Bank Math Master

Question: Express A = {2, 4, 6, 8, 10, 12, 14, 16, 18} in set builder form.

Solution:
A = {2, 4, 6, 8, 10, 12, 14, 16, 18}
All the elements of A is even Natural number and less than 20.
∴ The builder from is A = {x : x ∈ N, x is an even number < 20}

Now, If we take A = {x = 2n : n ∈ N and n < 20}
Then the set will be A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38}

∴ Option 1 is the correct answer.

Question: If you toss a coin twice, what is the probability that you will get heads the second time?

Solution:
When we toss a coin the sample outcomes are = {HH, HT, TH, TT}
Total number of outcome = 4

Heads in the second time {HH, TH} = 2

∴ Probability that I will get heads the second time = 2/4 = 1/2 = (1 × 100)/2 % = 50%

Question: If the median of 21 observations is 40 and if the observations greater than the median are increased by 5 then the median of the new data will be-

Solution:
• The median of a set of data is the middle value when the data is arranged in ascending order.

For 21 observations, the median is the 11th value.
According to the question, the original median is 40. Now, observations greater than the median are increased by 5. Since only the values greater than the median are increased, the position of the median (the 11th observation) remains unchanged, and the value of the median (40) itself does not change.

⇒ The median of the new data will still be 40.
∴ The median of the new data remains 40.

Question: Solve |2x + 5| < 14.

Solution:
We have,
|2x + 5| < 14
⇒ - 14 < 2x + 5 < 14
⇒ - 19 < 2x < 9
⇒ (- 19/2) < x < (9/2)

Question: In an arithmetic progression, the first term a = 5 and the common difference d = 3. What is the 10th term?

Solution:
Here,
a = 5
d = 3

10th trem = a + (10 - 1)d 
= 5 + 9 × 3
= 5 + 27
= 32

Question: On a six-sided die, each side has a number between 1 and 6. What is the probability of throwing a 3 or a 4?

Solution:
On a six-sided die, the probability of throwing any number is 1 in 6.
The probability of throwing a 3 = 1/6
The probability of throwing a 4 = 1/6

Therefore, the probability of throwing either a 3 or 4 is 1/6 + 1/6 = (1+1)/6 = 2/6 = 1/3

Question:

Solution:
Here,
Mean  = 20
Number of observation = 10

Question: Which of the following is true for x if 3(x - 4) > 2x + 5?

Solution:
3(x - 4) > 2x + 5
⇒ 3x - 12 > 2x + 5
⇒ 3x - 2x > 5 + 12
⇒ x > 17

Question: For a geometric sequence, the first term a = 6 and the common ratio r = 3. What is the sum of the first 3 terms?

Solution:
a = 6
r = 3 > 1
n = 3

S = {a(rⁿ - 1)}/(r - 1)
= {6(3³ - 1)}/(3 - 1)
= (6 × 26)/2
= 3 × 26
= 78

Question: If A = {a, b, r, t, s, y} and B = {e, d, f, r, s}, then B - A =

Solution:
B - A = {e, d, f, r, s} - {a, b, r, t, s, y}
= {e, d, f}

Question: A two-digit number is chosen at random. What is the probability that the chosen number is a multiple of 7?

Solution:
There are 90 two-digit numbers (all integers from 10 to 99).
Of those, there are 13 multiples of 7 ⇒ 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.

∴ The probability that the chosen number is a multiple of 7 is 13/90

Question: What is the mean of the range, mode and median of the data given below?
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Solution:
Arranging the given data in ascending order 2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here,
Most frequent data is 4
so Mode = 4

Total terms in the given data, (n) = 15 (It is odd)
Median = {(n + 1)/2}^th term
= {(15 + 1)/2}^th term
= (8)^th term
= 6

Now,
Range = Maximum value - Minimum value = 19 - 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3 = (17 + 4 + 6)/3 = 27/3 = 9

Question: If H > M = D > P and K = H > T > Z, then which of the following options is not correct?

Solution:
H > M = D > P and K = H > T > Z
By combining: Z < T< K = H > M = D > P

1) P < M True (as, M = D > P → M > P)

2) H > D True (as, H > M = D → H > D)

3) Z < H → True (as, Z < T< K = H → H > Z)

4) Z > K → False (as, Z < T< K, implies Z < K)

Question: If the sum of an infinite geometric series is 20 and the common ratio r = 1/2, what is the first term?

Solution:
Here,
r = 1/2 
a = ?

We know that,
S_∞ = a/(1 - r)
⇒ 20 = a/(1 - 1/2)
⇒ 20 = a/(1/2)
⇒ 20 = 2a
∴ a = 10

Question: If C = {a, b, x, y} and D = {m, n, o, p} then C union D is:

Solution:
C union D = C ∪ D = {a, b, x, y} ∪ {m, n, o, p}
= {a, b, m, n, o, p, x, y}

Question: A bag contains 14 blue, 6 red, 12 green, and 8 purple buttons. 25 buttons are removed from the bag randomly. How many of the removed buttons were red if the chance of drawing a red button from the bag is now 1/3?

Solution:
Total number of button = 14 + 6 + 12 + 8 = 40

If 25 buttons are removed, there are (40 - 25) = 15 buttons remaining in the bag.

If the chance of drawing a red button is now 1/3, then 5 of the 15 buttons remaining must be red.
The original total of red buttons was 6.
So, (6 - 5) = 1 red button was removed.

Question: Find the median of the given set of numbers 2, 6, 6, 8, 4, 2, 7, 9.

Solution:
Arranging the given data in ascending order 2, 2, 4, 6, 6, 7, 8, 9
Total terms in the given data, (n) = 8 (It is even)

Median = {(8/2)^th term + (8/2 + 1)^th term}/2
= (4^th term + 5^th term)/2
= (6 + 6)/2
= 12/2
= 6

Question: A regular deck of cards has 52 cards. Assuming that you do not replace the card you had drawn before the next draw, what is the probability of drawing three aces in a row?

Solution:
The probability of getting three aces in a row is the product of the probabilities for each draw.
For the first ace, that is 4/52 = 1/13;
for the second, it is 3/51 = 1/17;
and for the third, it is 2/50 = 1/25.

So the overall probability = (1/13) × (1/17) × (1/25) = 1/5525

Question: If n(A) = 16, n(B) = 18 and n(A ∩ B) = 7, then what is the value of n(A ∪ B)?

Solution:
We know that,
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 16 + 18 - 7
= 27