Bank Math

Let the denominator of the required fraction be X.
Given that,
the numerator of a fraction is 3 less than its denominator;
then its numerator becomes X-3.
Then the required fraction (X-3)/X ...(1)
If we add 10 to the numerator(X-3), original fraction (X-3)/X is increased by 1(3/7) (i.e., 10/7).
i.e.,
(X -3)/X + 10/7 = (X - 3 + 10)/X
(X - 3)/X + 10/7 = (X + 7)/X
(X + 7)/X - (X - 3)/X = 10/7
10/X = 10/7
X = 7.
Therefore, (X-3)/X = 4/7.
Hence the required fraction is 4/7.

There are 7 letters in the word DRAUGHT, the two vowels are A and U.
Since, the vowels are not to be separated; AU can be considered as one entity.
Therefore, the number of letters will be 6 instead of 7.
The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
The required number of arrangements = 2!.6! = 1440 ways.

Question: What is the ratio of simple interest earned on a certain amount at the rate of 12% p.a. for 9 years and that for 12 years?

Solution: 
As the amount and interest rate are the same, 
Ratio = 9/12
= 3/4

Question: The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

Solution:
Here,
The order of each letter in the dictionary is ABLORU.

Now,
with A in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

Similarly,
with B in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

With L in the beginning,
the first word will be LABORU, the second will be LABOUR.

Hence, the rank of the word LABOUR = 5! + 5! + 2
= 120 + 120 + 2
= 242

Question: A cistern can be filled by two taps X and Y in 15 hours and 20 hours respectively. The full cistern can be emptied by a third tap Z in 10 hours. If all the taps are turned on at the same time, in how much time will the empty cistern be filled completely?

Solution :

X’s 1 hour work = 1/15
Y’s 1 hour work = 1/20
Z’s 1 hour work = 1/10 (emptying → negative)

Net 1 hour work = 1/15 + 1/20 – 1/10

Find LCM of denominators (15, 20, 10) → 60

Convert:
1/15 = 4/60, 1/20 = 3/60, 1/10 = 6/60

Net work = 4/60 + 3/60 – 6/60 = 1/60

Time taken = 1 ÷ (1/60) = 60 hours.

P(A) = 2/5
P(B) = 4/7
E = {A and B both get selected}

P(E) = P(A) × P(B)
= (2/5) × (4/7)
= 8/35

Question: What is the nature of the roots of the equation 9x² + 12x + 4 = 0?

Solution:
Given that, 
9x² + 12x + 4 = 0  
Here,  
a = coefficient of x² = 9  
b = coefficient of x = 12  
c = constant term = 4  

Discriminant = b² - 4ac  
= (12)² - 4 × 9 × 4  
= 144 - 144  
= 0
When the discriminant = 0, the roots are real and equal.

Therefore, the roots are real and equal.

Note: 
- If b² - 4ac > 0 and a perfect square ⇒ roots are real, unequal and rational  
- If b² - 4ac > 0 but not a perfect square ⇒ roots are real, unequal and irrational  
- If b² - 4ac = 0 ⇒ roots are real and equal  
- If b² - 4ac < 0 ⇒ no real roots (complex roots)

Question: In January, the stock price went up by 50%. It then dropped by 20% in February, rose again by 25% in March, and declined by 10% in April. If Tk. 200 was invested initially and sold after April, calculate the net percentage change in price.

Solution:
At the end of January,
The value of the stock is = Tk. 200 + 50% of (Tk. 200)
= Tk. 200 + Tk. 100 = Tk. 300.

At the end of February,
The value of the stock is = Tk. 300 - 20% of (Tk. 300)
= Tk. 300 - Tk. 60 = Tk. 240.

At the end of March,
The value of the stock is = Tk. 240 + 25% of (Tk. 240)
= Tk. 240 + Tk. 60 = Tk. 300.

At the end of April,
The value of the stock is = Tk. 300 - 10% of (Tk. 300)
= Tk. 300 - Tk. 30 = Tk. 270.

Now, the percentage change in price is,
= (Change in price/Original price) × 100%
= (270 - 200)/200 × 100%
= (70/200) × 100%
= 35%

Question: find the value of sin²21° + cos²21°

Solution: 
sin²21° + cos²21°
= 1 [sin²θ + cos²θ = 1]

Question: 60% of a number is 30 less than three-fourth of the same number. Find the number.

Solution:
Let,
the number = x.

ATQ,
60% of x = (3/4) of x - 30
⇒ 60x/100 = (3x/4) - 30
⇒ 3x/5 = (3x/4) - 30
⇒ (3x/4) - (3x/5) = 30
⇒ (15x - 12x)/20 = 30
⇒ 3x/20 = 30
∴ x = 200

So, the number is 200.

