Daily Mail: Education

Increase ur solving mathematics

Probability

How probable is it to get probability questions ?

Probability questions are becoming increasingly common. They tend to be bundled among the difficult questions, so high scorers will commonly encounter 1, 2, or 3 of them. If you are a low scorer and are pressed for time, consider skipping most of the material past "Simple Probability
Do I have to be a genius to solve probability questions?

Absolutely not. Both this brief course and do not require any math knowledge beyond what you learned in your high school. You have not to be a genius either. Just be sure to try solving the problems and get a grip of the solution tools, and you'll crack it. To tell you the secret, the myth of the complexity of the probability theory is simply another way to secure the math instructors' wages.

What is probability?

Probability is a measure of how likely is an event to happen. It is measured in fractions from 0 to 1 (0 is impossible, 1 is unavoidable or certain). Sometimes it is denoted in percentages, again from 0% to 100%.

What are an event and an outcome?

Event is anything that happens. In probability theory we speak of events having outcomes or results. A coin flip (an event) has two possible outcomes—heads and tails. A die toss has six possible outcomes. When a coin is flipped (an event is tested), one of the outcomes is obtained. Either heads or tails.
How is probability used?
A probability is commonly denoted as p(Some Event). So, p(Heads) = 50% means that you have 1 chance in 2 to get heads in a coin flip. This also means that if you flip the coin 100 times, you'll get about 50 heads. But not exactly 50. You may get 49, or 63, or even no heads. But you're most likely to get such a number of heads that will be close to 50. This works for any probability. So, if the probability of getting married after going to the cinema is 3%, out of 1,000 movies you'll be married about 1,000 * 3% = 30 times. Maybe 26 or 34, but the average expectation is 30. That's what you use probability for, apart from cracking competitive exam.

A. Simple Probability: The F/T Rule

In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. This is known as the F/T Rule, and 90% of the problems are solved with this tool. No kidding.

Probability = (# of favorable outcomes) / (# of possible outcomes)

Example 1

What is the probability that a card drawn at random from a deck of cards will be an ace?

Solution
In this case there are four favorable outcomes

the ace of spades,

the ace of hearts,

the ace of diamonds,

the ace of clubs.

Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes. Therefore, the probability is 4/52 or 1/13. The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice.

Example 2

Two fair six-sided dice are rolled; what is the probability of having 5 as the sum of the numbers?

Solution
There are 36 possible outcomes when a pair of dice is thrown (six outcomes for the first die times six outcomes for the second one). Since four of the outcomes have a total of 5 [(1,4), (4,1), (2,3), (3,2)], the probability of the two dice adding up to 5 is 4/36 = 1/9.

Example 3

Two six-sided dice are rolled; what is the probability of having 12 as the sum of the numbers?

Solution
We already know the total number of possible outcomes is 36, and since there is only one outcome that sums to 12, (6,6—you need to roll double six), the probability is simply 1/36.

Dinosaur example

A blonde girl (G.W. Bush, your boss, or whoever you love too heartily) was asked once what is the probability of meeting a dinosaur in the street. The answer was: "50%. I either meet it or not." This is how you DON'T use the F/T rule. When counting the outcomes, make sure that:

all of them are equally likely to happen

you have not left out any possibilities when counting T

(quote) F and T are in the same currency, i.e. if F is combinations and T is permutations, you'll get an error.

B. Probability of Multiple Events

For questions involving single events, the F/T rule is sufficient. In fact, it is often sufficient for all other cases too. But, for questions involving multiple events, some other tools may be more appropriate. Even when the problem can be solved with F/T, these tools still may provide a more elegant solution. Here're the tools:

NOT tool

If you know that the probability of an event (or one of the outcomes) is p, the probability of this event NOT happening (or the probability of it NOT having this given outcome), is (1-p).

p(not A) + p(A) = 1

AND tool

If two (or more) independent events are occurring, and you know the probability of each, the probability of BOTH (or ALL) of them occurring together (event A and event B and event C etc) is a multiplication of their probabilities.

p(A and B) = p(A) * p(B)
p(A and B and C ... and Z) = p(A) * p(B) * p(C) * ... * p(Z)

Suppose I will only be happy today if I get an email and win the lottery. I've a 90% chance to get an email and 0.1% chance to win the lottery. What are my chances for happiness? Since email and lottery are independent (getting an email doesn't change my lottery chances, and vice versa), we can use the AND tool: p(email and lottery) = p(email) * p(lottery) = 90% * 0.1% = 0.09%; So I have 9 chances in 10,000... Not bad.

