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The numbers a, b and c form a sequence such that b is 3 more than a, a
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12 Dec 2023, 04:06
12 Dec 2023, 04:06
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Question Stats:
60% (01:56) correct
39% (01:38) wrong based on 23 sessions
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The numbers a, b and c form a sequence such that b is 3 more than a, and c is 9 more than b.
Also \(\frac{a}{b}=\frac{b}{c}\)
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.
Also \(\frac{a}{b}=\frac{b}{c}\)
Quantity A |
Quantity B |
b |
3/2 |
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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Re: The numbers a, b and c form a sequence such that b is 3 more than a, a
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12 Dec 2023, 09:55
12 Dec 2023, 09:55
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The answer is A.
b = a + 3 --> a = b - 3
c = b + 9
Since a/b = b/c --> ac = b^2
--> (b - 3)(b + 9) = b^2
--> b^2 + 6b - 27 = b^2
--> 6b = 27
--> b = 9/2
b = a + 3 --> a = b - 3
c = b + 9
Since a/b = b/c --> ac = b^2
--> (b - 3)(b + 9) = b^2
--> b^2 + 6b - 27 = b^2
--> 6b = 27
--> b = 9/2
The numbers a, b and c form a sequence such that b is 3 more than a, a
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29 Jan 2024, 22:20
29 Jan 2024, 22:20
\(a,b,c\)
\(a,3+a,9+b\)
\(a,3+a,9+3+a\)
\(a,3+a,12+a\)
we are given
\(\frac{a}{b}=\frac{b}{c}\)
LHS
\(\frac{a}{b} = \frac{a}{3+a}\)
RHS
\(\frac{b}{c} = \frac{3+a}{12+a}\)
therefore, \(\frac{a}{b}=\frac{b}{c}\) will be
\(\frac{a}{3+a} = \frac{3+a}{12+a}\)
\((3+a)^2 = a(12+a)\)
\(9+a^2+6a=12a+a^2\)
\(6a+9=12a\)
\(6a=9\)
\(a=\frac{9}{6} =\frac{3}{2}\). This is Quantity B.
Since we know that Quantity A is \(b\) and \(b\) is \(3\) times more than \(a\), automatically Quantity A, that is \(b\), is greater than Quantity B, which is the value of A.
Quantity A is the greatest.
\(a,3+a,9+b\)
\(a,3+a,9+3+a\)
\(a,3+a,12+a\)
we are given
\(\frac{a}{b}=\frac{b}{c}\)
LHS
\(\frac{a}{b} = \frac{a}{3+a}\)
RHS
\(\frac{b}{c} = \frac{3+a}{12+a}\)
therefore, \(\frac{a}{b}=\frac{b}{c}\) will be
\(\frac{a}{3+a} = \frac{3+a}{12+a}\)
\((3+a)^2 = a(12+a)\)
\(9+a^2+6a=12a+a^2\)
\(6a+9=12a\)
\(6a=9\)
\(a=\frac{9}{6} =\frac{3}{2}\). This is Quantity B.
Since we know that Quantity A is \(b\) and \(b\) is \(3\) times more than \(a\), automatically Quantity A, that is \(b\), is greater than Quantity B, which is the value of A.
Quantity A is the greatest.
