The post Practice Probability for CAT appeared first on Common Admission Test Blog.

]]>Two fair twelve-sided dice are thrown. What is the probability of getting a sum of 15?

A. 1/16

B. 5/72

C. 1/12

D. 7/24

In a certain precinct, 40% of the population identifies with the Red Party, 30% with the Blue Party, and the rest have no party affiliation. Of the Red Party, 50% are male; of the Blue Party, 60% are male; and of independents, 30% are male. What is the probability that a randomly selected female voter is neither Red Party nor Blue Party?

A. 20/41

B. 23/49

C. 19/52

D. 21/53

There are two bags of marbles on the table. The first bag has 6 blue and 3 yellow marbles, and the second bag has 5 blue and 3 yellow marbles. One bag is selected at random and two marbles chosen from the bag, one after the other without replacement. What is the probability that both marbles are yellow?

A. 5/56

B. 5/28

C. 7/54

D. 8/61

Two players take turns trying to eliminate the other in a game of chance. Player A has a 40% chance of eliminating Player B each turn. Player B has a 30% chance of eliminating Player A each turn. The game continues until one player is eliminated. If A goes first, what is the probability that A wins the game?

A. 19/29

B. 19/30

C. 20/29

D. 23/30

An ATM PIN number consists of 4 digits (0-9). What is the probability that a randomly assigned ATM PIN number has digits that add to 9?

A. 103/10000

B. 31/1000

C. 11/500

D. 21/1000

What is the probability that a 4 digit number that is a multiple of 11 will also be a multiple of 7?

A. 116/819

B. 1/7

C. 117/818

D. 13/90

In the card game of Poker, each player is dealt five cards from a standard pack of 52 cards. (In a standard deck, there are four different suits, and each suit contains 13 different denominations.) A three-of-a-kind is a Poker hand in which three cards have the same denomination while the other two are different denominations. What is the chance that you are dealt a three-of-a-kind on your first deal?

A. 88/4165

B. 97/6513

C. 13/1035

D. 15/3217

In how many ways can 5 distinct marbles be placed into 4 cups so that there is at least one marble in each cup?

A. 220

B. 240

C. 360

D. 480

Are you ready to check your work? First of all, here’s the list of correct answers.

**B****D****A****C****C****B****A****D**

First, there are 12^{2} = 144 possibilities for the outcome of rolling two 12-sided dice. Out of these, exactly 10 combinations result in a sum of 15:

3 + 12, 4 + 11, 5 + 10, 6 + 9, 7 + 8, 8 + 7, 9 + 6, 10 + 5, 11 + 4, and 12 + 3.

Thus the probability is 10/144 = 5/72.

Let *D*, *R*, and *I* be the events that a randomly selected voter is Red Party, Blue Party, and independent, respectively. Furthermore, let *M* and *F* be the events that the voter is male and female, respectively.

This problem asks for *P*(*I* | *F*), that is, the probability that a randomly selected voter is independent, given that the voter is female.

What we are given is: *P*(*D*) = 0.4, *P*(*R*) = 0.3, and *P*(*I*) = 1 – 0.4 – 0.3 = 0.3.

Moreover, we have the conditional probabilities, *P*(*F* | *D*) = 0.5, *P*(*F* | *R*) = 0.4, and *P*(*F* | *I*) = 0.7. (These are the complementary probabilities based on the given information about males.)

Now use Bayes’ Theorem.

First work out the probabilities for each bag separately.

Bag 1: *B* = 6 and *Y* = 3. The probability of getting a yellow first is 3/9. Then, there is one less yellow marble, so the chance of getting a second yellow would be 2/8. Multiply the probabilities to get 1/12.

Bag 2: *B* = 5 and *Y* = 3. The same method yields: 3/8 × 2/7 = 3/28.

Next, scale each probability by 1/2 to account for the fact that one of the two bags was chosen at random. Add the resulting fractions.

So, 1/24 + 3/56 = 15/168 = 5/56.

The key to solving this problem is to break it down into the rounds in which Player A might win. Either Player A eliminates the opponent (0.4 chance) or he/she does not (0.6 chance). In the event that Player A does not eliminate B in a given round, then we must also factor in the chance that Player B does **not** eliminate A on his/her turn (1 – 0.3 = 0.7 chance).

The following table summarizes the results in the first four rounds.

Round | Probability of Player A winning in this round | Partially simplified expression |
---|---|---|

1 | 0.4 | 0.4 |

2 | (0.6)(0.7)(0.4) | (0.42)(0.4) |

3 | (0.6)(0.7)(0.6)(0.7)(0.4) | (0.42)^{2}(0.4) |

4 | (0.6)(0.7)(0.6)(0.7)(0.6)(0.7)(0.4) | (0.42)^{3}(0.4) |

Notice that the probability of A winning in round *n* is exactly (0.4) × (0.42)^{n – 1}. Add up all of these individual terms to find the probability of winning overall.

*P*(A wins) = Σ_{n ≥ 1} (0.4) × (0.42)^{n – 1}

The sum can be computed using the formula for a sum of a Geometric Series.

First note that there are a total of 10^{4} = 10000 possible PIN numbers. Assume that each combination is equally likely.

Now we just need to compute the number of 4-digit combinations that add to 9. This is an example of a “ball-and-bin” problem (see Permutations and Combinations for CAT for a quick review).

The number 9 is the sum of nine ones. Imagine distributing these nine ones into four positions, corresponding to the four digits in each ATM. For example, we could place two ones in the first position, none in the second, six in the third, and one in the last position:

1+1 | | 1+1+1+1+1+1 | 1

This corresponds to the PIN number 2061.

Thus, with the 9 ones being “balls” and the 4 digit positions being the “bins,” the total number of ways to do this is ^{12}C_{3}.

Therefore, the answer is 220/10000 = 11/500

Let’s make a careful count of the multiples. The first 4-digit number, 1000, is not a multiple of 11. However, because 1000 ÷ 11 = 90.9, we know that 11 × 91 = 1001 is the first multiple of 11 to consider.

Then since 9999 ÷ 11 = 909, we can list all 4-digit multiples of 11:

11 × 91, 11 × 92, 11 × 93, …, 11 × 908, 11 × 909

There are 909 – 91 + 1 = 819 of these.

Next, we must locate all multiples of 7 within this list. Because 11 and 7 are **coprime**, that is, they have no common factors, the trick is to look for multiples of 7 in the numbers 91, 92, 93, …, 908, 909.

Now 91 = 7 × 13 happens to be a multiple of 7. The next one is 98, and so on. Check 909 by dividing: 909 ÷ 7 = 129.86. Therefore, 7 × 129 = 903 is the final multiple of 7 in the list.

So that gives a total of 129 – 13 + 1 = 117 multiples of 7 within the list of multiples of 11.

Thus, the fraction of multiples of 11 that are also multiples of 7 is equal to: 117/819 = 1/7.

This is a popular kind of problem which requires a careful step-by-step counting procedure.

First of all, the total number of five-card hands is: ^{52}C_{5}. This number is rather large:

Next, we must count the hands that qualify as three-of-a-kind.

- Choose one of the 13 denominations to be the repeated one: 13 choices.
- Of that denomination, pick 3 of the 4 cards to be in your hand:
^{4}C_{3}= 4 combinations. - Of the remaining 52 – 4 = 48 cards, choose two to fill out the hand:
^{48}C_{2}= (48 · 47)/(2 · 1) = 1128. However, we have actually over-counted in this step! From 1128, take away the choices that represent having the same denominations in those final two cards. Remember, there are now 12 denominations to work with, so that’s 12 ×^{4}C_{2}= 12 × 6 = 72. So that leaves 1128 – 72 = 1056.

