Question

Neelam and Rahul together have 10 pens both of them lost 2 each and the product of the numberof pens they Now have is 6.find how many pens they had to start with

Answer 1

Step-by-step explanation:

Let the initial number of pens be n and r

We have:

• n+r=10  => r=10-n

• (n-2)(r-2)=6

Replacing r in second equation with 10-n:

• (n-2)(10-n-2)=6
• (n-2)(8-n)=6
• 8n-n^2-16+2n=6
• -n^2+10n-16=6
• n^2-10n+22=0

n has no integer solution ...

So something must be wrong with given numbers. But this is how you would need to solve it. If numbers are different then just replace to get correct answer.

Answer 2

Step-by-step explanation:

let Neel has x pens and Rahul has y.

=> x+y = 10............(1)

again

(x-2) × (y-2) = 6

=> xy-2(x+y) +4 = 6

=> xy = 6-4+2(x+y)= 2+2×10 = 22

=> x(10-x) = 22

=> 10x-x² = 22

=> x²-10x+22 = 0

=> x= 5±√3

from (1)

if x= 5+√3 then y = 5-√3

=> Neelam has 5 pens,

=> Rahul has 4 pens

=> one pen is shered both of them.

verification:-

x+y= (5+4)+1 = 10

(x-2) (y-2) = (5-2)(4-2)= 3×2= 6

hope this helps you