there are 680 coins divided into 3 piles. The second pile has three times as many coins as the first. The third pile has 50 more coins than the second pile. Solve This problem using a tape diagram.
Dunno about tape diagrams, but if the 1st pile has x coins, then
x + 3x + 3x+50 = 680
x = 90
To solve this problem using a tape diagram, we can represent each pile as a section on a straight line, with the length of each section representing the number of coins in that pile.
Let's start by representing the first pile. We can let x be the number of coins in the first pile. So, the first section on the tape diagram will have a length of x.
Now, the second pile has three times as many coins as the first pile. Therefore, the length of the second section on the tape diagram will be 3x.
Lastly, the third pile has 50 more coins than the second pile. So, the length of the third section on the tape diagram will be 3x + 50.
To find the value of x, we need to consider the total number of coins. We know that the total number of coins is 680, so the sum of the lengths of all three sections on the tape diagram should be equal to 680.
Therefore, we can form the equation: x + 3x + (3x + 50) = 680.
Simplifying the equation, we get: 7x + 50 = 680.
Subtracting 50 from both sides: 7x = 630.
Dividing both sides by 7: x = 90.
So, the first pile contains 90 coins. The second pile contains 3 times as many coins, which is 3 * 90 = 270 coins. And the third pile contains 50 more coins than the second pile, which is 270 + 50 = 320 coins.