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Nata [24]
3 years ago
12

A. 109.45 B. 115.14 C. 117.85 D. 154.13

Mathematics
1 answer:
frozen [14]3 years ago
4 0

Answer: C

Step-by-step explanation: i dont know srry

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The sum of four siblings ages is 71. The second child’s age is twice the youngest, the third child’s age is twelve more than the
user100 [1]

Answer: 9, 18, 21, and 23 respectively

Step-by-step explanation:

To solve any problem like this, you want to assign a variable to each child because you dont know their age. So lets write an equation and assign variables to these children.

r = first child (youngest)

x = second child

y = third child

z = fourth child (oldest)

We know that r + x + y + z = 71

We also are given info on the comparisons of their ages.

r = r

x = 2r (2 times the youngest)

y = 12 + r (12 more than the youngest)

z = 3r - 4 (4 less than 3 times the youngest)

That was the easy part tbh, now we have to solve for 'r'. We can see that all the ages are a function of the same variable (r), and we know the sum of them is equal to 71. So lets solve for r and then plug that r value into each equation we just derived.

<u>Child #1 (youngest)</u>

r+2r+(12+r)+(3r-4)=71\\3r+12+r+3r-4=71\\7r+12-4=71\\7r+8=71\\7r=63\\r=9yrs

<u>Child #2</u>

x=2*9=18yrs

<u>Child #3</u>

y=12+r=12+9=21yrs

<u>Child #4 (oldest)</u>

z=3r-4=(3*9)-4=27-4=23yrs

<u>Fact Check:</u>

9+18+21+23=71

8 0
3 years ago
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