Question

Peter the optician sees 16 patients one Tuesday. 4 are neither short nor long sighted. The total number of short sighted patients is two more than the total number of long sighted patients. How many patients can be short sighted,how many can be long sighted and how many could require varifocal lenses?

Answer 1

Answer:

No of patients having longsightedness =  5

No of patient having shortsightedness = 7

Patients requiring varifocal lenses will be  =12                  

Step-by-step explanation:

Total no of patients = 16

As

No of patients  having neither long nor shortsightedness = 4

Let

No of patients having longsightedness = x

So

No of patient having shortsightedness = $x+2$

Equation becomes

$x+(x+2)+4= 16\\2x+6 = 16\\2x= 16-6\\2x= 10\\x= \frac{10}{2} \\x= 5$

Now

No of patients having longsightedness = x= 5

No of patient having shortsightedness = $x+2$      =$5+2$

                                                                 = 7

Patients requiring varifocal lenses will be = 5+7

                                                                         = 12