Quants 1

Two containers contain diluted acid of 0.5 and 0.75 concentration. A solution of 2 litres from each container contains acid and water in ratio

2 litres from the first container contains 2*0.5 = 1 litre of acid and 2-1=1 litres of water 2 litres from the second container contains 2*0.75 = 1.5 litres of acid and 2-1.5=0.5 litres of water The final solution contains 1+1.5=2.5 litres of acid and 1+0.5=1.5 litres of water Ratio = 2.5:1.5 = 5:3

n^5 - n^3 < 0 Compare n and n^2

given, n^5 < n^3 Note that this is possible only when n is b/w 0 and 1. Eg. n = 1/2; (1/2)^5 < (1/2)^3 => n^2 = 1/4 < n OR, n can be less than -1, Eg. -2 (-2)^5 < (-2)^3 => n^2 = 4 > n So, relationship cannot be determined.

Pam has a field in the shape of a polygon with 10 sides. Into how many triangular fields can she cut her field so that she can access each field from the same vertex?

n-2 Visualize that one vertex can be joined to n-3 other vertices however, will yield in (n-3) + 1 triangles. Try with a pentagon.

15 points on a plane of which 6 are collinear. How many pentagons can be drawn?

nCr(15, 5) - (nCr(6,3) + nCr(6,4) + 6) = 2962 (selecting any more than 2 points from the 6 collinear will not result in a pentagon)

How many necklaces can be made using at least 5 from 8 beads of different colours?

Circular (anti = clock) combinations taken r items at a time from n, = nPr/2r So, Ans = 8P5/2x5 + 8P6/2x6 + 8P7/2x7 + 8P8/2x8 = 7752 https://www.quora.com/How-many-necklaces-can-be-made-using-at-least-5-from-8-beads-of-different-colours

How many numbers between 1 and 1000 are either a multiple of 28 or a multiple of 32?

Do this normally, 28x <= 1000 and 32y <= 1000 add the numbers (x+y) and subtract the multiples of their LCM, viz, 224 -> 4 multiples. = 66 - 4 = 62.

The sum of the squares of the numbers is 37/6 times their product. The sum of the numbers is 35. Find the difference of the two numbers.

For this question, we know: (x+y)^2 = x^2 + y^2 + 2xy substitute the values in this and solve for xy finally, put x in terms of y in x+y = 35 (given) Solve the quadratic and get y = 5,30 so x = 30, y = 5 diff = 25

What are the remainders when 8^6 is divided by 7 and 8?

For this type: when 8^1 is divided by 7, rem=1; 8^2/7, rem=1; .. and so on.. 8^6/7, rem=1. And by 8, rem=0.

Find the value of (bx-ay), if x/a + y/b = 2 and ax-by = a^2 - b^2.

From eq. 1, bx + ay = 2ab. Now multiply this by 'b' and eq. 2 by 'a' and solve the eqns. It will result in x = a. Put in (1) and get y = b. So, bx-ay = ba-ab = 0 (Ans)

Additive inverse

In mathematics, the additive inverse of a number a is the number that, when added to a, yields zero. This number is also known as the opposite (number), sign change, and negation.

Which of the following are the LCM of 3!, 5! and 7!? a) 7! (b) 3! (c) 5! (d) 5040 (e) 12

LCM = 3!, 5x4, 7x6 = 7x6x5x4x3! = 7! = 5040 a, d

If a - b > a + b, where a and b are integers, which of the following must be true? Compare the signs of a, b and ab

Note that in this type, solve the eq. => 0 > 2b or b < 0 Hence, b is -ve. Nothing can be said about a and hence ab.

Given isosceles triangle. Exterior angle comparison problem.

Remember that the exterior angle is equal to the sum of the opposite two angles. https://www.dropbox.com/s/ikh6yhwk7a3cv1u/IsoscelesExteriorAngle.PNG?dl=0

-x^2 + 4x = -12 y^2 - 6y - 8 = 4 - 2y Compare x and y.

Remember that though the two equations evaluate to the same roots, the ans. is: the relationship cannot be determined because when x = 6, y can be -2 and v.v.

Type: When a/an (complex) equation is exactly divisible by another (simpler) equation, then find (b-a)..

This means that the roots of the simpler equations will be factors of the other (complex) equation. Simply put for x and then solve.

Select all the values of x which make the following true: 2x^4 = 7x^3 + 4x^2

This mistake done in this question was that after simplifying and solving the quadratic eqn., there were two values for x. However, there was also an option '0' that satisfied the eqn. Missed that!