OR tool

If two (or more) incompatible events are occurring, the probability of EITHER of them occurring (event A or event B or event C etc) is a sum of their probabilities.

p(A or B) = p(A) + p(B)
p(A or B or C ... or Z) = p(A) + p(B) + ... + p(Z)

Incompatible means that they can't happen together, i.e. p(A and B) = 0. In case of two compatible events, the OR tool looks a bit more complicated:

p(A or B) = p(A) + p(B) - p(A and B)

If we know that A and B are independent, we can apply AND tool to rewrite:

p(A or B) = p(A) + p(B) - p(A) * p(B)

Suppose I will now be happy in both cases - either getting an email or winning the lottery. What are my chances to happiness now? p(email or lottery) = p(email) + p(lottery) - p(email) * p(lottery) = 90% + 0.1% - 0.09% = 90.01%; My chances are 9,001 in 10,000 now. I'd rather choose this one.

Expressions/Brackets tool

When you're being asked for something complex, try reducing it to events and outcomes, and writing a formula. Use brackets to denote complex events, such as (A and B), or (A and (B or C)), etc. It is common to use AND as if it is multiplication and OR as if it is addition in the order preference, i.e. (A and B or C) = ((A and B) or C), but (A and (B or C)) <> (A and B or C). When you figure out the formula, it'll be easy to reduce it to simple arithmetic operations by using NOT, AND, and OR tools.

Elimination tricks

Given that 0 <= p(A) <= 1, you get the following rules:

p(A and B) <= p(A)

p(A or B) >= p(A)

p(A and B) <= p(A or B)

Thinking of these rules is often an excellent strategy for eliminating certain answer choices.

Example 4

If a fair coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails?

1/6

1/3

1/4

1/2

Solution
Suppose first toss is A, second is B. We know that p(A_heads) = 50% and that p(B_tails) = 50%. Also, A and B are independent. So, p(A_heads and B_tails) = p(A_heads) * p(B_tails) = 50% * 50% = 25% = 1/4. Answer is C.

Example 5

If a fair coin is tossed twice what is the probability that it will land either heads both times or tails both times?

1/8

1/6

1/4

1/2

Solution
Let first toss be A, second B.

p(Ah) = p(At) = p(Bh) = p(Bt) = 1/2
p(Ah and Bh) = p(Ah) * p(Bh) = 1/4
p(At and Bt) = p(At) * p(Bt) = 1/4
p((Ah and Bh) or (At and Bt)) = p(Ah and Bh) + p(At and Bt) = 1/4 + 1/4 = 1/2

Note that AND rule works because A and B are independent, and OR rule works because (Ah and Bh) and (At and Bt) are incompatible.

Alternatively, you may use F/T rule to solve this. Enumerate outcomes as (HH, HT, TH, TT). Favorable are HH and TT. So, p = 2/4 = 1/2. Although in this case F/T rule works more gracefully, the AND/OR approach is still helpful - you can learn it on such easy examples as this to prepare for the more difficult ones.

Example 6

A bowman hits his target in 1/2 of his shots. What is the probability of him missing the target at least once in three shots?

Solution

An optimal way to solve this is to think that (missing the target at least once) = 1 – (hitting it every time). So, p(hitting it every time) = p(shot1_hit and shot2_hit and shot3_hit) = p(shot1_hit) * p(shot2_hit) * p(shot3_hit) = 1/2 * 1/2 * 1/2 = 1/8; p(missing at least once) = 1 – p(hitting it every time) = 1 – 1/8 = 7/8.

Alternatively, use the F/T rule. The T are HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM. T = 8. The F are HHM, HMH, HMM, MHH, MHM, MMH, MMM. F = 7.