Sum of 6 numbers = (6 × 25) = 150.
Sum of 3 additional numbers = (3 × 22) = 66.
Sum of (6 + 3) =9 numbers = (150 + 66) = 216
∴ average of the combined 9 numbers = 216/9 = 24

Let the cost price = 100
Selling price = (100×117)/100 = 117
So, Marked price = (117×100)/90 = 130
∴ Profit Percentage without discount = {(130 - 100)/100} × 100 = 30%

Question: 40 workers can complete a work is 15 days. After six days 20 more workers joined them. How many days will they require to complete the remaining work?

Solution: 
15 days for the full work
∴ 6 day for = 6/15 = 2/5 of the work

∴ remaining work is = (1 - 2/5) = 3/5
And total workers = 40 + 20 = 60


∴ 40 men can do 1 work in 15 days
∴ 40 men can do 3/5 work in = (15 × 3/5) = 9 days
∴ 60 men can do 3/5 work in = (9 × 40)/60 days
= 6 days

Question: The average of 7 consecutive numbers is 21. The largest of these numbers is-

Solution:
Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,
Then,
(x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6)/7 = 21
⇒ 7x + 21 = 147
⇒ 7x = 126 
∴ x = 18

∴ Largest number = x + 6 = 18 + 6 = 24

Question: Find the greatest value of sin⁴A + cos⁴A.

Solution:
We know,
sin²A + cos²A = 1
⇒(sin²A + cos²A)² = 1²
⇒ (sin²A)² + (cos²A)² + 2sin²Acos²A = 1
⇒ sin⁴A + cos⁴A = 1 - 2sin²Acos²A
⇒ sin⁴A + cos⁴A = 1 - 2sin²90°cos²90° [since we need maximum value]
⇒ sin⁴A + cos⁴A = 1 - 2(1 × 0)
⇒ sin⁴A + cos⁴A = 1 - 0
∴ sin⁴A + cos⁴A = 1 

We are given that two-thirds of the 6 km was covered at 4 km/hr i.e. 4 km distance was covered at 4 km/hr.
Time taken to cover 4 km = 4 km/4 km/hr = 1 hr = 60 minutes.
Time left = 84 – 60 = 24 minutes

Now, the man has to cover the remaining 2 km in 24 minutes or 24/60 = 0.4 hours
Speed required for remaining 2 km = 2 km/0.4 hr = 5 km/hr

Let the length of the train be x metres.
Then, length of the platform = 2x metres.
Speed of the train = 90× (5/18) m/sec
= 25m/sec
∴(x+2x)/25 = 36
⇒ 3x= 900
⇒ x= 300
Hence, length of platform
= 2x= (2×300)m= 600m

Let the present age of the person be x years.

According to the question,
Then, 3(x + 3) - 3(x - 3) = x
⇒ 3x + 9 - 3x + 9 = x
⇒ 18 = x
Hence the present age of the person was 18 years.

Question: A car travels at a speed that is 3/4th the speed of a bike. The bike covers 240 km in 4 hours. How much distance will the car cover in 30 minutes?

Solution:
Speed of the bike = distance/time
= 240/4 = 60 km/h

And speed of the car = 3/4 of the speed of the bike
= (3/4) × 60 = 45 km/h

∴ Time for the car = 30 minutes
= 30/60 = 1/2 hours

∴ Distance covered by the car = speed × time
= 45 × (1/2) = 22.5 km

So the car will cover 22.5 km in 30 minutes.

Question: A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is:

Solution:
Given that,
Total bulbs = 20
Defective bulbs = 4
Non-defective bulbs = 20 - 4 = 16
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (16/20) × (15/19) = 240/380 = 12/19

And,
∴ P(at least one defective) = 1 - P(both non-defective)
= 1 - (12/19)
= (19 - 12)/19
= 7/19
∴ The probability that at least one of them is defective is 7/19

Number of shares = (14400/120) = 120
Face value = (100 × 120) = 12000
Annual income = (5/100 × 12000) = 600

Question: A man's salary was reduced by 50%, again the reduced salary was increased by 50%. Find the loss of in terms of percentage.

