Bank Math

Question: The fourth proportional to 5, 6, and 15 is- 

Solution: 
Let, The fourth proportional is x.

So, 5/6 = 15/x
⇒ 5x = 90
∴ x = 18

Question: Find the next number in the series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ?

Solution:
এই সিরিজটি হলো ফিবোনাচ্চি সিরিজ (Fibonacci sequence)। 
প্রতিটি সংখ্যা আগের দুটি সংখ্যার যোগফল।
0 + 1 = 1, 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, 21 + 34 = 55

সুতরাং পরের সংখ্যা হবে 55।

∴ সিরিজটি হলো- 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Question: The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?

Solution: 
Using options,
We find that four consecutive odd numbers are 37, 39, 41 and 43
The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.
Thus, 41 is one of the odd numbers

Question: The price of an item was increased by 20% and then was reduced by 25% the next day. Which of these can be the new price of the item in Tk if all prices are integer?

Solution: 
Let the original price be Tk. p  (an integer).
After a 20% increase,
New price = p + p of 20% = p + (20p/100)
= p + (p/5)
= 6p/5

And then reduced by 25%,
Final price = (6p/5) - (6p/5) of 25%
= (6p/5) - (6p/20) = (6p/5) - (3p/10)
= (12p - 3p)/10
= 9p/10
For the final price to be an integer, 9p/10 must be integer and p must be divisible by 10.

Now, check the options, 
ক) p = 300 × (10/9) = 333.33 (not integer)
খ) p = 325 × (10/9) = 361.11 (not integer)
গ) p = 450 × (10/9) = 500 (integer)
ঘ) p = 475 × (10/9) = 527.78 (not integer)

Only 450 works as a possible final integer price.
∴ The new price can be Tk 450.

Question: The price of sugar falls by 20%. By how much percent can a person increase consumption without increasing expenditure?

Solution:
Let,
Original price = 100 units
After 20% fall
New price = (100 - 20) = 80 units

With the same money,
Now we can buy = (100/80) × 100
= 1.25 × 100
= 125 units

Increase = 125 - 100
= 25 units

Percentage increase in consumption = (25/100) × 100%
= 0.25 × 100%
= 25%

∴ 25% consumption can be increased.

Question: A passenger sitting by the window of a train notices that he can count 16 electric poles in one minute. If the poles are known to be 30 meters apart, at what speed is the train traveling in km/h?

সমাধান:
মোট গুনে দেখা বৈদ্যুতিক খুঁটির সংখ্যা = 16 টি।
খুঁটিগুলির মধ্যবর্তী গ্যাপের সংখ্যা = (16 - 1) = 15 টি।

দুটি খুঁটির মধ্যবর্তী দূরত্ব = 30 মিটার।
∴ 1 মিনিট বা 60 সেকেন্ডে মোট অতিক্রান্ত দূরত্ব = (30 × 15) মিটার = 450 মিটার।

ট্রেনটির গতিবেগ = দূরত্ব/সময়
= 450/60 মিটার/সেকেন্ড
= 45/6 মিটার/সেকেন্ড
= 15/2 মিটার/সেকেন্ড
= (15/2 × 18/5) কিমি/ঘন্টা
= (3 × 9) কিমি/ঘন্টা
= 27 কিমি/ঘন্টা।

∴ ট্রেনটির গতিবেগ হলো 27 কিমি/ঘন্টা।

Question: If rSinθ = √3 and rCosθ = 1, then the value of (√3Cotθ + 1) = ?

Solution:
Given that,
rSinθ = √3 and rCosθ = 1
⇒ rCosθ/rSinθ = 1/√3
⇒ Cotθ = 1/√3
⇒ √3Cotθ = 1

(Add 1 both sides)
√3Cotθ + 1 = 1 + 1
∴ √3Cotθ + 1 = 2

Question: How many triangles can be formed with 18 points?

