Practice Advance math - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in ascending order ?
(a)
(b)
(c)
(d)
Any 3 numbers out of 9 can be selected in $^9C_3$ ways.
Now, these three numbers can be arranged among themselves in ascending order in only 1 way.
Hence, total no. of ways = $^9C_3$ × 1 = 84
Q-2) In a tournament, each of the participants was to play one match against each of the other participants. Three players fell ill after each of them had played three matches and had to leave the tournament. What was the total number of participants at the beginning, if the total number of matches played was 75 ?
(a)
(b)
(c)
(d)
Let the total no. of participants be 'n' at the beginning.
Players remaining after sometime = n – 3
Now, $^{n–3}C_2$ + (3 × 3) = 75
${(n - 3)!}/{2!(n - 5)!}$ + 9 = 75
$n^2$ – 7n – 120 = 0
(n + 8) (n – 15) = 0
neglecting n = – 8, n = 15
Q-3) In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position ?
(a)
(b)
(c)
(d)
Let us take books A and B as one i.e., they are always continuous.
Now, number of books = 4 – 2 + 1 = 3
These three books can be arranged in 3! ways and also A and B can be arranged in 2 ways among themselves.
So, number of ways when books A and B are always continuous = 2 × 3!
Total number of ways of arrangement of A, B, C and D = 4!
Hence, number of ways when A and B are never continuous = Total number of ways – number of ways when A and B always continuous
= 4! – 2 × 3! = 12
Q-4) A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if atleast one black ball is to be included in the draw ?
(a)
(b)
(c)
(d)
At least one black ball can be drawn in the following ways:
(i) one black and two other colour balls
(ii) two black and one other colour balls, and
(iii) all the three black balls
Therefore the required number of ways is
$^3C_1 × ^6C_2 + ^3C_2 × ^6C_1 + ^3C_3$ = 64.
Q-5) The number of ways of choosing a committee of 2 women and 3 men from 5 women and 6 men, if Mr. A refuses to serve on the committee if Mr. B is a member and Mr. B can only serve, if Miss C is the member of the committee, is
(a)
(b)
(c)
(d)
(i) Miss C is taken
(1) B included ⇒A excluded ⇒$^4C_1 . ^4C_2$ = 24
(2) B excluded ⇒$^4C_1 . ^5C_3$ = 40
(ii) Miss C is not taken
⇒B does not comes ; $^4C_2 . ^5C_3$ = 60⇒Total = 124
Q-6) The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices is
(a)
(b)
(c)
(d)
We have in all 12 points. Since, 3 points are used to form a triangle, therefore the total number of triangles including the triangles formed by collinear points on AB, BC and CA is $^{12}C_3$ = 220. But this includes the following :
The number of triangles formed by 3 points on AB = $^3C_3$ = 1
The number of triangles formed by 4 points on BC = $^4C_3$ = 4.
The number of triangles formed by 5 points on CA = $^5C_3$ = 10.
Hence, required number of triangles = 220 – (10 + 4 + 1) = 205.
Q-7) A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from each group. The number of ways in which he can choose the 7 questions is
(a)
(b)
(c)
(d)
No. of way to answer.
= $^6C_2 ^6C_5 + ^6C_3 ^6C_4 + ^6C_4 ^6C_3 + ^6C_5 ^6C_2$
= $2({^6}C_2 ^6C_5 + ^6C_3^6C_4) = 2({^6}C_2 ^6C_1 + ^6C_3 ^6C_2)$
= 2$({6 × 5}/2 × 6 + {6 × 4 × 5}/{3 × 2} × {6 × 5}/2)$
= 2 (90 + 300) = 780
Q-8) The probability that at least one of the events A and B occurs is 0.7 and they occur simultaneously with probability 0.2. Then P($\ov{A}$) + P($\ov{B}$) =
(a)
(b)
(c)
(d)
We have P (A ∪ B) = 0.7 and P (A ∩ B) = 0.2
Now, P(A ∪ B) = P(A) + P(B) - P (A ∩ B)
⇒P(A) + P(B) = 0.9⇒1-P($\ov{A}$) + 1 - P($\ov{B}$) = 0.9
⇒P($\ov{A}$) + P($\ov{B}$) = 1.1
Q-9) Each of two women and three men is to occupy one chair out of eight chairs, each of which numbered from 1 to 8. First, women are to occupy any two chairs from those numbered 1 to 4; and then the three men would occupy any, three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done ?
(a)
(b)
(c)
(d)
2 Women can occupy 2 chairs out of the first four chairs in $^4P_2$ ways. 3 men can be arranged in the remaining 6 chairs in $^6P_3$ ways.
Hence, total no. of ways = $^4P_2 × ^6P_3$ = 1440
Q-10) If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, the number of points in which they intersect each other is
(a)
(b)
(c)
(d)
Points are formed when two lines cuts each others.
No. of ways $^{20}C_2 = {20!}/{2!18!} = {20 × 19}/2$ = 190
Q-11) There are 6 different letters and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes ?
