Practice Model questions set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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-2) 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-3) 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-4) 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-5) 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-6) 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-7) 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-8) 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-9) 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-10) 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