The number of pairs of natural numbers the difference of whose squares is 45 will be ?
Options:
A . 2
B . 3
C . 6
D . 5
Answer: Option B Let the numbers be x and y According to question, $$\eqalign{ & \left( {x > y} \right) \cr & {x^2} - {y^2} = 45 \cr & \left( {x + y} \right)\left( {x - y} \right) = 45 \cr & {\text{Make factor of 45}} \cr & {\text{15}} \times 3 \cr & \,\,9 \times 5\,\,\,\,\,\,\,\,\,\,(3\,{\text{pairs}}) \cr & 45 \times 1 \cr} $$ These pairs gives the value of x and y which satisfy the given condition.
Submit Comment/FeedBack