Solution:
আমরা জানি,
একটি ত্রিভুজ গঠন করতে ৩ টি বিন্দু প্রয়োজন হয়।
তাহলে,
18 টি বিন্দু দিয়ে গঠিত ত্রিভুজের সংখ্যা = ¹⁸C₃
= 18!/{3! × (18 - 3)!}
= 18!/(3! × 15!)
= (18 × 17 × 16 × 15!)/(3 × 2 × 1 × 15!)
= (18 × 17 × 16)/(3 × 2 × 1)
= (18 × 17 × 16)/6
= 4896/6
= 816

∴ 18 টি বিন্দু দিয়ে মোট 816 টি ত্রিভুজ গঠন করা যাবে।

​প্রশ্ন: If x + (1/x) = 2, The value of x⁴⁹⁹⁹ + x⁵⁰⁰⁰  is:
​
​সমাধান:
দেয়া আছে, 
​x + (1/x) = 2
⇒ (x²+ 1)/x = 2
⇒ x²+ 1 = 2x
⇒ x²- 2x + 1 = 0
⇒ (x - 1)²= 0 
⇒ x - 1 = 0
∴ x = 1
​
​ x⁴⁹⁹⁹ + x^5000
​= 1 + 1
​= 2

Question: If tan(θ + 15°) = √3, what is the value of sinθ?

Solution:
Given that,
tan(θ + 15°) = √3
⇒ tan(θ + 15°) = tan 60°
⇒ θ + 15° = 60°
⇒ θ = 60° - 15°
⇒ θ = 45°

Now,
sinθ
= sin45°
= 1/√2

Question: A water tank is one-third full. Pipe A can fill the tank in 6 minutes, and pipe B can empty it in 12 minutes. If both pipes are open together, how long will it take to fill the tank completely?

Solution:
Let total tank = 1 unit.
Current water = 1/3

A’s 1 minute work = 1/6 (filling)
B’s 1 minute work = 1/12 (emptying → negative)

Net work per minute = 1/6 – 1/12 = (2 – 1)/12 = 1/12

Remaining to fill = 1 – 1/3 = 2/3

Time to fill = (2/3) ÷ (1/12) = (2/3) × 12 = 8 minutes

We know,
The ratio of Investment x Time = Ratio of Profit
∴ (A's investment x Time) : (B's investment x Time) = Profit of A : Profit of B

∴ (Robi's Investment x Time) : (Rasel's Investment x Time) = Robi's Profit : Rasel's Profit
∴ 40000 x 12 : 50000 x 8 = Robi's Profit : Rasel's Profit
∴ Robi's Profit : Rasel's Profit = 4,80,000 : 4,00,000 = 6:5
∴ Rasel's profit = (5/11) × 187000 = Tk. 85000

Question: If ⁶P_r = 120, what is the value of r?

Solution:
আমরা জানি, ⁿP_r = n!/(n - r)!
দেওয়া আছে,
⁶P_r = 120
⇒ 6!/(6 - r)! = 120
⇒ 720/(6 - r)! = 120 (কারণ 6! = 720)
⇒ (6 - r)! = 720/120
⇒ (6 - r)! = 6
⇒ (6 - r)! = 3!   (কারণ 3! = 3 × 2 × 1 = 6)
⇒ 6 - r = 3
⇒ r = 6 - 3
∴ r = 3

Let X hours be the time taken to fill a tank by P.
Let Y hours be the time taken to empty the tank by Q.
Then the time taken to fill the tank when P and Q are switched together : XY/(Y - X) hours.
Here, X = 16 minutes And Y = 32 minutes
Therefore,
Required time = (16 × 32)/(32 - 16)
= (32 × 16)/16
= 32 minutes.

Question: If

Solution:

C.P. of 56 kg rice = (26 x 20 + 30 x 36)
=(520 + 1080)
=1600.
S.P. of 56 kg rice = (56 x 30)
= 1680.
Gain = 80x100% = 5%.

Let,
The number is X
According to the question,
(60/100) × X + 120 = X
⇒ 3X/5 + 120 = X
⇒ X - 3X/5 = 120
⇒ (5X - 3X)/5 = 120
⇒ 2X = 600
⇒ X = 300.

Question: If 4x + 5y = 140 and 4x / 5y = 2 / 5, then find y - x.
 
