abhi47 wrote:

Shaggy has to learn the same 71 hiragana characters, and also has one week to do so; unlike Velma, he can learn as many per day as he wants. However, Shaggy has decided to obey the advice of a study-skills professional, who has advised him that the number of characters he learns on any one day should be within 4 off the number he learns on any other day.

(A) What is the least number of hiragana that Shaggy could have to learn on Saturday?

(B) What is the greatest number of hiragana that Shaggy could have to learn on Saturday?

Could someone please provide a simpler solution to this problem ? The actual solution provided is bit confusing.

I must admit the wording is pretty awkward. Anyway:

As "the number of characters he learns on any one day should be within 4 off the number he learns

on any other day", then the number of characters he learns on any day must be between \(x\) and \(x+4\) characters .

A. To minimize # of characters per day he should learn \(x\) (so min possible) characters on one day and \(x+4\) (so max possible) characters on all other 6 days, so we would have: \(x+6(x+4)=71\) --> \(x_{min}=\frac{47}{7}=6.something\), as \(x\) must be an integer than round up to 7 (it can not be less than 7 as 6.something is minimum);

B. To maximize # of characters per day he should learn \(x\) (min possible) characters on 6 days and \(x\) characters on all other 6 days, so we would have: \(6x+(x+4)=71\) --> \(x_{max}=\frac{67}{7}=9.something\), as \(x\) must be an integer than round down to 9 (it can not be more than 9 as 9.something is maximum), so max value for \(x+4\) is 9+4=13.

Answer: 7 and 13.

General rule for such kind of problems: to maximize one quantity, minimize the others;

to minimize one quantity, maximize the others.

Similar question to practice:

a-certain-city-with-population-of-132-000-is-to-be-divided-76217.htmlHope it helps.

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