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Joined: 23 Jan 2021
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In the first year, the sales of a store is a. Its sales incr
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26 Jun 2021, 23:57
26 Jun 2021, 23:57
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In the first year, the sales of a store is a. Its sales increase by x% every two years, and reach 1.4a in the fifth year.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
20% |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
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In the first year, the sales of a store is a. Its sales incr
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27 Jun 2021, 02:34
27 Jun 2021, 02:34
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kumarneupane4344 wrote:
In the first year, the sales of a store is a. Its sales increase by x% every two years, and reach 1.4a in the fifth year.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
20% |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
If we assume that the sale increases by some percent each year;
Sale First year = \(a\)
Sale Second year = \((1 + \frac{x}{100})a\)
Sale Third year > Sale Second year
Sale Fourth year = \((1 + \frac{x}{100})(1 + \frac{x}{100})a\)
Sale Fifth year = \(1.4a\)
Sale Fifth year > Sale Fourth year
\(1.4a > (1 + \frac{x}{100})^2a\)
\(1.4 > (1 + \frac{x}{100})^2\)
\((1.183)^2 > (1 + \frac{x}{100})^2\)
\((1 + 0.183)^2 > (1 + \frac{x}{100})^2\)
On comparison;
\(0.183 > \frac{x}{100}\)
\(18.3 > x\)
Col. A < Col. B
But, If we assume that sale increase by x% exactly after 2 years;
Sale First year = \(a\)
Sale Second year = \((1 + \frac{x}{100})a\)
Sale Third year = \((1 + \frac{x}{100})a\)
Sale Fourth year = \((1 + \frac{x}{100})(1 + \frac{x}{100})a\)
Sale Fifth year = \(1.4a = (1 + \frac{x}{100})(1 + \frac{x}{100})a\)
So, Sale Fifth year = Sale Fourth year
\((1.183)^2 = (1 + \frac{x}{100})^2\)
\((1 + 0.183)^2 = (1 + \frac{x}{100})^2\)
On comparison;
\(0.183 = \frac{x}{100}\)
\(18.3 = x\)
Col. A < Col. B
Hence, option B
Re: In the first year, the sales of a store is a. Its sales incr
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27 Jun 2021, 03:18
27 Jun 2021, 03:18
1
For the second one that you explained A=B, you have assumed that 1.4=(1.2)2 which is not the case. It is (1.18)2 so the expression would be 1+0.18 and then x would translate to 18% in which case B is greater than A even in this case. So B is the correct answer.
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