Red and blue LEGO blocks

Let $x$ be the number of blue blocks and let $y$ be the number of red blocks.

We are told that $\frac{1}{6}$ of the blocks are blue. This gives rise to the equation $\frac{x}{x + y} = \frac{1}{6}$ . Cross-multiplying and rearranging terms gives $y = 5x$ .

It must be that a blue block was drawn, because the proportion of blue blocks has decreased from $\frac{1}{6}$ to $\frac{1}{7}$ . This allows us to write the equation $\frac{x - 1}{x + y - 1} = \frac{1}{7}$ . Again, cross-multiplying and rearranging terms yields $y = 6x - 6$ . We can substitute the $y = 5x$ we obtained before into $y = 6x - 6$ to obtain $5x = 6x - 6 \longrightarrow x = 6$ .

We have 6 blue blocks, meaning we have $y = 5(6) = \boxed{30}$ red blocks.

Initially, for every blue block in the box there are $5$ red blocks. So if we start out with $n$ blue blocks we also have $5n$ red blocks, resulting in a probability of $\dfrac{n}{n + 5n} = \dfrac{1}{6}$ of drawing a blue block.

Now since after drawing one block the probability of drawing another blue block decreases, we can assume that a blue block was drawn. We thus have $n - 1$ blue blocks and $5n$ red blocks remaining, and so the given information implies that

$\dfrac{n - 1}{(n - 1) + 5n} = \dfrac{1}{7} \Longrightarrow 7(n - 1) = 6n - 1 \Longrightarrow n = 6.$

Thus there are $5n = \boxed{30}$ red blocks in the box.

I used a simple logic. Since initially the blue blocks are 1/6 of total blocks, the no of total blocks has to be in multiple of 6. The number on removing one block;has to be in multiple of 7. So I checked table of 6 and found that 6*6 = 36 and 36-1= 35 is multiple of 7. Therefore the initial no of blocks has to be 36.and Red blocks are 5/6 of that i.e. 30

let x = red + blue, then x/6 = blue and 5x/6 = red, because the propotion of blue decreases, so the block removed is blue. We have, (x-1)/7 = blue, 6(x-1)/7 = red. In this step, we have 2 ways to solve: [1]. (x-1)/7 = (x/6) -1 or [2]. 6(x-1)/7=5x/6 => x = 36 => red = 30.

Since initially the blue blocks are 1/6 of total blocks, the number of total blocks has to be in multiple of 6. The number on removing one block has to be a multiple of 7. So I found that 6*6 = 36 and 36-1= 35 is multiple of 7. Therefore the initial number of blocks has to be 36, and red blocks are 5/6 of that, i.e. = 30.

the blue blocks are 1/6 of total blocks, the number of total blocks has to be in multiple of 6. The number on removing one block has to be a multiple of 7. So I found that 6*6 = 36 and 36-1= 35 is multiple of 7. Therefore the initial number of blocks has to be 36, and red blocks are 5/6 of that, i.e. = 30.

If only 1 block gets removed, then we need to look for 2 numbers that satisfy the following: one of ´them is a multiple of 6, the other one is a multiple of 7, and the difference between the two numbers is 1. 35 and 36 are the first pair that satisfy these 3 criteria. Starting with 36: if 1/6 of blocks, are blue (ie. 6) then 30 are red. One gets removed, ie total of 35. 1./7 of blocks are blue, ie 5, leaving 30 red ones. Hence, 30 must be red. (What I haven't figured out or in is the following: How do we know which color the 1 block is that gets removed? Does this influence the result at all?)

let x be total number of blocks. So in the beginning there are x/6 blue blocks and 5x/6 red blocks. After picking one block the fraction of blue blocks reduce to 1/7. So the picked block must be blue. Now, the total number of blue blocks is (x-1)/7. the change in number of blue blocks occurred because of picking 1 blue block out. So, x/6 - (x-1)/7 =1 solving this we get x=36. 5/6 * 36=30

Let initially red lego is x, total will be 6x and blue will be 5x. First let's assume we remove blue lego, according to the question x-1=(6x-1)/7. Solve for x. x=6. Total red lego are 6*5=30

If you think about it, the total blocks will be 6x7-6 and 6x7-7 so it's pretty easy to know that there are 6 blue and 30 red to start Like 1/9 to 1/10 would be 90-9 and 90-10