Therefore, upon multiplying the number of choices at each step above, we obtain: 13 × 4 × 1056 = 54,912.

Finally, let’s put it all together. Divide the number of desirable hands by the total hands and reduce the fraction.

Ok, so this one isn’t really a probability problem. But it does display the kinds of subtle counting arguments that you often must perform when computing probabilities.

It helps to think about the situation before attempting to set up any formulas. The key is to first place 4 of the marbles into the 4 cups. There are ^{5}C_{4} = ^{5}C_{1} = 5 ways to pick these 4 marbles. Then there are 4! = 24 arrangements into 4 cups.

So far, we have 5 × 24 = 120 possibilities.

Finally, the fifth marble has to go into one of the 4 cups. There are clearly 4 choices for this step.

That gives 120 × 4 = 480 possibilities.

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]]>Permutations and combinations are just one small part of the CAT Quantitative section. For more tips, check out How to Study for the CAT Quantitative Ability Section.

Technically, a **permutation** is a way of rearranging items in a set, while a **combination** is a particular choice of some elements from a set. However, we see these concepts in problems, we usually mean the associated formulas and methods for counting items from a set.

A **permutation** is an arrangement of some number of distinct items (no repetitions) in which the order matters.

Suppose that you and your two closest friends decide to line up for a selfie. In how many ways can this be done?

For the sake of brevity, let’s name the three people A, B, and C. First of all, one of the three, A, B, or C, can choose to be on the left. That leaves two choices for the middle spot. Finally, once the first two people have been placed, the final person must go on the right.

Therefore, there are (3 choices) × (2 choices) × (1 choice) = 6 total permutations.

This generalizes to any number of items. The number of permutations of *n* distinct items is exactly *n*! (“*n* factorial”), where

Notice that 0! = 1. At first this may seem strange, but think about it like this: How many ways are there to arrange zero objects? Exactly *one* way!

Suppose ten people enter a race. How many possible arrangements of first, second, and third place winners?

It sounds like a permutation, but this time we don’t have to arrange all ten runners — just the top three. Nevertheless, the process starts the same.

- The first place winner must be one of the 10 runners (10 choices).
- After deciding first place, there are 9 runners to choose from to establish second place (9 choices).
- Once first and second are chosen, now there are 8 runners to pick from for third (8 choices).

Therefore, there are a total of 10 × 9 × 8 = 720 possible arrangements.

In general, the number of permutations of *r* items from a set of *n* distinct elements is denoted * ^{n}P_{r}*. The formula is:

Combinations are much like permutations, except that the order does not matter. For example, we might ask how many combinations of three runners could be chosen out of the 10 entrants. Here we do not order the runners by first, second, or third places.

The key here is to start with a permutation number, and then divide by the number of ways to rearrange the smaller set of elements. In our example, we would have:

^{10}*P*_{3} / 3! = 720 / 6 = 120.

Using the notation * ^{n}P_{r}* for the number of combinations of

The methods developed above generalize to rearrangements of sets with repetition of any numbers of items. Suppose a set has *n* items, and *r*_{1} of them are of one type, *r*_{2} of another type, *r*_{3} of another type, and so on. How many ways are there to make an arrangement of this set?

The key is to first find the total number of permutations of the entire set (*n*!). Then (just as in a combination), divide off by the number of rearrangements of any items that are considered the same. Thus, the total number will be:

Ok, now that you know the basics, let’s explore how the concepts of permutations and combinations might show up on the CAT!

How many distinct ways can the letters of the word MISSISSIPPI be rearranged?

This is a permutation with repetition. There are *n* = 11 letters total, with the following repetitions:

- Four I’s:
*r*_{1}= 4 - Four S’s:
*r*_{2}= 4 - Two P’s:
*r*_{3}= 2

(We could also include *r*_{4} = 1 for the single M, but it does not affect the value.)

As you work out the number, look for cancellations to make your life easier.

In how many ways can 8 identical balls be placed into three bins? Each bin may contain any number of the balls, including none at all.

The “ball-and-bin” problems are fairly common on the CAT, showing up in a variety of ways. Fortunately, there’s a standard trick for this. Use a symbol (like |) to separate the 8 balls. For example, here’s one way to fill the bins:

* * | * * * * * | * *(2, 5, and 1, respectively, in each bin) *

Now we answer the equivalent question: Of the 10 symbols (8 balls plus 2 dividers), which two will be chosen as the dividers? That’s a combination number:

^{10}*C*_{2} = (10)(9)/(2) = 45.

How many 5-digit numbers whose only digits are 4’s and 3’s are multiples of 12?

This is an interesting combination (*no pun intended!*) of number theory and counting.

A whole number is a multiple of 12 if and only if it is a multiple of 3 and a multiple of 4. Multiples of 3 have digits that add to a multiple of 3. On the other hand, the last two digits of every multiple of 4 must be a multiple of 4.

So we need a string of five 4’s and 3’s that

- add up to a multiple of 3,
*and* - whose last two digits are a multiple of 4.

To get a sum of a multiple of three, there are only a few possibilities:

- All 3’s
- Two 3’s and three 4’s

But the first scenario does not permit a multiple of 4 in the last two digits. That leaves scenario 2. In fact, the only possibility is that the final two numbers are 44.

So now we just have to count 3-digit strings using the digits 3, 3, 4. This one is easy enough to count directly: 3 total! (But if you *really* want to use permutations and combinations, the appropriate count at this step is ^{3}*C*_{1} = 3.)

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]]>The post Practice Trigonometry for CAT appeared first on Common Admission Test Blog.

]]>First, let’s review the fundamentals.

Suppose that triangle ABC is right, and assume that ∠ C is the right angle. To find the **sine**, **cosine**, and **tangent** of ∠ A, just remember: SOHCAHTOA!

The tangent can also be defined in terms of sine and cosine — as can the three other trigonometric ratios. Let’s call *θ* = ∠ A.

You should also be familiar with the **Unit Circle**. Each point (*x*, *y*) on the circle gives the value of cosine and sine of the corresponding angle.

For example, look for the angle 2π/3 (radians). Based on the unit circle diagram, you know:

- cos 120° = -1/2.
- sin 120° = √ 3 /2.

The six basic trig ratios can be extended to functions defined on all (or most) real numbers by interpreting arbitrary angles on the unit circle. The key is that each trig function becomes *periodic*, which means that values repeat in equal intervals.

It’s important to know the periods, domains, and ranges of each function.

Perhaps the most famous and useful equation in all of mathematics is the Pythagorean Theorem. (For this and other essentials about triangles, check out: Triangle Properties to Know for the CAT).

When applied to the unit circle, we get a trio of useful trigonometric identities:

Of course there are a myriad of other trigonometric identities, including sums or differences of angles, half and double angle formulas, products-to-sums, and sums-to-products, to name a few.

For a summary list, check out this table of trigonometric identities.

Now let’s test our knowledge and skills!

