Bank Math

Question: The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is = ?

Solution: 
Given that, 
LCM = 495, HCF = 5
And,
The sum of the numbers is, a + b = 100
The product of two numbers is equal to the product of their LCM and HCF. 
Product(a × b) = LCM × HCF = 495 × 5 = 2475

We know, 
(a - b)² = (a + b)² - 4ab
⇒ (a - b)² = 100² - 4 × 2475 ; [where, a + b = 100 and  ab = 2475 ]
⇒ (a - b)² = 10000 - 9900
⇒ (a - b)² = 100
⇒ a - b = √100 = 10
∴ a - b = 10 

So the difference between the numbers is  10.

Question: A shopkeeper sells goods at 25% profit. If he had sold for Tk. 200 more, the profit would have been 30%. What is the cost of that goods?

Solution:
Let the cost price = C Tk.

Selling price at 25% profit:
SP₁ = C + 25% of C = C × 1.25

Selling price at 30% profit:
SP₂ = C + 30% of C = C × 1.30

Difference:
SP₂ - SP₁ = 200
⇒ 1.30C - 1.25C = 200
⇒ 0.05C = 200
⇒ C = 200/0.05
⇒ C = 4,000 Tk.

∴ Cost Price = 4,000 Tk.

As, d days is greater than 7 days
So, charge for first week is c cents
∴ For (d - 7) days the charge is = f(d - 7) cents
So, total cost = c + f(d - 7)

Question: Hasan can paint 54 walls in 27 minutes. Rafiq can paint 25 walls in 10 minutes. Working together, how many walls can they paint in 20 minutes?

Solution:
Hasan can paint in 1 minute = 54 ÷ 27 = 2 walls

Rafiq can paint in 1 minute = 25 ÷ 10 = 25/10 = 5/2 walls

∴ Together they can paint in 1 minute = 2 + (5/2) walls
= (4 + 5)/2 = 9/2 walls

∴ In 20 minutes, they can paint = (9/2) × 20 walls
= (9 × 20)/2
= 90 walls

Let, total income = 100
Total expense food and education = 35 + 25
= 60
Remaining = (100 - 60) = 40
Expense on house rent = 40 × (80/100) = 32
Remaining after all expense = 40 - 32 = 8

Question: In a box of 5 eggs, 2 are rotten. What is the probability that two eggs chosen at random from the box are rotten?

Solution: 
Given,
Total number of eggs = 5
Number of rotten eggs = 2
Number of non-rotten (good) eggs = 3
Number of eggs to be chosen = 2

We can think of this as picking one egg and then picking another without replacing the first one.
The probability that the first egg picked is rotten = 2/5
  
Now, if the first egg was rotten, there is now 1 rotten egg left and 4 total eggs left.
The probability that the second egg picked is also rotten = 1/4
  
Hence, the probability of selecting 2 rotten eggs in a row = (2/5) × (1/4)
= 1/10

Question: In a box, the ratio of red marbles to blue marbles is 7 : 4. Which of the following could be the total number of marbles in the box?

Solution: 
18, 19, 21 কোনটিই 7 + 4 বা 11 দ্বারা নি:শেষে বিভাজ্য নয়। 
অতএব, সঠিক উত্তর 22

Question: Two unbiased coins are tossed. What is the probability of getting at least 1 tail?

Solution:
Total outcomes = {TT, TH, HT, HH} = 4
Favorable outcomes = {TT, TH, HT} = 3

So, the probability of getting at least 1 tail = Favorable outcomes/Total outcomes
= 3/4

 Question: A dozen Bombay peppers are purchased for Tk. 37.50 and sold for Tk. 39.75. What is the profit percentage?

Solution:
∴ লাভ = ৩৯.৭৫ - ৩৭.৫০ = ২.২৫ টাকা

∴ ৩৭.৫০ টাকায় লাভ হয় = ২.২৫ টাকা
∴ ১ টাকায় লাভ হয় = ২.২৫/৩৭.৫০ টাকা
∴ ১০০ টাকায় লাভ হয় = (২.২৫ × ১০০)/৩৭.৫০ টাকা
= (২২৫ × ১০০)/৩৭৫০ টাকা 
= ৬ টাকা বা ৬% 

Question: A man distributes Tk. 16,000 among his daughter, wife, and son such that the daughter’s and wife’s shares are in the ratio 1 : 2, and the son receives half of the total amount. What is the daughter’s share?