Solution:
We are given:
⇒ 4x / 5y = 2 / 5

We simplify:
x / y = (5 / 4) × (2 / 5)
⇒ x / y = 2 / 4 = 1 / 2
∴ x = y / 2

Also given:
4x + 5y = 140

Substitute x = y / 2:
⇒ 4 × (y / 2) + 5y = 140
⇒ (4y / 2) + 5y = 140
⇒ (2y + 5y) = 140
⇒ 7y  = 140
⇒ y = 20

Then x = y / 2 = 10
So, 20 - 10 = 10 

Question: The difference (in taka) between simple and compound interest at 4% per annum on a sum of Tk 1500 after 2 years is:

Solution:
Simple interest, I = Pnr/100 
= (1500 × 2 × 4)/100
= 120 Tk

Compound interest, A = P{1 + (r/100)}²
= 1500{1 + (4/100)}²
= 1500{1 + (1/25)}²
= 1500 × (26/25)²  
= 1622.4 Tk

Interest = 1622.4 - 1500 = 122.4 Tk
 ∴ Difference = 122.4 - 120 = 2.4 Tk

Question: A man completes a journey of 120 km by a car. He covers 60 km at 30 km/h, 40 km at 40 km/h, and the remaining 20 km at 20 km/h. What is his average speed for the whole journey?

Solution:
A man completes a journey of 120 km by a car.
First case, 
Distance covered: 60 km at 30 km/h
∴ Time = 60 ÷ 30 = 2 hours

Second case,
40 km at 40 km/h
∴ Time = 40 ÷ 40 = 1 hour

Third case,
the remaining 20 km at 20 km/h
 ∴ Time = 20 ÷ 20 = 1 hour

∴ Total distance = 60 + 40 + 20 = 120 km
∴ Total time = 2 + 1 + 1 = 4 hours

∴ Average speed = Total distance ÷ Total time = 120 ÷ 4 = 30 km/h

Question: A petrol tank is initially one-third full. After removing 5 gallons of petrol, the tank becomes one-fifth full. What is the total capacity of the tank in gallons?

Solution:
Let,
The capacity of the tank in gallons is x gallons.

According to question,
⇒ (x/3) - 5 = x/5
⇒ (x - 15)/3 = x/5
⇒ 5(x - 15) = 3x
⇒ 5x - 75 = 3x
⇒ 5x - 3x = 75
⇒  2x = 75
∴ x = 37.5 gallons

Let,
Speed = x km/h
So,
40x - 40x × (14/15) = 5
Or, 40x - 112x/3 = 5
Or, 120x - 112x = 15
Or, x = 15/8
∴ distance covered = (15 × 40)/8 = 75 km

Total match = 60 + 32 = 92
75% of 92 = 69
Match needed to win = 69 - 40
= 29.

Let Tap A take T minutes to fill the tank alone.
Since Tap A is 5 times faster than Tap B, Tap B takes 5 times more time.
So time taken by Tap B = 5T minutes
Also, 5T-T = 32 ----------- Given
∴ T = 8 minutes = Time taken by A
Time taken by B = 5 x 8 = 40 minutes.

In 1 min, A + B fills = 1/8 + 1/40 = 3/20 parts
So entire tank is filled in = 20/3 hours.

In whole mixture, there are:
Water = 3/(3 + 5) portion = 3/8 portion
And Milk = 5/(3 + 5) portion = 5/8 portion

Let, the portion of the mixture to be drawn off and replaced with water = x
So, in x portion mixture there are,
Water = 3/(3+5) portion of x = 3x/8 portion
And
Milk = 5/(3+5) portion of x = 5x/8 portion

As per the question,
(3/8 − 3x/8 + x) : (5/8 − 5x/8) = 1:1
Or, (3 − 3x + 8x)/8 = (5 − 5x)/8
Or, 3 + 5x = 5 − 5x
Or, 5x + 5x = 5 − 3
Or, 10x = 2
Or, x = 2/10
Or, x = 1/5

Answer: The 1/5 portion of the mixture to be drawn off and replaced with water, in order to get the mixture as half milk and half water.

Question: The area of the base of a rectangular tank is 6500 sq. cm and the volume of water contained in it is 2.6 cubic meters. The depth of water is?

Solution:
Let depth = D cm.
We know,
Volume Base Area x Depth
and, 1 cubic meter = 1,000,000 cubic centimeters

Then,
D × 6500 = 2.6 × 100 × 100 × 100 
∴ D = (2.6 × 100 × 100 × 100)/6500 cm 
= 400 cm
= 4 m

Number of shares = 20000
Face value of each share = Tk. 10
dividend per share = (10 × R/100) where R is the Rate of interest.
Total dividend = 20000 ×10 ×R/100
20000 ×10 ×R/100 = 24000
R = 24000/2000
= 12
Hence the dividend is 12%.