- You are standing on the corner of a square whose side length is 25 feet. Standing on the opposite corner from you is a tall tree. The angle of elevation from your position to the top of the tree is exactly 60°. How tall is the tree?
A. 25√ 2

B. 25√ 3

C. 25√ 6

D. 50√ 3

- Tweedledee and Tweedledum are 100 m from each other. Between them there is a tower. Tweedledee notes that the top of the tower is at
*x*°, while Tweedledum records that the top of the tower is at*y*°. Which expression below correctly computes the height of the tower?A. 100 tan

*x*° tan*y*° / (tan*x*° + tan*y*°)B. 100 (tan

*x*° + tan*y*°) / (tan*x*° – tan*y*°)C. 100 (tan

*x*° + tan*y*°) / (tan*x*° tan*y*°)D. 100 tan

*x*° tan*y*° / (tan*x*° – tan*y*°) - What is the maximum value of 8 sin
*θ*+ 6 cos*θ*?A. 9.5

B. 10

C. 10.3

D. 10.8

- Suppose the angle of elevation of the top of a flag pole changes from
*x*° to 45° as you walk 15 m toward it. Assuming that*x*< 45, find the height of the flag pole in terms of*x*.A. 30(1 + tan

*x*°) / tan*x*°B. 45 tan

*x*° / (1 + tan*x*°)C. 15 tan

*x*° / (1 + tan*x*°)D. 15 tan

*x*° / (1 – tan*x*°)

**C.**First find the distance of the diagonal

*d*along the ground from corner to corner. Using Pythagorean theorem with sides 25 and 25, we get:25

^{2}+ 25^{2}=*d*^{2}2 × 25

^{2}=*d*^{2}*d*= 25√ 2 .Then to obtain the height

*h*of the tree, use the tangent ratio with angle 60°.tan 60° =

*x*/ (25√ 2 )√ 3 =

*x*/ (25√ 2 )*x*= 25√ 2 × √ 3 = 25√ 6**A.**First let

*d*represent the distance from Tweedledee to the tower. Then the distance from Tweedledum to the tower must be 100 –*d*. Next, let*h*be the height of the tower. So, there are two right triangles, one determined by Tweedledee and the other by Tweedledum, sharing the same height. Finally, use trigonometry to write two equations:Next, solve the first equation for

*d*:*d*tan*x*° =*h*→*d*=*h*/ tan*x*°Now plug this result into the second equation and isolate

*h*.**B.**Here, the clue is that 6 and 8 form part of a Pythagorean Triple: 6-8-10.

Let

*f*= 8 sin*θ*+ 6 cos*θ*, and divide both sides by the constant 10 to get:*f*/10 = (8/10) sin*θ*+ (6/10) cos*θ*Because 10 is constant, it’s equivalent to find a maximum value for

*f*/10.Next, we can assume that

*θ*is a first-quadrant angle (otherwise, either its sine value or cosine value would be negative). So construct a right triangle with angle*θ*.Now the trick is to compare this triangle to the one whose sides are in ratio 6 : 8 : 10. In fact, in a triangle whose adjacent side is 8 and opposite side is 6, then the hypotenuse would be 10. So, calling the angle of this new triangle

*φ*, we can identify:Then, sin

*φ*= 6/10 and cos*φ*= 8/10.Fortunately, these ratios already show up in our equation. (That’s why I picked the 6-8-10 triangle to compare to!) So we may substitute for the trig ratios:

*f*/10 = cos*φ*sin*θ*+ sin*φ*cos*θ*Next, using the sum of angles formula for sine, we could write:

*f*/10 = sin(*φ*+*θ*)What is the maximum value of this function? Remember, the range of sin

*x*is from -1 up to 1. So the maximum occurs when sin(*φ*+*θ*) = 1. This will happen as long as*φ*+*θ*= 90°. In other words, the unknown angle*θ*must be complementary to the angle of a 6-8-10 triangle. That makes our unknown triangle into a 6-8-10 triangle as well!But the bottom line is that we can find such an angle, and so the maximum sine value (1) will be obtained. Finally, use

*f*/10 = 1 (maximum) to find that*f*= 10.**D.**Let

*h*be the height of the flag pole. When you reach the point at which the angle of elevation is exactly 45°, then you are at the base of an isosceles right triangle with height*h*. Therefore, at that point, you are*h*meters from the base of the flag pole. Thus your original position was 15 +*h*meters from the pole.This information allows us to set up a trigonometric equation:

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]]>The post What are HCF and LCM? appeared first on Common Admission Test Blog.

]]>(Check out this handy Math Syllabus for CAT Exam to see what other mathematical topics are covered on the test!)

In order to understand HCF and LCM, first you need to know a thing or two about factors and multiples.

A **factor** of a number is any number that divides into it with zero remainder. For example, the factors of 15 are: 1, 3, 5, and 15. The number 2, for example, is not a factor of 15, because 15/2 = 7.5 is not a whole number (there is a remainder). Factors are always less than or equal to the original number.

**Multiples** are numbers that have the original number as a factor. In other words, a multiple of *N* is simply the product of *N* by another whole number *k* (that is, *k* × *N*). For example, the multiples of 5 are: 5, 10, 15, 20, 25, 30, … Multiples are always greater than or equal to the original number, and there are infinitely many of them.

For any whole number *N*, the numbers 1 and *N* itself are always factors. Any number that has *only* those two factors is called **prime**. Prime numbers become very important when computing HCF and LCM, because the quickest way to compute these items is to compare prime factorizations.

There are infinitely many primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, … Fortunately, you may only need to know the first five or six of these for most applications.

Every whole number has a unique **prime factorization**, that is, a list of the primes or prime powers that are factors of the number. The easiest way to find the prime factorization of a number is to use a **factor tree**.

To build a factor tree for a number *N*, follow these steps:

- If the first prime, 2, is a factor, then divide by 2.
- Place 2 on the left branch and the quotient on the right branch.
- Repeat as needed until the quotient no longer has 2 as a factor.
- Now, do steps 1–3 again on the current quotient, but with the next prime, 3.
- Repeat, using larger and large primes until the last quotient is itself prime.
- The primes on the “leaves” of the tree make up the prime factorization. Write your answer as a product of powers of primes.

Let’s see how it works for the number 168.

Therefore, 168 = 2^{3} × 3 × 7.

Now you try it for the number 90. (You can check your answer by reading through the next section.)

Let’s start with just two numbers, *M* and *N*, though the methods work for three or more numbers just as easily.

- Find prime factorizations of both numbers.
- To compute
**HCF**of*M*and*N*: For each prime number, select the**smaller**power of the prime from the two prime factorizations. If one of the numbers does not have a particular prime, then that prime power will be zero in the HCF. - To compute the
**LCM**of*M*and*N*: For each prime number, select the**larger**power of the prime from the two prime factorizations.

Ok, so let’s use our steps to compute the HCF and LCM of 168 and 90.

- 168 = 2
^{3}× 3 × 7 - 90 = 2 × 3
^{2}× 5 (Did you get the right prime factorization?) - HCF(168, 90) = 2 × 3 = 6
- LCM(168, 90) = 2
^{3}× 3^{2}× 5 × 7 = 2520

Note: If there are more than two numbers, then just choose the smallest prime powers overall to find HCF, or the largest prime powers overall to find LCM.

Ok, so why do we care about the highest common factor or least common multiple of a pair of numbers?

The HCF helps in factoring. For example, if you need to factor the polynomial 168*x*^{7} – 90*x*^{5}, then you first find the HCF of 168 and 90 (which we did above — it’s 6). Then you find the smallest power of the variable(s) showing up (which should remind you of the process of finding HCF for numbers, by the way). Then divide all terms by those items and write your answer appropriately:

168*x*^{7} – 90*x*^{5} = 6*x*^{5}(28*x*^{2} – 15)

On the other hand, the LCM is useful in combining (adding/subtracting) fractions. Let’s see how it works!

Suppose you need to simplify 55/168 – 7/90. First find the LCM of the denominators, 168 and 90. Luckily, we’ve already done that step above and found that LCM(168, 90) = 2520. Now you also need to know how many times each denominator goes into the LCM, because those multipliers must be used on the numerators as well.

2520 ÷ 168 = 15

2520 ÷ 90 = 28

Now convert both fractions to equivalent ones with 2520 as denominator.

