Question

An algebra test contains 38 problems. Some of the problems are worth 2 points each. The rest of the questions are worth 3 points each. A perfect score is 100 points.
How many problems are worth 2 points? How many problems are worth 3 points?

Answer 1

Let 'x' be the problems worth 2 points.

Let 'y' be the problems worth 3 points.

Since, there are 38 total problems.

So, $x+y =38$ (equation 1)

x = 38-y

Since, a perfect score is 100 points.

So, $2x+3y = 100$ (equation 2)

Substituting the value of 'x', we get

$2(38-y)+3y=100$

$76-2y+3y=100$

$76+y = 100$

y = 24

x+y = 38

x = 38-24 = 14

So, 14 problems are worth 2 points and 24 problems are worth 3 points.

Answer 2

Total questions in the test = 38

Let 2 points questions be = x

Let 3 points questions be = y

As total questions are 38, so

$x+y=38$

From here, we can derive x as: $x=38-y$  ...... (i)

As given, few questions are 2 marks each and rest 3 marks each and perfect square is 100, so we get,

$2x+3y=100$   .......(ii)

Put the value of y from (i) in (ii)

$2(38-y)+3y=100$

$76-2y+3y=100$

y=24

Putting y in equation (i), we get

$x=38-24$

x=14

Hence, 2 points questions are = 14

3 points questions are = 24