Question

1. julius and walt are finding the product of 25 and 16.

part a
julius answer is incorrect. what did julius do wrong?
part b
what did walt do wrong?
part c
julius
25
× 16
150
+ 250
500
walt
25
× 16
200
50
120
+ 300
670

Explanation:

Part A

Step1: Recall multiplication steps

When multiplying \(25\times16\), we can break \(16\) into \(6 + 10\). So, \(25\times6 = 150\) (this is correct for Julius) and \(25\times10=250\) (also correct). But when adding, \(150 + 250\) should be \(400\), not \(500\). Julius made an addition error: \(150+250 = 400\), not \(500\).

Step2: Verify correct product

The correct product of \(25\times16\) is \(400\). Julius added \(150\) and \(250\) incorrectly (he got \(500\) instead of \(400\)).

Step1: Recall partial product method

When multiplying \(25\times16\) using partial products, we should multiply \(25\) by the units digit (\(6\)) and \(25\) by the tens digit (\(1\) in \(16\), which represents \(10\)). So, \(25\times6 = 150\) (not \(200\)) and \(25\times10 = 250\) (not \(50\), \(120\), or \(300\)). Walt used incorrect partial products. He didn't multiply \(25\) by \(6\) and \(25\) by \(10\) correctly. Instead, he used wrong intermediate values (\(200\), \(50\), \(120\), \(300\)) which don't come from \(25\times16\) breakdown.

Step2: Identify wrong partial products

Walt's partial products (\(200\), \(50\), \(120\), \(300\)) are not the correct partial products of \(25\times16\) (should be \(150\) (from \(25\times6\)) and \(250\) (from \(25\times10\))).

Answer:

Julius added \(150\) and \(250\) incorrectly (he calculated \(150 + 250 = 500\) instead of \(400\)).

Part B