- Home
- Standard 13
- Quantitative Aptitude
4.Average
hard
In a class, average height of all students is '$a$ ' cms. Among them, average height of $10$ students is '$b$' cms and the average height of the remaining students is '$c$' cms. Find the number of students in the class. (Here $a>c$ and $b>c$ )
A
$\frac{(a(b-c))}{(a-c)}$
B
$\frac{(b-c)}{(a-c)}$
C
$\frac{(b-c)}{10(a-c)}$
D
$\frac{10(b-c)}{(a-c)}$
Solution
$\frac{ T }{ N }=a$
$T =a N$
and $b=$ average of $\frac{10}{10}$
average of $10=10 b$
$c=\frac{(\text { average of } n-10)}{(n-10)}$
average of $n-10=(N-10) \times c$
and
T $=$ average of $(N-10)+$ average of 10
Substituting.
$a N =( N -10) \times c+10 b$
$a N-c N=10 \times(b-c)$
$(a-c) \times N=10(b-c)$
$N =10 \times\left(\frac{(b-c)}{(a-c)}\right)$
Standard 13
Quantitative Aptitude
