Advance Math Model Questions Set 1 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) ${15}/{184}$
(b) ${15}/{92}$
(c) ${91}/{276}$
(d) None
The correct answers to the above question in:
Answer: (b)
Exhaustive number of cases = $^{24}C_{14}$
Favourable cases = $^{22}C_{14}$
Probability = ${^{22}C_{14}}/{^{24}C_14} = {15}/{92}$
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 »Answer: (b)
Required no. of the ways = $^6C_3 × ^4C_2$ = 20 × 6 = 120
Question : 2
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 »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 : 3
3 digits are chosen at random from 1,2,3,4,5,6,7,8 and 9 without repeating any digit. What is the probability that their product is odd?
a) $5/108$
b) $5/42$
c) $2/3$
d) $8/42$
Answer »Answer: (b)
Let E be the event of selecting the three numbers such that their product is odd and S be the sample space. For the product to be odd, 3 numbers choosen must be odd.
∴ n(E) = $^5C_3$
= n(S) = $^9C_3$
∴ P(E) = ${n(E)}/{n(S)} = {^5C_3}/{^9C_3} = 5/{42}$
Question : 4
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) $3/5$
Answer »Answer: (b)
Required probability = $(1 - 1/6) × (1- 2/5) = 5/6 × 3/5 = 1/2$
Question : 5
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 »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 : 6
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 »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
= ${^4C_1}/{^9C_1} × {^7C_2}/{^{10}C_2} + {^5C_1}/{^9C_1} × {^3C_1 × ^7C_1}/{^{10}C_2}$
= $4/9 × 7/{15} + 5/9 × 7/{15} = 7/{15}$
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