Finally it’s your turn to practice the concepts. You can check your answers by scrolling down.

- A bowl full of candies can be redistributed into boxes each containing 4 candies, but there will be 1 left over. If put into boxes of 5 candies each, then there will still be 1 left over. Finally, if put into boxes of 6 candies each, there will again be 1 left over. What is the smallest number of candies that could have been in the bowl, assuming there is more than one piece of candy?
- Suppose two positive numbers add to 72. What is the largest possible HCF of the numbers?

**61**. Let*N*be the number of candies. The information implies that*N*– 1 is a multiple of 4, 5, and 6. Since we are looking for the least number possible, we need to compute LCM(4, 5, 6). We obtain*N*– 1 = 60, so*N*= 61.**36**. Let*x*+*y*= 72. Now if any common factors exits between*x*and*y*, then those factors must also be factors of 72. Because both numbers are positive, the largest such common factor must be half of 72, or*x*= 36 and*y*= 36. Finally, HCF(*x*,*y*) = HCF(36, 36) = 36.

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]]>The post CAT Interest Practice appeared first on Common Admission Test Blog.

]]>(Interest problems form an important part of the CAT Quantitative section. For more about what to expect on the CAT, check out this article: Math Syllabus for CAT Exam.)

Suppose you lend $100 to your friend, but you know that it will take her a few months to pay it back. To give her some incentive to pay back the money sooner, you tell her: “Every month that you have the money out, you have to add $5 to the payback amount.”

So, the *value* of the loan after one month is $105.

After two months, the value is $110; after three months, $115 — you get the picture.

The extra money ($5 per month that the loan is out) is called **interest** and gets added to the original amount (the **principle**) when the loan comes due. We call this **simple interest** because the interest grows at a constant rate over time. In our little example, there is a monthly simple interest rate of 5%.

Suppose the principle is *P* (dollars, pounds, euros, rupees, yen, or whatever the currency is in the problem). Let *t* stand for the time that the money is invested or lent (typically in years).

Finally, let *r* be the rate at which the interest grows. Make sure that *r* is expressed as a decimal rather than a percent. And check the units! If *t* is in years, then *r* must be an *annual* rate; if *t* is in months, then *r* should be a monthly rate.

Then we can compute the interest (*I*) and the total amount of the investment (*A*) using the following formulas:

- Interest:
*I*=*Prt* - Amount:
*A*=*P*+*Prt*=*P*(1 +*rt*)

Let’s work out a simple interest practice problem.

Suppose you lend $100 to your friend who agrees to pay you back with simple interest accruing at a monthly rate of 5%. If your friends does not pay any money back until a year later, what is the amount due at the end of that year?

Set up the simple interest formula. This problem asks for total amount due. Convert the rate of 5% to a decimal value: 0.05. The time will be *t* = 12 months.

*A* = *P*(1 + *rt*) = $100(1 + 0.05 · 12) = $100(1.6) = $160.

Simple interest may be fine for loans between friends, but no business or bank would ever use this formula when lending money. The problem is that simple interest does not grow fast enough to keep up with inflation.

In other words, there would be no incentive to pay back the amount in a timely manner — just wait until the value of money decreases enough!

Suppose that you loan another friend $100, again at 5% monthly interest, but this time you compute the amount due in a different manner. Every month that goes by without payment on the loan, you find the interest due and *add it back to the principle*.

For example, in the first month, both simple and compound interest loans will be the same: $105. Then in the second month, the first loan would just be $105 + $5 = $110, as we saw above. However, for the second friend, you would consider the new loan amount to be $105 and charge the amount of interest based on that new principle:

*A* = $105(1 + 0.05) = $105(1.05) = $110.25

It’s not a huge difference, $110.25 versus $110, but with monthly compounding, the amount will grow much quick over a long time period. The amount “snowballs” as higher balances will generate more and more interest over time.

Let’s see what the effect is after a full year.

End of Month Amount | Simple Interest | Compounding Interest |
---|---|---|

1 | $105 | $105 |

2 | $110 | $110.25 |

3 | $115 | $115.76 |

4 | $120 | $121.55 |

5 | $125 | $127.63 |

6 | $130 | $134.01 |

7 | $135 | $140.71 |

8 | $140 | $147.75 |

9 | $145 | $155.13 |

10 | $150 | $162.89 |

11 | $155 | $171.03 |

12 | $160 | $179.59 |

The difference is almost $20 by the end of the year. While that may not seem like very much, the real power of compounding shows up in long term loans. After five years, that same $100 loan would amount to $400 with simple interest and $1,867.92 using monthly compounding!

In contrast to the formula for simple interest, which only requires three parameters (principle, time, and rate), the formula for compound interest requires *four*. We also need to know how often compounding should take place. It could be monthly, quarterly, yearly, even daily!

Let’s define our parameters:

*P*— the principle*r*— the annual rate, or APR (sometimes labeled p.a.)*t*— the time, in years*n*— the number of compounding periods per year

Then the following two formulas compout the amount (*A*) of the loan after *t* years and the total interest (*I*) on the loan.

Ready for some compound interest practice?

Find the amount due in four years for a loan of $1,750 at 8% APR, compounded monthly. Then compute the total interest.

First, plug *P* = 1750, *r* = 0.08, *t* = 4, and *n* = 12 into the compounding interest formula to find the amount *A*. Note, we use *n* = 12 because there are 12 months in a year.

It’s very important to use as much precision as possible during the intermediate steps. *Do not round anything until the final step*.

So we have a total amount of $2,407.42. To compute the interest, simply take the principle amount away.

*I* = $2,407.42 – $1,750 = $657.42

Here’s a set of interest practice questions modeled from actual CAT problems. Scroll down to check your answers.

- What amount will be available after six years if $5200 is invested in an account bearing 4.5% interest compounded quarterly?
**A.**$6,801.55**B.**$6,942.12**C.**$7,209.07**D.**$7,551.19 - Suppose that a particular investment grows your money to 121% of its original value at the end of two years, using annual compounding. How many years will it take to double your money if invested into an account with the same APR but earning simple interest?
**A.**7**B.**9**C.**10**D.**11 - Sarah invests Rs.8000 for 10 years at 7% p.a. compound interest, where compounding occurs semi-annually. She decides to withdraw half of her accrued interest every six months for personal expenses, leaving the remaining amount invested. How much will her investment be worth at the end of the 10-year period?
**A.**Rs.10,981.37**B.**Rs.11,318.23**C.**Rs.11,527.99**D.**Rs.11,778.62 - Mike bought a new house and secured a mortgage from a bank for $250,000 to pay for it. Due to unforeseen circumstances, Mike could not begin paying the mortgage back until one year later. In the mean time, the mortgage accrued interest, compounded monthly. After the initial year, the mortgage balance was $259,667.22. What was the annual rate that this bank charged?
**A.**3.4%**B.**3.5%**C.**3.7%**D.**3.8% - Two sisters, Lisa (age 9) and Reesa (age 11), have the same birth day, which happens to be today. Their family sets up two investment accounts today for the girls totaling $12,000. If both accounts earn 5% interest compounded annually, how much money will each girl have if the money was split so that they would have the same amount at the age of 18?
**A.**$8,769.01**B.**$8,854.19**C.**$8,892.32**D.**$8,909.17

**A.**$6,801.55.Be sure to use the compounding interest formula. Here,

*P*= 5200,*r*= 0.045,*t*= 6, and*n*= 4 (quarterly means 4 times per year).

