Advance Math Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Advance Math topic of quantitative aptitude
(a) $1/2$
(b) $1/3$
(c) $2/3$
(d) $1/4$
The correct answers to the above question in:
Answer: (c)
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.
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
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) 24
b) 27
c) 21
d) 33
Answer »Answer: (b)
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
Question : 2
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) 6! × 1440
b) 18! × 2! × 1440
c) 18! × 1440
d) None of these
Answer »Answer: (c)
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
Question : 3
A Chartered Accountant applies for a job in two firms X and Y. The probability of his being selected in firm X is 0.7 and being rejected at Y is 0.5 and the probability of atleast one of his applications being rejected is 0.6. What is the probability that he will be selected in one of the firms ?
a) 0.8
b) 0.4
c) 0.2
d) 0.7
Answer »Answer: (a)
P(x) = 0.7
= The probability of selected in firm X
P(y) = 1 – 0.5 = 0.5
= probability of selected in firms Y
P(X ∪ Y) = 1 – 0.6 = 0.4
Probability that he selected in one of the firms
= 0.7 + 0.5 – 0.4 = 0.8
Question : 4
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) 8
b) 9
c) 7
d) More than 9
Answer »Answer: (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.
Question : 5
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) 60
b) 40
c) 48
d) 30
Answer »Answer: (c)
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
Question : 6
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) ${37}/{60}$
b) $1/5$
c) $7/{60}$
d) $1/8$
Answer »Answer: (a)
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
= $P (E_1 or E_2 or E_3) = P(E_1) + P(E_2) + P(E_3)$
= $1/4 + 1/5 + 1/6 = {37}/{60}$
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