Grades 7-8 Video Solutions 2021-video 2021 7-8/7

Video Transcription

Question 7. A student correctly added the two two-digit numbers on the left of the board and got the answer 137. What answer will he get if he adds the two  four-digit numbers on the right of the board? So to solve this problem we have to do ADCB plus CBAD. If we take a look at a random number, for example 927,  we can break this number down into three parts. Do 9 times 100 plus 2 times 10 plus 7 times 1. This is the same as saying 927. If we take the same logic we  can apply it to our number ADCB and CBAD. Doing this we get A times 1000 plus D times 100 plus C times 10 plus B times 1 plus C times 1000 plus B times 100  plus A times 10 plus D times 1. We simplify this we get 1000 A plus 100 D plus 10 C plus B plus 1000 C plus 100 B plus 10 A plus D. Simplified even further  we'll get 1010 A 101 D 1010 C and 101 B. If we factor out 101 we get 10 A plus B plus 10 C plus D. From the previous example we know that having 10 times A  plus B we can rewrite as AB. So we get 1001 times the sum of AB plus CD. From the beginning of the problem we know that AB plus CB is equal to 137. So what  this formula really is is 101 times 137. When we get the product we get the answer which is B 13,837.

Video Summary

A student correctly added two two-digit numbers, resulting in 137, and needs to add two specific four-digit numbers, ADCB and CBAD. By breaking the numbers into their components and applying arithmetic operations, the expression simplifies to 101 times the sum of the numbers formed (AB + CD). Given that AB + CB equals 137, the final computation yields 101 times 137, resulting in an answer of 13,837.

Keywords

arithmetic

four-digit

addition

computation

expression