**C.**10.This problem has two main parts. First we have to determine the rate of the original investment. Use the compounding formula with

*n*= 1 (annual compounding).Next, use

*r*= 0.1 to set up an equation involving the simple interest formula. Solve for the unknown time*t*. Again, the principle amount (*P*) cancels from the equation, so we never needed to know exactly how much money was invested in the first place.*P*(1 + 0.1*t*) = 2*P*1 + 0.1

*t*= 2*t*= 10**B.**Rs.11,318.23Semi-annual compounding implies that

*n*= 2 in the formula. However, each compounding period (6 months), half of the accrued interest is diverted out of the investment. So the effective rate of return would be:*r*= 7%/2 = 3.5% = 0.035. Now plug everything into the compounding interest formula.**D.**3.8%Generally, when there is an unknown rate in a compounding interest problem, you’ll have to use logarithms to solve the equation. If you need a refresher, check out: CAT Logarithm Practice.

Set up the compounding interest formula with all of the parameters given in the problem and solve for

*r*.Therefore, the rate is 3.8%.

**B.**$8,854.19This problem requires some algebra! Let

*x*be the amount that is initially invested in Lisa’s account. Then Reesa’s account would start with 12000-*x*. Furthermore, the time of investment will be different for each girl: Lisa’s account will be for 18-9 = 9 years, and Reesa’s will be for 18-11 = 7 years.First solve for

*x*to find out the initial investment.Finally, use

*x*= 5707.49 as the principle amount in Lisa’s account, for*t*= 9 years.

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]]>The post CAT Logarithm Practice appeared first on Common Admission Test Blog.

]]>**Logarithms** are basically the “opposite” (or *inverse*) of exponentiation. The little “*b*” in the following notation is called the **base** of the logarithm, and we always require *b* > 0 and *b* ≠ 1.

In other words, if you need to know what is the value of log_{2} 512, then what you’re really asking is this: What exponent do I need on 2 to get 512? Since 512 = 2^{9}, now you can say that log_{2} 512 = 9. See how the base of the logarithm (2) becomes the base of the exponentiation?

Of course not every logarithm problem is a straightforward as that. We’ll try our hand at some more challenging questions later.

There are a couple of bases that are incredibly important: 10 and *e*.

- Logarithm base 10 is called
**common logarithm**. The usual notation for common log is to omit the base completely. log*x*= log_{10}*x*. So if you don’t see a base mentioned, you can safely assume it’s a common log. - We use the base
*e*≈ 2.71828 in the**natural logarithm**. The notation for natural log is “ln.” So ln*x*= log_{e}*x*.

What’s so *natural* about *e*, you might be wondering? Well if you want to know more about this important constant, you might want to peruse this Wikipedia article.

Here are the graphs of common log and natural log alongside log base 2 for comparison.

Logarithms are useful for a number of reasons, not the least of which is that they have some amazing properties.

Here is a nice refresher for the properties of logarithms: ACT Math Logarithms: What You Need to Know. You could also check out this video to get you started.

In practice, the three most important rules involve logarithms of products, quotients, and powers.

**Product Property:**log_{a}(*x**y*) = log_{a}*x*+ log_{a}*y*.**Quotient Property:**log_{a}(*x*/*y*) = log_{a}*x*– log_{a}*y*.**Power Property:**log_{a}(*x*^{r}) =*r*log_{a}*x*.

Here is a more complete list of logarithm properties.

The logarithm function is also **one-to-one (injective)**, which means that logarithms can be “cancelled” from both sides of any equation.

In other words, if you know that log_{a} *M* = log_{a} *N*, then it must be true that *M* = *N*.

Finally, here are a few more important properties of the logarithm function, *f*(*x*) = log_{a} *x*.

- The domain of
*f*is all positive numbers (*x*> 0). - The range of
*f*is all real numbers (**R**). - The graph of
*f*has a vertical asymptote at*x*= 0. - If
*a*> 1, then*f*is always increasing. If 0 <*a*< 1, then*f*is always decreasing.

Now let’s practice what we’ve learned!

Find the largest value of *x* that satisfies log_{4}(3*x*^{2} + 44*x*) = 3.

(A) 3/5 (B) 4/3 (C) 7/4 (D) 8/3

If log_{a} 2 = 0.3562, log_{a} 3 = 0.5646, and log_{a} 5 = 0.8271, find log_{a} 21600.

(A) 5.129 (B) 4.9873 (C) 4.8661 (D) 3.9012

If *m* = *n*^{2} = *p*^{3}, then what is log_{m}(*mnp*)?

(A) 1 (B) 11/6 (C) 6 (D) Can’t be determined

If log(*x* – *y*) – log(*x* + *y*) = log(*y*/*x*), find the value of (*x*/*y*)^{2} + (*y*/*x*)^{2}

(A) 2 (B) 3 (C) 4 (D) 6

If three positive numbers, *a*, *b*, *c*, are in geometric progression, then which of the following is equivalent to ln *a* + ln *b* + ln *c*?

(A) 3 ln *a* (B) 3 ln *b* (C) 3 ln *c* (D) ln(*a* + *b* + *c*)

(B) 4/3

Use the definition of logarithm to rewrite the given equation as an exponential equation.

log_{4}(3*x*^{2} + 44*x*) = 3

3*x*^{2} + 44*x* = 4^{3} = 64.

3*x*^{2} + 44*x* – 64 = 0

(3*x* – 4)(*x* + 16) = 0

There are two solutions, *x* = 4/3 and *x* = -16. Of these two, 4/3 is the largest.

(A) 5.129

The key is to break down the number 21600 into its prime factors.

21600 ÷ 2 = 10800

10800 ÷ 2 = 5400

5400 ÷ 2 = 2700

2700 ÷ 2 = 1350

1350 ÷ 2 = 675

So far, we have a factor of 2^{5}.

675 ÷ 3 = 225

225 ÷ 3 = 75

75 ÷ 3 = 25

So the next factor is 3^{3}.

Finally, 25 = 5^{2}. Thus, the prime factorization is: 2^{5}3^{3}5^{2}.

Now use the product property of logarithms….

log_{a} 21600 = log_{a}(2^{5}3^{3}5^{2}) = log_{a}(2^{5}) + log_{a}(3^{3}) + log_{a}(5^{2})

….and the power property:

= 5 log_{a} 2 + 3 log_{a} 3 + 2 log_{a} 5

Finally, we substitute the given information and evaluate.

= 5(0.3562) + 3(0.5646) + 2(0.8271) = 5.129

(B) 11/6

Let’s start with the multiplicative property of logarithms.

log_{m}(*mnp*) = log_{m} *m* + log_{m} *n* + log_{m} *p*.

The first term is easy: log_{m} *m* = 1.

For the other terms, we have to use the given equations to rewrite *n* and *p* in terms of *m*.

Since *m* = *n*^{2}, we have: *n* = *m*^{1/2}.

Therefore, log_{m} *n* = log_{m} *m*^{1/2} = 1/2.

In a similar way, since *m* = *p*^{3}, we have: *p* = *m*^{1/3}.

Therefore, log_{m} *p* = log_{m} *m*^{1/3} = 1/3.

In total, we obtain 1 + 1/2 + 1/3 = 11/6.

(D) 6

This problem will definitely test our abilities to use the logarithm rules and properties correctly *and* strategically.

First, use the quotient property *in reverse* to combine the logarithms on the left side of the equation. Then you can use the cancellation property to get rid of the logs.

The next few steps involve careful algebraic manipulation. Perhaps a little trial and error is necessary along the way. The key is to get the left side of the equation to match exactly with (*x*/*y*)^{2} + (*y*/*x*)^{2}.

We *almost* have the correct expression on the left side! In order to make it match, square both sides, but be careful with that middle term.

