Bank Math Master

Question: If the sum of three consecutive odd integers is 141, what is the largest number?

Solution:
Let the three consecutive odd integers be:
x,  x + 2,  x + 4

Accordingly:
x + (x + 2) + (x + 4) = 141
⇒ 3x + 6 = 141
⇒ 3x = 141 - 6
⇒ 3x = 135
⇒ x = 135/3
⇒ x = 45

So the numbers are:
45, 47, 49(Largest)

∴ The largest number is 49.

Question: The ages of five persons are 54, 82, 45, 55, and 59 years. What should be the age of a sixth person so that the average age becomes 61 years?

Solution:
Let the age of the sixth person be x.
The average of six persons = Total age of 6 person/ 6

Accordingly:
(54 + 82 + 45 + 55 + 59 + x) / 6​ = 61
⇒ (295 + x) / 6 = 61
⇒ 295 + x = 366
⇒ x = 366 - 295
⇒ x = 71

Question: A group of 10 boxes has their average weight increased by 4 kg after replacing a 60 kg box with a new one. What is the weight of the new box?

Solution:
Let the weight of the new box be x kg.
Let the total weight of the original 10 boxes = W
Original average = W/10

After replacing the 60 kg box with x,
The total weight becomes: W - 60 + x
The new average = (W - 60 + x) / 10

Accordingly:
New average = Old average + 4
(W - 60 + x) / 10 = W/10 + 4
⇒ W - 60 + x = W + 40
⇒ x = 40 + 60
⇒ x = 100

Question: Two alarm clocks ring their alarms at regular intervals of 40 seconds and 60 seconds. If they first beep together at 10:30 AM, when will they next beep together for the first time?

Solution:
They will ring together after the LCM of 40 and 60 seconds.
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
∴ LCM = 2 × 2 × 2 × 3 × 5 = 120 seconds
= 120 ÷ 60 = 2 minutes [ 1 min = 60 sec. ]

∴ They will beep together again at 10:30 AM + 2 minutes = 10:32 AM

Question: If 2/3 of a number is 5 more than 1/4 of the number then 7/2 of the number is-

Solution:
Let,
the number be x

According to the question, 
⇒ (2x/3) - x/4 = 5
⇒ (8x - 3x)/12 = 5
⇒ 5x = 5 × 12
⇒ 5x = 60
⇒ x = 12

Then 7/2 of the number will be = x × 7/2
= (12 × 7)/2
= 42

Question:

Solution:

Question: A sum of Tk. 3300 is divided among A, B and C such that A gets 2/5 of what B gets and B gets 1/3 of what C gets. B’s share is: 

Solution:
Let,
C’s share = Tk. x
Then,
B’s share = Tk. x/3
A’s share = Tk. (2/5) × (x/3) = Tk. 2x/15

∴ 2x/15 + x/3 + x = 3300
⇒ (2x + 5x + 15x)/15 = 3300
⇒ 22x/15 = 3300
⇒ 22x = 3300 × 15
⇒ 22x = 49500
⇒ x = 49500/22
⇒ x = 2250

∴ B’s share = 2250/3 = 750 Tk.

Question: If √5ⁿ = 625, then the value of n is?

Solution:
Given that, √5ⁿ = 625 
⇒ √5ⁿ = 5⁴ 
⇒ (√5ⁿ)² = (5⁴)² 
⇒ 5ⁿ = 5⁸ 
∴ n = 8

Question: What is the difference between the biggest and the smallest fraction among 3/7, 4/9, 5/11 and 6/13?

Solution:
Converting each of the given fractions into decimal form, we get,
3/7 = 0.4286
4/9 = 0.4444
5/11 = 0.4545
6/13 = 0.4615

Since, 0.4615 > 0.4545 > 0.4444 > 0.4286 So, 6/13 > 5/11 > 4/9 > 3/7

∴ Required difference = 6/13 - 3/7
= (42 - 39)/91
= 3/91

Question: (0.1 × 0.01 × 0.001 × 10⁷) is equal to: 

Solution:
Given expression,
(0.1 × 0.01 × 0.001 × 10⁷)
= (1/10) × (1/100) × (1/1000) × 10⁷
= 10⁷/10⁶
= 10 ^{(7 - 6)}
= 10

Question: How many perfect squares lie between 120 and 300?