In cases like this it is evident that F/T rule soon becomes too hard to apply.

C. Event Types and Sets Analogy

Compatible vs. Incompatible (Mutually exclusive) Events

Sometimes you have to distinguish compatible and mutually exclusive events. Mutually exclusive are those events that can't happen together. Heads and tails are mutually exclusive events. Formally, two events are mutually exclusive if p(A and B) = 0. Otherwise, they are compatible. Note that mutually exclusive events are independent. (!)

Dependent vs. Independent Events

Most of the events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events if you put the cards back.

An example of a dependent event, one in which the probability of the second event is affected by the first, is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, there are one fewer aces and one fewer spades. This fact affects the F in the F/T rule.

What to do if you encounter dependent events? If possible, try to use F/T rule to the composite event of the two. In the cards example, you may consider counting all 2-card combinations you may draw (T), and then counting those that fit (F). This will be discussed in detail later. But sometimes the events can't be reduced to outcomes that can be counted. In these cases, use the sets analogy.

Sets Analogy

Remember the familiar problem type about students attending three language classes, say, French, German, and Chinese? There you had to calculate the number of students attending one of the classes, or number of students attending both French and German, but not Chinese, etc? The greatest way to solve such problems is to draw intersecting circles representing the three sets of students, and then to write there their numbers and try to find the answer.

What does it have to do with probability, one might wonder. But this is precisely the way to solve probability problems with dependent events. This charts you may have drawn for simple sets problems are called Venn diagrams in the probability theory. Perhaps to scare you away.

The logic is simple: each event is a language class, and each chance is a student in that class. And the probability of the event is the number of students (chances) attending it divided by the total number of students. Where the classes intersect is where two events happen at once. Mutually exclusive events do not intersect. Finally, independent events intersect in such an interesting way that, supposing French and German classes represent two independent events, the proportion of French students in the German class is the same as the proportion of French students in the school as a whole (100 students, 40 study German, 50 study French, and 20 study both: 20/40 = 50/100).

Conditional Probability

Conditional probability is a simple way to denote proportions you understand with the sets analogy. Simply put, p(A/B) is the probability of event A happening given that event B has already happened, or the number of students attending both A and B classes divided by the number of students attending B class.

So, for any two events, including dependent events, this statement hold:

p(A and B) = p(A) * p(B/A) = p(B) * p(A/B)

This statement, however scary, is self-evident. Look at it. It says that to find the number of students studying French and German you have to either multiply the number of those who study French by the proportion of German scholars in the French class (p(B/A)), or multiply the number of German students by the proportion of French students in the German class (p(A/B)). But that's self-evident, isn't it? So it is with events.

Independent events may, therefore, be defined as such that p(B/A) = p(B), p(A/B) = p(A).

Example 7

What is the probability that a card selected from a deck will be either an ace or a spade?

2/52

2/13

7/26

4/13

17/52

Solution

Let A stand for a card being an ace, and S for it being a spade. We have to find p(A or S). Are A and S mutually exclusive? No. Are they independent? Why, yes, because spades have as many aces as any other suit. Then,

p(A or S) = p(A) + p(S) - p(A) * p(S)

With simple F/T we get:

p(A) = 4/52 = 1/13
p(B) = 13/52 = 1/4

So,

p(A or S) = 1/13 + 1/4 - 1/52 = 16/52 = 4/13

Sets analogy can help you visualize the formula. Draw two intersecting circles—one for aces, the other for spades. To get the area (probability) of the figure formed by these two circles together (all chances that are either aces or spades), you add the areas of aces and spades and subtract the intersecting area, in order not to count it twice. What we subtract is the ace of spades that was counted twice.

Another way to think about the question is to just count aces and spades; that is, use the F/T rule. There are 13 spades in a deck and 3 aces other than the ace of spades already included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible outcomes, or 16/52 = 4/13.

Example 8

If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?

Solution

Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B/A) = 3/51 = 1/17, and therefore:

p(A and B) = p(A) * p(B/A) = 1/13 * 1/17 = 1/221

Example 9

If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw?

