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The following figure shows the data of students enrolled in $5$ years $(2014\;\text{to}\; 2018)$ for two schools $P$ and $Q$. During this period, the ratio of the average number of the students enrolled in school $P$ to the average of the difference of the number of students enrolled in schools $P$ and $Q$ is _______.

1. $8 : 23$
2. $23 : 8$
3. $23 : 31$
4. $31 : 23$

We can calculate the number of students enrolled in both schools during the given period of time,

• The total number of students enrolled in school $P\text{(in 5 years)} = (3 + 5 + 5 + 6 + 4) \times 1000 = 23000$
• Average number of students enrolled in school $P = \frac{23000}{5}$
• And, the total number of students enrolled in school $Q\text{(in 5 years)} = (4 + 7 + 8 + 7 + 5) \times 1000 = 31000$
• Average number of students enrolled in school $Q = \frac{31000}{5}$
• Average of the difference of the number of students enrolled in school $P$ and $Q = \frac{31000}{5} – \frac{23000}{5} = \frac{8000}{5}$

Now the required ratio $= \dfrac{\frac{23000}{8}}{\frac{8000}{5}} = \frac{23}{8}.$

$\therefore$ The ratio of the average number of the students enrolled in school $P$ to the average of the difference of the number of students enrolled in schools $P$ and $Q= 23:8.$

$\textbf{Short Method:}$  While calculating the ratio, common entities are cancelled.

• The total number of students enrolled in school $P=3 + 5+ 5 + 6 + 4 = 23$
• The total number of students enrolled in school $Q=4 + 7 + 8 + 7 + 5 = 31$

The difference between the number of students enrolled in school $P$ and school $Q=31 - 23 = 8$

Now the required ratio $= 23:8.$

So, the correct answer is $(B).$

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Total number of students enrolled in school P:   $3K+5K+5K+6K+4K=23K$

Total number of students enrolled in school Q:  $4K+7K+8K+7K+5K=31K$

Difference between the number of student enrolled in school P & school Q :  $31K-23K=8K$

Required ratio :  $\frac{23K}{8K}$

Option B is correct.
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