Solution:
We know that,
(11)² = 121 (Greater than 120 but less than 300)
(17)² = 289 (Greater than 120 but less than 300)
(18)² = 324 (Greater than 120 but not less than 300)

∴ We have 7 (11 to 17) numbers between 120 and 300 which are perfect squares.
121 = (11)²
144 = (12)²
169 = (13)²
196 = (14)²
225 = (15)²
256 = (16)²
289 = (17)² 

Question: A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in the month of June beginning with a Sunday?

Solution:
Given that,
The month begins with a Sunday, so there will be five Sundays in the month.
Number of the visitor in five sunday
⇒ 510 × 5 = 2550
Number of the other days visitor
⇒ 240 × 25 = 6000
Average = (2550 + 6000)/30 = 285

∴ The average number of visitors per day in a month of 30 days beginning with a Sunday is 285.

Question: Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31 and 51 respectively.

Solution:
If a number divides the given numbers with the mentioned remainders, then
964 - 41 = 923
1238 - 31 = 1207
1400 - 51 = 1349

Required number:
= HCF of (964 - 41), (1238 - 31) and (1400 - 51)
HCF of 923, 1207 and 1349 = 71

Question: Find the value of x if log_x 324 = 4.

Solution:
log_x324 = 4
⇒ x⁴ = 324
⇒ (x²)² = 18²
⇒ x² = 18²
⇒ x = √18
⇒ x = √3² × 2
⇒ x = 3√2

Question: If 0.13 ÷ p² = 13, then p equals:

Solution:
0.13 ÷ p² = 13
⇒ p² = 0.13/13
⇒ p² = 1/100
⇒ p = √1/100
⇒ p = 1/10
⇒ p = 0.1

Question: Among 80 students, the average marks in Mathematics is 65. If the 50 girls scored an average of 68, determine the average score of the remaining boys.

Solution:
Let,
the average marks of the boys = k
Total marks of 80 students = 80 × 65 = 5200 
Total marks of 50 girls = 50 × 68 = 3400

According to the question,
3400 + (80 − 50) × k = 5200
⇒ 3400 + 30k = 5200
⇒ 30k = 5200 − 3400
⇒ 30k = 1800
⇒ k = 1800 / 30
⇒ k = 60

∴ Average marks of the remaining 30 boys = 60

Question: Out of three numbers, the first is twice the second and is half of the third. If the average of the three number is 56, then the difference of first and third number is:

Solution:
Let, 
the second number be x.
Then first number = 2x, third number = 4x.

∴ 2x + x + 4x = 56 × 3
⇒ 7x = 168
⇒ x = 168/7
⇒ x = 24

Required difference:
= 4x - 2x
= 2x
= 2 × 24
= 48.

Question: P is five years younger than Q, who is twice as old as R. If the sum of their ages is 55, how old is Q?

Solution:
Let,
the age of R be = x years
Then,
the age of Q = 2x years
and age of P = (2x – 5) years

According to the question,
(2x – 5) + 2x + x = 55
⇒ 5x – 5 = 55
⇒ 5x = 60
⇒ x = 12

Hence, age of Q = 2x = (2 × 12) years = 24 years

Question: The average age of all the students of a class in 18 years. The average age of the boys of the class is 20 years and that of the girls is 15 years. If the number of girls in the class is 20, then find the number of boys in the class.

Solution:
Let,
the number of boys in the class be x.
Then, 18 × (x + 20) = 20x + (15 × 20)
⇒ 18x + 360 = 20x + 300
⇒ 20x + 300 = 18x + 360
⇒ 20x -18x = 360 -300
⇒ 2x = 60
⇒ x = 30

∴ The number of boys in the class is 30.