Bank Math

Question: When a person weighing 60 kg leaves a group of 50 people, the average weight of the remaining people increases by 0.3 kg. What is the new average weight of the remaining 49 people?

Solution:
let,
The average weight of 49 people is x kg
Total weight of 49 people = 49x
Total weight of 50 people = 49x + 60

ATQ,
50(x - 0.3) = (49x + 60)
⇒ 50x - 15 = 49x + 60
⇒ 50x - 49x = 60 + 15
∴ x = 75

∴ the new average weight of the remaining 49 people is 75 kg.

Question: If tanθ =√3 and sinθ = √3/2. Find the value of cosθ.

Solution:
Given,
tanθ =√3 and sinθ = √3/2

We know,
tanθ = sinθ/cosθ
⇒√3 = (√3/2)/cosθ
⇒ cosθ = (√3/2)/√3
∴ cosθ = 1/2   

Question: The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by:

Solution: 
Let, old length 100 m and breadth 100m
Old area = 100 × 100 m²
= 10000 m²

New length = 100 + 40 
= 140 m
New breadth = 100 + 30 
= 130 m 

New area = 140 × 130 m² 
= 18200 m²

The area of the resulting rectangle exceeds the area of the square by = {(18200 - 10000)/10000} × 100%
= (8200/10000)× 100%
= 83%

Laxmi present age = 51 - 8 = 43
So Madhu Age = (43 - 7) = 36
So, Diviya Present Age = 36 × 7/6 = 42
So Difference = (42 - 36) = 6 years

Question: If log_x(8/125) = - 3, then x = ?

Solution:
log_x(8/125) = - 3
⇒ x^{- 3} = 8/125
⇒ x^{- 3} = (2/5)³
⇒ x^{- 3} = (5/2)^{- 3}
⇒ x = 5/2
∴ x = 5/2

Question: A person buys certain number of gems at 20 per taka and an equal number at 30 per taka. He mixes them and sells them at 25 per taka. His gain or loss in the transaction is-

Solution:
The cost price of each gem is 20 per Tk. = Tk. 1/20
The price of each gem is 30 per Tk. = Tk. 1/30

∴ Average cost of gem = {(1/20) + (1/30)}/2
= (3 + 2)/(2 × 60)
= 5/120
= 1/24

The selling price of mixed gem = Tk. 1/25 

∴  Loss = (1/24) - (1/25) = 1/600

Cost price 1/24 then loss Tk. 1/600
Cost price 1 then loss Tk. (1/600)/(1/24)

∴ Cost price 1 then loss Tk. {(1/600)/(1/24)} × 100
= 4%

∴ There is a loss of 4%

Sum of their ages = 40 × 5 = 200 years
Sum of A & B’s ages = 35 × 2 = 70 years
Sum of C & D’s ages = 42 × 2 = 84
So, Age of E = 200 - (70 + 84) = 46

Question: There are 5 doors to a lecture room. Two are red, and the others are green. In how many ways can a lecturer enter the room and leave the room from different colored doors?

Solution:
There are 2 red and 3 green doors. We have two cases:
The room can be entered from a red door (2 red doors, so 2 ways) and can be left from a green door (3 green doors, so 3 ways): 2 × 3 = 6

The room can be entered from a green door (3 green doors, so 3 ways) and can be left from a red door (2 red doors, so 2 ways): 3 × 2 = 6 

Hence, the total number of ways = 6 + 6 = 12

Let investment of C be Tk. 100
So the investment of B = Three-fourths of C = 3/4 of Tk. 100 = Tk. 75
Investment of A = 20% more than B = 20% more than Tk. 75 = Tk. 90
Ratio of investment of A, B, and C = 90 : 75 : 100 = 18 : 15 : 20
Investment of C = 20/(18 + 15 + 20) × 3816
= Tk. 1440.

