The sum of three numbers is 98 . The third number is 8 less than the first. The second number is 3 times the third. What are the numbers?

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The sum of three numbers is 98 . The third number is 8 less than the first. The second number is 3 times the third. What are the numbers?

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Answer 1 · 2018-06-25T19:13:55

$n_{1} = 26$ $n_1 = 26$

$n_{2} = 54$ $n_2 = 54$

$n_{3} = 18$ $n_3 = 18$

Explanation:

Let the three numbers be denoted as $n_{1}$ $n_1$ , $n_{2}$ $n_2$ , and $n_{3}$ $n_3$ .

"The sum of three numbers is $98$ $98$ "

$[1] \Rightarrow n_{1} + n_{2} + n_{3} = 98$ $[1] => n_1+n_2+n_3 = 98$

"The third number is $8$ $8$ less than the first"

$[2] \Rightarrow n_{3} = n_{1} - 8$ $[2] =>n_3 = n_1 - 8$

"The second number is $3$ $3$ times the third"

$[3] \Rightarrow n_{2} = 3 n_{3}$ $[3] =>n_2 = 3n_3$

We have $3$ $3$ equations and $3$ $3$ unknowns, so this system may have a solution that we can solve for. Let's solve it.

First, let's substitute $[2] \to [3]$ $[2] -> [3]$

$n_{2} = 3 (n_{1} - 8)$ $n_2 = 3(n_1 - 8)$

$[4] \Rightarrow n_{2} = 3 n_{1} - 24$ $[4]=> n_2 = 3n_1 - 24$

We can now use $[4]$ $[4]$ and $[2]$ $[2]$ in $[1]$ $[1]$ to find $n_{1}$ $n_1$

$n_{1} + (3 n_{1} - 24) + (n_{1} - 8) = 98$ $n_1 + (3n_1-24) + (n_1-8) = 98$

$n_{1} + 3 n_{1} - 24 + n_{1} - 8 = 98$ $n_1 + 3n_1 - 24 + n_1 - 8 = 98$

$5 n_{1} - 32 = 98$ $5n_1 -32 = 98$

$5 n_{1} = 130$ $5n_1 = 130$

$[5] \Rightarrow n_{1} = 26$ $[5]=>n_1 =26$

We can use $[5]$ $[5]$ in $[2]$ $[2]$ to find $n_{3}$ $n_3$

$n_{3} = 26 - 8$ $n_3 = 26 - 8$

$[6] \Rightarrow n_{3} = 18$ $[6]=>n_3 = 18$

Lastly, we can use $[6]$ $[6]$ in $[3]$ $[3]$ to find $n_{2}$ $n_2$

$n_{2} = 3 (18)$ $n_2 = 3(18)$

$[7] \Rightarrow n_{2} = 54$ $[7]=>n_2 = 54$

Hence, our solution from $[5], [6], [7]$ $[5], [6], [7]$ is:

$n_{1} = 26$ $n_1 = 26$

$n_{2} = 54$ $n_2 = 54$

$n_{3} = 18$ $n_3 = 18$

Answer 2 · 2018-06-25T19:41:23

first no. =26; second no.=18;third no.=54

Explanation:

let a= first no.;b= second no. and c= third no.

now given(The third number is 8 less than the first. )
then, $c = a - 8$ $c=a-8$
also given(The second number is 3 times the third)
then, $b = 3 \cdot c; \Rightarrow 3 \cdot (a - 8); \Rightarrow 3 \cdot a - 24$ $b=3*c ;=>3*(a-8); =>3*a-24$
now adding
$a + b + c = 98$ $a+b+c=98$ (given)
putting the values of b and c

$\Rightarrow a + 3 \cdot a - 24 + a - 8 = 98$ $=>a+3*a-24+a-8=98$
$\Rightarrow 5 \cdot a - 32 = 98$ $=>5*a-32=98$
$\Rightarrow 5 a = 130; \Rightarrow a = 26$ $=>5a=130;=>a=26$

now $c = a - 8; \Rightarrow 26 - 8; \Rightarrow 18$ $c=a-8; =>26-8 ; =>18$
and $b = 3 \cdot c; \Rightarrow 3 \cdot 18; \Rightarrow 54$ $b=3*c ; =>3*18; =>54$

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The sum of three numbers is 98 . The third number is 8 less than the first. The second number is 3 times the third. What are the numbers?

2 Answers

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