Estimation:
23 --> rounded to 20
71 --> rounded to 70
70 + 20 = 90. I rounded both numbers down so the exact sum must be only a little greater than my estimate.
Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.
We are given that, the final solution should be 42 ml, this means that:
x + y = 42 ...........> equation I
This can also be written as:
x = 42-y .......> equation II
We are also given that the final concentration should be 45%, this means that:
5% x + 65% y = 45% (x+y)
0.05x + 0.65y = 0.45(x+y)
We have x+y = 42 from equation I, therefore:
0.05x + 0.65y = 0.45(42)
0.05x + 0.65y = 18.9 .........> equation III
Substitute with equation II in equation III as follows:
0.05x + 0.65y = 18.9
0.05(42-y) + 0.65y = 18.9
2.1 - 0.05y + 0.65y = 18.9
0.6y = 18.9 - 2.1
0.6y = 16.8
y = 28 ml
Substitute with y in equation II to get x as follows:
x = 42-y
x = 42 - 28
x = 14 ml
Based on the above calculations:
amount of 5% solution = x = 14 ml
amount of 65% solution = y = 28 ml
The correct choice is:
The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
Answer:
316.25
I hope I Got The Right Answer
3/9<3/8<3/7.
When we have fractions with same numerator, and and different denominators, the fraction that have the lowest denomiator will be the greatest while the one with the highest denomiator will be the smallest.
Answer:
This is the answer of your question.
