Bank Math

Question: A man travelled a distance of 1260 km in 17 hours. He travelled partly by car at a speed of 40 km/h, and partly by train at a speed of 80 km/h. What is the distance traveled by the train?

Solution: 
Given,
 Total distance = 1260 km
Total time = 17 hours
Speed of car = 40 km/h
Speed of train = 80 km/h

Let,
Distance travelled by car = x km
Distance travelled by train = (1260 - x) km

ATQ,
(x/40) + [(1260 - x)/80] = 17
⇒ 2x + 1260 - x = 17 × 80
⇒ x = 1360 - 1260
⇒ x = 100

∴ Distance traveled by the train = (1260 - 100) km
= 1160 km

Members in A = 71, B = 18, C = 53

Average spent per member: A = Tk. 397, B = Tk. 421 and C = Tk. 137

Now,
Total amount by A = 71 × 397 = 28187
Total amount by B = 18 × 421 = 7578
Total amount by C = 53 × 137 = 7261

∴ Total amount = 28187 + 7578 + 7261 = 43026

∴ Total members = 71 + 18 + 53 = 142

∴ Average = 43026/142 = 303

Total average amount spent per member = Tk. 303

Question: How much should be added to each term of 4 : 7 so that it becomes 2 : 3?

Solution:
Given that,
Ratio of two numbers is 4 : 7
Let the number added to denominator and numerator be 'x' 

Now according to the question,
(4 + x) : (7 + x) = 2 : 3
⇒ (4 + x)/(7 + x) = 2/3
⇒ 12 + 3x = 14 + 2x
∴ x = 2 

∴ 2 will be added to make the term in the ratio of 2 : 3.

Given,
Q = 2/3(P + R)
Or, 3Q = 2(P + R)
Again,
P = (Q + R)/2
Or, 2P = Q + R
Now,
P + Q + R = 366
Or, P + 2P = 366
Or, 3P = 366
∴ P = 122

Question: Two trains running in opposite directions cross a man standing on the platform in 17 seconds and 7 seconds respectively and they cross each other in 13 seconds. The ratio of their speeds is-

Solution: 
Let, the speed of the trains are x m/sec and y m/sec respectively

We know, Distance/Length = Speed × Time
∴ Length of first train = 17x meter
And, length of second train = 7y meter

Now, Total time = Total length/Total speed
(17x + 7y)/(x + y) = 13
⇒ 17x + 7y = 13x + 13y 
⇒ 17x - 13x = 13y - 7y
⇒ 4x = 6y
⇒ x/y = 3/2
∴ x ∶ y = 3 ∶ 2

Question: In a geometric progression, the 4^th term is 16 and the 7^th term is 128. Find the 10^th term.

Solution:
Let the first term = a
Common ratio = r
We know,
n-term = ar^{n - 1}

Then,
4^th term, ar³ = 16  ........(1)  
7^th term, ar⁶ = 128  ........(2)

Now, divide equation (2) by equation (1) then we get,
(ar⁶)/(ar³) = 128/16  
⇒ r³ = 8  
⇒ r³ = 2³  
∴ r = 2  

Then substitute r = 2 into equation (1) 
a (2)³ = 16   
⇒ a × 8 = 16  
∴ a = 2

Now, 10^th term
= ar⁹  
= 2 × 2⁹ 
= 2 × 2⁹  
= 2¹⁰ 
= 1024

∴ The 10th term is 1024

Question: A train running at the speed of 60 kmph crosses a 200 m long platform in 27 seconds. What is the length of the train?

Solution: 
Given that,
Time = 27 sec
∴ Speed = (60 × 5/18)m/sec = 50/3 m/sec

Let the length of the train be x metres.
Then,
(x + 200)/27 = 50/3
⇒ x + 200 = (50/3) × 27 
⇒ x + 200 = 450
⇒ x = 450 - 200
∴ x = 250 metres

So the length of the train is 250 meters.

Question: If (x/y) > 0, which of the following must be true?

