Sum of the numbers | AMC-10A, 2001 | Problem 16

Try this beautiful problem from Algebra based on Sum of the numbers....

Sum of the numbers - AMC-10A, 2001- Problem 16

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

$5$
$20$
$30$
$25$
$36$

Key Concepts

algebra

Mean

Median

Check the Answer

Answer: $30$

AMC-10A (2001) Problem 16

Pre College Mathematics

Try with Hints

Mean of three numbers means average of three numbers......

Let $x$ be the mean of three numbers then we can say that the least of the numbers is $m-10$ and the greatest is $m + 15$

Can you now finish the problem ..........

Given that The median of the three numbers is $5$. Now "median" means the middle of the three numbers

so we can write $\frac{1}{3}[(m-10) + 5 + (m + 15)] = m$,

$\Rightarrow m=10$

Therefore The sum of three numbers are $3(10)=30$

Medians of triangle | PRMO-2018 | Problem 10

Try this beautiful problem from Geometry based on medians of triangle

Medians of triangle | PRMO | Problem 10

In a triangle ABC, the medians from B to CA is perpendicular to the median from C to AB. If the median from A to BC is 30,determine $(BC^2 +AC^2+AB^2)/100$?

$56$
$24$
$34$

Key Concepts

Geometry

Medians

Centroid

Check the Answer

Answer:$24$

PRMO-2018, Problem 10

Pre College Mathematics

Try with Hints

We have to find out $(BC^2 +AC^2+AB^2)/100$. So, we have to find out $AB, BC, CA$ at first. Now given that, the medians from B to CA is perpendicular to the median from C to AB and the median from A to BC is 30,

So clearly $\triangle BGC$,$\triangle BGF$,$\triangle EGC$ are right angle triangle.Let $CF=3x$ & $BE =3y$ then clearly $CG=2x$ & $BG= 2y$ given that $AD=30$ SO $AG=20$ & $DG =10$ (as $G$ is centroid, medians intersects at 2:1). Therefore from pythagoras theorem we can find out $BC,BF,CE$ i.e we can find out the value $AB,BC,CA$

Can you now finish the problem ..........

$CE^2=(2x)^2+y^2=4x^2 +y^2$

$BF^2=(2y)^2+x^2=4y^2+x^2$

Also, $CG^2+BG^2=BC^2$ $\Rightarrow 4x^2 + 4y^2={20}^2$ $\Rightarrow x^2+y^2=100$

$AC^2=(2CE)^2=4(4x^2+y^2)$

$AB^2=(2BF)^2=4(4y^2+x^2)$

Can you finish the problem........

$(BC^2 +AC^2+AB^2)=20(x^2+y^2)+20^2=2400$

so, $(BC^2 +AC^2+AB^2)/100$=24

Sum of the numbers | AMC-10A, 2001 | Problem 16