The greatest of the 31 positive integers in a set, is 20. The median of the set is 15. What is the greatest possible average (arithmetic mean) of the 31 integers?

To maximize the avg. let all numbers above median be 20 and all below be 15. So, sum = 15(20) + 15(15) + 15 = 540 So, avg = 540/31

Cards numbered 1 through 10 are placed in an urn. One ticket is drawn at random. In how many chances out of 20 shall the card have a prime number written on it?

Total primes = 2,3,5,7 So, P(prime) = 4/10 or, 8/20 => 8 out of 20 chances.

What is the area enclosed by the points (1,4), (4,-2) and (9,-12)?

Try it! (This will confirm the subtraction sequence, which is important in this case).

In a kilometer race, A can give B a start of 50 metres and B can give C a start of 40 metres. How much start can A give C in a 2 km race?

Use the 'if a units costs y dollars then 1 unit will cost y/a dollars' method. Substitute 1 unit of B to 1 unit of A and get C. You did it that way and got the ans as 176 m.

Find (r-x) if 2P(5,3) = P(r,4) and P(9,5) + 5P(9,4) = P(10,x)

(2 x 5!)/2! = r(r-1)(r-2)(r-3) ----- kyunki (r-4)! ko udaa diya yahaan se. now, 5! = r(r-1)(r-2)(r-3) --- so if 5! is a product of 4 consecutive nos. then they must be 2.3.4.5 => r = 5 Solving the second eqn., after a lot of simplification, (2 x 9!)/4! = (10 x 9!)/(10-x)! or, 1/4! = 5/(10-x)! or, (10-x)! = 4! x 5 or, 10-x = 5 or, x = 5 so, r-x = 0

Which two of the following numbers have the quotient greater than 3? a) -8 (b) -4 (c) -2 (d) 4 (e) 8

-8/-2 = 4

A) 30 cents per ounce B) $5 per pound

1 pound = 16 ounce so, 30 x 16 cents per pound or 480/100 = 4.8 dollars per pound is (A) which is less than (B).

What are rational numbers?

A rational number is one which can be written as a fraction of integers, eg. -3 = -3/1 4/3 0.5 = 1/2 sqrt(16) = 4/1, etc.

Find the number of words formed by permuting all the letters in the word INDEPENDENCE such that the E's do not come together.

Ans => total no. of words - no. of words when all E's are together. Total no. of words = 12!/(3!2!4!) = 1663200 no. of words when all E's are together = consider all 4 E's as a group hence as a single alphabet = 9!/(3!2!) = 30240 hence, ans = 1663200 - 30240 = 1632960

If x/y > 1 and x + y < 0, which of the following must be true? (Indicate all that apply) 1) |x| > |y| 2) x < y 3) x > y 4) x^2 > y^2

Ans: 1,2,4 For such question, it is better to conjure an example. viz, -3/-2 > 1 and -3 + (-2) < 0 another one, x = -1/2, y = -1/3 Observing closely, it turns out that considering the two conditions, it is possible only when x and y are both -ve.

Two chords of lengths 32 cm and 112 cm lie in a circle of radius 65cm. Find the maximum and minimum possible distances b/w them.

Applying Pythagoras Theorem, The maximum distance is possible when the two chords are on opposite sides of the centre and the least distance is when the two chords are on the same side. The maximum distance = 63+33=96cm The minimum distance = 63-33=30cm

ab != 0 & a^x = b^x Compare a and b.

As neither a nor b is 0, a and b have several possibilities. If x is an even integer, a can be -ve and b the same +ve. Next a and b can be any number with x = 0. So, the relationship cannot be determined from the information given.

Find the probability that a leap year will contain either 53 Tuesdays or Wednesdays.

Assuming that the year is equally likely to start on any day of a week, there will be 53 Tues if the year starts on Mon or Tue and 53 Wed if it starts on Tue and Wed. So, P(53T U 53W) = P(53T) + P(53W) - P(53T and 53W) = 2/7 + 2/7 - 1/7 = 3/7

Given a>1, b>1, Compare: 1/a + 1/b AND 1/(1/a + 1/b)

At first look, I thought quantity B will always be greater, but.. there's one outlier. a=b=2. Hence, relationship cannot be determined.

Type: If factorial eq. like, (n+1)! = 12(n-1)! is given, then the quadratic equation, upon simplification, will result in two values of n...

