Question

Given below is the picture of box method for calcualtion of $29\times 15$.

The missing numbers are:

• A. $A=200$ and $B=345$
• B. $A=90$ and $B=445$
• C. $A=200$ and $B=435$
• D. $A=90$ and $B=435$

Answer 1

The correct option is C $A=200$ and $B=435$


In the box method, we first split the numbers into tens and ones.
We have to find the value of $29\times 15$.

Using place value we can split $29$ and $15$ as:
$29=20+9$
$15=10+5$.

So, we can write the muliplication result as:
$(29\times 15)=(20+9)\times (10+5)$.

Putting the respective values in the box above, we get:


The numbers inside the boxes are the products of the numbers on the sides of the boxes.

Observing and matching the values with the image given in question, we get:
$A=20\times 10=200$.


Now, to calculate the final product $29\times 15$, we add all the partial products.

So, final product will be:
$=29\times 15$
$=200+90+100+45$
$=435$.

∴ $B=435$.

The box diagram finally looks like:


Hence, the missing numbers are: $A=200$ and $B=435$.