Practice Advance math - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A,B,C and D are four towns any three of which are noncollinear. Then the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is
(a)
(b)
(c)
(d)
To construct 2 roads, three towns can be selected out of 4 in 4 ×3×2 = 24 ways. Now if the third road goes from the third town to the first town, a triangle is formed, and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.
Q-2) Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2 ?
(a)
(b)
(c)
(d)
There can be 3 cases :
I. When one dice shows 2.
II. When two dice shows 2.
III. When three dices shows 2.
Case I : The dice which shows 2 can be selected out of the 3 dices in $^3C_1$ ways.
Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = $^5C_1$ each, so no of ways when one dice shows 2 = $^3C_1 × ^5C_1 × ^5C_1$ .
Case II : Two dices, showing 2 can be selected out of the 3 dices in $^3C_2$ ways and the rest one can have any 5 numbers except 2, so number of ways for the remaining 1 dice = 5.
So, number of ways, when two dices show
2 = $^3C_2$ × 5
Case III : When three dices show 2 then these can be selected in $^3C_3$ ways.
So, number of ways, when three dices show
2 = $^3C_3$ = 1
As, either of these three cases are possible.
Hence, total number of ways
= (3 × 5 × 5) + (3 × 5) + 1 = 91
Q-3) The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat and the terminal digits are even is
(a)
(b)
(c)
(d)
The first and the last (terminal) digits are even and there are three even digits. This arrangement can be done in $^3P_2$ ways. For any one of these arrangements, two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits (leaving out the terminal digits) may be arranged using these 5 digits in $^5P_4$ ways. The required number of numbers is $^3P_2 × ^5P_4$ = 6 × 120 = 720.
Q-4) In a chess tournament, where the participants were to play one game with another, two chess players fell ill, having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was
(a)
(b)
(c)
(d)
Let there be n participants in the beginning. Then the number of games played by (n – 2) players = $^{n - 2}C_2$
∴ $^{n - 2} C_2 + 6 = 84$
(Two players played three games each)
⇒$^{n - 2}C_2$ = 78⇒(n - 2)(n - 3) = 156
⇒$n^2$ - 5n - 150 = 0
⇒$n^2$ - 15n + 10n = 150 = 0
⇒n(n - 15) + 10(n - 15) = 0
(n - 15)(n + 10) = 0
n = 15, –10
n cannot be –ve
Therefore n = 15.
Q-5) There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown below such that balls of the same colour do not occupy any two consecutive cells ?
(a)
(b)
(c)
(d)
Let us start with Red colour
Where, R = Red, B = Black, W = White
| R | B | R | W |
| R | W | R | B |
| R | B | R | B |
| R | W | R | W |
| R | B | W | R |
| R | W | B | R |
| R | B | B | R |
| R | W | W | R |
There are eight such arrangements, if we start with Red ball. Similarly, there are 8 arrangements, if we start with black or white ball.
Hence, No. of arrangements = 8 + 8 + 8 = 24
Q-6) Three dice are rolled. The number of possible outcomes in which at least one die shows 5 is
(a)
(b)
(c)
(d)
Required number of possible outcomes
= Total number of possible outcomes – Number of possible outcomes in which 5 does not appear on any dice. (hence 5 possibilities in each throw)
= $6^3 – 5^3$ = 216 – 125 = 91
Q-7) A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a)
(b)
(c)
(d)
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.
∴ No. of choices available to the student
= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.
Q-8) To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection ?
(a)
(b)
(c)
(d)
Required no. of the ways = $^6C_3 × ^4C_2$ = 20 × 6 = 120
Q-9) A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour ?
(a)
(b)
(c)
(d)
Six balls can be selected in the following ways: one red balls and 5 blue balls or
Two red balls and 4 blue balls
Total number of ways
= $^3C_1 × ^7C_5 + ^3C_2 × ^7C_4$
= 3 × ${7 × 6}/{2 × 1} + 3 × {7 × 6 × 5}/{3 × 2 × 1}$
= 63 + 105 = 168
Q-10) ABCD is a convex quadrilateral. 3, 4, 5 and 6 points are marked on the sides AB, BC, CD and DA respectively . The number of triangles with vertices on different sides is
(a)
(b)
(c)
(d)
Triangles with vertices on AB, BC and CD are
3×4×5 = 60
Triangles with vertices on AB, BC and DA are
3×4×6 = 72
Triangles with vertices on AB, CD and DA are
3×5×6 = 90
Triangles with vertices on BC, CD and DA are
4×5×6 = 120
∴ Total no. of triangles = 60 + 72 + 90 + 120 = 342
Q-11) If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word is
(a)
(b)
(c)
(d)
Starting with the letter A, and arranging the other four letters, there are 4! = 24 words. These are the first 24 words. Then starting with G, and arranging A, A, I, and
N in different ways, there are ${4!}/{2!1!1!} = {24}/2$ = 12 words.