But remember, n cannot be negative. So jettison the negative value.

Centroid of a triangle given the vertices

C = [(x1+x2+x3)/3, (y1+y2+y3)/3] Which is the intersection of the lines joining each vertex to the midpoint of the opposite side.

C can complete the work in 18 days working alone. B takes 3 days less than C and A takes 5 days less than B to complete the same work. On which day will they complete the work if C works for the initial two days only?

C's 1d work = 1/18 B's " " = 1/15 A's " " = 1/10 A,B,C's 1 d work = 1/18 + 1/15 + 1/10 = 6/27 = 2/9 A,B,C's 2 d work = 4/9 Remaining work = 1 - 4/9 = 5/9 A,B's 1 day work = 1/15 + 1/10 = 1/6 So, # more days required = 5/9 by 1/6 or (5*6) / 9 = 3.33 Total days spent = 3.33 + 2 = 5.33 So, the work will be completed on the 6th day.

3% chance of 1 egg broken and .5% chance of all 12 eggs broken, in a crate. How many eggs can be expected broken in a 1000 crate delivery?

E(x) = Σ(x.P(x)) so, .03 x 1 + .005 x 12 = .09 for 1000, E = 1000 x .09 = 90 eggs

6 students standing in a circle, playing a game, passing a ball to each other. How many such passes possible?

Every child has (6-1) options/choices to pass. So, 5^6 = 15625 PS: Neither is it 6C2 nor 6*(6-1).

Children of a school have to stand in different rows with each row having a flag of the same color. There are 25 with red, 35 with green and 100 children with blue flags. What is the least no. of rows that should be formed for this arrangement such that each has equal no. of children.

Find the HCF (basically, just take the LCM and select the greatest factor that divides all). = 5 So, # rows = total children / HCF = (25+35+100)/5 = 32 (Ans)

Which of the following points are on the straight line through (1,4) and (3,-2)?

For this type of question, construct the eqn. of the line, viz. y-y1 = m(x-x1) m = (y2-y1)/(x2-x1) = -6/2 = -3 so, line through (1,4): y-4 = -3(x-1) y = -3x + 7 Now, put the abscissa of the given points and check if it results in the respective ordinate.

At a certain school, the ratio of students to teachers is 3:5 and teachers to administrators is 4:5. Then it is asked which of the following could be the potential values of the number of students in the school.

For this type, always combine the ratios => 12/20 = 20/25 => 12:20:25 => # students has to be 12y or any multiple of 12 is a potential candidate.

Type: Time taken by A to complete a work is 20 days. B is 25% faster than A, and C is 60% faster than B.

For this type, apply percentages on one day works and not the full work (days). Ex. B's 1 day work is 125% * 1/20 and C's 160% of that. (Ans: 16, 10)

a^3/b^3 + b^3/a^3 = ? given, a^2 + b^2 = 5ab

For this type, bring the given equation to a familiar form: divide both sides by ab a/b + b/a = 5 Now take cubes and solve normally. Use (a+b)^3 = a^3 + b^3 + 3ab(a+b) Also, (a-b)^3 = a^3 - b^3 - 3ab(a-b)

P(Whole class passes in Maths) = .8 P("..".. in Hindi) = .7 P("..".. in either M or H) = .95 Find the P of not getting any failures in M and H in the whole class.

Given, P(M U H) = .95 as, P(M U H) = P(M) + P(H) - P(M and H) P(M and H) = .8 + .7 - .95 = 1.5 - .95 = .55 (Ans)

S = k/T; If S increases by 50%, by what percent will the value of T decrease?

Given, T = k/S. If, S becomes 1.5xS, T = k/1.5S or, new T = 1/1.5 times old T = 2/3 times T So, T decreased by 1 - 2/3 = 1/3 = 33.33%

The least common multiple of two numbers is 44 and the highest common factor of the numbers is 264. How many such pairs of numbers exist? (Doubt)

I don't understand this but they did something like: The product of the two nos. = 264/44 = 6 so nos. are (1,6), (2,3). Crazy!

Type: If the roots of the equation ax^2 + bx + c = 0 are equal, then... blah

If the roots are equal, D is zero. or b^2 - 4ac = 0. Also, if D is -ve, then the roots are imaginary. Also, Sum of roots = b/a Produc of roots = c/a

ABC inc. has a total of 450 employees who work for a total of 13500 hours per week. If the number of weekly work hours per person has normal distribution and the standard deviation is 7 hours, then how many employees work more than 37 hours per week.

