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Question: On Monday morning, Mr. Smith had a certain amount of money that he planned to spend during the week....

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?

Explanation

Solution

To solve this question, we will use the GP (Geometric Progression) nth{{n}^{th}} term of a GP whose first term is ‘a’ and the common ratio is ‘r’ and is given by Tn=arn1.{{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 M4.\dfrac{M}{4}.
Money on Monday=M\text{Money on Monday}=M
Money on Tuesday=M4\text{Money on Tuesday}=\dfrac{M}{4}
Money on Wednesday=M4(14)=M16\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,M4,M16,....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 nth{{n}^{th}} term of a GP is given by Tn=arn1.{{T}_{n}}=a{{r}^{n-1}}. We have our GP as first term a = M and r=14.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.
T6=1\Rightarrow {{T}_{6}}=1
Using the formula of the nth{{n}^{th}} term of a GP, we have,
T6=ar61\Rightarrow {{T}_{6}}=a{{r}^{6-1}}
T6=ar5\Rightarrow {{T}_{6}}=a{{r}^{5}}
We have, a = M and r=14.r=\dfrac{1}{4}.
T6=1=M(14)5\Rightarrow {{T}_{6}}=1=M{{\left( \dfrac{1}{4} \right)}^{5}}
1=M45\Rightarrow 1=\dfrac{M}{{{4}^{5}}}
M=45\Rightarrow M={{4}^{5}}
M=1024\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.
Money on Monday=M\text{Money on Monday}=M
Money on Tuesday=M4\text{Money on Tuesday}=\dfrac{M}{4}
Money on Wednesday=M16\text{Money on Wednesday}=\dfrac{M}{16}
Money on Thursday=M64\text{Money on Thursday}=\dfrac{M}{64}
Money on Friday=M256\text{Money on Friday}=\dfrac{M}{256}
Money on Saturday=M1024\text{Money on Saturday}=\dfrac{M}{1024}
Now, money on Saturday = 1 dollar
M1024=1\Rightarrow \dfrac{M}{1024}=1
M=1024\Rightarrow M=1024
Hence, our answer is the same.