Hello,
We don't need algebra to solve this.
I suppose the ages are 1,2,3,4
Their sum is 1+2+3+4=10.
To obtain 94 we must add 94-10=84 for 4 persons =>84/4=21 to each
Ages are
1+21=22
2+21=23
3+21=24
4+21=25
But if really want an algebric resolution:
Let's x the age of the youngest
x+1,x+2,x+3 the others.
x+(x+1)+(x+2)+(x+3)=94
==>4x+6=94
==>4x=88
==>x=22
Ages are 22,23,24,25.
The oldest friend is 25 years old.
X=number of boys
y=number of girls.
We can suggest the following system of equations:
x+y=33
x/y=3/8
We solve this system of equations by substitution method.
x/y=3/8 ⇒x=3y/8
(3y/8)+y=33
least common multiple=8
3y+8y=33*8
11y=33*8
y=(33*8)/11
y=3*8
y=24
x+y=33
x+24=33
x=33-24
x=9
number of boys /total number of students=9/33=3/11
There are 9 boys and 24 girls in the class.
Answer: C: 9. since 3/11 of the students are boys.
Answer:
120 = 28 + 12(½) + X
Where X is the no. of pages of articles
120 = 28 + 6 + X
X = 120 - 34
X = 86
15 tickets = $105
105 ÷ 15 = 7
1 ticket costs $7.
$7 x 12 = $84
12 tickets cost $84
Answer: $84 (Answer C)
Note that a^-3 = 1/a^3
9^3 = 729
1/9^3 = 1/729
Solution: 1/729
