Bank Math Master

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

Solution:
The events when three unbiased coins are tossed = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of events 8

The events of getting at least two heads {HHH, HHT, HTH, THH}

Number of expected events = 4

∴ The probability of getting at least two heads is = 4/8 = 1/2

Question: In an AP, The 7th term is 3 times of 2nd term. If the 5th term is 11, what is the 20th term?

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

ATQ,
3(a + d) = a + 6d
⇒ 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 20th term is: a + 19d
= 3 + 19 × 2
= 3 + 38
= 41

Question: In a certain language ICEBERG is coded as GCEBERI, how is ICHNITE coded in that language?

Solution:
ICEBERG ⇔ GCEBERG
I ⇒ C ⇒ E ⇒ B ⇒ E ⇒ R ⇒ G 
1 ⇒ 2 ⇒ 3 ⇒ 4 ⇒ 5 ⇒ 6 ⇒ 7
G ⇒ C ⇒ E ⇒ B ⇒ E ⇒ R ⇒ G

Here,
Letters of 1st and 7th position inter-change their position.
Letters of 3rd and 5th position also inter-change their position.
And the letters of even number position remain same.

I ⇒ C ⇒ H ⇒ N ⇒ I ⇒ T ⇒ E 
1 ⇒ 2 ⇒ 3 ⇒ 4 ⇒ 5 ⇒ 6 ⇒ 7
E ⇒ C ⇒ I ⇒ N ⇒ H ⇒ T ⇒ I

∴ ICHNITE ⇔ ECINHTI

Question: If |2x - 3| < 1 and p < 3x - 2 < q, then find the values of p, q.

Solution:
Given that,
|2x - 3| < 1
⇒ - 1 < 2x - 3 < 1
⇒ - 1 + 3 < 2x - 3 + 3 < 1 + 3
⇒ 2 < 2x < 4
⇒ 1 < x < 2
⇒ 3 < 3x < 6
⇒ 3 - 2 < 3x - 2 < 6 - 2
∴ 1 < 3x - 2 < 4

∴ p = 1 and q = 4

Question: A person first moves 8 meters towards north, then 12 meters towards east and finally 8 meters towards north. How far did he go along from the first position?

Solution:

DE = EC + CD = 8 + 8 = 16 m 
AD = √(AE² + DE²) 
= √(12² + 16²)
= √(144 + 256)
= √(400)
= 20 m

Question: In a simultaneous throw of two dice, what is the probability of getting a total of 9?

Solution:

The total number of events ⇒ 36.
The number of events of getting total 9 ⇒ 4.

∴ The probability of getting a total of 9 is 4/36 = 1/9.

Question: Which number will complete the series: 127, 63, 31, 15, __?

Solution:
(127 - 1)/2 = 126/2 = 63
(63 - 1)/2 = 62/2 = 31
(31 - 1)/2 = 30/2 = 15
(15 - 1)/2 = 14/2 = 7

Question: What is the difference between the 5^th and the 4^th terms of the sequence 2, 4, 7, ____ whose n^th term is n + 2^{(n - 1)}?

Solution:
Given that, a_n = n + 2^{n - 1} 

Thus:
a₄ = 4 + 2^{4 - 1} = 12;
a₅ = 5 + 2^{5 - 1} = 21;

∴ The difference is = 21 - 12 = 9.

Question: If HEAD = 18 then HAIR =?

Solution:

HEAD
H + E + A + D 
8 + 5 + 1 + 4
= 18
Now,
H + A + I + R
8 + 1 + 9 + 18
= 36
∴ HAIR = 36

Question: Tickets numbered 1 to 30 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 2 and 3?

Solution: 
Number which are multiple of 2 and 3 between 1 to 30  = {6, 12, 18, 24, 30} 

∴ The probability is =  5/30 = 1/6

Question: If x = (√7/2) + (√3/2), then x + 1/x =?