Solution:
The daughter's share and the wife's share are of 1 : 2 ratio
daughter's share = x
wife's share = 2x

As the son gets half of the total amount
∴ son's share = 16000/2
= 8000 Tk

∴ Rest of the share = 3x
∴ According to the question,
Or, 3x = 8000
Or, x = 8000/3
∴ x = 2666.66

∴ daughter's share = x = 2666.66 Tk

Question: A person subscribing to Netflix for a year pays BDT. 1560. If the monthly subscription is BDT. 150, how much discount does a yearly subscriber get?

Solution: 
Yearly subscription rate = BDT. 1560
Charge for 12 month as rate of BDT. 150 per month = 12 × 150
= BDT. 1800

∴ Discount = 1800 - 1560 = BDT. 240

Hence, discount % = (240 × 100)/1800 
= 13.33%

Let the constant be x,
so putting the first scenario in equation
A = x × (B/C)
⇒ 6 = x × 2/3
⇒ 2x = 18
⇒ x = 9
We can find out A in the second scenario by putting the value of x as 9
A = 9 × 8/6
⇒ A = 12.

100 years contain 5 odd days.
∴ Last day of 1st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Question: If Kamal gves 20 marbles to Shuvo, then, both of them will have equal numbers of marbles in their possessions. If Shovo gives 40 marbles to Kamal then Kamal will have twice the number of marbles that Shuvo will retain. What is the number of marbles that Kamal has?

Solution:
কামালের মার্বেল আছে = x টি
শুভর মার্বেল আছে = y টি

১ম শর্তমতে
x - 20 = y + 20
x - y = 20 + 20 
x - y = 40 ......................(1)

২য় শর্তমতে
x + 40 =2(y - 40)
x + 40 = 2y - 80
2y - x = 40 + 80 
2y - x = 120 ......................(2)

(1) × 2 + (2) ⇒ 
2x - 2y + 2y - x = 80 + 120
x = 200

কামালের মার্বেল আছে = 200 টি

Question: A train crosses a platform 300 meters long in 50 seconds and another platform 150 meters long in 35 seconds. What is the length of the train?

Solution:
Let the length of the train = x meters

Then,
For the first platform, the distance covered by the train = (x + 300) meters
And,
For the second platform, the distance covered by the train = (x + 150) meters

According to the question,
(x + 300)/50 = (x + 150)/35
⇒ 50(x + 150) = 35(x + 300)
⇒ 50x + 7500 = 35x + 10500
⇒ 50x - 35x = 10500 - 7500
⇒ 15x = 3000
⇒ x = 3000/15 
⇒ x = 200

∴ The length of the train is 200 meters

Let,
The speed of the stream is x km/hr.
Then,
Downstream Speed = (12 + x) km/hr
Upstream Speed = (12 - x) km/hr
The boat covers, 40 km downstream in 7 hours
Then we have,
[ 40/(12 + x) ] + [ 40/(12 - x) ] = 7
⇒ [{40(12 - x) + 40(12 + x)}/{(12 + x)(12 - x)}]
⇒[40(12 - x) + 40(12 + x)] = 7(144 - x²)
⇒ 480 - 40x + 480 + 40x = 1008 - 7x²
⇒ 960 = 1008 - 7x²
⇒ 7x²= 1008 - 960
⇒ 7x² = 48
⇒ x² = 48/7
⇒ x² = 6.92
⇒ x = 2.63
Hence, the speed of the stream is 2.63 km/hr.

Question: If the price of a commodity increases by 50%, by what fraction should the quantity consumed be decreased to maintain the same total expenditure?

Solution:
Let the initial price of the commodity be 100.

After a 50% increase in price,
The increased price will be 150.
Now, we have to reduce the consumption to keep expenditure at 100.

Increase in price = 150 - 100 = 50

∴ We have to reduce the consumption
= (50/150) × 100
= 33.33% or 1/3

Question: The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 4 times the father's age at that time. What are the present ages of the father and son?

