Question 324444
Let R be the number of correct problems.
Let W be the number of incorrect problems.
1.{{{R+W=26}}}
2.{{{8R-5W=0}}}
From eq. 1,
{{{R=26-W}}}
Substitute into eq. 2,
{{{8(26-W)-5W=0}}}
{{{208-8W-5W=0}}}
{{{13W=208}}}
{{{highlight(W=16)}}}
Then from eq. 1,
{{{R+16=26}}}
{{{highlight(R=10)}}}
.
.
.
10 correct and 16 incorrect.