Question

In a question on division, the divisor is 7 times the quotient and 3 times the remainder. If the remainder is 28, then the dividend is: $$$$
A.1008$$$$
B.1516$$$$
C.1036$$$$
D.2135$$$$

Answer 1

Hint: We make equations $d=7\times q=7q,d=3\times r=3r$ with the information given in the question where $d$ is the divisor, $q$ is the quotient and $r$ is the remainder. We put $r=28$ in the second equation to find $d$ and put the obtained value of $d$ in the first equation to get $q$. We put $d,q,r$ in the Euclid’s lemma of division to find dividend $n=dq+r$ .$$$$

Complete step by step answer:
We know that in arithmetic operation of division the number we are going to divide is called dividend, the number by which divides the dividend is called divisor. We get a quotient which is the number of times the divisor is of dividend and also remainder obtained at the end of division. $$$$If the number is $n$, the divisor is $d$, the quotient is $q$ and the remainder is $r$, they are related by the following equation,
$$n=dq+r$$
Here the divisor can never be zero. The above relation is called Euclid’s division Lemma.
We are given in the question that the divisor is 7 times the quotient and 3 times the remainder. We know that in mathematics ‘times’ means multiplication. We have,
$$\begin{array}{rl}& d=7\times q=7q........\left(1\right)\\ & d=3\times r=3r.........\left(2\right)\end{array}$$
We are further given in the question the remainder is 28. Let us put $r=28$ in equation (2) and have the dividend as
$$d=3r=3\times 28=84$$
We put $d=84$ in equation and have,
$$84=7q$$
We divide both side of the above equation by 7 to have,
$$\begin{array}{rl}& {\displaystyle \frac{84}{7}}={\displaystyle \frac{7q}{7}}\\ & \Rightarrow 12=q\end{array}$$
We put the values of obtained divisor$d=84$, quotient $q=12$ and remainder $r=28$ in Euclid’ division lemma to have the dividend as
$$n=dq+r=84\times 12+28=1008+28=1036$$

So, the correct answer is “Option C”.

Note: We note that ‘lemma’ means a small result accompanying a theorem. We can alternatively assume the divisor as an unknown variable $x$ and solve for $x$. We also note that in Euclid’s lemma $n=dq+r$ we always have $0\le r<\left|b\right|$ where $\left|b\right|$ is the absolute value of $b$.