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In a cup of saline water (salt water), the mass ratio of salt to water is 1:10. After adding 22g of salt, the mass ratio of salt to water becomes 2:9. how many grams of saline water were there originally.

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Let's assume the original mass of the saline water (salt water) is x grams. According to the given information, the mass ratio of salt to water is initially 1:10.

This means the mass of salt in the saline water is (1/11) * x grams, and the mass of water is (10/11) * x grams.

After adding 22 grams of salt, the mass ratio of salt to water becomes 2:9.

This means the mass of salt is now (2/11) * (x + 22) grams, and the mass of water is (9/11) * (x + 22) grams.

Since the mass of salt remains the same in both scenarios, we can set up an equation:

(1/11) * x = (2/11) * (x + 22)

Simplifying the equation:

x = 2 * (x + 22)

x = 2x + 44

x - 2x = 44

-x = 44

x = -44 (disregard this negative value)

Since the value of x is negative, it is not a valid solution in this context.

It seems there may be an error or inconsistency in the given information or calculations. Please double-check the values and ratios provided to ensure accuracy.

User Gianni Carlo
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