Time is taken by A to cover 100 meters = 60 seconds
A gives a start of 4 seconds then time takes by B = 72 seconds
B takes 72 seconds to cover 96 meters
Speed of B = 96/72 = 1.33 m/s

Question: Two trains with lengths 126 m and 119 m respectively are moving towards each other. Their speeds are 12 m/s and 23 m/s, respectively. What will be the time needed by the trains to cross each other?

Solution:
To cross each other completely, the two trains must cover a total distance equal to the sum of their lengths.
∴ Total distance to be covered = 126 + 119 = 245 meters

Since they are moving towards each other, their relative speed is the sum of their individual speeds.
∴ Relative speed = 12 + 23 = 35 m/s

∴ Time taken to cross each other = Total distance/Relative speed
= 245/35
= 7 seconds

∴ The trains will take 7 seconds to completely cross each other.

Question: A worker was hired for 5 days. Each day, he was paid Tk. 20 more than what he was paid for the previous day of work. The total amount he was paid in the first 3 days of work equaled the total amount he was paid in the last 2 days. What was his starting pay?

Solution: 
Let
Starting payment was = x

Salary for the 1st 3 days:
x, x + 20, x + 40 

Salary for the last 2 days:
x + 60, x + 80

Now
sum of the salary of the 1st 3 days = sum of the salary of the last 2 days
x + x + 20 + x + 40 = x + 60 + x + 80
⇒ 3x + 60 = 2x + 140
⇒ 3x - 2x = 140 - 60
⇒ x = 80

∴ The starting payment was Tk. 80

Question: A party consists of a grandmother, father, mother, four sons and their wives, and one son and two daughters of each of the sons. How many females are there in all?

Solution: 
Grandmother: 1 female
Mother: 1 female
Four sons' wives: 4 females
Two daughters for each of the four sons: 4 × 2 daughters = 8 females

1 (Grandmother) + 1 (Mother) + 4 (Wives) + 8 (Daughters) = 14 females
So, there are a total of 14 females in the party.

প্রশ্ন: If a and b are whole numbers such that a^b = 81, what is the value of (a + 1)^{b - 1}?

Solution:
We know that 81 = 3⁴
∴ a = 3 and b = 4

Now,
(a + 1)^{b - 1}
= (3 + 1)^{4 - 1}
= 4³ = 64

Question: In how many different ways can the letters of the word ‘BALLOON’ be arranged?

Solution:
Number of letters in the word = 7

Repeated letters:
L = 2 times
O = 2 times
The remaining letters (B, A, N) are different.

∴ Number of arrangements
= 7!/(2! × 2!)
= 5040/4
= 1260

Question: The area of a trapezium is 96 square cm. The length of one of the parallel sides is 12 cm, and the distance between the parallel sides is 8 cm. Find the length of the other parallel side.

Solution:
Given,
Area of the trapezium = 96 cm²
One parallel side a = 12 cm
Distance between the parallel sides h = 8 cm

Let
the other parallel side = b cm

We know,
The area of a trapezium = (1/2) × (a + b) × h
⇒ 96 = (1/2) × (12 + b) × 8
⇒ 96 = (12 + b) × 4
⇒ (12 + b) = 96/4
⇒ 12 + b = 24
⇒ b = 24 - 12
∴ b = 12

∴ The other parallel side is 12 cm.

Question: A farmer borrowed Tk. 3600 at 15% simple interest per annum. At the end of 4 years, he cleared this account by paying Tk. 4000 and a cow. The cost of the cow is:

Solution:
P = 3600 tk, R = 15%, T = 4 yrs

S.I = PRT/100
= (3600 × 15 × 4)/100
= 2160 Tk.

Hence,
amount after 4 years = (3600 + 2160) = 5760 Tk.
∴ Cost of the cow = (5760 – 4000) = 1760 Tk.

Face value of each share = Tk. 5
Total dividend received by Jobayed = {100 × 5 × (12/100)}
= Tk. 60
Let market value of 100 shares = Tk. x
x × (10/100) = 60
x = 600
ie, Market value of 100 shares = Tk. 600
Hence, Market value of each share = Tk. 6

Let first discount = x
91% discount of (100 – x) % of 120 = 90
⇒ (91/100) × {(100 – x)/100} × 120 = 90
⇒ (100 - x) = (90 × 100 × 100)/(120 × 91) = 82.42
⇒ x = (100 – 82.42) = 17.58
Therefore, first discount = 17.58%