Solution: 
Let, The wages of man = 100 Tk.
Then wages decreased by 50%
So wages = (100 - 50) = 50 Tk
The reduced wages were increased by 50%
Then wages = 50 × (150/100) = 75 Tk.
So, the percentage of wages loss = (100 - 75) tk
= 25 Tk

Question: A, B and C jointly thought of engaging themeselves in a business venture. It was agreed that A would invest TK. 8000 for 7 months, B, TK. 10,000 for 4 months and C, TK. 12,000 for 3 months. A wants to be the working member for which, he was to receive 10% of the profits. The profit earned was TK. 6600. Calculate the share of C in the profit.

Solution:
Ratio of their investments = (8000 × 7) : (10,000 × 4) : (12,000 × 3)
= 56,000 : 40,000 : 36,000
= 14 : 10 : 9

sum of ratio = (14 + 10 + 9) = 33

Here,
The profit earned was TK. 6600

For working A received = 10% of TK. 6600
= TK. (10 × 6600)/100
= TK. 660

Remaining profit = TK. (6600 - 660) = TK. 5940

∴ C's share = TK. {5940 × (9/33)
= TK. 1620

Given,
A fills the tank in 22.5 minutes and A remains open for 18 minutes.
So total tank filled by A = Tank filled in 1 min × 18 minutes = 1/(22.4) × 18 = 4/5
This is the entire work done by A.

So whatever is the remaining work, it is done by only B
Let B be open for T minutes.
Total tank filled by B = 1 - (4/5) = 1/5 = Tank filled in 1 min × T minutes = (1/15) × T
1/5 = (1/15) × T

∴ T = 3 minutes = B should be closed after this much time.

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

Let there be x pupils in the class.
The total increase in marks = (x × 1/2) = x/2
∴ x/2 = (83 - 63)
⇒ x/2 = 20
⇒ x = 40
Answer: 40

Question: (256)^0.16 × (256)^0.09 = ?

Solution:
(256)^{0.16 + 0.09}
= (256)^0.25
= (256)^25/100
= (256)^1/4
= (4⁴)^1/4 (because 256 = 4⁴)
= 4^{4 × 1/4}
= 4¹
= 4

Question: The area of a regular hexagon of side 3√2 cm is:

Solution:
A regular hexagon consists of 6 equilateral triangle
Area of regular hexagon
= 6 × (√3/4) × (side)²
= 6 × (√3/4) × (3√2)²
= 27√3 cm²

Question: You are looking at a billboard 40m away with an angle of elevation of 30°. At what height is the billboard?

Solution:


ধরি,
উচ্চতা = XZ = h
অতিভুজ = XY = 40 মিটার

প্রশ্নমতে,
Sin30° = লম্ব/অতিভুজ
⇒ 1/2 = h/40
∴ 2h = 40 মিটার
h = 20

Question: X’s age is 20% more than Y’s age. If the sum of their ages if 20 more than the difference of their ages, find their respective ages.

Solution: 
Let X's age be x and Y's age be y

First condition,
X’s age is 20% more than Y’s age.
 ⇒ xx = y + 0.20y
 ⇒ x = 1.20y
 ∴ x = (6/5)y.......(1)

Second condition,
The sum of their ages is 20 more than the difference of their ages.
x + y = (x - y) + 20
⇒ x + y = x - y + 20
⇒ 2y = 20
⇒ y = 20/2
∴ y = 10 years

Now that we have Y's age, substitute  y = 10 into Equation 1
⇒ x = (6/5)y = (6/5) × 10 = 12
∴ x = 12 years

X's age is 12 years and Y's age is 10 years.

Question: In a 729 liters mixture of milk and water, the ratio of milk to water 7 : 2. To get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is- 

Solution: 
Quantity of milk in 729 litre of mixture
= 7 × 729/9 = 567 litres

Quantity of water
= 729 - 567
= 162 litres

Let x litre of water be added to become ratio 7 : 3

According to the question,
7/3 = 567 / (162 + x)
Or, 162 × 7 + 7x = 567 × 3
Or, 7x = 1701 - 1134 = 567
Or, x = 567 / 7 = 81
Therefore, 81 litres of water is to be added.

Let,
Habib’s age is = 6x
And Shikha’s age = 4x
ATQ,
(6x - 5)/(4x - 5) = 5/3
⇒ 18x - 15 = 20x - 25
⇒ 2x = 10
⇒ x = 5
So, Habib’s age is = 6×5 = 30 year

Question: If sinθ/cosθ = 1, then find θ.

Solution:
sinθ/cosθ = 1
⇒ tanθ = 1
⇒ tanθ = tan45°
∴ θ = 45°

Sum of ages of husband, wife and child= (24 × 3) + 9 = 81 year
Sum of ages of wife and child 
=> 25 × 2 + 10 = 50 + 10 = 60 year
Age of husband = 81 - 60 = 21 year