Advance Math Model Questions Set 1 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 : Ram and Shyam appear for an interview for two vacancies in an organisation for the same post. The probabilities of their selection are 1/6 and 2/5 respectively. What is the probability that none of them will be selected ?

(a) $1/5$

(b) $1/2$

(c) $5/6$

(d) $3/5$

The correct answers to the above question in:

Answer: (b)

Required probability = $(1 - 1/6) × (1- 2/5) = 5/6 × 3/5 = 1/2$

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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers

Question : 1

A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighbouring places are empty is

a) ${15}/{184}$

b) ${15}/{92}$

c) ${91}/{276}$

d) None

Answer: (b)

Exhaustive number of cases = $^{24}C_{14}$

Favourable cases = $^{22}C_{14}$

Probability = ${^{22}C_{14}}/{^{24}C_14} = {15}/{92}$

Question : 2

To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection ?

a) 132

b) 120

c) 1,490

d) 60

Answer: (b)

Required no. of the ways = $^6C_3 × ^4C_2$ = 20 × 6 = 120

Question : 3

A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective, is:

a) $7/{19}$

b) ${12}/{19}$

c) $4/{19}$

d) ${21}/{95}$

Answer: (a)

P (None is defective) = ${^{16}C_2}/{^{20}C_2} = ({16 × 15}/{2 × 1} × {2 × 1}/{20 × 19})$

= ${12}/{19}.$

P (at least one is defective) = 1 - ${12}/{19} = 7/{19}$

Question : 4

The probability of getting 10 in a single throw of three fair dice is :

a) $1/8$

b) $1/9$

c) $1/6$

d) $1/5$

Answer: (a)

Exhaustive no. of cases = $6^3$

10 can appear on three dice either as distinct number as following (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and each can occur in 3! ways. Or 10 can appear on three dice as repeated digits as following (2, 2, 6), (2, 4, 4), (3, 3, 4) and each can occur in ${3!}/{2!}$ ways.

∴ No. of favourable cases = 3 × 3! + 3 × ${3!}/{2!}$ = 27

Hence, the required probability = ${27}/{6^3} = 1/8$

Question : 5

A bag has 4 red and 5 black balls. A second bag has 3 red and 7 black balls. One ball is drawn from the first bag and two from the second. The probability that there are two black balls and a red ball is :

a) ${11}/{45}$

b) $7/{15}$

c) ${14}/{45}$

d) $9/{54}$

Answer: (b)

Required probability

= Probability that ball from bag A is red and both the balls from bag B are black + Probability that ball from bag A is black and one black and one red balls are drawn from bag

= ${^4C_1}/{^9C_1} × {^7C_2}/{^{10}C_2} + {^5C_1}/{^9C_1} × {^3C_1 × ^7C_1}/{^{10}C_2}$

= $4/9 × 7/{15} + 5/9 × 7/{15} = 7/{15}$

Question : 6

In shuffling a pack of cards three are accidentally dropped. The probability that the missing cards are of distinct colours is

a) ${165}/{429}$

b) ${162}/{459}$

c) ${169}/{425}$

d) ${164}/{529}$

Answer: (c)

The first card can be one of the 4 colours, the second can be one of the three and the third can be one of the two. The required probability is therefore

4 × ${13}/{52} × 3 × {13}/{51} × 2 × {13}/{50} = {169}/{425}$.

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