Question

The city library donated some children’s books to Mr. Clark’s first-grade class. If each student takes 4 books, there will be 20 books left. If 3 students do not take a book and the rest of the students take 5 books each, there will be a number of books left. How many books were donated to the class?
A.120
B.140
C.160
D.175

Answer 1

Hint: First of all, let the number of students be $x$ and number of books be $y$. Form the equations using the given conditions. Solve the resultant equations by substitution method to find the values of $x$ and $y$.

Complete step-by-step answer:
Let the number of students be $x$ and the number of books be $y$.
It is given that if each student takes 4 books, there will be 20 books left
That is, $4x$ books were taken by students and there were 20 books still left.
We can write this as
$4x + 20 = y$ (1)
Similarly, it is given that if 3 students do not take a book and the rest of the students take 5 books each, there will be no books left.
We can say that total books are 5 times \[x - 3\] where $x$ represents the total number of students .
Here, we will have $5\left( {x - 3} \right) = y$
On solving the bracket we get,
$5x - 15 = y$ (2)
Now, we have two equations in two variables.
We will solve both the equations using the substitution method.
Substitute the value of $y$ from equation (1) in equation (2)
That is,
$4x + 20 = 5x - 15$
Rearrange by bringing constant terms to one side and terms containing $x$ on one side and solve it further.
$
  4x - 5x = - 15 - 20 \\
   \Rightarrow - x = - 35 \\
   \Rightarrow x = 35 \\
$
Now substitute the value of $x$ in equation (1)
$4\left( {35} \right) + 20 = y$
On solving the equation, we get,
$
  140 + 20 = y \\
   \Rightarrow y = 160 \\
$
Hence, the total number of books donated to the class were 160.
Thus, option C is correct.

Note: Here, we have solved linear equations in two variables using substitution method. We can also use elimination method and cross-multiplication method. The equations are linear because the degree of variables is 1.