Brazilian Grapes

total number of grapes be x. first brother got grapes= 1/5 x second brother got = 1/5(x-1/5x)= 4x/25 remaing grapes to be divided in three brothers = x- 1/5x - 4x/25 =16x/25 to divide 16x/25 between 3 brothers i,e 16x/(25*3)or 16x / 75 , 16x should be exactly divisible by 75 so we put largest value of x as 75 , because total number of grapes can't be more than 100. thus, number of grapes is 75

Though this problem was presented under the Algebra Category but can easily be solved without one with a simple concept of Number theory. Considering the fact that the number was initially divided by 5 in succession followed by 3, so the number should be divisible by

5 x 5 x 3 = 75

So that is the minimum number of grapes in the box

Note

1. This will work only if all of the divisors are not co-prime to 4
2. At any moment anyone realizes that someone may have already taken the grapes, all of the remaining should realize and share it equally, else it would be a chaotic problem to handle

The problem can be solved by analyzing the clue words given in the problem. mother instructed her children to "equally" divide among 5 siblings so the total number of grapes should exactly divisible by 5,and "no one got more than 20" so total number should not exceed 100.another clue word was when andrea woke up "thinking she was firdt to wake up" .this means that the remaining grapes after the first sibling took his share is also exactly divisible by 5. and after she took her share. the remaining grapes should then be exactly divided between the 3 remaining siblings. so to solve this problem we must think of a number divisible by 5 that after taking away 1/5 the difference is also exactly divisible by 5, then the remaining "difference" should be exactly divided between the 3 remaining siblings. 75- 1/5= 75-15=60; 60/5=12= 60-12=48: 48/3=16
15+12+16=16=16=75

let, k was the total number of grapes. now A took one-fifth, B took one-fifth of the remains and rest 3 divide equally among three. Now, A took k/5, B took 4k/25, C took 16k/25, D took 16k/25, E took 16k/25. check: $\frac{k}{5}+\frac{4k}{25}+\frac{16k}{75}+\frac{16k}{75}+\frac{16k}{75}=k$ now $\frac{1}{5}=\frac{15}{75}\;and\;\frac{4}{25}=\frac{12}{75}$ $=>\frac{16k}{75}>\frac{k}{5}>\frac{4k}{25}$ $=> \frac{16k}{75}\leq20$ $=> k\leq93.75$ but it should be an +ve integer => take 93. $$ since it is divisible by 5, so last digit should be 5 or 0. so maximum 90 grapes possible. Again, number of grapes for all should be a positive integer. so k is also at least divisible by 75, 5 and 25. LCM of these three is 75. $$ now k should be multiple of 75 and could have a value of maximum 90. $$ So take 75. $$

simply let the grapes left for the last three brothers be $3x$ , then the total number of the grapes given by their mother will be $3x\times\frac{5}{4}\times\frac{5}{4}$ or $\frac{3\times 5^{2} x}{16}$ . Since the number of grapes is and integer, $x$ must be divisible by 16, and also $x$ must be less than 20. Therefore the only possible value for $x$ is 16. By pluck in $x$ = 16, we will get the total number of grape be 75.

The grapes are indivisible. thus it should be a whole number <20 each child ABCD and E are the children.

A got 20% of the total grapes

B got 20% of the 80% remaining then B took only 16% of the total grapes

thus CD and E got the highest share of 21.33% thus the we should try to start over here

100 could not be the number because the 3 remaining bros will have 21.33 and it should be a whole number and on top of that it exceeded the ceiling maximum grape per brother,

thus find a number that will satisfy a whole number if the formula is used

Formula is Total Grapes=.2T + .2(.8T)+ ((T-.2T- .2T(.8T))/3)

WE will find it thru trial and error. ===75

Let x be the initial amount of grapes.

A gets (1/5)x.

B gets (1/5)(4/5)x = (4/25)x

C, D, E each get y, which totals 3y.

The five together get (1/5)x + (4/25)x + 3y, which is also x.

So, (1/5)x + (4/25)x + 3y = x.

Solve for y... y = (16/75)x

The smallest x that gives an integer for y is 75.

So there were 75 grapes.

Checking...A gets 15, B gets 12, C, D, E each get 16. the sum is 75 and each is less than 20.

75 grapes were in the bowl

5 x 5 x 3 = 75

It would be easier to start by taking x as the number of grapes each of the last three brothers had. Thus, they took 3x grapes. Before Beatrice took, there must have been (3x)(5/4) grapes. Before Albert, there must have been (3x)(5/4)(5/4) grapes. This is 75x/16, which is less than 100, as the last clue states. Thus 75x/16 is an integer and less than 100. To cancel 16, x=16, and the total number of grapes becomes 75