2 3 of the pens in a box are blue 1 3 of the remainder are black and the rest are red if there are 34 red pens in the box how many pens are there in the box altogether?
Let's say there are x pens in the box.
There are 2/3 * x = (2/3)x blue pens in the box.
The remainder, which is x - (2/3)x = (1/3)x pens, is not blue.
There are 1/3 * (1/3)x = (1/9)x black pens in the box.
The total number of red, black, and blue pens in the box is (1/9)x + (1/3)x + (2/3)x = 34 pens.
Combining like terms, we get (1/9 + 1/3 + 2/3)x = 34 pens.
Simplifying the left side of the equation gives (4/9 + 2/3)x = 34 pens.
Combining like terms again, we get (10/9)x = 34 pens.
Dividing both sides by (10/9), we get x = (34 * 9)/10 = 306/10 = <<34=34>>34 pens. Answer: \boxed{34}.
To determine the total number of pens in the box, we need to follow the given information step by step.
Let's assume the total number of pens in the box is "x".
1/3 of the pens in the box are black, which means (1/3)x pens are black.
The remaining pens in the box, after subtracting the black pens, are (x - (1/3)x) = (2/3)x pens.
2/3 of these remaining pens are blue, which gives (2/3) * (2/3)x = (4/9)x blue pens.
Adding the black and blue pens together, we get ((1/3)x) + ((4/9)x) = (7/9)x pens.
Since we know that there are 34 red pens in the box, we can write (7/9)x + 34 = x.
To solve this equation, we can multiply both sides by 9 to eliminate the denominator:
7x + 306 = 9x.
Next, subtract 7x from both sides:
306 = 2x.
Finally, divide both sides by 2 to solve for x:
x = 306 / 2.
Therefore, x = 153.
Hence, there are 153 pens in the box altogether.
Let's solve this step-by-step:
Step 1: Find the number of red pens in the box.
Given that 1/3 of the remainder after blue pens are taken out is black pens, and the remaining pens are red, we can say that 2/3 of the remainder is red pens. Therefore, (2/3) * (1/3) of the total number of pens equals 34.
(2/3) * (1/3) * P = 34
2/9 * P = 34
Multiply both sides of the equation by 9/2:
P = (34 * 9) / 2
P = 153
Answer: There are 153 pens in the box altogether.