hello
to solve this question, we can write out two sets of equation and solve them
let the short answers be represented by x
let the multiple-choice questions be represented by y
we know that the test has 60 points
multiple-choice carries 2 points
short answers carries 5 points
![2x+5y=60](https://img.qammunity.org/2023/formulas/mathematics/college/imadogdesdwxg3ezhckfn2bfsd3slcexq5.png)
now we have a total of 15 questions which comprises of multiple-choice questions and short answers
![x+y=15](https://img.qammunity.org/2023/formulas/mathematics/college/sixxorckiqwr7gu5ehufuwly78x3h2wguj.png)
now we have two set of equations which are
![\begin{gathered} 2x+5y=60\ldots\text{equ}1 \\ x+y=15\ldots\text{equ}2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o66odvw3p2a5gudt7jx76awlm80tv4pi4b.png)
now let's solve for x and y
from equation 2, let's make x the subject of formula
![\begin{gathered} x+y=15 \\ x=15-y\ldots\text{equ}3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vvvsvcjbnr10nbykuyqr0k021skzv436l1.png)
put equation 3 into equation 1
![\begin{gathered} 2x+5y=60 \\ x=15-y \\ 2(15-y)+5y=60 \\ 30-2y+5y=60 \\ 30+3y=60 \\ \text{collect like terms} \\ 3y=60-30 \\ 3y=30 \\ \text{divide both sides by the coeffiecient of y} \\ (3y)/(3)=(30)/(3) \\ y=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kys7dfcrkd5jqnpb4jtzmoq4guqkcf8g5x.png)
now we know the value of y which is the number of multiple-choice question. we can use this information to find the number of short answer through either equation 1 or 2
from equation 2
![\begin{gathered} x+y=15 \\ y=10 \\ x+10=15 \\ \text{collect like terms} \\ x=15-10 \\ x=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bcw642e4eqle0knsdfjdqac2sqr3ux51zq.png)
from the calculations above, the number of short answers is equal to 5 and multiple-choice questions is equal to 10.
The answer to this question is option C