Hence, total 36 words.
Next, the 37th word starts with I. There are 12 words starting with I. This accounts up to the 48th word. The 49th word is NAAGI.
The 50th word is NAAIG.
Q-12) If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is :
(a)
(b)
(c)
(d)
Three vertices can be selected in $^6C_3$ ways.
The only equilateral triangles possible are $A_1 A_3 A_5 \text"and" A_2 A_4 A_6$
p = $2/{^6C_3} = 2/{20} = 1/{10}$
Q-13) A and B play a game where each is asked to select a number from 1 to 5. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is
(a)
(b)
(c)
(d)
| 1,1 | 1,2 | 1,3 | 1,4 | 1,5 |
| 2,1 | 2,2 | 2,3 | 2,4 | 2,5 |
| 3,1 | 3,2 | 3,3 | 3,4 | 3,5 |
| 4,1 | 4,2 | 4,3 | 4,3 | 4,5 |
| 5,1 | 5,2 | 5,3 | 5,4 | 5,5 |
No. of total events = 25.
Chance of winning in one trial = $5/25 = 1/5$
Hence, chance of not winning = $1 - 1/5 = 4/5$
Q-14) The probabilities of four cricketers A, B, C and D scoring more than 50 runs in a match are $1/2 , 1/3 , 1/4$ and $1/{10}$ . It is known that exactly two of the players scored more than 50 runs in a particular match. The probability that these players were A and B is
(a)
(b)
(c)
(d)
Let $E_1$ be the event that exactly two players scored more than 50 runs then P$(E_1) = 1/2 × 1/3 × 3/4 × 9/{10} + 1/2 × 2/3 × 1/4 × 9/{10} + 1/2 × 2/3 × 3/4 × 1/{10} + 1/2 × 1/3 × 1/4 × 9/{10} + 1/2 × 1/3 × 3/4 × 1/{10} + 1/2 × 2/3 × 1/4 × 1/{10} = {65}/{240}$
Let $E_2$ be the event that A and B scored more than 50 runs, then P$(E_1 ∩ E_2) = 1/2 × 1/3 × 3/4 × 9/{10} = {27}/{240}$
∴ Desired probability
= $P(E_2/E_1) = {P(E_1 ∩ E_2)}/{P(E_1)} = {27}/{65}$
Q-15) A woman goes to visit the house of some friend whom she has not seen for many years. She knew that besides the two married adults in the household, there are two children of different ages. But she does not knew their genders. When she knocks on the door of the house, a boy answers. What is the probability that the younger child is a boy ?
(a)
(b)
(c)
(d)
The total possible pairs of children (B, B), (B, G), (G, B). Now the one child is boy, is confirmed, but we don't know whether he is youngest or elder one. So the three ordered pairs could be the one describing the children in this family. So the probability of the younger children to be boy = 2/3.
Q-16) A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign, then the probability that they belong to different institutions is:
(a)
(b)
(c)
(d)
Required probability = ${^2C_1 × ^3C_1 × ^4C_1}/{^9C_3}$
= ${2 × 3 × 4}/{{9!}/{6!3!}} = 2/7$
Q-17) Six teachers and six students have to sit round a circular table such that there is a teacher between any two students. The number of ways in which they can sit is
(a)
(b)
(c)
(d)
Six students $S_1 , S_2 , ........, S_6$ can be arranged round a circular table in 5 ! ways. Among these 6 students there are six vacant places, shown by dots (•) in which six teachers can sit in 6 ! ways.
Hence, number of arrangement = 5 ! × 6 !
Q-18) A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged ?
(a)
(b)
(c)
(d)
Two tallest boys can be arranged in 2! ways. Rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways of arrangement = 2! × 18! × 6!
= 18! × 2 × 720 = 18! × 1440
Q-19) If A and B are two independent events and P(C) = 0, then A, B, C are :
(a)
(b)
(c)
(d)
Since, A and B are independent events.
∴ P(A ∩ B) = P(A)P(B)
Further since, A ∩ C, B ∩ C, A ∩ B ∩ C are subsets of C, we have
P(A ∩ C) ≤ P(C) = 0
P(B ∩ C) ≤ P(C) = 0
and P(A ∩ B ∩ C) ≤ P(C) = 0
⇒ P(A ∩ C) = 0 = P(A)P(C)
P(B ∩ C) = 0 = P(B)P(C)
P(A ∩ B ∩ C) = 0 = P(A)P(B)P(C).
Clearly A, B, C are pairwise independent as well as mutually independent. Thus, A,B,C are independent events.
Q-20) Suppose six coins are tossed simultaneously. Then the probability of getting at least one tail is :
(a)
(b)
(c)
(d)
If six coins are tossed, then the total no. of outcomes = $(2)^6$ = 64
Now, probability of getting no tail = $1/{64}$
Probability of getting at least one tail = 1 - $1/{64} = {63}/{64}$