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Dr. Xman is diluting concentrated HCl acid by mixing water in it. The quantity of mixture is 32 l. The ratio of concentrated HCl to water is 7 ∶ 1. If 3 l of water is added to the mixture, the ratio of HCl to water changes to new ratio. If he want to change back the new ratio to the original value, how much concentrated HCl is to be added?

Dr. Xman is diluting concentrated HCl acid by mixing water in it. The quantity of mixture is 32 l. The ratio of concentrated HCl to water is 7 ∶ 1. If 3 l of water is added to the mixture, the ratio of HCl to water changes to new ratio. If he want to change back the new ratio to the original value, how much concentrated HCl is to be added? Correct Answer 21 L

Given∶

Quantity of mixture = 32 L.

Ratio of Conc. HCl to water = 7 ∶ 1

Extra 3L of water is added to mixture.

Formula used∶

Simple Concept of Ratios

a ∶ b can be written as ax and bx where x is a positive integer.

Calculation∶

Initial ratio of conc. HCl ∶ water = 7 ∶ 1

Let the quantity of water be x and HCl be 7x.

According to Question,

7x + x = 32

⇒ 8x = 32

⇒ x = 4

In original mixture,

Quantity of HCl = 7x = 7 × 4 = 28 L

Quantity of water = x = 4 L

After 3L of water is added to the mixture,

Quantity of HCl = 28 L

Quantity of water = 4 + 3 = 7 L

New ratio of HCl ∶ water = 7 ∶ 28 = 1 ∶ 4

Original ratio = 7 ∶ 1

Let x be added to the new quantity of HCl

Thus,

New quantity of HCl / new quantity of water = 7/1

(28 + x)/7 = 7/1

⇒ x = 49 - 28 = 21 

∴ 21 L of HCl be added.

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