Today we have two algebra questions suitable for most students from the fifth grade and above – Algebra for all ages!
Solve for the variables A through F in the equations below, using the digits from 0 through 5. Every digit should be used only once. A variable has the same value everywhere it occurs, and no other variable will have that value.
A =3 B = 2 C = 0 D = 4 E = 1 F = 5
First, let’s look for the identity properties.
Remember x + 0 = x. In our problem, the second equation is B + C = B, so C must be 0.
Also, x ∙ 1 = x. In our problem, the third equation is D * E = D, so E = 1.
**************
Next, let’s look at the first equation: A + A + A = A².
In other words, 3A = A ∙ A. Divide both sides by A and A = 3.
**************
Now that we know A and E, we can solve the fourth sentence, A – E = B.
3 – 1 = 2. B = 2.
**************
The 5th equation says, B2 = D. 22 = 4, so D = 4.
**************
Finally, D + E = F. F = 4 + 1 = 5.
A =3 B = 2 C = 0 D = 4 E = 1 F = 5
Alicia was paid $125 for babysitting five days after school for the Smith family. Each day Mrs. Smith paid her $3 more than the day before. How much money did she earn on the first day?
$19
On the first day Alicia earned $x.
On the second day, she earned $(x + 3).
On the third day, she earned $(x + 6)
On the fourth day, she earned $(x + 9)
On the fifth day, she earned $(x + 12)
5x + 30 = 125
5x + 30 – 30 = 125 – 30
5x = 95
5x ÷ 5 = 95 ÷ 5
x = 19
Alicia earned $19 on the first day.
Check: (19 + 22 + 25 + 28 + 31 = 125)
Do you have a question to ask the math teacher? Our math teacher, Mary Lou loves puzzles and riddles! Please send your request to info@mytutorlesson.com You just may see it posted here on a future date.
