Practice Model questions set 2 - 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) 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-3) How many different letter arrangements can be made from the letter of the word EXTRA in such a way that the vowels are always together ?
(a)
(b)
(c)
(d)
Considering the two vowels E and A as one letter, the total no. of letters in the word 'EXTRA' is 4 which can be arranged in $^4P_4$ , i.e. 4! ways and the two vowels can be arranged among themselves in 2! ways.
∴ reqd. no. = 4! × 2! = 4 × 3 × 2 × 1 × 2 × 1 = 48
Q-4) Each of the 3 persons is to be given some identical items such that product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done ?
(a)
(b)
(c)
(d)
Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, There can be 5 cases :
Case I : When one of them is given 30 items and rest two 1 item each.
So, number of ways for (30 × 1 × 1) = ${3!}/{2!}$ = 3
(As two of them have same number of items)
Case II : Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III : Number of ways for (15 × 2 × 1) = 3! = 6
Case IV : Number of ways for (6 × 5 × 1) = 3! = 6
Case V : Number of ways for (5 × 3 × 2) = 3! = 6
Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
Q-5) Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?
(a)
(b)
(c)
(d)
For each person to know all the secrets the communication has to be between the Englishmen (who knows say E1 French) and one Frenchmen (say F1). The other two in each case will communicate with E1 & F1 respectively. So for minimum no. of calls, E2 gives information to E1 & receives it after E1 interacts with F1. So 2 calls for each of the four E2, E3, F2 and F3, i.e., 8 calls +1 call (between E1 & F1). Hence 9 calls in all.
Q-6) In how many ways can four children be made to stand in a line such that two of them, A and B are always together ?
(a)
(b)
(c)
(d)
Take, A and B to be always together as a single entity.
Now, total no. of children = 4 – 2 + 1 = 3
These can be arranged in 3! ways and A, B can be arranged among themselves in 2! ways.
Hence, no. of arrangements such that A and B are always together = 3! × 2! = 3 × 2 × 2 = 12
Q-7) The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is
(a)
(b)
(c)
(d)
Let the sides of the game be A and B. Given 5 married couples, i.e., 5 husbands and 5 wives. Now, 2 husbands for two sides A and B can be selected out of 5 = $^5C_2$ = 10 ways.
After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we are to choose 2 wives out of remaining 5 – 2 = 3 wives i.e., $^3C_2$ = 3 ways.
Again two wives can interchange their sides A and B in 2! = 2 ways.
By the principle of multiplication, the required number of ways = 10 × 3 × 2 = 60
Q-8) 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-9) 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-10) 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