Question

On Monday morning, Mr. Smith had a certain amount of money that he planned to spend during the week. On each subsequent morning, he had one fourth the amount of the previous morning. On Saturday morning, 5 days later, he had 1 dollar. How many dollars did Mr. Smith originally start with on Monday morning?

Answer 1

Hint: To solve this question, we will use the GP (Geometric Progression) \[{{n}^{th}}\] term of a GP whose first term is ‘a’ and the common ratio is ‘r’ and is given by \[{{T}_{n}}=a{{r}^{n-1}}.\] We will first form a sequence (G.P) of the amount of money Mr. Smith had in consequent weekdays and then try to find the sixth term (Saturday) from the above formula. Finally, equate it to 1 dollar to get the result.

Complete step-by-step solution
Let us first assume the initial amount of money that Mr. Smith had at the starting of the week that is on Monday. Let M be the initial amount of the money that Mr. Smith had on Monday. According to the given condition in the question, if the money on Monday is M, then on Tuesday it is \[\dfrac{M}{4}.\]
\[\text{Money on Monday}=M\]
\[\text{Money on Tuesday}=\dfrac{M}{4}\]
\[\text{Money on Wednesday}=\dfrac{M}{4}\left( \dfrac{1}{4} \right)=\dfrac{M}{16}\]
Similarly, we have a series of the terms given as \[M,\dfrac{M}{4},\dfrac{M}{16},....\]
This series forms a geometric progression (GP). Let us now define a GP. GP is a sequence of numbers in which each number is obtained from the previous one by multiplying by a constant. For example, 1, 2, 4, 8, 16, 32, … is a GP with a common ratio where r is 2, and the first term ‘a’ as 1.
The \[{{n}^{th}}\] term of a GP is given by \[{{T}_{n}}=a{{r}^{n-1}}.\] We have our GP as first term a = M and \[r=\dfrac{1}{4}.\] Now, we are given that the sixth day Saturday (Starting from Monday), he has 1 dollar. Therefore, the sixth term of GP is 1.
\[\Rightarrow {{T}_{6}}=1\]
Using the formula of the \[{{n}^{th}}\] term of a GP, we have,
\[\Rightarrow {{T}_{6}}=a{{r}^{6-1}}\]
\[\Rightarrow {{T}_{6}}=a{{r}^{5}}\]
We have, a = M and \[r=\dfrac{1}{4}.\]
\[\Rightarrow {{T}_{6}}=1=M{{\left( \dfrac{1}{4} \right)}^{5}}\]
\[\Rightarrow 1=\dfrac{M}{{{4}^{5}}}\]
\[\Rightarrow M={{4}^{5}}\]
\[\Rightarrow M=1024\]
Therefore, the amount of money he had starting the week on Monday is 1024.

Note: Another possible way to solve this question is without using GP and directly calculating the money he had each day.
\[\text{Money on Monday}=M\]
\[\text{Money on Tuesday}=\dfrac{M}{4}\]
\[\text{Money on Wednesday}=\dfrac{M}{16}\]
\[\text{Money on Thursday}=\dfrac{M}{64}\]
\[\text{Money on Friday}=\dfrac{M}{256}\]
\[\text{Money on Saturday}=\dfrac{M}{1024}\]
Now, money on Saturday = 1 dollar
\[\Rightarrow \dfrac{M}{1024}=1\]
\[\Rightarrow M=1024\]
Hence, our answer is the same.