Jayant gets 3 marks for each right sum and loses 2 marks for each wrong sum. he attempts 30 sums and obtains 40 marks. the number of sums attempted correctly is
Answers
x= correct ans
y= wrong ans
x+y=30
y=30-x
therefore
40=3x-2(30-x)
40=3x-60+2x
100=5x
x=20
The number of sums he attempted correctly are 20 sums
Explanation:
Given:
1. Jayant gets 3 marks for each right sum
2. Jayant loses 2 marks for each wrong sum
3. He attempts 30 sums and obtains 40 marks
To find:
The number of sums he attempted correctly
Solution:
==> Let, x for Correct Answer
==> y for Wrong Answer
==> Total sum he attended is 30
==> x+y = 30 ==>1
==> Jayant gets 3 marks for correct sum
==> so, it can be write as 3x
==> Jayant loses 2 marks for each wrong sum
==> So,-2y
==> He obtain 40 marks
==> 3x-2y = 40 ==>2
==> Using Elimination Method
==> Multiply by -3 in equation 1
==> -3(x+y=30)
==> -3x-3y=-90 ==>3
==> Equating equation 2 and 3
==> 3x-2y= 40
==> -3x-3y=-90
--------------------
-5y = -50
--------------------
==> 5y=50
==> y=10
==> Substitute y in equation 1
==> x+y = 30
==> x+10 =30
==> x = 30-10
==> x=20
==> The number of sums he attempted correctly is 20 sums
==> Check whether the answers are correct are not
==> Apply the x and y values in equation 1 and 2
==> x+y =30
==> 20+10 =30
==> 30=30
==> Hence proved
==> 3x-2y=40
==> 3(20)-2(10)=40
==> 60-20=40
==> 40=40
==> Hence proved