Solution:
যেহেতু (x/y) > 0 তাই
x ও y এর দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক হবে। 
আমরা জানি
 দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক সংখ্যার গুণফল ধণাত্মক হয়।
( - x) ( - y) = xy
x × y = xy

i) xy > 0 অবশ্যই ধণাত্মক

Question: Rima's height is 5.4". Eva is taller than Rima, but she is not taller than Promi. Promi is shorter than her cousin Rakib, but she is not shorter than Rima. Who is the tallest in the group?

Solution: 
Rima's height = 5'4"

Eva > Rima
Eva's height > 5'4"

Eva is not taller than Promi.
So, height of Promi > 5'4" + x

Let
Promi's height = (5'4" + x ) + y
Promi < Rakib.

Rakib's height >( 5'4" + x) + y

Rakib > Promi > Eva > Rima
Therefore, Rakib is the tallest.

15min = 1/4hrs
1 hr → 5 kms
1/4hr → 5/4 kms
So, length of the bridge=5/4 km = 1250 metres

Question: In how many different ways can the letters of the word "JASHORE" be arranged so that the vowels always come together?

Solution:
The word J A S H O R E = 7 letters
There, Vowels = A, O, E (3 vowels)
Consonants = J, S, H, R (4 consonants)

Now,
Treat the 3 vowels (A, O, E) as a single unit.
So now we have, [A O E]  J  S  H  R
Total units to arrange = 5 (1 vowel block + 4 consonants)

Number of ways to arrange these 5 units = 5! = 120 ways

And
The 3 vowels (A, O, E) can be arranged among themselves in = 3! = 6 ways

Total number of arrangements = (ways to arrange the 5 units) × (ways to arrange vowels inside the block)
= 120 × 6
= 720

So there are 720 different ways to arrange the letters of 'JASHORE' such that the vowels always come together.

Question: The average weight of 12 students in a class is 45.5 kg. What should be the weight of a 13th student so that the average weight of all 13 students becomes 47.2 kg?

Solution:
Average weight of 12 students = 45.5 kg
∴ Total weight of 12 students = (45.5 × 12) kg
= 546 kg

Again,
Average weight of 13 students = 47.2 kg
Total weight of 13 students = (47.2 × 13) kg
= 613.6 kg

∴ Weight of the 13th student = (613.6 - 546) kg
= 67.6 kg

Therefore, the weight of the 13th student should be 67.6 kg.

• Question: In an examination, 36% are pass marks. If an examine gets 17 marks and fails by 10 marks, what is the maximum mark?

Solution:
Pass mark = (17 + 10) = 27

Let maximum marks be x

Then 36% of x = 27
Or, 36x/100 = 27
Or, 36x = 2700

Hence, x = 75

∴ The maximum mark is 75

Question: If x² + yz + zx + xy is divided by x + z, the result is-

Solution:
x² + yz + zx + xy
= x² + xy + zx + yz
= x(x + y) + z(x + y)
= (x + y)(x + z)

∴ divided by (x + z) the result = (x + y)(x + z)/(x +z) = (x + y)

Question: The sum of the squares of three numbers is 120, and the sum of their products taken two at a time is 140. What is the sum of the three numbers?

Solution:
Let the numbers be a, b, c.
Given:
a2+ b2 + c2 = 120, 
ab + bc + ca = 140,
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 120 + 2 × 140 = 120 + 280 = 400

∴ a + b + c = √400 = 20

মনে করি,
z এর আয় = p টাকা
y এর আয় = 4p টাকা
x এর আয় = 2.5×4p টাকা = 10p টাকা

∴ x : y : z = 10p : 4p : p
= 10 : 4 : 1
তাহলে, z এর আয় = 1/(10 + 4 + 1) × 1800
= 1/15 × 1800
= 120 টাকা

Pillar above the water is 12 feet = 1 - (1/2 + 1/3) = 1/6
So, the length of the pillar is = 12 × 6 = 72 feet

Question: At the rate of 15% simple interest per annum, how much profit will be earned on Tk. 900 as principal in 7 years?

Solution:
Given,
Sum P = Tk. 900
Time n = 7 years
Rate r = 15%
= 15/100
= 3/20

We know,
Profit = Prn
= 900 × 3/20 × 7
= 945

Question: The average of 60 numbers is 25. If two numbers, namely 30 and 40, are discarded, what is the average of the remaining numbers?