In Normal Distribution, mean represents 50% of the distribution and one SD on either side, 34% of the remaining distribution. Between 1 - 2 SD, there is 13.5% distribution and beyond that, 2%. With that story, given u = 13500/450 = 30 (mean) 50% 1 SD = 30 + 7 = 37 So, employees with weekly hours less than or equal to 37 = 50% + 34% (above mean) = 84% (Not sure why we ignore the other side of the bell curve) So, employees who worked more than 37 hrs/week = 16% = .16 x 450 = 72

Type: 5 men can complete a work in 2 days, 3 women in 3 days, etc. Then asked how many days can 1 man and woman working together complete it?

In this type, simply find 1 man's 2 days work (1/5). Then find 1 man's 1 day work (1/(5*2)). Similarly for woman and then apply the 1/T = 1/Work1 + 1/Work2 formula. A: 4

Bob's weight is 30 pounds more than Sara's. If both gain 30 pounds then Bob's weight becomes 25% greater than Sara's. How much does Bob weigh right now?

Incisively frame the equation: Given, B = S+30 Now, B + 30 = 1.25(S + 30) or, S+30 + 30 = 1.25xS + 1.25x30 or, 60 - 37.5 = 1.25S - S or, S = 22.5/0.25 = 90 So, B = 90+30 = 120

A shopkeeper sold 200 tables at a loss. His loss was equal to the selling price of 4 tables. Which of the following are true? 1) His loss was Rs 2000 2) His loss% was 100/51% 3) His selling price was Rs 500 per table 4) The selling price cannot be determined 5) The total loss cannot be determined

Let SP 1 table = x, Loss = 4x CP = 200x + 4x = 204x So, loss is what percentage of CP? 4x = k/100 * 204x or 100/51 = k% Correct options: 2,4,5

40% of a 2:3 solution of A and B is replaced with B. What is the new ratio of A and B?

Let the initial qtty. by 10 L. So, qtty. A = (2*10)/5 = 4 L B = 10 - 4 = 6 L Now, 40%, or 4 L removed that contains 8/5 = 1.6 L of A and 4-1.6 = 2.4 L of B New solution contains 4-1.6 = 2.4 L of A and 6 - 2.4 (B removed) + 4 (B added) = 7.6 L of B So, A/B = 2.4/7.6 = 6:19

Two nos. are in the ratio 6:13. The LCM of the two nos. is 312. Find the sum of the two nos.

Let the nos. be 6k, 13k. Given, LCM = 6 x 13 x k = 312 so, k = 4 Nos. => 6 x 4 = 24; 13 x 4 = 52 so, 24+52 = 76.

A boat takes 6 hours to go downstream and upstream a certain distance. It takes 4 hours to travel twice the distance downstream. Which of the following is the ratio of the speed of the boat in still water to the speed of the river?

Let the speed of the boat in still water be x and the speed of the river be y Speed upstream = x-y Speed downstream = x+y Let the distance travelled be D Time = Distance/Speed D/(x+y) + D/(x-y) = 6 and 2D/(x+y) = 4...(1) Putting D/(x+y)=2 in the first relation, we get 2 + D/(x-y) = 6 D/(x-y) = 4 ...(2) From (1) and (2), we get (x+y)/(x-y) = 4/2 = 2 x+y=2x-2y x-2x=-2y-y -x=-3y x/y = 3/1

A child runs along the four sides of a square playground at speeds 200, 300, 600 and 800 m/hr. Find the average speed of the child.

Let x be the side of the square. So, T = x/200 + x/300 + x/600 + x/800 = x * 9/800 So, Avg. Speed = Total d/Total t = (4x * 800) / 9x = 355.55 m/hr

If a + b > 0 and b - c < 0, which of the following statements must be true? a + c > 0 c > a b > 0

Notice their something common in the two eqns., b. And because they look similar, they may be solved simultaneously. The key to solving this problem is to combine the two inequalities by summing them vertically: a + b > 0 + b - c < 0 ======== ? ======== Inequalities, however, can be added vertically only when the inequality signs face the same direction. To overcome this, we can 'flip the sign' of the second equation by multiplying both sides by -1: - (b - c) > 0 = c - b > 0 Now our inequalities can be summed: a + b > 0 c - b > 0 ======== a + c > 0 ======== Thus Choice A must be true.

A point P divides a line joining (a,b) and (c,d) in the ratio 1:6. Find the ordinate and abscissa of the point.

