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) $p_1 > p_2$
(b) $p_1 < p_2$
(c) $p_1 = p_2$
(d) None
The correct answers to the above question in:
Answer: (a)
$p_1 = {^{15}C_2}/{^{42}C_4} = {15 × 14 × 4!}/{2! × 42 × 41 × 40 × 39} = 1/{41 × 26}$ and
$p_2 = {^{30}C_4}/{^{84}C_8} = {15 × 14 × 13 × 12 × 8!}/{4! × 84 × 83 × 82 × ...... × 77}$
= ${15 × 14 × 13 × 12 × 8 × 7 × 6 × 5}/{84 × 83 × 82 × 81 × 79 × 78 × 77} < p_1$
⇒$p_1 > p_2$.
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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
= $P (E_1 or E_2 or E_3) = P(E_1) + P(E_2) + P(E_3)$
= $1/4 + 1/5 + 1/6 = {37}/{60}$
Question : 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) 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 : 4
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.
Question : 5
Six teachers and six students have to sit round a circular table such that there is a teacher between any two students. The number of ways in which they can sit is
a) 5 ! × 6 !
b) 5 ! × 5 !
c) 6 ! × 6 !
d) None of these
Answer »Answer: (a)
Six students $S_1 , S_2 , ........, S_6$ can be arranged round a circular table in 5 ! ways. Among these 6 students there are six vacant places, shown by dots (•) in which six teachers can sit in 6 ! ways.
Hence, number of arrangement = 5 ! × 6 !
Question : 6
A class consists of 80 students, 25 of them are girls and 55 are boys. If 10 of them are rich and the remaining poor and also 20 of them are intelligent then the probability of selecting an intelligent rich girl is
a) ${25}/{128}$
b) $5/{512}$
c) $5/{128}$
d) None of these
Answer »Answer: (b)
Total 80, Girls = 25, Boys = 55
10 R, 70 P, 20 I
$1/4 × 1/8 × {25}/{80} = 5/{512}$
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