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1 year ago

How much water do you add to make 12% sugar syrup using 5g sugar?

Mathematics

2 answers:

Alchen [17]1 year ago

8 0

so the syrup is just water and sugar, we know that the sugar percent is just 12%, and that happens to be 5grams.

now 100% - 12% is 88%, so the 88% leftover will just be water.

if 12% is 5g, how much is 88%?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 5&12\\ x&88 \end{array}\implies \cfrac{5}{x}=\cfrac{12}{88}\implies \cfrac{5}{x}=\cfrac{3}{22}\implies 110=3x \\\\\\ \cfrac{110}{3}=x\implies 36\frac{2}{3}=x$

Gnoma [55]1 year ago

6 0

Answer:

$36\dfrac{2}{3}$ g of water.

Step-by-step explanation:

You know that 5 g of sugar form 12%. Let x g be the amount of water needed. The total amount of syrup will be x+5 g. Then

5 g - 12%,

x+5 g - 100%.

Mathematically, you can write a proportion:

$\dfrac{5}{x+5}=\dfrac{12}{100}\Rightarrow 500=12(x+5),\ 500=12x+60,\ 12x=440, \\ \\x=\dfrac{440}{12}=\dfrac{110}{3}=36\dfrac{2}{3}.$

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Gary spent $48 on a watch. He spent 1/3 of the remainder on a pen if he still had 1/2 of his money left,how much money did he ha

xeze [42]

T= total
Money spent is: 
48+ 1/3 (T-48) = 1/2 T
One equation for one unknown, solve:

48 + 1/3 T -1/3*48= 1/2 T
48 - 16 =1/2 T- 1/3 T
32= 1/6 T
T= 6*32
T=192

192-48=144 money left after watch
1/3*144=48 third of leftover, spent on pen
144-48=96 left which is half of 192.

6 0

2 years ago

Read 2 more answers

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .