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) 190
(b) 250
(c) 220
(d) 120
The correct answers to the above question in:
Answer: (a)
Points are formed when two lines cuts each others.
No. of ways $^{20}C_2 = {20!}/{2!18!} = {20 × 19}/2$ = 190
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Question : 1
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
$1/4 × 1/8 × {25}/{80} = 5/{512}$
Question : 2
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 : 3
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 : 4
The alphabets of word ALLAHABAD are arranged at random. The probability that in the words so formed, all identical alphabets are found together, is
a) $16/17$
b) $5!/9!$
c) $1/63$
d) None of these
Answer »Answer: (c)
(AAAA), (LL), HBD
P = ${5!}/{{9!}/{4!2!}} = {5! × 4! × 2!}/{9!}$
= ${24 × 2}/{9 × 8 × 7 × 6} = 1/{63}$
Question : 5
A man and his wife appear for an interview for two posts. The probability of the husband's selection is $1/7$ and that of the wife's selection is $1/5$ . The probability that only one of them will be selected is
a) $4/{35}$
b) $6/{35}$
c) $6/7$
d) $2/7$
Answer »Answer: (d)
Probability that only husband is selected
= P(H) P($\ov{W}$) = $1/7(1 - {1/5}) = 1/7 × 4/5 = 4/{35}$
Probability that only wife is selected
= P($\ov{H}$) P(W) = $(1 - {1/7})(1/5) = 6/7 × 1/5 = 6/{35}$
∴ Probability that only one of them is selected
=$4/{35} + 6/{35} = {10}/{35} = 2/7$
Question : 6
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) 132
b) 1440
c) 40
d) 3660
Answer »Answer: (b)
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
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