Solution:
ধরি,
বর্তমানে বাবার বয়স = x বছর
বর্তমানে ছেলের বয়স = (45 - x) বছর

5 বছর আগে,
বাবার বয়স = x - 5
ছেলের বয়স = (45 - x) - 5 = 40 - x

প্রশ্নমতে,
(x - 5)(40 - x) = 4(x - 5)
⇒ 40 - x = 4
⇒ x = 40 - 4 = 36

∴ বাবার বর্তমান বয়স = 36 বছর
ছেলের বর্তমান বয়স = 45 - 36 = 9 বছর

Question: A person has to completely put each of three liquids: 403 litres of petrol, 465 litres of diesel and 496 litres of Mobil Oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?

Solution: For the least number of bottles, the capacity of each bottle must be maximum.
The capacity of each bottle = HCF of 403 liters, 465 liters, and 496 liters
= 31 liters.

For petrol, required number of bottles = 403/31 = 13
For diesel, required number of bottles = 465/31 = 15
For mobil, required number of bottles = 496/31 = 16

Hence,
required number of bottles = 13 + 15 + 16 = 44

Here, a + b + c = 12 & a + b = 4
So, c = 8
As, a + c = 7
∴ a = 7 - 8 = -1

Let A's age 10 year ago = x year. 
Then, B's age 10 year ago = 2x year.
(x + 10) / (2x+ 10) = 3/4 
=> x = 5.
So, the total of their present ages 
=(x + 10 + 2x + 10) 
= (3x + 20) 
= 35 year.

Question: A seller bought 120 pens for 600 taka. How many pens does he need to sell for 600 taka to make a profit of 20%?

Solution:
Cost price of 120 pens = Tk. 600 Cost price per pen = 600 ÷ 120
= Tk. 5

∴ Selling price per pen for 20% profit = 5 × (120/100)
= 5 × 1.2
= Tk. 6

∴ Number of pens to be sold for Tk. 600
= 600 ÷ 6
= 100 pens

1) Find the cost price

Let C.P. of book = x and S.P. = Tk. 450
S.P. of book = C.P. – (10% of C.P.)
S.P. = x – (0.10x)
450 = 0.9 x
x i.e cost price = Tk. 500

2) Find Selling Price to gain 10%.

Now, we are asked to find a selling price to gain 10% profit.
Selling price = (100 + Gain%)/100 × C.P.
Selling price = (100 + 10)/100 × 500
Selling price = (110/100) × 500
= Tk. 550

Question: The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

Solution:
As the ratio of their present ages is 3 : 4 ,
let their present ages be 3x and 4x.
So, 5 years ago, as the ratio of their ages was 5 : 7,

According to the question,
(3x - 5) : (4x - 5) = 5 : 7
⇒ (3x - 5)/(4x - 5) = 5/7
⇒ 21x - 35 = 20x - 25
⇒ x = 10

Hence, their present ages are 3x = 30 and 4x = 40

Question: A dice is cast twice, and the sum of the appearing numbers is 10. The probability that the number 5 has appeared at least once is-

Solution:
A dice is cast twice
Sum of the numbers = 10
Find probability that number 5 appears at least once

Now,
All possible pairs (first die, second die) whose sum = 10 is,
(4, 6), (5, 5), (6, 4)
∴ Total outcomes = 3

And,
Outcomes with at least one 5
From the above pairs we get only 1 outcome has at least one 5.

 ∴ Probability = Favorable Outcome/Total outcomes = 1/3

Question: The ratio of X : Y is 3 : 5, and the ratio of Y : Z is 6 : 7. If X = 18, what is the value of Z?

Solution:
Given that,
X : Y = 3 : 5
Y : Z = 6 : 7
And X = 18

Now,
X : Y = 3 : 5
⇒ X/Y = 3/5
⇒ 5X = 3Y
⇒ Y = (5 × 18)/3 ; [X = 18]
∴ Y = 30

And,
Y : Z = 6 : 7
Y/Z = 6/7
⇒ 30/Z = 6/7
⇒ Z = (30 × 7)/6
∴ Z = 35

So the value of Z is 35.

As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time.
10% = 10 min
100% = 100 min.

Now,
Distance traveled by him,
= (100/60) × 18
= 30 km.

