Advance Math Mcq Pdf Model Questions & Answers, Practice Test for ssc constables capfs nia ssf ar tier 1 2024
ssc constables capfs nia ssf ar tier 1 2024 SYLLABUS WISE SUBJECTS MCQs
Number System
Ratio & Proportion
Percentage
Time & Work
Mensuration
Advance Math
Trigonometric Ratios & Identities
Average
Profit & Loss
There are 6 positive and 8 negative numbers. Four numbers are chosen at random and multiplied. The probability that the product is a positive number is:
Answer: (b)
Total no. of numbers = 6 positive + 8 negative = 14
n(S) = $^{14}C_4$
The product of four numbers could be positive when,
(a) all the four numbers chosen are positive or
(b) all the four numbers chosen are negative or
(c) two of the chosen numbers are positive and two are negative.
Required Prob. = ${^6C_4}/{^{14}C_4} + {^8C_4}/{^{14}C_4} + {^6C_2 × ^8C_2}/{^{14}C_4}$
= ${505}/{1001}$
In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact centre of the squares and only one ball is to be placed in one square) shown in the figure given below such that they do not lie along the same straight line ?
Answer: (a)
3 balls can be placed be in any of the 12 squares in $^{12}C_3$ ways.
Total number of arrangements = $^{12}C_3$ = 220
Now, assume that balls lie along the some line.
There can be 3 cases :
Case I: When balls lie along the straight horizontal line.
3 balls can be put in any of the 4 boxes along the horizontal row in $^4C_3$ ways.
Now, since there are 3 rows, so number of ways for case I = $^4C_3$ × 3 = 12
Case II : When balls lie along the vertical straight line 3 balls can be put in any of the 3 boxes along the vertical row in $^3C_3$ ways.
Now, as there are 4 vertical rows, so number of ways for Case II = $^3C_3$ × 4 = 4
Case III : Balls lie along the 2 diagonal lines towards the left and 2 diagonal lines towards the right.
Number of ways = 2 + 2 = 4
Number of ways, when balls lie along the line = 12 + 4 + 4 = 20
Number of ways when balls don't lie along the line = Total number of ways – number of ways when balls lie along the line.
= 220 – 20 = 200.
In a question paper, there are four multiple choice type questions. Each question has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct ?
Answer: (b)
Since, every question has five options, so no. of choices for each question = 5
∴ total no. of choices = 5 × 5 × 5 × 5 = 625
Now, no. of choices of all correct answer = 1
Hence, no of choices for all the four answers not correct
= total no. of choices – no. of choices of all correct answer = 625 – 1 = 624
The probability of raining on day 1 is 0.2 and on day 2 is 0.3. What is the probability of raining on both the days ?
Answer: (b)
Required probability
= P (A) × P (B) = 0.2 × 0.3 = 0.06
The probability that a student passes in mathematics is 4/9 and that he passes in physics is 2/5. Assuming that passing in mathematics and physics are independent of each other, what is the probability that he passes in mathematics but fails in physics ?
Answer: (c)
Probability of passing in mathematics = $4/9$
Probability of passing in physics = $2/5$
Probability of failure in physics = 1  $2/5 = 3/5$
Given that both the events are independent.
Required probability = $4/9 × 3/5 = 4/{15}$
ssc constables capfs nia ssf ar tier 1 2024 IMPORTANT QUESTION AND ANSWERS

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