Solution

So, there are 12 red and 18 blue marbles. We are asked to draw twice and return the marble after each draw. Therefore, the first draw does not affect the probability of the second draw. We return the marble after the draw, and therefore, we return the situation to the initial conditions before the second draw. Nothing is altered in between draws; therefore, the events are independent.

p(drawing a red marble) would be 12/30 = 2/5. The same is true for the second draw. Then p(First_Red and Second_Red) = p(First_Red) * p(Second_Red) = 2/5 * 2/5 = 4/25.

Example 10

Now consider the same question with the condition that you do not return the marbles after each draw.

Solution

The probability of drawing a red marble on the first draw remains the same, 12/30 = 2/5. The second draw, however, is different. The initial conditions have been altered by the first draw. We now have only 29 marbles in the jar and only 11 red. So, p(Second_Red/First_Red) = 11/29. Using the dependent event formula,

p(First_Red and Second_Red) = p(First_Red) * p(Second_Red/First_Red) = 2/5 * 11/29 = 22/145

To summarize, if you return every marble you select, the probability of drawing another marble is unaffected; the events are INDEPENDENT. If you do not return the marbles, the number of marbles is affected and therefore DEPENDENT.

D. Learning the Advanced Tools

Detailed discussion of advanced solution tools is out of scope of this course, but here're some considerations to get you started:

Combinations. Good understanding of CT formulas (n!, nAk, nCk) is essential to solving complex F/T problems, where both F and T are so large you can't enumerate them manually, but only with a formula. See our Combinations Course ?.

Expectations. Some probability problems deal with money, gains, and bets. Often you have to calculate which bet will be better, or how much it will be worth. The tool that deals with this is Expectation. E = G * p, where G is gain, and p is probability. So, a 10% chance to get $100 is worth (has E) of $100 * 10% = $10. Therefore, it is better than to get $8 for granted, but worse than a 5% chance to get $300 (E = $300 * 5% = $15). Complex expectation works similarly: E1 = E * p, i.e. a 10% chance to get a 25% chance to get $100 is worth 10% * (25% * $100) = $2.5; This is how Expectations work.

Distributions. The three types of distributions are Binominal, Hypergeometric, and Poisson distributions. These are just handy formulas for solving 3 very specific kinds of problems, like these:

If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)

There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)

Each hour an average of ten cars arrive at the parking lot. The lot can handle at most fifteen cars per hour. What is the probability that at a given hour cars will not be accepted? (Poisson Distribution)

As you may have noticed, Poisson and Binominal Distribution problems are alike. In fact, these Distributions are two methods of solving the same kind of problems. The difference is that BD provides accurate but costly (many calculations) method, and PD provides and elegant approximation, and is therefore used only on large numbers.

While BD and HD are quite likely to appear on GMAT, PD is not. For GMAT Club's members it is an open question whether one can in fact encounter PD on GMAT. In any case, there won't be two questions on PD.

See the Appendix A below for an explanation of BD and HD.

An Afterword

We hope this course was helpful. If you can think of any improvements we could make, please let us know at the forum. Just think how great it would be to have a FREE clear and concise guide to Probability on GMAT. Your feedback is what hopefully will bring this to life.

Appendix 1: Binomial and Hypergeometric Distributions

Binomial Distribution

BD formula (also know as Bernoulli’s formula) is:

nPk = nCk * p^k * (1-p) ^ (n-k)

where nPk denotes the probability of an event having a given outcome exactly k times in n tests, p is the probability of the event having this outcome in each single test, and nCk comes from Combinations Theory.

