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) ${37}/{60}$
(b) $1/5$
(c) $7/{60}$
(d) $1/8$
The correct answers to the above question in:
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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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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) $1/2$
b) $1/3$
c) $2/3$
d) $1/4$
Answer »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.
Question : 4
From a pack of 52 cards, two cards are drawn, the first being replaced before the second is drawn. What is the probability that the first is a diamond and the second is a king ?
a) $4/{13}$
b) $1/{52}$
c) $1/4$
d) $4/{15}$
Answer »Answer: (b)
Probability of getting a diamond, P(D) = ${13}/{52} = 1/4$
and probability to king, P(K) = $4/{52} = 1/{13}$
So, required probability = P(D).P(K)
= $1/4 × 1/{13} = 1/{52}$
Question : 5
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) $p_1 > p_2$
b) $p_1 < p_2$
c) $p_1 = p_2$
d) None
Answer »Answer: (a)
$p_1 = {^{15}C_2}/{^{42}C_4} = {15 × 14 × 4!}/{2! × 42 × 41 × 40 × 39} = 1/{41 × 26}$ and
$p_2 = {^{30}C_4}/{^{84}C_8} = {15 × 14 × 13 × 12 × 8!}/{4! × 84 × 83 × 82 × ...... × 77}$
= ${15 × 14 × 13 × 12 × 8 × 7 × 6 × 5}/{84 × 83 × 82 × 81 × 79 × 78 × 77} < p_1$
⇒$p_1 > p_2$.
Question : 6
Twenty-seven persons attend a party. Which one of the following statements can never be true ?
a) Each person in the party has a different number of acquaintances
b) There is a person in the party who has an odd number of acquaintances
c) There is a person in the party who is acquainted with all the twenty-six others
d) In the party, there is no set of three mutual acquaintances
Answer »Answer: (a)
It can be clearly established that the choices (a) , (c) and (d) may or may not be true . Statement (b) can never be true because every person cannot have a different number of acquaintances.
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