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30 Jun 2021, 03:45
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D
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Difficulty:
45%
(medium)
Question Stats:
65% (01:40) correct
35%
(01:47)
wrong
based on 322
sessions
History
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The rate of chemical reaction X is inversely proportional to the square of the rate of chemical reaction Y. If the rate of chemical reaction Y increases by 100%, by what percent does the rate of chemical reaction X decrease?
A. 50%
B. 60%
C. 70%
D. 75%
E. 80%
A. 50%
B. 60%
C. 70%
D. 75%
E. 80%
04 Jul 2021, 23:16
Kudos
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Rate of reaction X = \(\frac{1}{Y^2}\)
After 100% increase in Y =>
Rate of reaction X = \(\frac{1}{(2Y)^2}\)
Decrease => \(\frac{1}{Y^2}-\frac{1}{(2Y)^2}\) = \(\frac{3}{4Y^2}\)
Percentage Decrease => 100 * \(\frac{3}{4Y^2}\) / \(\frac{1}{Y^2}\) = 75%
IMO D.
After 100% increase in Y =>
Rate of reaction X = \(\frac{1}{(2Y)^2}\)
Decrease => \(\frac{1}{Y^2}-\frac{1}{(2Y)^2}\) = \(\frac{3}{4Y^2}\)
Percentage Decrease => 100 * \(\frac{3}{4Y^2}\) / \(\frac{1}{Y^2}\) = 75%
IMO D.
08 Jul 2021, 04:44
Kudos
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Bunuel
Solution -
The rate of chemical reaction X is inversely proportional to the square of the rate of chemical reaction Y –
So, this can be represented as:
Rate of Chemical Reaction \(X = \frac{1}{Y^2} \)
Rate of Chemical Reaction Y increases by 100% which implies, Y + 100% of Y = 2Y
Since the rate of chemical reaction X and rate of chemical reaction Y are inversely proportional, we can represent the new rate as:
Rate of Chemical Reaction \(X = \frac{1}{〖(2Y)〗^2}\)
So, the decrease in the rate of chemical reaction X = \(\frac{1}{Y^2} - \frac{1}{(2Y)^2} = \frac{3}{〖4Y〗^2} \)
Percentage decrease in the rate of chemical reaction \(X = \frac{(3⁄〖4Y〗^2 )}{(1⁄Y^2 )} ×100 =\frac{3}{4}×100=75 percent \)
Answer Choice-D
