Algebra Quiz
In the following problems, try to find ways to apply algebra that minimize the work needed to solve for each solution. Look for places where variables can be reused to represent multiple unknowns or where equations can be combined to save valuable time. Answers on the bottom of the page. Q1.
The mass of a full jar of honey is 600 grams.
With half of the honey, the total mass is 350 grams.
What is the mass of the empty jar in grams? a. 100 b.150 c. 200. d.250
Q2.
It takes 5 cooks 5 hours to bake 5 pies. Assuming no change of rate, how many hours will it take 10 cooks to bake 10 pies?
a.1 b.2 c.5 d.10
Q3.
What is the cost of buying a pizza, a donut, and a piece of cake together?
Q4
Which is the largest?
a. 1110/1111 b. 2221/2223. c.3331/3334
Q5.
Let a, b , c and d be four numbers with
a < b c < d
The average of a and b is c. The average of c and d is b
If d - a = 60, what is the value of b - c ?
Explanation Q1:
The full jar of honey has a mass of 600 g and the half-full jar has a mass of 350 g. Therefore, the mass of honey necessary to fill half the jar would be 600 g−350 g=250 g.
Therefore, the jar must have mass 350 g - 250 g = 100 g.
Explanation Q2:
The number of pies baked is directly proportional to both the number of cooks and the number of hours, so if either of those quantities is scaled, the number of pies baked will be scaled by the same amount. It is important to note that only one of them has to change the number of pies. It takes 5 cooks 5 hours to bake 5 pies. If we double the number of cooks then we'll double the number of pies made in the same amount of time. So it takes 10 cooks 5 hours to bake 10 pies.
Explanation Q3:
Rather than solve for the price of each food item individually, we can solve for the price of all three together at once. In total, each of the food items appears in the three known equations twice. So if we add all three together, we find that the sum of two pizzas, two donuts, and two pieces of cake is equal to 111+ 113 + 30 = 254 Rs So the cost of one of each food item is 254/2 = 127Rs.
Explanation Q4:
We can rewrite each of the fractions.
The question now becomes which one is the smallest.
We can compare the first two fractions. 1/1111 = 2/2222 > 2/2223, so the first fraction is definitely not the smallest.
Now we compare the last two fractions. 2/2223 = 6/6669 and 3/3334 = 6/6668. Since 6/6669 < 6/6668, we have 2/2223 < 3/3334, so 2/2223 is the smallest of the fractions.
That means 2221/2223 is the largest fraction.
Explanation Q5:
We can represent the values of a b, c, and d on the number line to help us determine their differences.
Since c is the average of a and b, it will lie halfway between them on the number line.
Since b is the average of c and d, it will lie halfway between them on the number line. Since we already placed b and c on the number line, we can arrange this by placing d the same distance from b as c is, but on the opposite side of it.
We can now see that a, b, c, and d are equally spaced on the number line. Since the distance d - a = 60, we can divide this by 3 to see that the spacing between any two numbers immediately next to each other is 20.
So the value of b - c is the positive distance from b to c, which is 20.
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