So, the probability that the given event:

will have the outcome n times: nPn = p^n

will not have the outcome: nP0 = (1-p)^n

will have the outcome at least 1 time: P = 1 - (1-p)^n

will have the outcome no more than k times: P = nP0 + nP1 + ... + nPk

will have the outcome no less than k times: P = nPk + nP(k+1) + ... + nPn

Let's apply it to the example given in the text above: for a coin flip p = 50%; 5 flips means that n = 5. The probability to find is for the coin to have heads at least 3 times. The answer is:

P = 5P3 + 5P4 + 5P5 = 10/32 + 5/32 + 1/32 = 16/32 = 1/2

5P3 = 5C3 * 50% ^ 3 * (1 - 50%) ^ (5 - 3) = 10 * 1/8 * 1/4 = 10/32
5C3 = 5! / (3! * 2!) = 10

5P4 = 5C4 * 50% ^ 4 * (1 - 50%) ^ (5 - 4) = 5 * 1/32 = 5/32
5C4 = 5! / (4! * 1!) = 5

5P5 = 50% ^ 5 = 1/32

So, the answer is 1/2.

Hypergeometric Distribution

The HD formula is:

p = aCa' * bCb' * cCc' * ... * zCz' / (a+b+c+..+z)C(a'+b'+c'+...+z')

where a, b, ... z are show how many times the outcome was obtained in each sub-event, and a', b', ... z' show how many times the corresponding sub-event was tested. nCk stands for combinations formula.

Let us again apply it to the example given in the text above: we have 3 ball colors, or 3 sub-events: g for green, r for red, b for blue. We know that:
g' = 2
r' = 3
b' = 2
g = 1
r = 2
b = 1

Now, let's do the calculations:

p = 2C1 * 3C2 * 2C1 / (2+3+2)C(1+2+1) = 2C1 * 3C2 * 2C1 / 7C4

7C4 = 7! / (4! * 3!) = 5 * 6 * 7 / 6 = 35
2C1 = 2
3C2 = 3! / (2! * 1!) = 3

p = 2 * 2 * 3 / 35 = 12/35

So, the answer is 12/35.

Numbers System

1) 2^2n-1 is always divisible by 3
2^2n-1 = (3-1)^2n -1
= 3M +1 -1
= 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2?
ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6
Funda : if a number 'n' is represented as
a^x * b^y * c^z ....
where, {a,b,c,.. } are prime numbers then

Quote:

(a) the total number of factors is (x+1)(y+1)(z+1) ....
(b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)....
(c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)....
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x*y*...)

3) what is the highest power of 10 in 203!ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually
203/2 = 101
101/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5
203/5 = 40
40/5 = 8
8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT 2002 has similar question )
ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5

5) In how many ways can 2310 be expressed as a product of 3 factors?
ANS: 2310 = 2*3*5*7*11
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways

6) In how many ways, 729 can be expressed as a difference of 2 squares?
ANS: 729 = a^2 - b^2
= (a-b)(a+b),
since 729 = 3^5,
total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27.
So 4 ways
Funda is that, all four ways of expressing can be used to findout distinct a,b values,
for example take 9*81
now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.

7) How many times the digit 0 will appear from 1 to 10000
ANS: In 2 digit numbers : 9,
In 3 digit numbers : 18 + 162 = 180,
In 4 digit numbers : 2187 + 486 + 27 = 2700,
total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between 10 and 20 with denominator as 3?
ANS :
sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66…….
= 21 + 23 + ……
= 300

9) if n = 1+x where x is the product of 4 consecutive number then n is,
1) an odd number,
2) is a perfect square
SOLN : (1) is clearly evident
(2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number 'n' the reminder is also same, what is that number if the number is a odd prime number?
ANS : since both leave the same reminder, let the reminder be 'r',
then, 987 = an + r
and 643 = bn + r and thus
987 - 643 is divisible by 'r' and
987 - 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence 'r' is 43

11) when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?
ANS : whenever such problem is given,
we need to write the numbers in top row and rems in the bottom row like this

11 7 4
| \ \
5 6 3

( coudnt express here properly

)
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5
that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2
75 mod 7 = 5
10 mod 11 = 10

12) a^n - b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc
EXAMPLE: 40^3-17^3-23^3 is divisble by
since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.

