About Soroban Expertise I have a systematic and orderly way to organize the facts in a word problem, which (usually) leads clearly to the necessary equation. I think I can help with all types of word problems.
Experience 38 years of teaching college-level math, mostly at a two-year college.
Education/Credentials BS and MS in mathematics, SUNY Albany
Question You have 20 coins, 4 piles. The 1st pile is 4 more than the 2nd pile. The 2nd pile is 1 less than the 3rd pile and the 4th pile is twice as much as the 1st pile. How many coins are in each pile?
Answer Hello, Mandy!
There must be a typo in the problem.
The answers are not whole numbers.
I'll change the 20 to 28.
You have 28 coins, 4 piles.
The 1st pile is 4 more than the 2nd pile.
The 2nd pile is 1 less than the 3rd pile.
The 4th pile is twice the 1st pile.
Let's call the pile A, B, C, D (in that order).
Let x = number in pile C: C = x
Pile B is has one less: B = x - 1
Pile A has 4 more than B: A = x + 3
Pile D has twice A: D = 2(x + 3)
We are told that: A + B + C + D = 28
There is our equation! (x + 3) + (x - 1) + x + 2(x + 3) = 28
and we have: 5x +8 = 28 5x = 20 x = 4
Therefore:
A has 7 coins
B has 3 coins
C has 4 coins
D has 14 coins
.