(a)
(b)
(c)
(d)
As there are 6 letters and envelopes, so if exactly 5 are into correctly addressed envelopes, then the remaining 1 will automatically be placed in the correctly addressed envelope. Thus, the probability that exactly 5 go into the correctly addressed envelope is zero.
Q-12) The probability that the 13th day of a randomly chosen month is a Friday, is
(a)
(b)
(c)
(d)
Probability of selecting a month = $1/{12}$.
$13^{th}^$ day of the month is friday if its first day is sunday and the probability of this = $1/7$.
∴ Required probability = $1/{12}.1/7 = 1/{84}$.
Q-13) A programmer noted the results of attempting to run 20 programs. The results showed that 2 programs ran correctly in the first attempt, 7 ran correctly in the second attempt, 5 ran correctly in the third attempt, 4 ran correctly in the fourth attempt and 2 ran correctly in the fifth attempt. What is the probability that his next programme will run correctly on the third run ?
(a)
(b)
(c)
(d)
Total number of attempts = 20
Favourable no. of attempts = 5
Required probability (running the program correctly in the third run) = $5/{20} = 1/4$
Q-14) One bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, determine the probability that one ball is white and another is black.
(a)
(b)
(c)
(d)
Probability that first ball is white and second black
= (4/6) × (5/8) = 5/12
Probability that first ball is black and second white
= (2/6) × (3/8) = 1/8
These are mutually exclusive events hence the required probability
P = $5/{12} + 1/8 = {13}/{24}$.
Q-15) 3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd?
(a)
(b)
(c)
(d)
Let E be the event of selecting the three numbers such that their product is odd and S be the sample space. For the product to be odd, 3 numbers choosen must be odd.
∴ n(E) = $^5C_3$
= n(S) = $^9C_3$
∴ P(E) = ${n(E)}/{n(S)} = {^5C_3}/{^9C_3} = 5/{42}$
Q-16) If the probability that A and B will die within a year are p and q respectively, then the probability that only one of them will be alive at the end of the year is
(a)
(b)
(c)
(d)
Only one of A and B can be alive in the following, mutually exclusive ways.
$E_1$ A will die and B will live
$E_2$ B will die and A will live
So, required probability = $P(E_1) + P (E_2)$
= p(1 - q) + q (1 - p) = p + q - 2pq.
Q-17) A bag has 4 red and 5 black balls. A second bag has 3 red and 7 black balls. One ball is drawn from the first bag and two from the second. The probability that there are two black balls and a red ball is :
(a)
(b)
(c)
(d)
Required probability
= Probability that ball from bag A is red and both the balls from bag B are black + Probability that ball from bag A is black and one black and one red balls are drawn from bag
= ${^4C_1}/{^9C_1} × {^7C_2}/{^{10}C_2} + {^5C_1}/{^9C_1} × {^3C_1 × ^7C_1}/{^{10}C_2}$
= $4/9 × 7/{15} + 5/9 × 7/{15} = 7/{15}$
Q-18) A bag has 13 red, 14 green and 15 white balls, $p_1$ is the probability of drawing exactly 2 white balls when four balls are drawn. Then the number of balls of each colour are doubled. Let $p_2$ be the probability of drawing 4 white balls when 8 ball are drawn, then
(a)
(b)
(c)
(d)
$p_1 = {^{15}C_2}/{^{42}C_4} = {15 × 14 × 4!}/{2! × 42 × 41 × 40 × 39} = 1/{41 × 26}$ and
$p_2 = {^{30}C_4}/{^{84}C_8} = {15 × 14 × 13 × 12 × 8!}/{4! × 84 × 83 × 82 × ...... × 77}$
= ${15 × 14 × 13 × 12 × 8 × 7 × 6 × 5}/{84 × 83 × 82 × 81 × 79 × 78 × 77} < p_1$
⇒$p_1 > p_2$.
Q-19) In a given race the odds in favour of three horses A, B, C are 1 : 3; 1 : 4; 1 : 5 respectively. Assuming that dead head is impossible the probability that one of them wins is
(a)
(b)
(c)
(d)
Suppose $E_1 , E_2$ and $E_3$ are the events of winning the race by the horses A, B and C respectively
∴ $P(E_1) = 1/{1 + 3} = 1/4, P(E_2) = 1/{1 + 4} = 1/5$
$P(E_3) = 1/{1 + 5} = 1/6$
∴ Probability of winning the race by one of the horses A, B and C
= $P (E_1 or E_2 or E_3) = P(E_1) + P(E_2) + P(E_3)$
= $1/4 + 1/5 + 1/6 = {37}/{60}$
Q-20) A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighbouring places are empty is
(a)
(b)
(c)
(d)
Exhaustive number of cases = $^{24}C_{14}$
Favourable cases = $^{22}C_{14}$
Probability = ${^{22}C_{14}}/{^{24}C_14} = {15}/{92}$