Question

A small jar contained water, lime, and sugar in the ratio of 90:7:3. A glass contained only water and sugar in it. Contents of both (small jar and glass) were mixed in a bigger jar and the ratio of contents in the bigger jar was 85:5:10 (water, lime, and sugar respectively). Find the percentage of water in the bigger jar?

• A. 70
• B. 75
• C. 80
• D. 72.5
• E. 85

Answer 1

Answer: (E)

To solve this problem, we need to find the percentage of water in the bigger jar when the contents of the small jar and the glass are combined.

Let's break down the problem step-by-step:

Initially, the contents of the small jar are given in the ratio \(90:7:3\) for water, lime, and sugar, respectively.

The glass contains only water and sugar. When both the small jar and the glass are mixed in the bigger jar, the ratio becomes \(85:5:10\) for water, lime, and sugar, respectively.

To find the percentage of water in the bigger jar, we will determine the composition of the original quantities of each component and calculate their new proportions.

Let the volume of the small jar be \(x\) units, then:

• Water in small jar = \(\frac{90}{100}x\)
• Lime in small jar = \(\frac{7}{100}x\)
• Sugar in small jar = \(\frac{3}{100}x\)

Let the volume of the glass be \(y\) units, then it contains only water and sugar.

• Total water contribution from both containers is the sum of the water from the small jar and the water in the glass.
• Total sugar contribution from both containers is the sum of the sugar from the small jar and sugar in the glass, whereas lime contribution remains constant.

According to the final mixture ratio in the bigger jar \(85:5:10\):

Let W, L, S be the respective quantities of water, lime, and sugar. Hence, the equation system is:

• \(\frac{W}{L} = \frac{85}{5}\)
• \(\frac{S}{L} = \frac{10}{5}\)

From the lime ratio, solve for total weight proportions:

\(W + L + S = 100\)

Since \(\frac{W}{L} = \frac{85}{5}\) or \(W = 17L\), also, substitute \(S = 2L\) from the sugar ratio.

Now we have:

\(17L + L + 2L = 100\)

\(20L = 100\)

Thus, \(L = 5\), hence find W:

\(W = 17 \times 5 = 85\)

Therefore, the percentage of water in the bigger jar is:\(\frac{85}{100} \times 100 = 85\%\).

Thus, the correct answer is 85%.

Answer 2

The problem involves mixing contents from two sources to determine the percentage of water in a final mixture. We are given information about the initial ratios and the final ratio after mixing. We need to solve for the percentage of water in the final mixture labeled as the "bigger jar".

First, let's consider the ratios. In the small jar, the ratio of water, lime, and sugar is 90:7:3. Let the common multiple be x. Therefore, the quantity of water in the small jar is 90x, lime is 7x, and sugar is 3x.

In the glass, we have only water and sugar. Since the contents are mixed such that the final ratio in the bigger jar becomes 85:5:10, no lime from the glass is needed. Hence, the entire lime part comes from the small jar, making it 7x.

Let the quantities in the glass be represented by y (water) and z (sugar). Thus, the equations based on the final ratio are as follows:

1. Water: 90x + y
2. Lime: 7x
3. Sugar: 3x + z

These need to add up to the total as per the final ratio 85:5:10.

The final ratio represents:
(90x + y) as 85 parts
7x as 5 parts
(3x + z) as 10 parts

We start with the lime component:7x = 5 parts

Solve for x:x = (5 / 7)

Now, substitute x in the ratios for water and sugar:

1. For water:90x + y = 85 parts
   90(5/7) + y = 85
   450/7 + y = 85
   y = 85 - 450/7
   y = (595/7) - (450/7)
   y = 145/7
2. For sugar:3x + z = 10 parts
   3(5/7) + z = 10
   15/7 + z = 10
   z = 10 - 15/7
   z = 70/7 - 15/7
   z = 55/7

Now, calculate the percentage of water:

Total parts = 85 + 5 + 10 = 100.

Percentage of water = (Water parts / Total parts) * 100% 
Percentage of water = (85 / 100) * 100% = 85%

Thus, the percentage of water in the bigger jar is 85%.