Question

In a basket of apples 12% are rotten and 66 are in good condition . Find the total number of apples in the basket\[\]
A.75\[\]
B.70\[\]
C.71\[\]
D.80\[\]

Answer 1

\[\] Hint: We assume the total amount of apples in the basket as $x$ and the number of rotten apples in terms of $x$ using the given percentage. We add the number of rotten apples and the number of apples in good condition and then equate to $x$. We solve the equation to get the required result.\[\]

Complete step by step answer:
We know that when we say $p\%$ it means there are $p$ number of items out of every 100 items. When we say $p\%$ of $a$ it means there are $\dfrac{p}{a}\times 100$ number of items. \[\]
Let us suppose the total amount of apples in the basket is $x$. We are given that 12% of the apples are rotten. So the number of rotten apples are $x\times \dfrac{12}{100}=\dfrac{12x}{100}$. We are also given that the rest of the apples are in good condition and there are 66 good apples. If we add the number of rotten apples and good apples we shall total the number of apples. So we have
\[\dfrac{12x}{100}+66=x\]
Let us multiply 100 in all the terms of the above equation and get ,
\[\begin{align}
  & \dfrac{12x}{100}+66=x \\
 & \Rightarrow 12x+6600=100x \\
\end{align}\]
 We subtract $12x$ from both side of the equation and get,
\[\Rightarrow 6600=88x\]
We divide 88 both side and get,
\[\begin{align}
  & \Rightarrow \dfrac{6600}{88}=\dfrac{88x}{88} \\
 & \Rightarrow x=\dfrac{6600}{88}=\dfrac{600}{8}=75 \\
\end{align}\]
So there are total 75 number of apples and hence the correct option is A.\[\]
Alternatively,
The total percentage is 100%. If 12% of apples are rotten, the percentage of apples in good condition is $100-12=88$. Let us assume the total number of apples is $x$. We are given that the number of apples in good condition is 66. So we have
\[\begin{align}
  & x\times \dfrac{88}{100}=66 \\
 & \Rightarrow 88x=6600 \\
 & \Rightarrow x=\dfrac{6600}{88}=75 \\
\end{align}\]

Note:
We note that we use this strategy when there are two items say X , Y and the total number of items is the sum of the number of items in X type say $n\left( X \right)$ and in Y say $n\left( Y \right)$ type. If $n\left( X \right)+n\left( Y \right)$ is not equal to the total number of items we cannot use the strategy. We also note that percentage on sum of two numbers is equal to sum of percentage of each number , in symbols $\% p\left( x+y \right)=\% p\times x+\% p\times y$ .