14) There is a seller of cigerette and match boxes who sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found that there are no takers. So he reduced the price of cigarette and managed to sell all the cigerattes, realising Rs. 77.28 in all. What is the number of cigerattes?

a) 49
b) 81
c) 84
d) 92

ANS : (d)
since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer

Quote:

i have given this question to make the funda clear

15) What does 100 stand for if 5 X 6 = 33
ANS : 81
SOLN : this is a number system question,
30 in decimal system is 33 in some base 'n', by solving we will have n as 9
and thus, 100 will be 9^2 = 81

16) In any number system 121 is a perfect square,
SOLN: let the base be 'n'
then 121 can be written as n^2 + 2*n + 1 = (n+1)^2
hence proved

17) Most of you ppl know these, anyways, just in case

Quote:

(a) sum of first 'n' natural numbers - n*(n+1)/2
(b) sum of the squares of first 'n' natural numbers - n*(n+1)*(2n+1)/6
(c) sum of the cubes of first 'n' natural numbers - n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 - 25

18 ) See Attachment to know how to find LCM, GCF of Fractions

Quote:

CAT 2002 has 2 questions on the above simple concept

19) Converting Recurring Decimals to Fractions

let the number x be 0.23434343434........

thus 1000 x = 234.3434343434......
and 10 x = 2.3434343434.........
thus, 990 x = 232
and hence, x = 232/990

20) Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n
EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with

9? ( never seen such question though

)
the reminder would be (6 + 8 + 1) % 9 = 6

(b) (a*b*c) % n = (a%n * b%n * c%n) %n
EXAMPLE: What is the remainder left when 1073 * 1079 * 1087 is divided by 119 ? ( seen this kinda questions alot

)
1073 % 119 = ?
since 1190 is divisible by 119, 1073 mod 119 is 2
and thus, "the remainder left when 1073 * 1079 * 1087 is divided by 119 " is 2*8*16 mod 119 and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that is 18

Glossary : % stands for reminder operation

find the number of zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100

the expresion can be rewritten as (100!)^100 / 0!* 1!* 2!* 3!....99!

Now the numerator has 2400 zeros

the formular for finding number of zeros in n! is

[n/5]+[n/5^2]...[n/5^r]
where r is such that 5^r<=n<5^(r+1)

and [..] is the grestest integer function

for the numerator find the number of zeros using the above formulae..

for 0!...4! number of zeros ..0
5!...9!.number os zeros ..1
9!...14!... 2
15!..19!..................3
20!..24!..................4!
now at 25! the series makes a jump to 6
25!...29!.................6
30!...34!.................7
this goes on and again makes a jump at 50!
and then at 75!

so the number of zeros is...

5(1+2....19) + 25+ 50+ 75

the last 3 terms 25 50 and 75 are because of the jumps..

this gives numerator has 1100 zeros

now total number of zeros in expression is no of zeros in denominator - no of zeros in numerator
2400 - 1100

the Answer 1300

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

To find the number of factors of a given number, express the number as a product of powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (2⁴ * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The sum of first n natural numbers = n (n+1)/2

The sum of squares of first n natural numbers is n (n+1)(2n+1)/6

The sum of first n even numbers= n (n+1)

The sum of first n odd numbers= n^2

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

To find the squares of numbers near numbers of which squares are known

To find 41^2 , Add 40+41 to 1600 =1681

To find 59^2 , Subtract 60^2-(60+59) =3481

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
eg: x^4+3x^2+2x+6=0 has no positive roots .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = - b/a
sum of the product of the roots taken two at a time = c/a
product of the roots = -d/a
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = - b/a
sum of the product of the roots taken three at a time = c/a
sum of the product of the roots taken two at a time = -d/a
product of the roots = e/a
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if
x=y(=k/2). The maximum product is then (k^2)/4

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If for two numbers x*y=k(=constant), then their SUM is MINIMUM if
x=y(=root(k)). The minimum sum is then 2*root(k) .

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

|x| + |y| >= |x+y| (|| stands for absolute value or modulus )
(Useful in solving some inequations)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Product of any two numbers = Product of their HCF and LCM .
Hence product of two numbers = LCM of the numbers if they are prime to each other

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For any regular polygon , the sum of the exterior angles is equal to 360 degrees
hence measure of any external angle is equal to 360/n. ( where n is the number of sides)

For any regular polygon , the sum of interior angles =(n-2)180 degrees

So measure of one angle in

Square =90

Pentagon =108

Hexagon =120

Heptagon =128.5

Octagon =135

Nonagon =140

Decagon = 144

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If any parallelogram can be inscribed in a circle , it must be a rectangle.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sides equal).