Question:  A group has 8 men and 7 women. In how many ways can a committee of 5 people be formed if the number of women is at least 3?
​
​Solution:
​Given that,
​Total committee size = 5

And for women ≥ 3, possible distributions:
Three ways to formed the committee
1. 3 women + 2 men
2. 4 women + 1 man
3. 5 women + 0 men
​
​Now, 1st case- 3 women + 2 men
​Choose 3 women from 7, (⁷C₃) = 35
​Choose 2 men from 8,  (⁸C₂) = 28
​∴ Total ways = 35 × 28 = 980
​
​2nd case- 4 women + 1 man
​Choose 4 women from 7, (⁷C₄) = 35
Choose 1 man from 8, (⁸C₁) = 8
 ​∴ Total ways = 35 × 8 = 280
​
​And 3rd case-​5 women + 0 men
Choose 5 women from 7,  (⁷C₅) = 21
No men to choose
​∴ Total ways = 21
​
​∴ ​Total ways = 980 + 280 + 21= 1281

Cost of second hand car = Tk. 2,00,000
Repainting cost = Tk. 2000
Cost of threading per tyre = Tk. 200
Cost of threading for all tyres = Tk. 200 x 4 = Tk. 800
Total costs involved = Actual cost + Repainting costs + Threading costs
= 2,00,000 + 2000 + 800
= 202800
Profit Percentage = 10% = (SP - CP)/CP x 100%
Or, 10% = (SP - 202800)/202800 x 100%
Or, 10 = (SP - 202800)/2028
Or, 20280 = SP - 202800
Or, SP = 20280 + 202800
= 223080.
∴ Selling Price = 223080

The average of 6 players = X
Average increases by 10, when the seventh player makes a score of 112 runs.
Therefore, average of 7 players = X + 10
Average =Sum of Scores/Number of Players -------(1)
Here, average = X, number of players = 6
Hence,
Sum of scores = 6X
Score of 7 players = (Score of 6 players + score of 7 player) = (6X + 112)------- (2)
Total average = (X + 10) --------- (3)
Substitute (2) and (3), in (1)
(X + 10) = (6X + 112)/7
⇒ 7X + 70 = 6X + 112
⇒ 7X - 6X = 112 - 70
⇒ X = 42.

ধরি, C এর বয়স x বছর
তাহলে B এর বয়স = 2x বছর এবং A এর বয়স = (2x + 2) বছর
প্রশ্নমতে, x + 2x + 2x + 2 = 27
⇒ 5x = 27 - 2
⇒ x = 25/5
⇒ x = 5
∴ B এর বয়স (2×5) বছর = 10 বছর

Since 1/3 (boys) > 1/4 (girls); have to maximize number of boys

The number of girls has to be a multiple of 4 since we cannot have half girls and number of boys multiple of 3

Working backwards.

36 - 4 girls = 32 boys not divisible by three
36 - 8 girls = 28 boys not divisible by three
36 - 12 girls = 24 boys which is divisible by three

So there are 24 boys and 12 girls

∴ total of 24×1/3 + 12×1/4 = 11 students walk to school

প্রশ্ন: বার্ষিক ৫% মুনাফায় ১৬,০০০ টাকার ২ বছরের সরল ও চক্রবৃদ্ধি মুনাফার পার্থক্য কত হবে?

সমাধান:
এখানে,
মূলধন, P = ১৬০০০ টাকা
সুদের হার, R = ৫%
সময়, T = ২ বছর

সরল মুনাফা, I = (PRT)/১০০
= (১৬০০০ × ৫ × ২)/১০০
= ১৬০০০০/১০০
= ১৬০০ টাকা

চক্রবৃদ্ধি মুনাফার ক্ষেত্রে,
A = P × {১ + (R/১০০)}^T
= ১৬০০০ × {১ + (৫/১০০)}^২
= ১৬০০০ × (১.০৫)^২
= ১৬০০০ × ১.১০২৫
= ১৭৬৪০ টাকা

∴ চক্রবৃদ্ধি মুনাফা, CI = A − P
= ১৭৬৪০ − ১৬০০০
= ১৬৪০ টাকা

∴ পার্থক্য = ১৬৪০ − ১৬০০ = ৪০ টাকা

Question: Which number will complete the series:
3, 7, 15, 31, 63, ?

Solution:
The pattern is alternate multiplication by 2 and adding 1:
n = 2n + 1, where n = 3, 7, 15, 31...

The pattern:
(2 × 3) + 1 = 6 + 1 = 7
(2 × 7) + 1 = 14 + 1 = 15
(2 × 15) + 1 = 30 + 1 = 31
(2 × 31) + 1 = 62 + 1 = 63

So, the missing term = (2 × 63) + 1 = 126 + 1 = 127.