be the litres of 30% pure solution

Also;

Given that our total volume of solution is  100 litres

x+y =100  ---- (1)

The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.

(0.55)(x)+(0.30)(y) = 0.45(100) ---- (2)

From equation (1)

Let ; y = 100 - x

Replacing the value for y = 100 - x into equation (2)

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0.55x + 30 - 0.30x = 45

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x = 15/0.25

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From ; y = 100 - x

y = 100 - 60

y = 40  litres of 30% solution.

Therefore, the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

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