As per statement (a - b) is 6 more than (c + d).
a - b = (c + d) + 6 ............ (1)
As per statement (a + b) is 3 less than (c - d)
a + b = (c - d) - 3 ........... (2)

Adding both equations Eq(1) and Eq(2)
(a - b) + (a + b) = (c + d) + 6 + (c - d) - 3
⇒ a - b + a + b = c + d + 6 + c - d - 3
⇒ 2a = 2c + 3
⇒ 2a - 2c = 3
⇒ 2 (a - c) = 3
⇒ a - c = 3/2
∴ a - c = 1.5

Question: If 40 workers working at 80% efficiency can complete a task in 15 days, how many workers working at 100% efficiency would be needed to complete the same task in 10 days?

Solution: 
Case-1: 
W = 40 × 0.8 × 15
= 480

Case-2:
Let the required number of workers be x.
W = x × 1 × 10
480 = 10x
∴ x = 48

Question: Today is Wednesday. After 45 days, it will be:

Solution:
We know that each day of the week is repeated after 7 days.

45 ÷ 7 = 6 (remainder 3)

So, after (7 × 6) = 42 days it will be Wednesday.
∴ after 45 days it will be (Wednesday + 3 days) = Saturday

Let length = x and breadth = y. Then,
2(x + y)= 46 or
x + y = 23 and
x² + y² = (17)²
= 289.

Now, (x + y)² = (23)²
⇒ (x² + y²) + 2xy = 529
⇒ 289 + 2xy = 529
⇒ xy = 120.

∴ Area = xy = 120 cm²

Question: The difference between the average of all prime numbers between 30 and 60 and the average of all prime numbers between 15 and 30 is-

Solution:
The prime numbers between 30 and 60 are 31, 37, 41, 43, 47, 53 and 59.

∴ The average of all prime numbers between 30 and 60 is,
= (31 + 37 + 41 + 43 + 47 + 53 + 59)/7
= 311/7
= 44.43

And,
The prime numbers between 15 and 30 are 17, 19, 23 and 29.

∴ The average of all prime numbers between 15 and 30 is,
= (17 + 19 + 23 + 29)/4
= 88/4
= 22

∴ Required difference = (44.43) - 22 = 22.4

Total distance travelled = (39 + 25)
= 64 km
Total time taken = (45 + 35)
= 80 min.
= (80/60) hr.
= (4/3) hr.
∴ Average speed = {64 × (3/4)} km/hr
= 48 km/hr.
Hence, the average speed of the car is 48 km/hr.

Let, Tanks capacity is x gallons
ATQ,
5x/8 - x/4 = 6
⇒ 3x/8 = 6
∴ x = (6×8)/3 = 16 gallons

Question: If A's income is 20% less than that of B, then by what percent is B's income more than that of A?

Solution:
Let B’s income be Tk. x
∴ A's income = x - 20% of x
= x - 0.20x 
= 0.80x
So, A's icome = 0.80x

Difference between B's and A's income,
= x - 0.80x
= 0.20x

∴ Percentage = (Difference/A's income) × 100%
= (0.20x/0.80x) × 100%
= (1/4) × 100%
= 25%

Question: If 5 ≥ p ≥ - 1 and q ≥ - 1, which of the following cannot be a value of p - q?

Solution: 
Here, 5 ≥ p ≥ - 1 and q ≥ - 1

Now,
i) If, p = - 1 and q = - 1 then, p - q = - 1 - (- 1) = -1 + 1 = 0 
ii) If, p = 2 and q = 1 then, p - q = 2 - 1 = 1 
iii) If, p = 5 and q = 0 then, p - q = 5 - 0 = 5 
iv) If, p = 5 and q = -1 then, p - q = 5 - (- 1) = 5 + 1 = 6

All the given values are possible because the maximum value of p - q is 6, and 0, 1, 5, and 6 are all ≤ 6.

Correct answer: ঙ) None of these

Question: Tickets are numbered from 1 to 30. One ticket is drawn at random. What is the probability that the number is a multiple of 4 or 6?

