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ekwok
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 28 Dec 2023, 21:35
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

(A) \((1 + \frac{1}{100})(1 + \frac{1}{25})\)

(B) \((1 - \frac{1}{100})(1 - \frac{1}{25})\)

(C) \((1 + \frac{1}{10})(1 + \frac{1}{5})\)

(D) \((1 - \frac{1}{10})(1 - \frac{1}{5})\)

(E) \(1\)
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 28 Dec 2023, 22:24
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ekwok
Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

A) (1 + \(\frac{1}{100}\))(1 + \(\frac{1}{25}\))
B) (1 - \(\frac{1}{100}\))(1 - \(\frac{1}{25}\))
C) (1 + \(\frac{1}{10}\))(1 + \(\frac{1}{5}\))
D) (1 - \(\frac{1}{10}\))(1 - \(\frac{1}{5}\))
E) 1
Show more

Given:

  • B = 90% of A
  • C = 110% of B
  • D = 80% of C
  • E = 120 % D


E =\( \frac{120}{100} * \frac{80}{100} * \frac{110}{100} * \frac{90}{100} * A\)

Show Spoiler
E = \(\frac{120}{100} * D\)

E = \(\frac{120}{100} * \frac{80}{100} * C\)

E = \(\frac{120}{100} * \frac{80}{100} * \frac{110}{100} * B\)

E = \(\frac{120}{100} * \frac{80}{100} * \frac{110}{100} * \frac{90}{100} * A\)

\(\frac{E}{A} = \frac{120}{100} * \frac{80}{100} * \frac{110}{100} * \frac{90}{100}\)

\(\frac{E}{A} = \frac{12}{10} * \frac{8}{10} * \frac{11}{10} * \frac{9}{10}\)

\(\frac{E}{A} = \frac{6}{5} * \frac{4}{5} * \frac{11}{10} * \frac{9}{10}\)

\(\frac{E}{A} = \frac{6*4}{5*5} * \frac{11*9}{10*10}\)

\(\frac{E}{A} = \frac{24}{25} * \frac{99}{100}\)

From the options given below

Option B ⇒ (1 - \(\frac{1}{100}\))(1 - \(\frac{1}{25}\))

\((\frac{100-1}{100})(\frac{25-1}{25}) = (\frac{99}{100})(\frac{24}{25})\)

This matches the value obtained above

Option B
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 28 Dec 2023, 23:02
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ekwok
Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

A) (1 + \(\frac{1}{100}\))(1 + \(\frac{1}{25}\))
B) (1 - \(\frac{1}{100}\))(1 - \(\frac{1}{25}\))
C) (1 + \(\frac{1}{10}\))(1 + \(\frac{1}{5}\))
D) (1 - \(\frac{1}{10}\))(1 - \(\frac{1}{5}\))
E) 1
Show more

Algebraic

If you realize that we have two sets of number equidistant from 100, we should be tempted to use \((a-b)(a+b)=a^2-b^2\)
\(E=(1+\frac{20}{100})D\)
\(D=(1-\frac{20}{100})C\)
\(C=(1+\frac{10}{100})B\)
\(B=(1-\frac{10}{100})A\)

\(E=(1+\frac{20}{100})(1-\frac{20}{100})(1+\frac{10}{100})(1-\frac{10}{100})A\)
\(E=(1^2-(\frac{2}{10})^2)(1^2-(\frac{1}{10})^2)A\)
\(E=(1^2-(\frac{1}{5})^2)(1^2-(\frac{1}{10})^2)A\)
\(E=(1-\frac{1}{25})(1-\frac{1}{100})A\)
\(\frac{E}{A}=(1-\frac{1}{25})(1-\frac{1}{100})\)

OR

E=1.2D=1.2*0.8*C=1.2*0.8*1.1B=1.2*0.8*1.1*0.9A=0.96*0.99A=(1-0.04)(1-0.1)A
\(E=(1-\frac{4}{100})(1-\frac{1}{100})A\)
\(\frac{E}{A}=(1-\frac{1}{25})(1-\frac{1}{100})\)

