Bank Math Master

Question: The average of 4 positive integers is 59. The highest integer is 83 and the lowest integer is 29. The difference between the remaining two integers is 28. Which of the following integers is higher of the remaining two integers?

Solution:
Sum of four integers = 59 × 4 = 236
Let the required integers be x and x - 28
Then, x + (x - 28) = 236 - (83 + 29)
⇒ 2x - 28 = 124
⇒ 2x = 152
⇒ x = 76
Hence, required integer = 76

Question: What is the difference between the simple interest on a principal of Tk. 500 being calculated at 5% per annum for 3 years and 4% per annum for 4 years?

Solution:
I₁ = (500 × 3 × 5)/100 = 75

I₂ = (500 × 4 × 4)/100 = 80

∴ Difference = 80 - 75 = 5

Question: 7⁶ⁿ - 6⁶ⁿ, where n is an integer > 0, is divisible by-

Solution:
7⁶ⁿ - 6⁶ⁿ

Assume that n = 1
= 7⁶ - 6⁶
= (7³)² - (6³)²
= (7³ - 6³)(7³ + 6³)
= (343 - 216) × (343 + 216)
= 127 × 559
= 127 × 13 × 43

Clearly, it is divisible by 127, 13 as well as 559

Question: The price of a car depreciates in the first year by 25% in the second year by 20% in third year by 15% and so on. The final price of the car after 3 years, if the present cost of the car is Tk. 10,00,000-

Solution:
The final price of the car after 3 years = 1000000 × 0.75 × 0.80 × 0.85
= Tk. 5,10,000

Question: How many terms are there in the GP 5, 20, 80, 320,..........., 20480?

Solution:
Common ratio, r = 20/5
= 4

Last term or nth term of GP = ar^{n - 1}
⇒ 20480 = 5 × (4^{n - 1})
⇒ 4^{n - 1} = 20480/5
⇒ 4^{n - 1} = 4096
⇒ 4^{n - 1} =  4⁶
So, comparing the power,
Thus, n - 1 = 6
∴ n = 7

Number of terms = 7

Question: A trader sells goods to a customer at a profit of k% over the cost price, besides it he cheats his customer by giving 880 g only instead of 1 kg. Thus his overall profit percentage is 25%. Find the value of k?

Solution:
1 kg = 1000 g
Let,
The cost price of per gram Tk. 1
∴ The cost price of 1000g Tk. 1000
∴ The cost price of 880 Tk. 880

Selling price of 880 gram = 1000 + k% of 1000 = 1000 + {(1000k)/100}
= 1000 + 10k

∴ Profit = 1000 + 10k - 880
= 120 + 10k

Overall % Profit = {(120 + 10k)/880} × 100

∴ {(120 + 10k)/880} × 100 = 25
⇒ (120 + 10k)/880 = 25/100
⇒ 120 + 10k = (25 × 880)/100
⇒ 120 + 10k = 220
⇒ 10k = 100
∴ k = 10

Question: A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in  6 seconds. The speed of the second train is-

Solution:
Distance covered = (108 + 112)
= 220 meter.
Time = 6 seconds.

Relative speed = 220/6 = 110/3 m/s.
= (110 × 3600)/(3 × 1000) km/hr
= 132 km/hr.

Now,
50 + Speed of second train = 132 km/hr.
∴ Speed of second train = (132 - 50) km/hr.
= 82 km/hr.

Question: X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?

Solution:
Work done by X in 8 days = (1/40) × 8 = 1/5

Remaining work = 1 - (1/5) = 4/5

Now,
4/5 work is done by Y in 16 days
Whole work will be done by Y in = (16 × 5)/4
  = 20 days

∴ X's 1 day's work = 1/40
∴ Y's 1 day's work = 1/20

(X + Y)'s 1 day's work
= 1/40 + 1/20
= 3/40

Hence, X and Y will together complete the work in
= 40/3 = 13.33 days

Question: In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

Solution:
ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.
One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged.
The number of possible rearrangements is 4! = 24

The group of 3 vowels contains two a s and one u
The 3 vowels can rearrange amongst themselves in 3!/2! = 3

Hence, the total number of rearrangements in which the vowels appear together are = 24 × 3 = 72

Question: To gain 10% on selling sample of milk at the cost price of pure milk, the quantity of water to be mixed with 100 kg of pure milk is -

Solution:
Let the quantity of water mixed be x kg.
Let,
CP of 1 kg of pure milk = Tk. 1
CP of 100 kg of pure milk = Tk. 100

Hence,
% gain = (x/100) × 100
⇒ 10 = x

Question: If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

Solution:
Let,
The side of original square is x
∴ The area of original square is x²

The side of new square is x + 50% of x = x + x/2 = 3x/2
∴ The area of new square is (9x²)/4

∴ The ratio between the new area and the original area of the square = (9x²)/4 : x²
= (9/4) : 1
= 9 : 4

Question: The area of the four walls of a room is 120 square meters and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is -

Solution: 
Let the breadth = x meters and
length = (2x) metres

Area of 4 walls = 2(2x + x) × 4
= 24x

ATQ,
∴ 24x = 120
⇒ x = 5

So, length = 10 m, and breadth = 5 m

Area of the floor = 10 × 5 = 50 square meters

Question: The surface area of a sphere is the same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

Solution: 
The surface area of sphere = 4πr₁²
The curved surface area of cylinder =2πr₂h
diameter = 12 cm
radius r₂ = 6 cm

⇒ 4πr₁² = 2πr₂h
⇒ r₁² = (6 × 12)/2
⇒ r₁² = 36
⇒ r₁ = 6

radius of the sphere 6 cm

Question: Find out the wrong number in the series:
455, 445, 465, 435, 485, 415, 515, 455

Solution:
The given series is made of two different series:

First: 455, 465, 485, 515, and

Second: 445, 435, 415, 455

The sequence in first series is + 10, + 20, + 30, and
The sequence in second series is - 10, - 20. So, the number 455 is wrong as it should be 415 - 30 = 385

Question: Which number is the odd one in oval A and B respectively?


Solution:
In figure A,
42 is odd because except 42 all the other numbers are divisible by 12

In figure B,
52 is odd because except 52 all the other numbers are divisible by 6

Question: X's age 3 years ago was three times the present age of Y. At present Z's age is twice the age of Y. Also Z is 12 years younger than X. What is the present age of Z?

Solution:
Let the present age of Y be a years
Three years ago X's age = 3a years
Then, present age of X is (3a + 3)
Z's present age = 2a

According to question
(3a + 3) - 2a = 12
a = 9 year

∴ Present age of Z = 2a = 2 × 9 = 18 years