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) 72
(b) 288
(c) 144
(d) 720
The correct answers to the above question in:
Answer: (d)
The first and the last (terminal) digits are even and there are three even digits. This arrangement can be done in $^3P_2$ ways. For any one of these arrangements, two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits (leaving out the terminal digits) may be arranged using these 5 digits in $^5P_4$ ways. The required number of numbers is $^3P_2 × ^5P_4$ = 6 × 120 = 720.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If the probability that A and B will die within a year are p and q respectively, then the probability that only one of them will be alive at the end of the year is
a) p + q – 2pq
b) p + q – pq
c) p + q
d) p + q + pq.
Answer »Answer: (a)
Only one of A and B can be alive in the following, mutually exclusive ways.
$E_1$ A will die and B will live
$E_2$ B will die and A will live
So, required probability = $P(E_1) + P (E_2)$
= p(1 - q) + q (1 - p) = p + q - 2pq.
Question : 2
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
a) 280
b) 196
c) 140
d) 346
Answer »Answer: (b)
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.
∴ No. of choices available to the student
= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.
Question : 3
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 »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 × ${13}/{52} × 3 × {13}/{51} × 2 × {13}/{50} = {169}/{425}$.
Question : 4
A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is $1/2$ , $1/3$ and $1/4$ . Probability that the problem is solved is
a) $1/2$
b) $2/3$
c) $3/4$
d) $1/3$
Answer »Answer: (c)
$P(E_1) = 1/2, P(E_2) = 1/3 and P(E_3) = 1/4;$
P$(E_1 ∪ E_2 ∪ E_3)$ = 1 - P($\ov{E_1}$) P ($\ov{E_2}$) P ($\ov{E_3}$)
= 1 - $(1 - 1/2)(1 - 1/3)(1 - 1/4) = 1 - 1/2 × 2/3 × 3/4 = 3/4$
Question : 5
There are 6 different letters and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes?
a) $1/6$
b) $1/2$
c) Zero
d) $5/6$
Answer »Answer: (c)
As there are 6 letters and envelopes, so if exactly 5 are into correctly addressed envelopes, then the remaining 1 will automatically be placed in the correctly addressed envelope. Thus, the probability that exactly 5 go into the correctly addressed envelope is zero.
Question : 6
Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2 ?
a) 81
b) 91
c) 36
d) 116
Answer »Answer: (b)
There can be 3 cases :
I. When one dice shows 2.
II. When two dice shows 2.
III. When three dices shows 2.
Case I : The dice which shows 2 can be selected out of the 3 dices in $^3C_1$ ways.
Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = $^5C_1$ each, so no of ways when one dice shows 2 = $^3C_1 × ^5C_1 × ^5C_1$ .
Case II : Two dices, showing 2 can be selected out of the 3 dices in $^3C_2$ ways and the rest one can have any 5 numbers except 2, so number of ways for the remaining 1 dice = 5.
So, number of ways, when two dices show
2 = $^3C_2$ × 5
Case III : When three dices show 2 then these can be selected in $^3C_3$ ways.
So, number of ways, when three dices show
2 = $^3C_3$ = 1
As, either of these three cases are possible.
Hence, total number of ways
= (3 × 5 × 5) + (3 × 5) + 1 = 91
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