Bank Math Master

Question: Four unbiased coins are tossed. What is the probability of getting at most two heads?

Solution:

The total number of events = 16
The numbers of event with at most two heads = 11
∴ The probability of getting at most two heads = 11/16 

Question: If |x - 2| < 3 and m < 3x + 5 < n, then find the values of m, n.

Solution:
Given that,
|x - 2| < 3
⇒ - 3 < x - 2 < 3
⇒ - 3 + 2 < x - 2 + 2 < 3 + 2
⇒ -1 < x < 5
⇒ - 3 < 3x < 15
⇒ - 3 + 5 < 3x + 5 < 15 + 5
∴ 2 < 3x + 5< 20

∴ m = 2 and n = 20

Question: In a simultaneous throw of two dice, what is the probability of getting a total of 8?

Solution:

The total number of events ⇒ 36.
The number of events of getting total 8 ⇒ 5.

∴ The probability of getting a total of 8 is 5/36.

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Solution:
Total number of tickets = 20

The numbers which are multiple of 3 or 5 are {3, 5, 6, 9, 10, 12, 15, 18, 20}
∴ Total expected events = 9

∴ The probability = 9/20 

Question: If 3x - 4y > 2x + 3y, then which of the following must be true?

Solution: 
3x - 4y > 2x + 3y
⇒ 3x - 2x > 3y + 4y
⇒ x > 7y 
From that we surely say that x > y.

Question: If |x - 2| > 1, then what is the following should be correct?

Solution:
|x - 2| > 1

If (x - 2) is positive then,
x - 2 >1
⇒ x > 1 + 2
∴ x > 3 

If (x - 2) is negative then,
- (x  - 2) > 1
⇒ x - 2 < - 1
⇒ x < - 1 + 2
∴ x < 1

∴ x > 3 or x < 1

Question: - 2 < (6 - 2x)/3 < 4, find the value of x.

Solution:
- 2 < (6 - 2x)/3 < 4
⇒ - 6 < 6 - 2x < 12
⇒ - 6 -  6 < - 2x < 12 - 6
⇒ - 12 < -  2x < 6
⇒ 6 > x > - 3 
∴ - 3 < x < 6

Question: A student scored 60 marks in the first test and 45 marks in the second test of the terminal examination. How many minimum marks should the student score in the third test get a mean of least 62 marks?

Solution:
Let,
The marks scored in the third test be x marks.

(60 + 45 + x)/3 ≥ 62
105 + x ≥ 186
x ≥ 81
Therefore, the student must score 93 marks to maintain a mean of at least 62 marks.

Question: 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 them is defective is-

Solution:
Total bulbs 20
Number of defective bulbs 4
∴ Number of non-defective bulbs (20 - 4) = 16.

The probability of non-defective bulbs is ¹⁶C₂/²⁰C₂ = 120/190 = 12/19

∴ The probability of at least 1 bulb is defective = 1 - (12/19) = 7/19 

Question: A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least one woman?

Solution:
Total member = 3 + 2 = 5

Committee can be form with at least 1 one woman:
1 woman, 2 men : ²C₁ ×  ³C₂ = 2 × 3 = 6
2 women, 1 man: ²C₂ × ³C₁ = 1 × 3 = 3
∴ The total number of ways to make committe with at least 1 one woman: 6 + 3 = 9

The total number of ways to make committe with all members = ⁵C₃ = 10

∴ The probability that the committee has at least woman = 9/10

Question: Write the solution set of the equation x² - 4 = 0 in roster form.

Solution: 
Given that,
x² - 4 = 0
⇒ x² = 4
∴ x = ± 2

∴ The set will be {- 2, 2}

Question: What is the 12th term of the sequence - 2, - 4, - 6, ...... , - 100?

Solution:
Here,
- 4 - (- 2) = - 4 + 2 = - 2
- 6 - (- 4) = - 6 + 4 = - 2
∴ d = - 2
a = - 2
n = 12

∴ The 12th term of the sequence = a + (n - 1)d
= - 2 + (12 - 1)(- 2)
= - 2 + 11(- 2)
= - 2 - 22
= - 24 

Question: If set A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then write the universal set for all three sets.

Solution:
Let U is the universal set for sets A, B and C, 
Here,
U = A ∪ B ∪ C
U = {1, 3, 5} ∪ {2, 4, 6} ∪ {0, 2, 4, 6, 8}
U = {0, 1, 2, 3, 4, 5, 6, 8}

Question: The 6th term of an AP is 6 and the 16th term is 14. What is the 27th term?

Solution:
The 6th term is a + 5d = 6 ..............(1)
The 16th term is a + 15d = 14 ............(2)

from (2) - (1) we get,
a + 15d - a - 5d = 14 - 6
⇒ 10d = 8
⇒ d = 8/10
∴ d = 4/5

Put the value of d in (1) we get,
a + 5 × (4/5) = 6
⇒ a + 4 = 6
∴ a = 2 

∴ The 27th term is (a + 26d) = 2 + 26 × (4/5)
= 2 + (104/5)
= 114/5

Question: If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}. Find A ∩ (B ∪ C).

Solution:
B ∪ C = {7, 9, 11, 13} ∪ {11, 13, 15}
= {7, 9, 11, 13, 15}

A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}

Question: In an AP, the ratio of the 2nd term to the 7th term is 1/3. If the 5th term is 11, what is the 15th term?

Solution:
The 2nd term is a + d.
The 7th term is a + 6d 

ATQ,
(a + d)/(a + 6d) = 1/3
⇒ 3a + 3d = a + 6d
∴ 2a = 3d
∴ a = (3d)/2

The 5th term is  a + 4d = 11
⇒ (3d)/2 + 4d = 11
⇒ 3d + 8d = 22
⇒ 11d = 22
∴ d = 2

∴ a = (3 × 2)/2 = 3
∴ The 15th term is: a + 14d
= 3 + 14 × 2
= 3 + 28
= 31

Question: One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will not be an ace.

Solution:
Total number of Ace is 4.
∴ The probability of a card will be 4/53 = 1/13

∴ The probability that the card will not be an ace = 1 - (1/13)
= (13 - 1)/13
= 12/13 

Question: In an AP, the sum of the first 3 terms is - 36 and that of the last 3 is 27. If there are 10 terms, what is the 1st term?

Solution:
Let,
the first term of AP is a.
The common different, d 
∴ The AP will be, a, a + d, a + 2d, ..................., a + 7d, a + 8d, a + 9d

ATQ,
a + a + d + a + 2d = - 36 
⇒ 3a + 3d = - 36
⇒ a + d = - 12 
∴ d = - 12 - a

And,
a + 7d + a + 8d + a + 9d = 27
⇒ 3a + 24d = 27
⇒ 3a + 24(- 12 - a) = 27
⇒ 3a - 288 - 24a = 27
⇒ - 21a = 315
⇒  a = 315/(- 21)
∴ a =  - 15