P = [(a x 6 + c x 1)/(1 + 6), (b x 6 + d x 1)/(1+6)] And, P(abscissa,ordinate)

A bag contains 5 R and 5 B balls. In how many ways can 5 balls be drawn randomly such that there are atleast 2 balls of each color?

PS: Notice 'atleast'. The only two possiblities are draw 2 balls of one color and then 3 of the other (and the other way round). So. = 5C2 x 5C3 + 5C3 x 5C2 = 2 x C(5,2)C(5,3).

If a and b are integers and a = (3x5x7x9x11x13)/33b. Find b.

Reduce the fraction. 11x3/33 gone. Now because a is an integer, b must divide the numerator. Use the calc and divide by all the available options.

If n = pr^2, and p and r are distinct prime nos., how many factors does n have?

Remember when you take the LCM, by taking an example: 12 = 3x2^2, the factors are, 1,2,3,4,6,12 or 1,p,r,r^2,pr,pr^2

The cost of one orange is equal to the cost of 4 bananas and the cost of three bananas is equal to the cost of 5 eggs. Find the ratio of the price of one orange to that of one banana to one egg.

Simple, O:B:E = 4B:B:3B/5 or, 4:1:3/5 (multiplying all be 5) = 20:5:3 Alternate method: You can also do it the rudimentary way as O/B = B/E now, 4B ------------ 1 O so O/B = 4/1 and, 3B ------------ 5 E so B/E = 5/3 so, 4/1 = 5/3 (make ratio equal) 4 x 5 / 1 x 5 = 5/3 or 20/5 = 5/3 or 20:5:3

If the probability of shooting two white rabbits in a jungle where only white and black rabbits thrive is 2/11 and there are 5 white rabbits, then what is the total no. of rabbits in the jungle?

Simple, let black rabbits be x, then given, P(shooting 2 white) = 5/(5+x) * 4/(4+x) = 2/11 or x^2 + 9x - 90 = 0 or x = 6 (Ans)

For what value of k is the line (k-3)x - (4-k^2)y + k^2-7k+6 = 0, parallel to the x-axis?

Simply bring all to the form y = mx + c and then put m = 0. Don't worry about the other k terms lingering in 'c'. (Ans: 3)

If a>b>k>0, compare a/b and (a+k)/(b+k).

Simply plug in values and analyze.

Type: Find the value of 'a' if (x-a) is the GCD of (eq 1) and (eq 2).

Simply put x=a in the equations and equate them. Solve for a.

Type: For what value of k will the following eqns. have infinitely many solns.

Simply solve the equations for x and then put k that makes the result 0.

9 letters: 5 consonants and 4 vowels. Three letters are chosen at random. P(choosing more than one vowels) = ?

So, P(2 or 3 vowels) = ? n(E): 2 vowels = 4C2 * 5C1 = 30 so, P(2 v) = 30 / 9C3 = 30/84 n(E): 3 vowels = 4C3 = 4 so, P(3 v) = 4/84 so, P(2 or 3 v) = 34/84 = 17/42 (Ans)

The mixture contains 5 parts of substance A for every 2 parts of substance B. Substance A consists of 5 parts of sugar, 2 of salt and 4 of other ingredients. Substance B consists of 3 parts of sugar, 2 parts of salt and 3 parts of other ingredients. How many parts of sugar does the mixture contain for every 248 parts of salt?

Substance A contains 5 parts of sugar, 2 of salt and 4 of other ingredients. Divide substance A into 11 parts. 1 part of substance A contains 5/11 parts of sugar, 2/11 parts of salt and 4/11 parts of other ingredients. 5 parts of substance A contain (5x5/11= ) 25/11 parts of sugar, (5x2/11= ) 10/11 parts of salt and (5x4/11=) 20/11 parts of other ingredients. Substance B contains 3 parts of sugar, 2 of salt and 3 of other ingredients. Divide substance B into 8 parts. 1 part of substance B contains 3/8 parts of sugar, 2/8 parts of salt and 3/8 parts of other ingredients. 2 parts of substance B contain (2x3/8=) 6/8 parts of sugar, (2x2/8=) 4/8 parts of salt and (2x3/8=) 6/8 parts of other ingredients. Total sugar in the mixture = 25/11+6/8 = (25x8+6x11)/(11x8) = 266/88 Total salt in the mixture = 10/11+4/8 = (10x8+4x11)/(11x8) = (80+44)/88 = 124/88 248/(124/88) = 2x88 = 176 parts of total mixture (266/88)x176 = 266x2 = 532 The mixture contains 532 parts of sugar for every 248 parts of salt in it.