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a regular hexagon : root(3)*3/2*(side)*(side)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For any 2 numbers a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)

(GM)^2 = AM * HM

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
For any positive integer n

2<= (1+1/n)^n <=3

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.

a^4+b^4+c^4+d^4 >=4abcd

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(n!)^2 > n^n (! for factorial)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum
if a/p = b/q = c/r = d/s .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Consider the two equations

a1x+b1y=c1
a2x+b2y=c2

Then ,
If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to )
If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is
0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,
the minute hand describes 6 degrees /minute
the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .

The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight.
(This can be derived from the above) .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a triangle
1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2
=a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .

In any triangle
a=b*CosC + c*CosB
b=c*CosA + a*CosC
c=a*CosB + b*CosA

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to
(k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to
(a1+a2+a3+............./b1+b2+b3+..........)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(7)In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
2 < e < 3

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

In a GP the product of any two terms equidistant from a term is always constant .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.

(m+n)! is divisible by m! * n! .

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f)
is((a+c+e)/3 , (b+d+f)/3) .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a parallelogram = base * height

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

APPOLLONIUS THEOREM:

In a triangle , if AD be the median to the side BC , then
AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

for similar cones , ratio of radii = ratio of their bases.

The HCF and LCM of two nos. are equal when they are equal .

+++++++++++++++++++++++++++++++++++++++++++++++++++++

Volume of a pyramid = 1/3 * base area * height

+++++++++++++++++++++++++++++++++++++++++++++++++++++

In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

In any triangle the angular bisector of an angle bisects the base in the ratio of the
other two sides.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The quadrilateral formed by joining the angular bisectors of another quadrilateral is
always a rectangle.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

|a|+|b| = |a+b| if a*b>=0
else |a|+|b| >= |a+b|

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

2<= (1+1/n)^n <=3

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
WINE and WATER formula:

If Q be the volume of a vessel
q qty of a mixture of water and wine be removed each time from a mixture
n be the number of times this operation be done
and A be the final qty of wine in the mixture

then ,
A/Q = (1-q/Q)^n

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a hexagon = root(3) * 3 * (side)^2

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(1+x)^n ~ (1+nx) if x<<<1

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Some pythagorean triplets:

3,4,5 (3^2=4+5)
5,12,13 (5^2=12+13)
7,24,25 (7^2=24+25)
8,15,17 (8^2 / 2 = 15+17 )
9,40,41 (9^2=40+41)
11,60,61 (11^2=60+61)
12,35,37 (12^2 / 2 = 35+37)
16,63,65 (16^2 /2 = 63+65)
20,21,29(EXCEPTION)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height
where median is the line joining the midpoints of the oblique sides.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Let W be any point inside a rectangle ABCD .
Then
WD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn inside
this triangle touching each other then each's radius = a/(2*(root(3)+1))

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

when you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

To find the squares of numbers from 50 to 59

For 5X^2 , use the formulae

(5X)^2 = 5^2 +X / X^2

Eg ; (55^2) = 25+5 /25

=3025

(56)^2 = 25+6/36

=3136

(59)^2 = 25+9/81

=3481

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

many of u must b aware of this formula, but the ppl who don't know it must b useful for them.
a+b+(ab/100)

this is used for succesive discounts types of sums.
like 1999 population increses by 10% and then in 2000 by 5%
so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999

and if there is a decrease then it will be preceeded by a -ve sign and likeiwse

Numbers

Quote:

i have given this question to make the funda clear

Quote:

18 ) See Attachment to know how to find LCM, GCF of Fractions

Quote:

CAT 2002 has 2 questions on the above simple concept

Glossary : % stands for reminder operation

find the number of zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100

Daily Mail

Pages

Education

Increase ur solving mathematics

No comments:

tweet

Total Pageviews