Question

In how many ways can 5 pounds be paid in shillings and sixpences, including zero solutions (positive integer solutions)?
(a) 100
(b) 101
(c) 102
(d) 103

Answer 1

Hint: Assume that 5 pounds is distributed in such a way that the number of shillings be x and the number of sixpences be y. The summation of x shillings and y sixpences is equal to 5 pounds. Convert shillings into pounds and sixpences and pounds using the relation \[\text{1 pound = }\!\!~\!\!\text{ 20 }\!\!~\!\!\text{ shillings}\] and \[\text{1 pound = }\!\!~\!\!\text{ 40 }\!\!~\!\!\text{ sixpences}\] . Divide the equation \[2x+y=200\] , by 2. As x and 100 both are integers, so \[\dfrac{y}{2}\] must also be an integer. Let that integer be p. So, \[y=2p\] . Now, put the value of y in the equation \[2x+y=200\] and get the value of x. Since x is a positive integer and also includes zero so, \[x\ge 0\] . Now, solve it further and get the integral values of p. The number of solutions of p will be the same as the number of solutions of x and y.

Complete step-by-step answer:
We know the relation between pound and shillings. That is,
\[\text{1 pound = }\!\!~\!\!\text{ 20 }\!\!~\!\!\text{ shillings}\] …………….(1)
We know the relation between pounds and sixpences. That is,
\[\text{1 pound = }\!\!~\!\!\text{ 40 }\!\!~\!\!\text{ sixpences}\] ……………………(2)
Let us assume that 5 pounds is distributed in such a way that the number of shillings be x and the number of sixpences be y.
x shillings + and sixpences = 5 pounds …………………(3)
Using equation (1), Converting x shillings into pound, we get
x shillings = \[\dfrac{x}{20}\] pounds ……………….(4)
Using equation (2), Converting y sixpences into pound, we get
y sixpences = \[\dfrac{y}{40}\] pounds ……………….(5)
Now, from equation (3), equation (4), and equation (5), we get
\[\begin{align}
  & \dfrac{x}{20}+\dfrac{y}{40}=5 \\
 & \Rightarrow \dfrac{2x+y}{40}=5 \\
\end{align}\]
\[\Rightarrow 2x+y=200\] ………………………(6)
Now, we have to find the integral solution of x and y for the equation \[2x+y=200\] .
First of all, we need to simplify this equation into simpler form so that we can directly predict the positive integral values of x and y. It means that x and y both are positive integers.
\[\begin{align}
  & 2x+y=200 \\
 & \Rightarrow x+\dfrac{y}{2}=100 \\
\end{align}\]
\[\Rightarrow \dfrac{y}{2}=100-x\] ………………….(7)
Since x is an integer and 100 is also an integer so \[\dfrac{y}{2}\] must also be an integer. Let the integer \[\dfrac{y}{2}\] be p. So,
\[\Rightarrow \dfrac{y}{2}=p\]
\[\Rightarrow y=2p\] ……………………(8)
Now, from equation (6) and equation (8), we get,
\[\begin{align}
  & \Rightarrow 2x+2p=200 \\
 & \Rightarrow 2x=200-2p \\
 & \Rightarrow x=100-p \\
\end{align}\]
Since x is a positive integer and we also have to include zero solutions so,
\[\begin{align}
  & 100-p\ge 0 \\
 & 100\ge p \\
\end{align}\]
Here, we can say that p should be less than equal to 100.
The range of p is \[\left[ 0,100 \right]\] .
We have assumed that \[\dfrac{y}{2}\] is an integer and \[\dfrac{y}{2}\] is equal to p. So, p is also an integer.
So, we have to get the integral values of p in the range \[\left[ 0,100 \right]\] .
We can say that p can be 0, 1, 2, 3, 4,……………, 100.
Therefore, the number of possible integral values of p is 101.
Hence, the correct option is B.

Note: In this question, one might get confused how to approach this question because the relation between pounds and shillings, and pounds and sixpences is not given in the question. So, remember that, \[\text{1 pound = }\!\!~\!\!\text{ 20 }\!\!~\!\!\text{ shillings}\] and \[\text{1 pound = }\!\!~\!\!\text{ 40 }\!\!~\!\!\text{ sixpences}\] .
Also, one can forget to include the value of x and p equal to zero in the solution. In the question, it is mentioned to include zero as a solution. One can conclude \[\left[ 0,100 \right]\] as an answer. This is not correct. We only have to take the integral values and have to ignore the decimal point values.