(B) 3 ln *b*

**A word of caution:** As tempting as it looks, letter choice (D) is **not** correct! Logarithms do not preserve addition — that is,

log_{a} (*x* + *y*) ≠ log_{a} *x* + log_{a} *y*.

Instead, we have to analyze the given information for clues. Because *a*, *b*, and *c* are in a geometric progression, that means that there is a *common ratio*, r, such that *b* = *ar* and *c* = *ar*^{2}.

Then we can use the product property of logarithms *in reverse*.

ln *a* + ln *b* + ln *c* = ln(*abc*) = ln((*a*)(*ar*)(*ar*^{2})) = ln(*a*^{3}*r*^{3}) = ln((*ar*)^{3})

From this step, use the power property of logarithms to pull the exponent down. Then use the fact that *b* = *ar* to complete the problem.

ln((*ar*)^{3}) = 3 ln(*ar*) = 3 ln *b*

Logarithms are inverse to exponentiation. They satisfy many remarkable properties.

Some of the most important properties involve multiplication, division, and exponents:

**Product Property:**log_{a}(*x**y*) = log_{a}*x*+ log_{a}*y*.**Quotient Property:**log_{a}(*x*/*y*) = log_{a}*x*– log_{a}*y*.**Power Property:**log_{a}(*x*^{r}) =*r*log_{a}*x*.

The logarithm function is one-to-one, has domain all positive numbers, and range all real numbers.

With these fundamental rules and properties at hand, you’ll be well-prepared to tackle even the most challenging CAT logarithm questions!

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]]>The post CAT Remainder Practice Problems appeared first on Common Admission Test Blog.

]]>For more about mathematics topics on the CAT, check out the Math Syllabus for CAT Exam.

It all begins with division.

In elementary and middle school, you learned a process called **long division**.

One easy and useful thing to remember is that the remainder must always be *less than* the divisor. For example, if you divide any number *N* by the divisor 2, then your remainder can only be 0 (if *N* is *even*) or 1 (if *N* is *odd*).

In fact, a number *N* is **divisible** by *d* if and only if the remainder of the division *N* ÷ *d* is equal to 0.

If you take the remainders of consecutive whole numbers with respect to a fixed divisor *d*, then you end up with consecutive remainders… most of the time. The only exception is that the next remainder after *d*-1 must be 0.

*Confused?* Let’s look at an example.

The first row of the table shows consecutive whole numbers from 1 to 10. The second row shows their remainders upon division by *d* = 4. See how the remainders also increase by 1 until you hit *d* – 1 = 3?

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

1 | 2 | 3 | 0 | 1 | 2 | 3 | 0 | 1 | 2 |

Long division can be a pain in the neck. Fortunately, there are ways to avoid this step sometimes when finding remainders.

Remainders work pretty well with respect to addition, subtraction, and multiplication. Just as long as you’re working with the same divisor throughout.

- If you add two numbers and then take the remainder, it’s the same as if you first took remainders and then added.
- The same is true for subtraction. If you subtract two numbers and then take the remainder, it’s the same as if you first took remainders and then subtracted.
- And it works for multiplication too! If you multiply two numbers and then take the remainder, it’s the same as if you first took remainders and then multiplied.

The only caveat is that you still might have to take remainders again at the end of each operation. We call this process **modular arithmetic**. Then we call the divisor the **modulus**, and finding remainders by a modulus *d* will often be called **reduction (modulo d)**.

For example, what is the remainder when 250 is divided by 7? Instead of long division, just think of 250 as 25 × 10. The remainder for 25 ÷ 7 is 4 (since 25 = 7(3) + 4), and for 10, the remainder is 3. Multiply remainders to get 4 × 3 = 12. Then reduce again by 7 to find the remainder 5 to answer the problem.

By the way, because modular arithmetic works with multiplication, that means you can take whole number powers as well!

For example, what is the remainder of 2^{301} ÷ 3? First, note that 2^{2} = 4 reduces to 1 modulo 3. Rewrite 2^{301} = 2^{2(150) + 1} = 4^{150}× 2. Then by the rules for modular arithmetic, the remainder will be the same as for 1^{150} × 2, which is equal to 2.

Finding remainders of large powers divided by large divisors can be very time-consuming. Fortunately, there is a mathematical “trick” that can speed the process along.

**Euler’s Theorem.** If *m* and *n* are relatively prime (no common factors except 1), then there is a special number called φ(*n*) such that the remainder of the division *m*^{φ(n)} ÷ *n* is always 1. The value of φ(*n*) is given by the formula,

An example showing the use of this advanced formula will be given in the practice problems.

Now you can try your hand at each of these CAT remainder practice problems! Solutions appear at the end.

Find the sum of all two-digit numbers that leave a remainder of 3 when divided by 7.

(A) 503 (B) 676 (C) 777 (D) 832

What is the remainder of 3^{2017} upon division by 82?

(A) 3 (B) 15 (C) 27 (D) 51

What is the remainder of 3^{2053} upon division by 83?

(A) 3 (B) 16 (C) 27 (D) 51

Suppose *x* = *k*(*k*+1)(*k*+2), where *k* ε **N**. What must be true about the number *x*?

(A) *x* is even and divisible by 7.

(B) *x* is not divisible by 5.

(C) *x* is odd and divisible by 3.

(D) *x* is divisible by 6.

A two-digit number *n* has remainders 1, 1, and 6 when divided by 3, 5, and 7, respectively. Find the sum of digits of *n*.

(A) 9 (B) 11 (C) 13 (D) 14

Are you ready to check your work?

(B) 676

Because remainders of consecutive whole numbers form a repeating consecutive list, we know that all numbers having a certain remainder will form an **arithmetic sequence**. This sequence starts as follows (remember, we just want two-digit numbers):

10, 17, 24, 31, …

To find out the last number in the sequence, find the remainder of 99 ÷ 7, which is 1. So, working backward, the last number in the sequence whose remainder will be 3 must be 94.

Then you can use the formula for a sum of an arithmetic sequence. How many terms will be there? well if *a*_{1} = 10 = 7(1) + 3, and *a _{n}* = 7(

Sum = (*n*/2)(*a*_{1} + *a _{n}*) = (13/2)(10 + 94) = 676.

(A) 3

Remainders and modular arithmetic show up in two different ways in this problem.

First, let’s explore some of the small powers of 3. 3^{1} = 3, 3^{2} = 9, 3^{3} = 27, 3^{4} = 81. The fact that 81 is only 1 less than the modulus 83 is a big clue. This means that 81 is equivalent to -1 modulo 82.

Now let’s find out how many groups of 4 are in 2017. Using long-division, we find that 2017 = 4(504) + 1. So we can rewrite the original value:

3^{2017} = (3^{4})^{504} × 3^{1},

which is equivalent to: (-1)^{504} × 3 = 1 × 3.

Therefore the remainder is 3.

(C) 27

If you think you’ve seen double, take a closer look! This time the divisor is 83. If you tried the same procedure as the previous problem, then you immediately hit a snag. 3^{4} is equivalent to -2 modulo 83, and it’s a lot harder to deal with powers of -2 then with powers of -1.

This is a job for Euler’s Theorem. Because *n* = 83 is prime, it only has a single prime factor, and we get:

φ(83) = 83(1 – 1/83) = 83 – 1 = 82.

The theorem states that 3^{82} will have remainder 1 when divided by 83. So now let’s reduce the exponent 2053 modulo 82 by long division. The remainder happens to be 3.

So, 3^{2053} is equivalent to 3^{3} = 27.

(D) *x* is divisible by 6.

A little knowledge of remainders helps out. Notice that *x* is the product of three *consecutive* natural numbers: *k*, *k*+1, and *k*+2. (Recall, **N** is the set of natural numbers, 1, 2, 3, 4, ….)

