Advance Math Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 2 EXERCISES

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The following question based on Advance Math topic of quantitative aptitude

Questions : 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

The correct answers to the above question in:

Answer: (c)

(AAAA), (LL), HBD

P = ${5!}/{{9!}/{4!2!}} = {5! × 4! × 2!}/{9!}$

= ${24 × 2}/{9 × 8 × 7 × 6} = 1/{63}$

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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers

Question : 1

If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, the number of points in which they intersect each other is

a) 190

b) 250

c) 220

d) 120

Answer: (a)

Points are formed when two lines cuts each others.

No. of ways $^{20}C_2 = {20!}/{2!18!} = {20 × 19}/2$ = 190

Question : 2

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: (b)

Total 80, Girls = 25, Boys = 55

10 R, 70 P, 20 I

$1/4 × 1/8 × {25}/{80} = 5/{512}$

Question : 3

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: (a)

advance-math-aptitude-mcq

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 : 4

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: (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 : 5

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: (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

Question : 6

The probabilities of four cricketers A, B, C and D scoring more than 50 runs in a match are $1/2 , 1/3 , 1/4$ and $1/{10}$ . It is known that exactly two of the players scored more than 50 runs in a particular match. The probability that these players were A and B is

a) $5/6$

b) $1/6$

c) ${27}/{65}$

d) None of these

Answer: (c)

Let $E_1$ be the event that exactly two players scored more than 50 runs then P$(E_1) = 1/2 × 1/3 × 3/4 × 9/{10} + 1/2 × 2/3 × 1/4 × 9/{10} + 1/2 × 2/3 × 3/4 × 1/{10} + 1/2 × 1/3 × 1/4 × 9/{10} + 1/2 × 1/3 × 3/4 × 1/{10} + 1/2 × 2/3 × 1/4 × 1/{10} = {65}/{240}$

Let $E_2$ be the event that A and B scored more than 50 runs, then P$(E_1 ∩ E_2) = 1/2 × 1/3 × 3/4 × 9/{10} = {27}/{240}$

∴ Desired probability

= $P(E_2/E_1) = {P(E_1 ∩ E_2)}/{P(E_1)} = {27}/{65}$

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