CAT Algebra Practice Problems

CAT Algebra Practice Problems

CAT algebra practice problems are an important part of CAT prep. After all, algebra is about 25% of your grade in CAT Quant.

The 5 Types of Algebra Problems on the Current IIM-CAT Exam

In the most recent official CAT mock test online, algebra questions fell into 5 general categories:

  • Rate of speed
  • Work rate
  • Solving for a single variable
  • Ratios and proportions
  • Function problems

Below, we’ll do one practice problem for each type of CAT algebra question. Then at the end of the post, there will be an answer key and answer explanations.

(1) Rate of speed

If a truck drives at 50 km/hr, it will get to its destination by 7pm. However, if the truck instead drives at 75 km/hr, it will reach its destination by 5pm. At what speed, in km/hr, must the truck travel in order to get to its destination by 6pm?

Enter your answer here: ___________

(2) Work rate

8 farmers can construct a barn in 7 days. In how many days could 14 farmers construct the same barn?

A) 5
B) 7
C) 4
D) 3
E) None of these

(3) Solving for a single variable

20% of 80% of 1/5 of X = 200

A) 6250
B) 6180
C) 6075
D) 6200

(4) Ratios and proportions

A painter mixes 30 ounces of water with 60 ounces of pigment. After using one-third of the paint, he adds in more water to replenish the amount of mixture he used. What is the current proportion of water to pigment?

A) 3:4
B) 2:3
C) 1:2
D) 5:4

(5) Function problems

Let f(x) be a function satisfying the following for all real x,y: f(x)f(y) = f(xy). If f(3)=6, then what is the value for f(1/3)?

A) 0
B) 1/6
C) 1/3
D) 1

Answer key, with explanations

1) 60

Explanation:

  • Original text of the problem:
    If a truck drives at 50 km/hr, it will get to its destination by 7pm. However, if the truck instead drives at 75 km/hr, it will reach its destination by 5pm. At what speed, in km/hr, must the truck travel in order to get to its destination by 6pm?

To solve this, remember the formula for distance problems. Distance = rate of speed * time, or D = rt.

We have the “r” variable already. Two “r” variables, in fact. If the truck is going 50 km/hr, D = 50(t). And if the truck is going 75 km/hr, D = 75(t). Now, we also know that at 75 km/hr, the truck arrives 2 hours earlier than it does at 50 km/hr. And we know that the truck is traveling the same distance regardless.

So, the distance is the same regardless of speed, but at 75 kph the travel time is two hours less. This means we can say that D = 50(t), and also that D = 75(t-2). And since these two distances are equal, we can also say that 50(t) = 75(t-2). Solve for t, and you can know how long it took the truck to reach its destination at 50 kph:

50(t) = 75(t-2)
50(t) = 75t – 150
50(t) + 150 = 75t
150 = 25t
6 = t
t = 6

OK, so now we know the time. If the truck travels at a speed of 50 kph, it goes the needed distance in 6 hours, arriving at 7pm. But what we want to know is what speed it will take to get the truck there at 6pm, one hour earlier.

So to solve for that speed, AKA rate, AKA variable “r”, we rearrange D = rt and plug our numbers in:

D = rt
r = D/t
r = D (t-1)
r = (rt)/(t-1) *Remember, D = rt. We don’t have the value for Distance, but we do have the values for rate of speed and time. So in this step, we replace D with rt so that we can plug in those values.
r = (50t)/(t-1)
r = (50*6)(6-1)
r = 300/5
r = 60 km/hr

So there’s your answer!

2) C

Explanation

  • Original text of the problem:
    8 farmers can construct a barn in 7 days. In how many days could 14 farmers construct the same barn?

If 8 farmers can construct a barn in 7 days, it would take one farmer 8 times as long. 8*7 = 56. If 56 is the work rate of a single farmer, you can divide 56 by 14 to get the work rate of 14 farmers. 56/14 = 4 days.

3) A

  • Original text of the problem:
    20% of 80% of 1/5 of X = 200

Here, you need to turn the percentages into fractions, and you need to recognize that the word “of” means “multiplied by.” (“Of” always indicates multiplication in English language story problems.)

So we change 20% to 1/5 and 80% to 4/5. Then we change the equation to: (1/5)*(4/5)*(1/5)*X = 200. From there, here are the operations:
(1/5)*(4/5)*(1/5)*X = 200
(4/125)X = 200
4X/125 = 200
4X = 200(125)
4X = 25000
X = 6250

4) D

  • Original text of the problem:
    A painter mixes 30 ounces of water with 60 ounces of pigment. After using one third of the paint, he adds in more water to replenish the amount of mixture he used. What is the current proportion of water to pigment?

You solve this in a three-step process. What you need to pay attention to is the exact amount of liquid the painter has: (1) at the start, (2) after he or she uses 1/3 of the paint, and (3) after the amount of solution is replenished. The ratio of water to pigment is 1:2.

At (1), the painter has 90 ounces of solution (30 ounces of water plus 60 ounces of pigment).

Then at (2) when a third of the paint is gone, the painter has lost 1/3 of the water, which is 10 ounces. And 1/3 of the pigment, or 20 ounces has been lost. This is a total loss of 30 ounces from the solution. There were originally 90 ounces. Take away 30 to get 60 ounces. The ratio is still the same. 20 ounces of water to 40 ounces of pigment is still a ratio of 1:2.

Now we’re at (3). To get the 60 ounces of solution back up to 90 ounces, the painter adds 30 ounces of water. There were 20 ounces of water in the solution in (2). Add an extra 30 ounces, and you have 50 ounces of water. No pigment has been added since step (2), so there are still 40 ounces of pigment. The ratio of water to pigment is now 50:40, which can be simplified to 5:4.

5) B

  • Original text of the problem:
    Let f(x) be a function satisfying f(x)f(y) for all real x,y. If f(3)=6, then what is the value for f(1/3)?

Remember that when you set a function–f(3)=6 in this case–the number of the function will always lead to the result after the equal sign. So in this problem, whenever f is set to 3, which is expressed as f(3), the result will always be 6 no matter what other variables you add in the equation.

In other words, if you hold additional f values as also true, they will interact with f(3) to result in 6, regardless.

So we’re introduced to an additional f value, f(1/3). We know that f(3) equals 6. But we’re not told what f(1/3) equals. What must it equal, in order to always lead to an answer of 6 when it’s used in an equation with f(3)?

We can find this when we look at the other stated property of f in this problem. We’re told that f(x)f(y) = f(xy).

Since we’ve been given two values of f, 3 and 1/3, we can plug those values in as x and y for f(x)f(y) = f(xy). That looks like:

f(3)f(1/3) = f[3*(1/3)] = f(1)

So, when we apply apply f to the equation, we get f(1). But that’s not the true end result. Remember, the real solution needs to be 6, because we’ve been told that anytime 3 is a true value for f, the end result must be 6. In other words, f(1) must have a result of 6, because it must lead to 6 when used with f(3), as seen above. We can find the true value of f(1) using– you guessed it– algebraic operations:

f(1) = 6
f(1/6) = 6/6
f(1/6) = 1
f(1) = 1/6

So we found that f(1/3), when multiplied by f(3) in the equation, equals f(1). Then we found that f(1), if it is used alone in this set of equations, will equal 1/6. In this way, if you plug in the numbers, f(3) can always equal 6, as the set requires.

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