Solution: 
Given that,
x = (√7/2) + (√3/2)
x = (√7 + √3)/2
Now,
1/x = 2/(√7 + √3)
= {2(√7 - √3)}/{(√7 + √3)(√7 - √3)}
= {2(√7 - √3)}/(7 - 3)
= {2(√7 - √3)}/4
= (√7 - √3)/2 

∴ x + 1/x = (√7 + √3)/2 + (√7 - √3)/2
= (√7 + √3 + √7 - √3)/2
= (2√7)/2
= √7

Question: In a group of 100 students, 72 students can play Football and 43 can play Cricket. How many can play both Football and Cricket?

Solution:
Let,
x people can play both Football and Cricket.
∴ People can only play football 72 - x.
People can only play cricket 43 - x 

Here,
72 - x + 43 - x + x = 100
⇒ 115 - x = 100
∴ x = 15 
 

Question: Four boys are sitting on a bench. Rasel is to the left of Tofail. Rakib is to the left of Hasib. Rakib is between Hasib and Tofail. Who would be second from the left on the bench?

Solution:
Rasel is to the left of Tofail.
Rasel ⇔ Tofail

Rakib is to the left of Hasib.
Rakib ⇔ Hasib

Rakib is between Hasib and Tofail.
Tofail ⇔ Rakib ⇔ Hasib

∴ Rasel ⇔ Tofail ⇔ Rakib ⇔ Hasib.

Tofail would be second from the left on the bench.

Question: Two cards are drawn from a pack of 52 cards. What is the probability that both are black cards?

Solution:
number of black cards = 26

probability both card black = ²⁶C₂/⁵²C₂
= 325/1326
= 25/102

Question: A committee of 5 members is to be formed by selecting out of 6 men and 7 women. What is the probability that the committee has exactly 2 men and 3 women?

Solution:
Total member = 6 + 7 = 13
2 men can be selected out of 6 men in  ⁶C₂ ways
3 women can be selected out of 7 women in ⁷C₃ ways
Required number of ways = ⁶C₂ × ⁷C₃ = 15 × 35 = 525

The total number of ways to make committee with all members = ¹³C₅ = 1287

∴ The probability that the committee has exactly 2 men and 3 women = 525/1287
= 175/429

Question: If (x + 3)² = 225 then what is the value of x - 1?

Solution:
(x + 3)² = 225
⇒ x + 3 = ± 15
Take the negative value wet get,
x + 3 = - 15
⇒ x = - 15 - 3
∴ x = - 18

∴ x - 1 = - 18 - 1 = - 19

Question: The test scores for a class are 86, 94, 70, 81, 92, 74, 75, 89, 76, and 97. What is the median of the data set?

Solution:
Arrenge the data set in accending order: 70, 74, 75, 76, 81, 86, 89, 92, 94, 97 

Total number of data is 10. Which is even number.

∴ The median is = [(10/2)th element + {(10/2) + 1}th element]/2
= (5th element + 6th element)/2
= (81 + 86)/2
= 167/2
= 83.5 

Question: If f(2a) = 2f(a) and f(6) = 11, Then f(24) =?

Solution:
f(24) 
= f(2 × 12)
= 2 × f(12)
= 2 × f(2 × 6)
= 2 × 2 × f(6)
= 4 × 11 
= 44

Question: Find the harmonic mean of 4 and 8.

Solution:
To calculate the harmonic mean of given numbers, you would divide the number of observations by the reciprocal of each number. For example, a, b, c are given numbers. So the number of observations is 3.
∴ The harmonic mean of a, b, c =


∴ The harmonic mean of 4 and 8 = 

Question: If LOSS = 1355 and GAIN = 2468 then 84615 =?

 Solution:
LOSS = 1355, GAIN = 2468
∴ 84615 = NAILS

Question: Solve the inequality 3 ≤ - 6 - 5x < 12.

Solution:
3 ≤ - 6 - 5x < 12
⇒ 3 + 6 ≤ - 5x < 12 + 6
⇒ 9 ≤ - 5x < 18
⇒ - 9 ≥ 5x > - 18
⇒ - 9/5 ≥ x > - 18/5
⇒ - 1.8 ≥ x > - 3.6 
∴ - 3.6 < x ≤ - 1.8