Solution:
Total tickets = 30

Now, 
Multiples of 4 is-  4, 8, 12, 16, 20, 24, 28 = 7
multiples of 6 is- 6, 12, 18, 24, 30 = 5
And multiples of both (12, 24) = 2

∴ Favorable outcomes = 7 + 5 - 2 = 10

∴ Probability = favorable outcomes/total outcomes 
= 10/30
= 1/3

So the probability is 1/3.

6% হার সুদে 360 টাকার 5 বছরের সুদ = (360 × 5 × 6)/100 = 108 টাকা
∴ 360 টাকার 4 বছরের সুদ = 540 - (360 + 108) = 72 টাকা
∴ প্রথম 4 বছরে সুদের হার ছিল = (100 × 72)/(360 × 4) = 5% [এখানে, r = (100 × 1)/(p × n) সূত্র ব্যবহার করে]

প্রশ্ন: If 5 - 3x ≤ 14, then what is the value of x?

Solution:
5 - 3x ≤ 14
⇒ - 3x ≤ 14 - 5
⇒ - 3x ≤ 9
⇒ 3x ≥ -9 [উভয় পক্ষকে -1 দ্বারা গুণ করলে]
⇒ x ≥ - 9/3
∴ x ≥ - 3

​এই সমাধানটিকে ব্যবধি (interval) আকারে প্রকাশ করলে হয়: [- 3, ∞)
​এখানে তৃতীয় বন্ধনী [ দ্বারা বোঝায় যে - 3 সমাধান সেটের অন্তর্ভুক্ত, এবং ∞ এর পাশে প্রথম বন্ধনী ) বোঝায় যে এটি অসীম পর্যন্ত বিস্তৃত।

Work done = 5/8
Balance work =(1 - 5/8) = 3/8
Let the required number of days be x.
Then,(5/8 : 3/8)= :: 10 : x ⇔ 5/8 × x = 3/8 × 10
x = 3/8 × 10 × 8/5
⇒ x = 6.

Question: Kamal invested Tk. 500 in EBL for 2 years and Tk. 300 in SIBL for 4 years. At the end of the period, he received a total of Tk. 220 as simple interest from both banks. What was the rate of interest per annum?

Solution:
Let the common rate of interest be r % per annum.

We know,
SI = (P × r × t)/100

Interest from EBL,
(500 × r × 2)/100 = (1000r)/100 = 10r

And, 
Interest from SIBL,
(300 × r × 4)/100 = (1200r)/100 = 12r

∴ Total interest received = Interest from EBL + Interest from SIBL
⇒ 220 = 10r + 12r 
⇒ 220 = 22r
⇒ r = 220/22
∴ r = 10

So the common rate of interest is 10% per annum.

Question: X, Y, and Z share Tk. 1680 in such a way that X has 3.5 times as much as Y, and Y has 2 times as much as Z. How much money does Z receive?

Solution:
Given,
X = 3.5Y
Y = 2Z

∴ X = 3.5 × 2Z = 7Z

So, the ratio of X, Y, Z = 7 : 2 : 1

∴ Z receive = [1680 × (1/10)] Tk.
= 168 Tk.

Question: If x + y = 21 and xy = 110, find the value of x² + y² = ?

Solution:
Given that,
x + y = 21 and xy = 110

We know that,
(x + y)² = x² + y² + 2xy
⇒ x² + y² = (x + y)² - 2xy
⇒ x² + y² = 21² - 2 × 110
⇒ x² + y² = 441 - 220
∴ x² + y² = 221

Question: If x is equal to 2 more than the product of 4 and z, and y is equal to 3 less than the product of 5 and z, then 3x is how much greater than 2y when z is 3?

Solution: 
Given:
x = 4z + 2
y = 5z - 3

When z = 3,
∴ x = 4 × 3 + 2 = 12 + 2 = 14
∴ y = 5 × 3 - 3 = 15 - 3 = 12

Now,
∴ 3x = 3 × 14 = 42
∴ 2y = 2 × 12 = 24

∴ Difference = 3x - 2y = 42 - 24 = 18
∴ 3x is 18 greater than 2y.