B


Approximate
\(E=1.2D=1.2*0.8*C=1.2*0.8*1.1B=1.2*0.8*1.1*0.9A=0.96*0.99A \)
\(=> 0.90\leq \frac{E}{A}\leq 1\)
Discard A and C as they are greater than 1, and discard E as it is 1.
D is 0.9*0.8 = 0.72...Discard
Only B left
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 28 Dec 2023, 22:21
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ekwok
Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

A) (1 + \(\frac{1}{100}\))(1 + \(\frac{1}{25}\))
B) (1 - \(\frac{1}{100}\))(1 - \(\frac{1}{25}\))
C) (1 + \(\frac{1}{10}\))(1 + \(\frac{1}{5}\))
D) (1 - \(\frac{1}{10}\))(1 - \(\frac{1}{5}\))
E) 1
Show more

Let A = 100
B= 90
C= 99
D= 79.2
E = 1.2*79.2 = 95.04

\(\frac{E}{A}\)=0.9504

Eliminate Option A, C and E ( all these are greater than or equal to 1)

D) (1 - \(\frac{1}{10}\))(1 - \(\frac{1}{5}\)) = 0.88 <\(\frac{E}{A}\)
Remaining Option B is the answer.

Choice B
B) (1 - \(\frac{1}{100}\))(1 - \(\frac{1}{25}\))
\((\frac{99}{100})*(\frac{24}{25})\)
\((\frac{99}{100})*(\frac{96}{100})\)
On solving this, we get 0.9504 = \(\frac{E}{A}\)
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 02 Feb 2025, 07:49
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[E][/A]
E= 120% x 80% x 110% x 90% A

in the division [E][/A] A will be cancel and we will have 120% x 80% x 110% x 90%
= (1+20%)(1-20%)(1-10%)(1+10%)

applying exponents theory

= (1- (20%)^2) (1- (10%)^2)
= (1- 20% x 20%) (1- 10% x 10%)
= (1-4%) (1-1%)

reexpressing in fractions:
= (1- [4][/100]) (1-[1][/100])
= (1- [1][/25]) (1-[1][/100])

Answer choice B
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 12 May 2025, 20:55
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

Reducing a number by 10 percent and then increasing the resulting smaller number by 10 percent results in a final number that is smaller than the original.

Similarly, reducing a number by 20 percent and then increasing the resulting smaller number by 20 percent results in a final number that is smaller than the original.

So, the four steps must result in a final number E that is smaller than A. Thus, only choices (B) and (D) are possible.

Choice (D) reduces the number by 10 percent and then 20 percent without the increases of 10 percent and 20 percent included among the four steps to E.

So, (D) is incorrect, and the correct answer must be (B).

(A) \((1 + \frac{1}{100})(1 + \frac{1}{25})\)

(B) \((1 - \frac{1}{100})(1 - \frac{1}{25})\)

(C) \((1 + \frac{1}{10})(1 + \frac{1}{5})\)

(D) \((1 - \frac{1}{10})(1 - \frac{1}{5})\)

(E) \(1\)

Correct answer: B
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Let A be a positive real number. If B is 90% of A, C is 110% of B, D 13 May 2025, 05:35
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ekwok
Let A be a positive real number. If B is 90% of A, C is 110% of B, D is 80% of C, and E is 120% of D, which of the following is equal to \(\frac{E}{A}\)?

(A) \((1 + \frac{1}{100})(1 + \frac{1}{25})\)

(B) \((1 - \frac{1}{100})(1 - \frac{1}{25})\)

(C) \((1 + \frac{1}{10})(1 + \frac{1}{5})\)

(D) \((1 - \frac{1}{10})(1 - \frac{1}{5})\)

(E) \(1\)
Show more

Let, A = 1000
then B = 90% of A = 900
then C = 110% of B = 1.1*900 = 990
then D = 80% of C = 0.8*990 = (4/5)*990
then E = 120% of D = 1.2*(4/5)*990 = (6/5)*(4/5)*990

\(\frac{E}{A} = {(6/5)*(4/5)*990} / 1000 = (\frac{99}{100})*(\frac{24}{25})\)

Answer: Option B which is \((1 - \frac{1}{100})(1 - \frac{1}{25}) = (\frac{99}{100})*(\frac{24}{25})\)

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