Question 29 & 30 Sum of squares of three numbers is 138. Sum of their products with each other is 131. Find the sum of numbers. Two candidates contested an election .7500 votes were cast in an election. Out of these 20% were invalid. 55% votes were scored by one candidate. Find the number of votes scored by each candidate.

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Question 29 & 30 Sum of squares of three numbers is 138. Sum of their products with each other is 131. Find the sum of numbers.

Solution

Let the numbers be ‘x’, ‘y’ and ‘z’;

According to first condition sum of squares is 138;

$\Rightarrow x^{2}+ y^{2}+ z^{2}=138----(1)$

According to 2nd condition, sum of products of two numbers taken at a time is equal to 131, so;

$\Rightarrow xy+yz+xz=131----(2)$

To solve this we take sum of numbers and then take square of their algebraic sum!

$\Rightarrow (x+y+z)^{2}= x^{2 }+ y^{2}+ z^{2}+2(xy+yz+zx)-----(3)$

Put values from eq (1) and (2) in eq (3);

$\Rightarrow (x+y+z)^{2}=138+2(131)$

$\Rightarrow (x+y+z)^{2}=138+262$

$\Rightarrow (x+y+z)^{2}=393$

Taking square root on both sides;

$\sqrt{(x+y+z)^{2}}=\sqrt{393}$

$\Rightarrow x+y+z = 19.82$

rounding off

$\Rightarrow x+y+z = 20$

so, Their sum is 20

Question 30 : Two candidates contested an election .7500 votes were cast in an election. Out of these 20% were invalid. 55% votes were scored by one candidate. Find the number of votes scored by each candidate.

Solution:

Here,

$Total Votes=7500$

$Invalid Votes = \frac{20}{100}\times 7500$

$Invalid Votes = \frac{1}{5}\times 7500$

$Invalid Votes = 1500$

So,

$Valid Votes = 7500-1500=6000$

$Votes By 1st Candidate= \frac{55}{100}\times 6000$

$Votes By 1st Candidate= 3300$

$Valid Votes By 2nd Candidate= 6000-3300=2700$

So, Valid Votes by 2nd candidate were 2700

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