Prove the following
Question:

If $27 \div A=33$, then find the value of $\mathrm{A}$

Solution:

We observe that, 4 x 3 can never be a single digit number 2, so 4 x 3 must be a two-digit number, whose ten’s digit is 2 and unit’s digit is the number

less than or equal to 4. Therefore, the value of 4 can be 9, as the values of 4 from 1 to 8 do not fit.

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