Aptitude

Question

Given $\displaystyle \cfrac{\;\cfrac{a}{b} + \cfrac{b}{a}\;}{\cfrac{a}{b} - \cfrac{b}{a}} = 1$

If $a$ and $c$ are positive integers, then how many ordered pairs are possible for $(a,c)$, where: $a + 4b^2 + c \leq 8$?

• A. 45
• B. 28
• C. 17
• D. 18

Answer 1

Ans will be 28, 

((a^2+b^2)/ab)/((a^2-b^2)/ab)=1

=>(a^2+b^2)/(a^2-b^2)=1

=>(a^2+b^2)=(a^2-b^2)

=>2b^2=0

=>b=0

a+c<=8   Because a and c are +ve integers.

=>a=7          c=1                         =1 ordered pair

=>a=6         c=1,2                      =2 ordered pair    

...............

=>a=1          c=1,2.......,7           =7 ordered pair

no, of ordered pair 1+2+.......7= 28 

Answer 2

Ans will be 45

((a^2+b^2)/ab)/((a^2-b^2)/ab)=1

=>(a^2+b^2)/(a^2-b^2)=1

=>(a^2+b^2)=(a^2-b^2)

=>2b^2=0

=>b=0

a+c<=8

=>a=8          c=0                            =1 ordered pair

=>a=7          c=0,1                         =2 ordered pair

=>a=6         c=0,1,2

...............

=>a=0          c=0,1,2.......,8               =9 ordered pair

no, of ordered pair 1+2+..........9=45 

Comments

−3 < −2 < −1 < 0 < 1 < 2 < 3 < … An integer is positive if it is greater than zero and negative if it is less than zero. Zero is defined as neither negative nor positive.

I think that answer should be 28. What do u think?

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@Lord_Krishna

https://www.mathsisfun.com/whole-numbers.html

see it

0 is also an integer

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yes. but "0 is not a positive integer."  In the question its given that :

If a and c are positive integers.....

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ok yes

Answer 3

This is an easy question.

after getting b=0, we'll have to do little observation.

suppose i ask you to find how many pairs give sum equal to 2. answer is 1 way

again, how many pairs give sum equal to 3, answer is 2 ways

how many pairs give sum equal to 4 , answer is 3 ways.

similarly, how many pairs give sum equal to 8 , answer is 7 ways.

therefore, it is nothing but the sum of first 7 natural number starting from 1 which gives (7*8)/2= 28. answer.