a box of red,blue and green pencils contains 20 in all.there are 6 times as many blue pencilas green ones,and fewer rd pencils than blue ones. how many red pencilsare in he box
There are 7 or 14 blue and green pencils. Blue ones, respectively, 6 or 12, and green ones 1 or 2.
In all there are 20 pencils, hence, there are two possibilities for the reds:
either there are 20−7=13,
or 20−14=6.
But the reds are less than the blue ones, so the only possible answer is 12 blue pencils, 2 green pencils and 6 red pencils.
let x be the number of green ones
blue one = 6x
Well, it seems like this box of pencils is a colorful party! If we let the number of red pencils be represented by 'r', the number of blue pencils by 'b', and the number of green pencils by 'g', we can set up a system of equations to solve the problem.
From the information given, we know that:
1) r + b + g = 20 (because there are 20 pencils in total)
2) b = 6g (since there are 6 times as many blue pencils as green ones)
3) r < b (because there are fewer red pencils than blue ones)
Let's substitute equation 2 into equation 3 to eliminate 'b':
r < 6g
Now, since we need to solve for 'r', let's substitute equation 2 and the updated inequality into equation 1:
r + 6g + g = 20
r + 7g = 20
Solving the system of equations, we find that there are many potential combinations of red and green pencils that could satisfy the given conditions. Without further information, we can't determine a single numerical value for 'r'.
But hey, let's give this problem a little more color... I mean, humor! Why don't we imagine that the box of pencils is like a clown car, full of surprises? So, the number of red pencils could be anywhere between 0 and 20 (inclusive), leaving us with a colorful range of possibilities.
To determine the number of red pencils in the box, let's break down the information given and solve the problem step by step.
Let's represent the number of red pencils as "x," the number of blue pencils as "y," and the number of green pencils as "z."
We are given the following information:
1. The total number of pencils is 20: x + y + z = 20.
2. There are six times as many blue pencils as green ones: y = 6z.
3. There are fewer red pencils than blue ones: x < y.
Now let's solve the equations to find the value of x:
Substitute the value of "y" from equation 2 into equation 1:
x + (6z) + z = 20
x + 7z = 20 -- Equation 3
Since we don't have the exact values of "x" and "z," we need another equation to solve for them.
From equation 2, we have:
6z > x -- Equation 4
Let's consider some possible values for z (green pencils) and find corresponding values for x (red pencils) that satisfy the given conditions.
Let z = 1
From equation 3: x + 7(1) = 20
x + 7 = 20
x = 13
However, this solution doesn't satisfy equation 4 (6z > x), as 6(1) is not greater than 13.
Let z = 2
From equation 3: x + 7(2) = 20
x + 14 = 20
x = 6
This solution satisfies both equations 3 and 4.
Therefore, when there are 2 green pencils in the box, there are 6 red pencils.
Hence, the box contains 6 red pencils.