Advance Math Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts
ADVANCE MATH PRACTICE TEST [2  EXERCISES]
Advance Math Model Questions Set 1
Advance Math Model Questions Set 2
Question : 1
Suppose six coins are tossed simultaneously. Then the probability of getting at least one tail is :
a) ${53}/{54}$
b) ${63}/{64}$
c) ${71}/{72}$
d) $1/{12}$
Answer »Answer: (b)
If six coins are tossed, then the total no. of outcomes = $(2)^6$ = 64
Now, probability of getting no tail = $1/{64}$
Probability of getting at least one tail = 1  $1/{64} = {63}/{64}$
Question : 2
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
Question : 3
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 : 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
The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat and the terminal digits are even is
a) 72
b) 288
c) 144
d) 720
Answer »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.
IMPORTANT QUANTITATIVE APTITUDE EXERCISES
Advance Math Model Questions Set 1 Online Quiz
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