Solution:
ATQ,
The average of 60 numbers is = 25
The sum of 60 numbers is = 25 × 60 = 1500
The two numbers discarded = 30 + 40 = 70
The sum of the remaining 58 numbers = 1500 - 70 = 1430

∴ The new average = 1430/58 = 24.66 (approximately)

Question: The average of a non-zero number and its square is 11 times the number. The number is-

Solution:
Let the number be x (x ≠ 0).

According to the question,
The average of the number and its square is 11 times the number.
⇒ (x + x²)/2 = 11x
⇒ x + x² = 22x
⇒ x² + x - 22x = 0
⇒ x² - 21x = 0
⇒ x(x - 21) = 0
So, x = 0  or  x = 21
But the number is non-zero, so we discard x = 0.

Therefore, the number is 21.

Number of red balls = 4
Number of yellow ball = 5
Number of pink ball = 6
Total number of balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls = ¹⁵C₂
= 15!/2!(15 - 2)!
= 15!/(2! × 13!)
= (15 × 14)/(1 × 2)
= 105.
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
¹⁰C₂
= 10!/2!(10 - 2)!
= 10!/2!8!
= (10 × 9)/(1 × 2)
= 45.
∴ Required Probability = 45/105
= 3/7

L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
=22×22×32×52
=3600

Question: If sec²θ + tan²θ = 5/3, then what is the value of tan2θ?

Solution:
We know,
sec²θ = 1 + tan²θ

Given that,
⇒ sec²θ + tan²θ = 5/3
⇒ 1 + tan²θ + tan²θ = 5/3
⇒ 2tan²θ = 2/3
⇒ tanθ = 1/√3
⇒ θ = 30°

∴ tan(2θ) = tan(2 × 30°) = tan60° = √3

Question: If log_x (64/125) = - 3, then x = ?
(Janata RC 2022 অনুযায়ী) 

Solution:
log_x (64/125)  = - 3
⇒ x^{- 3} = 64/125
⇒ x^{- 3} = (4/5)³
⇒ x^{- 3} = 1/(5/4)³
⇒ x^{- 3} = (5/4)^{- 3}
⇒ x = 5/4

Question: 


Solution:

Question: A triangular garden has sides measuring 36 meters, 40 meters, and 32 meters. A fence is to be constructed around the garden, with pillars placed at intervals of 4 meters. How many pillars will be required to completely surround the garden?

Solution:
Perimeter of the triangle = 36 + 40 + 32 = 108 meters
Distance between pillars = 4 meters

Number of pillars required = Perimeter ÷ Distance
= 108 ÷ 4 
= 27
= 27 pillars

∴ 27 pillars are needed to surround the triangular garden.

Question: Pipe B takes 12 hours to fill a tank by itself, while pipe A works three times faster. If both pipes are opened together, how many hours will they need to fill the tank?

Solution:
B নল দ্বারা চৌবাচ্চা পূর্ণ হয় = 12 ঘণ্টায় 
∴ 1 ঘণ্টায় পূর্ণ হয় = 1/12 অংশ

A নল দ্বারা পূর্ণ হয় = 12/3 = 4 ঘণ্টায় 
∴ 1 ঘণ্টায় পূর্ণ হয় = 1/4 অংশ

দুইটি নল দ্বারা একত্রে 1 ঘণ্টায় পূর্ণ হয় = (1/12) + (1/4)
= (1 + 3)/12
= 4/12
= 1/3 অংশ 

দুইটি নল দ্বারা একত্রে,
1/3 অংশ পূর্ণ হয় = 1 ঘণ্টায়
∴ 1 অংশ পূর্ণ হয় = 1 × 3 = 3 ঘণ্টায় 

Question: Two containers contain milk and water in the ratios 5 : 2 and 9 : 5. What ratio should the mixtures be combined in to achieve a final ratio of 2 : 1 milk to water?

Solution:
Let,
P unit of the first mixture is added to Q unit of the second mixture.