So if you took remainders of the three factors with respect to the divisor 2, you’d get either 0, 1, 0, or 1, 0, 1. Then when you multiply them, you’d get 0 as a remainder of *x* by 2.

Thus, *x* is even. But that’s not all!

Taking remainders of *k*, *k*+1, and *k*+2 by the divisor 3 would result in a list like 0, 1, 2, or 1, 2, 0, or 2, 0, 1. No matter what, 0 will show up in the list. So when we multiply those remainders together we end up with 0.

Thus, *x* is divisible by 3.

Any number divisible by both 2 and 3 must be divisible by 6.

(C) 13

Let’s work backward.

Because *n* has remainder 1 when divided by 3, we know that *n* = 3*a* + 1 for some *a* ε **N**.

Next, we reason that 3*a* must be divisible by 5 (so that *n* = 3*a* + 1 has the required remainder of 1 upon division by 5). Therefore, since 3 is relatively prime to 5, we know that 5 has to be to be a factor of *a*. We can now replace *a* by 5*b* (for some *b* ε **N**). That is,

*n* = 3(5*b*) + 1 = 15*b* + 1.

Finally, reduce the number modulo 7. Noting that 15 = 2(7) + 1, we find that *n* is equivalent to *b* + 1 mod 7. Then *b* = 5 is the smallest possibility that would give the required remainder of 6 mod 7.

To find *n*, simply plug in *b* = 5 to get *n* = 15(5) + 1 = 76. Finally, the sum of digits is 7 + 6 = 13 to answer the problem.

Now that you’ve studied a few practice problems and their solutions, you should be a little better prepared for the test.

Although CAT remainder problems can be challenging, it’s important to remember the basics.

- Know how to use long division to find quotients and remainders.
- Understand the patterns in how remainders appear for consecutive whole numbers.
- Use the additive, subtractive, and multiplicative properties of remainders (modular arithmetic).
- Look for those mathematical methods and theorems (such as Euler’s Theorem) that will allow you to work out remainders of a large power of a number.

With these tips in mind, you’ll be sure to succeed on the CAT!

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]]>The post Area Formulas to Know for the CAT appeared first on Common Admission Test Blog.

]]>You will need to know the area formulas for each of the following basic shapes.

- Triangles
- Rectangles and Parallelograms
- Circles

Moreover, you should know how to break down a more complicated shape into basic shapes to find its total area.

The area of a triangle is: *A* = (1/2)*bh*, where *b* is the length of the base, and *h* is the height (length of the altitude).

Often you’ll have to find a suitable altitude in a triangle first.

Alternatively, if you know the lengths of each of the sides, you could use **Heron’s Formula** for area.

The quantity “*s*” is called the **semiperimeter** of the triangle.

By the way, you can check out Triangle Properties to Know for the CAT for more about triangles!

The basic rule for both rectangles and parallelograms is that *Area* equals *Base* times *Height*. That is,

*A* = *bh*

The area of a circle of radius *r* is: *A* = π*r*^{2}. For most purposes, you can approximate π ≈ 3.14 (or, even better, 3.14159). Remember, if the problem gives you a diameter for a circle, then its radius is half the diameter length.

Other shapes may be broken down into rectangles and triangles.

For example, a trapezoid consists of a rectangle together with a triangle on either side.

But if you want to memorize another formula, you can use *A* = (*h*/2)(*a* + *b*), where *h* is the height, and the top and bottom bases have length *a* and *b*, respectively.

A **regular** polygon has all sides and angles equal. You can break up a regular polygon into *n* congruent isosceles triangles, where *n* is the number of sides. Then if you can find the area of one such triangle, simply multiply that result by *n* to obtain the area of the whole polygon.

On the other hand, if you’re good at memorizing formulas, then here’s one for the regular polygon of *n* sides with side length *s*:

Let’s review!

- Triangles:
*A*= (1/2)*bh*(or use Heron’s formula) - Rectangles:
*A*=*bh* - Parallelograms:
*A*=*bh* - Circles:
*A*= π*r*^{2} - Break up complicated shapes into simpler pieces to find area, or use a formula specific to that shape (if such a formula exists)

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]]>The post CAT Arithmetic Practice Problems appeared first on Common Admission Test Blog.

]]>Here are some of the arithmetic topics that you may run across on the CAT:

- Mean, median, and mode (basic statistics)
- Percentage, ratios, and proportions
- Simple and compound interest
- Applications in profit & loss, speed, time & distance, and rates

The study of statistics has to do with data and how it is distributed. Perhaps the simplest statistical measure is the **mode** of a data set.

The **mode** is the number that occurs most often in a data set. If no number occurs more than once, then there is no mode. On the other hand, a set can have more than one mode — that’s when more than one number has the maximum number of occurrances.

For example, the set (3, 1, 4, 1, 5, 9, 2, 6, 5) has two modes: 1 and 5.

Next, the **median** is the “middle” number. In other words, if you arranged all of the data in increasing order, then the one in the middle of the list is your median. But this only works if there are an odd number of data points. For an even number, you have to take the two middle numbers, say *a* and *b*, and average them together to get the median: (*a* + *b*)/2.

And that brings us to the **mean**. The **arithmetic mean** of a set is equal to the sum of the data divided by the number of items. We also call this the **average** value.

So, the mean (average) of (*a*_{1}, *a*_{2}, *a*_{3}, …, *a _{n}*) is equal to: (

You’ll also need to know how to find **weighted averages**. This time, each value gets a *weight*, which could represent the number of times that value occurs, or the percentage of a population that corresponds to that value, or other useful interpretations.

Suppose the data set (*a*_{1}, *a*_{2}, *a*_{3}, …, *a _{n}*) has weights (

For example, suppose your class has 15 boys and 18 girls. The test average for the boys was 78, while girls’ average was 81. What was the class average for this test?

The data points are 78 with weight 15, and 81 with weight 18. So we obtain: (15 × 78 + 18 × 81)/(15 + 18) = 2628/33 = 79.6.

We use percents, ratios, and proportions to help understand fractional relationships among quantities.

A **percentage** is measure of how frequent an item would be out of a hundred. So, *P*% really stands for the fraction *P*/100.

The *percentage change* between two values is equal to the *difference* divided by the *original*.

In order to find the new amount after applying a percentage *increase* or *decrease*, use the following formulas:

*P*% increase of*a*is: (1 +*P*/100) ×*a*.*P*% decrease of*a*is: (1 –*P*/100) ×*a*.

Ratios and proportions are other ways to express fractional relationships. The ratio *A* : *B* can be interpreted as a fraction *A*/*B*. If two ratios are supposed to be the same, then this sets up a proportion.

*A* : *B* = *C* : *D* means *A*/*B* = *C*/*D*. Equivalently, *AD* = *BC*.

Interest can be computed in a number of different ways. For the CAT arithmetic section, you’ll need to know simple and compound interest.

**Simple interest** varies directly with the time of the investment. If my interest payment is $100 for a period of one month, then it would be $200 for a total of two months. Simple, right? The formula for computing simple interest is likewise very simple.

If *P* is the principle, *r* is the interest rate as a decimal, and *t* is the time in years, and *I* stands for the computed simple interest, then

*I* = *Prt*

Then to find the total amount of the investment, simply add:

*P* + *I* = *P* + *Prt* = *P*(1 + *rt*).

Compounding interest means that the interest accrued in any given time period is then added back to the principle. That way, the interest computation for the next month becomes slightly larger. The process continues for the entire time of the investment. So compound interest generates much more in the long run than simple interest at the same rate.