There are three teams A, B and C containing 5, 4 and 5 equally skilled members respectively. If teams A, B and C can complete a set of tasks in 2, 3 and 3 days respectively, then how many days will it take a fourth team to complete the work when it comprises of three members, one from each of the three teams?

Team A's 1 member will take 5*2 days. Similarly, B's 1 mem --- 4*3 days C's 1 mem --- 5*3 days And, 1/A + 1/B + 1/C = 1/T or, 1/T = 1/10 + 1/12 + 1/15 = 450/1800 = 1/4 => 4 days will be taken by the new team.

Car sold median question

The price ranges are already in sorted order, and we know that the median is at 15th car that lies b/w 7 and 10 cars. https://www.dropbox.com/s/iax7mn4y2gah6lf/CarSoldMedianQ.PNG?dl=0

Type: Finding area of a triangle when the co-ordinates of its three vertices are given.

This is given by something called, Shoelace Formula. So, Area of triangle (x1,y1),(x2,y2),(x3,y3) = | 1/2 [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)] |

Given a triangle ABC and a line connecting the midpoints of two of its adjacent sides. Also given, the area of the smaller triangle formed by the middle line = 8. Asked, the area of triangle ABC.

This is simple. The two triangles are similar as they share the common angle, ex. angle C and their sides are in the ratio 1:2. Now the area of the smaller triangle = 1/2 x (bh). Area of the bigger triangle (ABC) = 1/2 x (2b.2h) = 2bh From smaller triangle, bh = 8x2 = 16 so, ABC = 2 x 16 = 32

There are 10 flags of different colours. Tim has to prepare 2 different arrangements, each using 4 flags of different colours. How many different arrangements can he possibly make if the flags of one set cannot be used in the other?

Watch out, this is a permutation problem because flags are different. So choosing AB is not the same as choosing BA. nPr(10,4) * nPr(6,4)

The mth term of an AP is 1/n and the nth term is 1/m. Find the first term and common difference.

by def., Tn = a + (n-1)d Now create two eqns. Tn and Tm, equate them to 1/m & 1/n respectively. Then subtract eq. 2 from 1 and find d = 1/mn. Substitute in any of 2 or 1. Solve for a = 1/mn or a = d

3 dice are rolled, i) Find the P (even on one and 5 on other two) ii) P (total of 6)

i) n(E) => {2,4,6}, {5}, {5} = 3 * 3!/2! (similar to permutation of a word with duplicate letters) So, P = 9/6^3 ii) n(E) => 1,2,3 = 3! 1,1,4 = 3!/2! 2,2,2 = 1 So, n(E) = 6 + 3 + 1 So, P = 10/6^3

n and q are +ve integers. Compare q^(n-5) and q^(n/5).

put q=1 and n=5, both quantities = 1. put q=2 and n=10, first qtty. is greater. So the relationship cannot be determined.

Two dice are tossed once. The probability of getting an even number on the first die or a total of 8 is

total possible outcomes for two dice: 36 case 1 prob of getting an even number on the first die first die can have 2,4,6 second die can have any of 1,2,3,4,5,6 or favorable outcomes 3x6 = 18 probability= 18/36 case 2 getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4) probability=5/36 now both of the above cases have some cases common to them i.e. when the first die has an even number and the sum is also 8 there are 3 cases of this kind (2,6) (6,2) (4,4) prob=3/36 also P(A or B)=P(A) + P(B) - P(A & B) so we have P(even or sum of 8) = 18/36 + 5/36 - 3/36 20/36 http://gmatclub.com/forum/two-dice-are-tossed-once-the-probability-of-getting-an-even-126477.html OR My way: Case 1: ===== {2,4,6} x {...6 nos...} = 18 Case 2: ===== (2,6) --- 2 nos. (Already included in case 1) (3,5) --- 2 nos. (4,4) --- 1 nos. (Already included in case 1) So, n(E) = 18 + 2 = 20 n(S) = 36 so, P = 20/36 = 5/9

Three unbiased dice are rolled, find the P of getting a factor of 8 greater than 1 on first, 1 on second and 3 on third dice.

{2,4},{1},{3} so, n(E) = 2 x 3! so, P = (2x2x3)/6^3 = 2/6^2 or P = 1/18