There were five people in group A, with a mean lQ of 100, and five in group B, with a mean IQ of 74. When Jack left group A for group B, both means increased, and the total of the two new means was 180. What Is Jack's IQ?

Aug 29, 2016

Jack's IQ is $80$.

Explanation:

Suppose that, Jack's IQ is $j$, and those of the rest of the $4$

persons in group A, be, ${x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}$.

Accordingly, $\frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} + j}{5} = 100. \ldots \ldots \ldots . \left(1\right)$.

After, leaving the group by Jack, the new mean for the group A is

$= \frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}}{4.} \ldots \ldots \ldots \ldots \ldots \ldots . \left(1 '\right)$.

Let ${y}_{i} , 1 \le i \le 5 ,$ be the IQs of $5$ persons of group B. Since, the

average IQ of Group be is $74$, we have,

$\frac{{y}_{1} + {y}_{2} + {y}_{3} + {y}_{4} + {y}_{5}}{5} = 74. \ldots \ldots \ldots . . \left(2\right)$

Finally, after Jack's joining the group B, since it has now $6$ members, the new mean becomes,

$\frac{{y}_{1} + {y}_{2} + {y}_{3} + {y}_{4} + {y}_{5} + j}{6.} \ldots \ldots \ldots \ldots \ldots \ldots \left(2 '\right)$

Using $\left(1 '\right) \mathmr{and} \left(2 '\right)$ in what is given, we get,

$\frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}}{4} + \frac{{y}_{1} + {y}_{2} + {y}_{3} + {y}_{4} + {y}_{5} + j}{6} = 180$, or,

3((x_1+x_2+x_3+x_4)+2(y_1+y_2+y_3+y_4+y_5+j)=2160...(3)

Here, $\left(1\right) \Rightarrow \left({x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}\right) = 500 - j$, and,

$\left(2\right) \Rightarrow \left({y}_{1} + {y}_{2} + {y}_{3} + {y}_{4} + {y}_{5}\right) = 370$.

Utilising these in $\left(3\right)$,

$3 \left(500 - j\right) + 2 \left(370 + j\right) = 2160$

$\therefore 1500 - 3 j + 740 + 2 j = 2160$, i.e.,

$j = 80$.

Hence, Jack's IQ is $80$.

Enjoy maths.!