So, in the P unit of the first mixture,
Amount of milk present = (5/7) × P = 5P/7
Amount of water present = (2/7) × P = 2P/7

In the Q unit of the second mixture,
Amount of milk present = (9/14) × Q = 9Q/14
Amount of water present = (5/14) × Q = 5Q/14

ATQ,
{(5P/7) + (9Q/14)}/{(2P/7) + (5Q/14)} = 2/1
⇒ {(10P + 9Q)/14}/{(4P + 5Q)/14} = 2
⇒ 10P + 9Q = 8P + 10Q
⇒ 2P = Q

∴ P : Q = 1 : 2

- 2 + (-2) - {-(2)} - 2
= - 2 - 2 + 2 - 2
= - 4

Here,                    A   :   B    :   C
Ratio of Profit → 2   :   3    :    7
Average gain= (2+3+7)/3 = 4 units
According to the question,
4 units= Tk. 8000
1 unit= Tk. 2000
3 units=3×2000= Tk. 6000
∴Share of B= Tk. 6000

Question: If the n^th terms of an arithmetic progression is 4n + 1, then the common difference is-

Solution:
The n^th terms of an arithmetic progression is 4n + 1
n = 1 then, T₁ = 4 . 1 + 1 = 5
n = 2 then, T₂ = 4 . 2 + 1 = 9
n = 3 then, T₃ = 4 . 3 + 1 = 13
n = 1 then, T₄ = 4 . 4 + 1 = 17
.....................................................
.....................................................

The common difference 
 T₂ - T₁ = 9 - 5 = 4
 T₃ - T₂ = 13 - 9 = 4
 T₄ - T₃ = 17 - 13 = 4

∴ The common difference is 4.

Question: The sum of the squares of three numbers is 123, and the sum of their products taken two at a time is 119. What is the sum of the three numbers?

Solution:
Let the three numbers be a, b, and c
Then,
a² + b² + c² = 123
ab + bc + ca = 119

Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 123 + (2 × 119) 
⇒ (a + b + c)² = 361
⇒ (a + b + c) = √361
∴ (a + b + c) = 19

Question: The investment ratio of two partners, P and Q, is 7 : 9, while their profit ratio is 14 : 27. Given that P invested his funds for 6 months, determine the duration of Q's investment.

Solution:
Let P invested Tk 7x for 6 months
Q invested Tk 9x for y months

Now,
(7x × 6) : (9x × y) = 14 : 27
⇒ (7x × 6)/(9x × y) = 14/27
⇒ 42x/9xy = 14/27
⇒ 42 × 27 = 14 × 9y
⇒ 1134 = 126y
⇒ y = 1134/126
∴ y = 9

So, Q invested for 9 months.

Question: The number of distinct permutations of the letters of the word "AMERICA" is how many times that of the word "CANADA"?

Solution:
In the word AMERICA there are 7 letters in total,
with A appearing 2 times and all other letters appearing 1 time each.

∴ Number of distinct permutations = 7!/2!
= (7 × 6 × 5 × 4 × 3 × 2 × 1)/2
= 7 × 6 × 5 × 4 × 3
= 2520

Again,
In the word CANADA there are 6 letters in total,
with A appearing 3 times and all other letters appearing 1 time each.

∴ Number of distinct permutations = 6!/3!
= (6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 6 × 5 × 4
= 120

Therefore, the number of arrangements of AMERICA is 2520/120 = 21 times the number of arrangements of CANADA.

So the number of distinct permutations of 'AMERICA' is 21 times the number of distinct permutations of 'CANADA'.

Question: Mr. Ali is a trader. He mixes 26 kg of rice at Tk. 20 per kg with 30 kg of rice of other variety at Tk. 36 per kg and sells the mixture at Tk. 32 per kg. His profit percent is-

Solution:
Cost Price of 56 kg rice = {(26 × 20) + (30 × 36)}
= (520 + 1080)
= 1600 taka

Selling Price of 56 kg rice = (56 × 32)
= 1792 taka

∴ Profit = 1792 - 1600
= 192 taka

∴ Profit percentage = (192/1600) × 100%
= 12%

By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)
Hence, his time originally must have been 30 minutes.
Thus required distance = 20 kmph × 0.5 hours = 10 km.