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CAT 2017 Question Paper | Quants Slot 2

CAT Previous Year Paper | CAT Number Theory Questions | Question 28

Looking out to strengthen your hold on the Number Theory part of the Quantitative Aptitude section in the CAT Exam? You've come to the right place! CAT previous year paper are the best way to go about achieving expertise in any topic and here we have a beautiful question from Number Theory taken from CAT Question paper 2017 Slot 2. Try out this question and see if you can solve it without looking at the solutions provided!
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Question 28 : How many different pairs (a, b) of positive integers are there such that a ≤ b and \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9}) ? [TITA]


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Explanatory Answer

Method of solving this CAT Question from Number Theory

Given that \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9})
Let us assume that a ≤ b.
For first case, let a = b, then
\\frac{1}{18}) + \\frac{1}{18}) = \\frac{1}{9})
Hence (18 , 18) is one pair of possibility.
In all the cases, both a and b should be greater than 9 to satisfy the equation.
So, we can take a = 10 , 11 , 12 , ..... , 18 and check for the different possibilities.
For example, let us take a = 15
⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{15})
⟹ \\frac{5 - 3}{45}) = \\frac{2}{45})
Hence a = 15 is not possible.
Similarly when a = 17, 16, 14, 13, and 11 does not work either.
Now let us take a = 12,
⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{12})
⟹ \\frac{4 - 3}{36}) = \\frac{1}{36})
Hence, a = 12 and b = 36 is the other pair of possibility.
When a = 10,
⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{10})
⟹ \\frac{10 - 9}{90}) = \\frac{1}{90})
Hence, a = 10 and b = 90 is the other pair of possibility.
Therefore (18,18) , (12,36) and (9,90) are the three different pairs of positive integers which satisfy the condition

The question is "How many different pairs (a, b) of positive integers are there such that a ≤ b and \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9}) ? [TITA]"

Hence, the answer is 3

 

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