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13 Feb 2023, 03:18
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
75% (02:45) correct
25%
(03:03)
wrong
based on 75
sessions
History
Date
Time
Result
Not Attempted Yet
A square is divided into four identical rectangles and a smaller square. The rectangles have length a and width b, with a > b. If the area of the larger square is 16 times of the area of the smaller square, what is the ratio of a to b ?
A. 1:3
B. \(\sqrt{3}:\)
C. 3:5
D. 5:3
E. 3:1
Attachment:
2023-02-13_14-14-15.png [ 7.94 KiB | Viewed 4222 times ]
13 Feb 2023, 08:23
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What is the ratio of a to b?
The rectangles have length a and width b, with a > b.
Because a > b --> options A and C are Incorrect.
Option B is incomplete so lets see if we require to know the value given in option B.
Clearly, the side of larger square is a + b and
side of smaller square is a - b
Approach 1
Since, the area of the larger square is 16 times of the area of the smaller square
\((a + b)^2 = 16*(a - b)^2\)
=> \(a^2 + b^2 + 2ab\) = \(16 (a^2 + b^2 - 2ab)\)
=> 15 \((a^2 + b^2) \)= 34ab
=> \(\frac{(a^2 + b^2)}{ab }\) = \(\frac{34}{15}\)
=> \(\frac{a}{b} + \frac{b}{a} = \frac{34}{15} \) ............eqn 1
Now we can substitute the options D and E in eqn 1 and see if it satisfies the eqn.
We can find that D is the right answer.
Approach 2
Since, the area of the larger square is 16 times of the area of the smaller square
\((a + b)^2 = 16*(a - b)^2\)
=> \({\frac{(a + b)}{(a - b)}}^2 \)= 16 = \(4^2\)
=> \(\frac{(a + b)}{(a - b)} = + 4 \)
On solving we get \(\frac{a}{b} = \frac{5}{3 } \) .... which is correct since a > b
or \(\frac{(a + b)}{(a - b)} = -4\)
On solving we get \(\frac{a}{b}\) = \(\frac{3}{5}\). This can be rejected as we know that a>b
therefore Answer is D
The rectangles have length a and width b, with a > b.
Because a > b --> options A and C are Incorrect.
Option B is incomplete so lets see if we require to know the value given in option B.
Clearly, the side of larger square is a + b and
side of smaller square is a - b
Approach 1
Since, the area of the larger square is 16 times of the area of the smaller square
\((a + b)^2 = 16*(a - b)^2\)
=> \(a^2 + b^2 + 2ab\) = \(16 (a^2 + b^2 - 2ab)\)
=> 15 \((a^2 + b^2) \)= 34ab
=> \(\frac{(a^2 + b^2)}{ab }\) = \(\frac{34}{15}\)
=> \(\frac{a}{b} + \frac{b}{a} = \frac{34}{15} \) ............eqn 1
Now we can substitute the options D and E in eqn 1 and see if it satisfies the eqn.
We can find that D is the right answer.
Approach 2
Since, the area of the larger square is 16 times of the area of the smaller square
\((a + b)^2 = 16*(a - b)^2\)
=> \({\frac{(a + b)}{(a - b)}}^2 \)= 16 = \(4^2\)
=> \(\frac{(a + b)}{(a - b)} = + 4 \)
On solving we get \(\frac{a}{b} = \frac{5}{3 } \) .... which is correct since a > b
or \(\frac{(a + b)}{(a - b)} = -4\)
On solving we get \(\frac{a}{b}\) = \(\frac{3}{5}\). This can be rejected as we know that a>b
therefore Answer is D
13 Feb 2023, 10:51
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Bunuel
Solution:
- In the diagram above, the dimensions look like this:Attachment:
squareinsquare.png [ 9.97 KiB | Viewed 3269 times ]
- Area of larger square \(=side^2=(a+b)^2\)
- Area of smaller square \(=side^2=(a-b)^2\)
- According to the question, \((a+b)^2=16\times (a-b)^2\)
\(⇒\frac{(a+b)^2}{(a-b)^2}=16\)
\(⇒(\frac{a+b}{a-b})^2=16\)
\(⇒\frac{a+b}{a-b}=4\) (we are ignoring -4 because \(\frac{a+b}{a-b}≠-4\))
\(⇒a+b=4a-4b\)
\(⇒-3a=-5b\)
\(⇒a:b=5:3\)
Hence the right answer is Option D