Here is the formula for the total amount *A*, if there are *n* compounding periods per year. (Remember, your rate *r* must be the decimal equivalent of your percentage rate.)

There are numerous applications of arithmetic that show up on the CAT. Here are just a few.

- Profit = Revenue – Costs. If Profit < 0, then it is a loss. Revenue = (Price per unit) × (units sold).
- Distance = Speed × Time. Or, Speed = Distance / Time.
- Rates are measures of change in one quantity per change in another. Therefore, a rate is like a ratio, or fraction. Rate of A with respect to B = (Change in Quantity A) / (Change in Quantity B).

Now let’s see if you can apply what you know about arithmetic on the following problems! Solutions will be given at the end.

The number of part-time employees at a large chain of retail stores changes from month to month. Use the following chart to determine by what percent the part-time workforce increased from March to April.

Month | Jan. | Feb. | March | April | May |

Part-time Employees | 320 | 308 | 358 | 402 | 357 |

(A) 9.3% (B) 10.1 % (C) 12.3% (D) 15.7%

A company has two factories, A and B, both producing the same model of smart phones. The ratio of production between factories A and B is 2 : 5, and the ratio of costs generated between the two factories is 3 : 4. What is the ratio of cost per unit between factories A and B?

(A) 2: 7 (B) 18 : 13 (C) 6 : 11 (D) 15 : 8

There were three assignments in Writing Class, a 5-page paper worth 20% of the total course grade, a 10-page paper worth 30% of the course grade, and a final 20-page paper worth 50% of the course grade. If your scores were 82% for the 5-page paper, 65% for the 10-page paper, and 91% for the 20-page paper, what was your final course score to the nearest percentage?

(A) 81% (B) 79% (C) 77% (D) 75%

Three brothers, Alan, Bobby, and Charles, work for a house-painting business. If working alone, Alan can complete one house job in 15 days, Bobby can do it in 20 days, and Charles takes 30 days to complete it. For a particular job, they work together for some number of days, after which Charles gets sick and cannot work, leaving Alan and Bobby to complete the project. The total amount of money received by the crew was $4,500, and Charles got $1500 less than either of his brothers. Assuming the daily rate of pay was the same per person, find out how many days the job took in total.

(A) 4 (B) 6 (C) 8 (D) 10

(C) 12.3%

This is a job for the *percentage change* formula. Take the difference of the March and April numbers: 402 – 358 = 44. Then divide by the original (March) value: 44/358 = 0.1229. Finally, convert to a percentage to answer the problem. Roughly 12.3%, which is letter (C).

(D) 15 : 8

This problem is extra challenging due to the apparent lack of information. The problem does not give any concrete numbers for total production or cost, only their ratios. However, there is just enough information to answer the question!

First, for production, we can say that the total production for factory A is 2*x* and for factory B is 5*x*. What is *x*? Surprisingly, it doesn’t matter… just think of “*x*” as an unspecified number of “parts.”

We do the same for costs, only we should use a different letter for the number of “parts.” Costs for factory A: 3*y*, and for factory B: 4*y*.

Now cost per unit is equal to the total cost divided by the total production. For A, we get: (3*y*)/(2*x*). For B, we find: (4*y*)/(5*x*). The problem asks for the ratio of these two quantities, so let’s treat the ratio as a fraction.

The answer is (D) 15 : 8.

(A) 81%

This problem requires a weighted average. Here, the weights add up to 1 (20% + 30% + 50% = 100%), so we just have to compute the top part of the weighted average formula:

(82)(0.20) + (65)(0.30) + (91)(0.50) = 16.4 + 19.5 + 45.5 = 81.4.

Therefore, to the nearest percent, the course score is 81%.

(C) 8

By far, this is the most complicated question on this practice set! There are numerous quantities to keep track of, and it isn’t clear how they all might be related.

First, think of the individual job completion times in terms of rates. The rate equals 1 job / *x* days. So Alan’s rate is 1/15, Bobby’s rate is 1/20, and Charles’ rate is 1/30.

Next, we must distinguish the work done when all three brothers were working (say, a fraction *p* of the total house job) from the when only two worked (1 – *p* of the total job).

Now let’s talk about the payments. Because Charles lost out on $1500 compared to each of his brothers, that means that the two other brothers earned $3000 for those days when Charles was missing, which leaves $1500 in payments when all the brothers were working, or $500 per brother. Given the ratio of $500 : $1500 for per-worker pay, we know that the two remaining brothers were on the job for three times the number of days as when all three were working.

Let *T* be the number of days when all three worked. Then Charles was absent for 3*T* days beyond that time.

Lastly, we must put everything together. Using Quanity = Rate × Time, we may observe:

- Before Charles left:
*p*= (1/15 + 1/20 + 1/30) ×*T*= (3/20)*T* - After Charles left: 1 –
*p*= (1/15 + 1/20) × 3*T*= (7/20)*T*

Then using Equation (1) in Equation (2), we find:

1 – (3/20)*T* = (7/20)*T*

1 = (10/20)*T*

*T* = 20/10 = 2

Therefore, the total time on the job was *T* + 3*T* = 2 + 6 = 8 days.

For more practice, try this: Profit and Loss Practice Problems: CAT Quant Sample Question and Answer.

If you’d like to learn more about what is covered on the CAT, check out What is the CAT Syllabus?

Happy studying!

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]]>The post Triangle Properties to Know for the CAT appeared first on Common Admission Test Blog.

]]>Let’s review the basics. It may help to refer to the following diagram.

- The sum of
**interior angles**is always 180 degrees. That is,*x*+*y*+*z*= 180. **Triangle Inequality**. The sum of any two sides is always greater than the third side, and the difference is always less than the third side. So we have*a*+*b*>*c*, and |*a*–*b*| <*c*.- The side opposite the largest angle is the largest side, and the side opposite the smallest angle is the smallest side. That is, if
*a*<*b*<*c*, then*x*<*y*<*z*. - A perpendicular line drawn from a vertex to the base is called an
**altitude**and measures the*height*of the triangle. Once you know the height, you can find**area**using Area = (1/2) Base × Height, or more concisely,*A*=*bh*/2. - The
**perimeter**of the triangle is the sum of its sides, or*P*=*a*+*b*+*c*. - Two triangles are
**similar**if their corresponding angles are the same. If Triangle ABC is similar to Triangle DEF, then the ratios AB : DE, AC : DF, and BC : EF are all the same.

A number of properties apply only to specific kinds of triangles.

- An
**equilateral triangle**has all three sides congruent. As a consequence, all angle are congruent, and they each measure 60 degrees. The area of an equilateral is (√3)/4 ×*s*^{2}, where*s*is the side length. - An
**isosceles triangle**has two sides congruent. The angles opposite the equal sides must also be equal. - A
**scalene triangle**has no congruent sides.

A **right triangle** must have a right angle (90 degrees). The side opposite the right angle is the **hypotenuse**, and the other two sides are called **legs**.

**Pythagorean Theorem.** In a right triangle, *a*^{2} + *b*^{2} = *c*^{2}, where a and b are the leg lengths and c is the length of the hypotenuse.

There are two special right triangles that are useful to remember.

The **30-60-90 triangle** is essentially half of an equilateral triangle. It’s side lengths are in ratio 1 : 2 : √ 3.

The **45-45-90 triangle** is the only triangle that is both right and isosceles. It’s side lengths are in ratio 1 : 1 : √ 2.

For more about right triangles, check out this article from Magoosh: Conquering Right Triangles and the Pythagorean Theorem.

You’ll find a good number of geometry problems on the CAT, so it’s important to understand the fundamentals, including triangle properties. With enough study, you’ll ace all of the geometry problems on the CAT!

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