Question

Two types of liquids having the rates of ₹ 8/kg and ₹ 10/kg respectively are mixed in order to produce a mixture having the rate of ₹ 9.20/kg. What should be the amount of the second type of liquid if the amount of the first type of liquid in the mixture is 20 kg?

Answer 1

Answer:

$\huge \tt \colorbox{green}{✍Solution}$

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Given , First Liquid (L1) = ₹ 8/kg

And , Second Liquid (L2) = ₹ 10/kg

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Amount of first type of liquid in the mixture (M1) = 20kg

Amount of M2 = ?

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So , By balancing method:-

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L1×M1 + L2×M2 = Mixture rate × (M1 + M2)

=> 8×20 + 10× M2 = 9.2(20+M2)

=> 160 + 10M2 = 184 + 9.2 M2

=> 10M2 - 9.2M2 = 184-160

=> 0.8M2 = 24

=> M2 = 24/0.8 ×10

=> M2 = 30 kg

Therefore, the amount of the second type of liquid is 30kg .

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Answer 2

Answer:

The amount of the second type of liquid in the mixture is 30 kg.

Explanation:

Given:

The price of the first liquid, $L_{1}$ = ₹ $8/kg$

The price of the second liquid, $L_{2}$ = ₹ $10/kg$

The amount of the first type of liquid in the mixture, $M_{1}$ = $20 \ kg$

To find:

We have to find the amount of the second type of liquid in the mixture.

$M_{2}$ =?

Solution:

By using the method of balancing, we get:

$L_{1} \times M_{1} + L_{2} \times M_{2} = Mixture \ rate \times (M_{1} + M_{2} )$

Substituting the values in the above equation we get,

$8 \times 20 + 10 \times M_{2} = 9.2 (20+M_{2} )$

$160 + 10M_{2} = 184 + 9.2M_{2}$

$10M_{2} - 9.2M_{2} = 184-160$

$0.8M_{2} = 24$

$M_{2} = \frac{24}{0.8} \times 10$

$M_{2} = 30 \ kg$

Final Answer:

The amount of the second type of liquid in the mixture is 30 kg.

Link:

https://brainly.in/question/14581630?referrer=searchResults

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