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6)
So, n (S) = 6
E is the event of getting an odd number.
So, n (E) = 3
Probability of getting an odd number P (E) = (Total number of favorable outcomes)/(Total number of outcomes)
n(E)/n(S) = 3/6
= 1/2

Question: Which of the following fractions is greater than 2/5 and less than 3/4 ?

Solution:
2/5 = 0.40
3/4 = 0.75

Than,   
1/3 = 0.333   
3/8 = 0.375
5/8 = 0.625
4/5 = 0.8

Clearly, 0.625 lies between 0.40 and 0.75

∴ 5/8 lies between 2/5 and 3/4.

Question: ∠P and ∠Q are complementary to each other. If ∠P = 20° + 4x and ∠Q = 6x, find the value of ∠Q.

Solution:
Here,
∠P = 20° + 4x and ∠Q = 6x

For complementary angles,
∠P + ∠Q = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 4x + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
∴ x = 7°

So, ∠Q = 6 × 7° = 42°

Question: A sells an article which costs him Tk. 400 to B at a profit of 20%. B then sells it to C, making a profit of 10% on the price he paid to A. How much dose C pay to B?

Solution:
A এর 20% লাভে বিক্রয়মূল্য = 400 + 400 এর 20%
=  400 + 400 এর 20/100
= 400 + 80
= 480 

A এর বিক্রয়মূল্য = B এর ক্রয়মূল্য

B এর 10% লাভে বিক্রয়মূল্য = 480 + 480 এর 10%
= 480 + 480 এর 10/100
= 480 + 48
= 528

B এর বিক্রয়মূল্য = C এর ক্রয়মূল্য = 528 টাকা

Question: Rifat is twice as old as his friend Sifat. Sifat is 5 years older than Rafi. In 5 years, Rifat will be three times as old as Rafi. How old is Sifat now?

Solution:
Let
Rifat’s age = R
Sifat’s age = S
Rafi’s age = F

Now,
Rifat is twice as old as Sifat
∴ R = 2S
Sifat is 5 years older than Rafi
S = F + 5
∴  F = S - 5

In 5 years, Rifat will be three times as old as Rafi.
That means after 5 years,
R + 5 = 3(F + 5)
⇒ 2S + 5 = 3(S - 5 + 5)
⇒ 2S + 5 = 3S
∴ S = 5

So Sifat is 5 years old now.

Question: If A : B = 3 : 4, B : C = 5 : 6, and C : D = 7: 9 then A : D is equal to-  

Solution:
Given that,
A : B = 3 : 4, B : C = 5 : 6, and C : D = 7: 9
Then,
A/D = (A/B) × (B/C) × (C/D)
= (3/4) × (5/6) × (7/9)
= 35/72
= 35 : 72
∴ A : D = 35 : 72

None of the above
Radius of the circle = r
So hypotenuse = 2r
So, 2r = √(2.4² + 1.8²)
Or, 2r = √9
Or, 2r = 3
the circumference of the circle = 2πr = 3π

Question: Rice and wheat are in a mixture in the ratio 5 : 3. If 16 kg wheat is added to this mixture, the ratio of rice to wheat changes to 5 : 7. How much wheat is in new mixture?

Solution: 
Let the initial amount of rice = 5x kg
Initial amount of wheat = 3x kg

After adding 16 kg wheat,
Rice remains = 5x kg
And wheat becomes = 3x + 16 kg

New ratio of rice : wheat = 5 : 7
So,
⇒ 5x/(3x + 16) = 5/7
⇒ 7 × 5x = 5 × (3x + 16)
⇒ 35x = 15x + 80
⇒ 35x - 15x = 80
⇒ 20x = 80
⇒ x = 80/20
∴ x = 4

Now, wheat in the new mixture = 3x + 16
= 3 × 4 + 16
= 12 + 16
= 28 kg

So, 28 kg of wheat is in the new mixture.