Bank Math

Question: If 1 < p < 4 and 2 < q < 6, which of the following best describes p - q?

Solution: দেয়া আছে:
1 < p < 4 -------------(1)
2 < q < 6 -------------(2)

এখন, আমরা p - q এর সীমা বের করতে চাই।

(2)⇒
2 < q < 6
⇒ - 2 > -q > - 6 (যদি -1 দ্বারা গুণ করি)
⇒ - 6 < -q < - 2 ----------(3)

(1) এবং (3) যোগ করি,
⇒ (1 + (- 6 )) < p - q < (4 + (- 2))
⇒ - 5 < p - q < 2

Let,
The original number of artisans be x.
More artisans, Less days [Indirect proportion]
∴ (x + 8) : x :: 16 : 12
⇒ 12 (x + 8) = 16x
⇒ 12x + 96 = 16x
⇒ 16x - 12x = 96
⇒ 4x = 96
⇒ x = 96/4
⇒ x = 24.

Place value of 6 = 6000
Face value of 6 = 6
Difference = 6000 - 6
= 5994.

Question: A is thrice as good a workman as B and so takes 40 days less than B for doing a job. The time in which they can do the job together is - 

Solution:
Let A alone takes x days 
and B alone takes 3x days to complete the job.

ATQ,
3x - x = 40
⇒ 2x = 40
⇒ x = 20

So, A alone takes 20 days and B alone takes 3 × 20 = 60 days to complete the job.

(A + B)'s 1 days work = (1/20 + 1/60) = 1/15 part

∴ A and B together can do the work in 15 days.

Question: What number should replace the question mark?


Solution:
In the first figure:
(4 × 8) + 7 = 39

In the second figure:
(6 × 3) + 9 = 27

In the third figure:
(9 × 7) + 4 = 67

∴ The number is 67

Question: A sum of money at simple interest amounts to Tk. 815 in 3 years and to Tk. 854 in 4 years. The sum is-

Solution:
Simple interest for 1 year = Tk. (854 - 815)
= Tk. 39

∴ Simple interest for 3 years = Tk.(39 × 3)
= Tk. 117

∴ Sum = (815 - 117)
= Tk. 698

Question: If 12 men can do a piece of work in 14 days, the time taken by 14 men to do the same piece of work will be - 

Solution: 
from 12 men can do a piece of work in 14 days, we get - 
M₁ = 12
D₁ = 14

The time taken by 14 men to do the same piece of work will be
M₂ = 14
D₂ = ?

we know that,
M₁ × D₁ = M₂ × D₂
D₂ = (12 × 14)/14
∴ D₂ = 12 days

Let the speed of the man in still water = x
Given that, speed of the stream = 3 km/h
So, Speed downstream = (x + 3) km/h
and, Speed upstream = (x – 3) km/h
ATQ,
(x + 3) × 1 = (x - 3) × 1.5
Or, x + 3 = 1.5x – 4.5
Or, 0.5x = 7.5
∴ x = 15

Question: A, S and D Started a business with the investment of TK 5000, TK 4000 and TK 3000 respectively. At the end of the year, the profit received by S and D together is TK 2000 more than  the profit received by A. What is the total profit of the business?

Solution:

বিনিয়োগের অনুপাত, A : S : D = 5000 : 4000 : 3000
= 5 : 4 : 3

মনে করি,
মোট লাভ p টাকা
∴ A এর লাভ = 5p/12

S এবং D এর লাভ একত্রে = (4P/12) + (3P/12) = 7p/12

প্রশ্নমতে,
7p/12 = (5p/12) + 2000
or, (7p/12 ) - (5p/12) = 2000
or, (7p - 5p)/12 = 2000
or, 2p/12 = 2000
∴ p = 12000

মোট লাভ 12000 টাকা 

Let x be the required number.
Given, x is as much greater than 36 as is less than 86.
ATQ, x - 36 = 86 - x
Or, x + x = 86 + 36
Or, 2x = 122
Or, x = 122/2 = 61

Question: A fort has enough food for 360 soldiers for 120 days. After 40 days, 120 soldiers leave. For how many more days will the remaining food last.

Solution:
After 40 days: 360 soldiers had food for 80 days. Suppose 240 soldiers had food for x days.
Now, fewer soldiers, more days (Indirect Proportion):
240 : 360 : : 80 : x
⇒ 240/360 = 80/x
⇒ 240x = 360 × 80
⇒ x = (360 × 80)/240
∴ x = 120

Question: An outgoing pipe is attached to a tank that can empty it in 18 hours. An ingoing pipe is connected to the tank that can fill it in 9 hours. How much time will it take to fill the half-full tank?

Solution: 
ingoing pipe in one hour can fill = 1/9
outgoing pipe in one hour can empty = 1/18

in one hour, the total fill up = 1/9 - 1/18 = 1/18

∴ To fill the tank it will take = 18 hours.
∴ To fill half the tank it will take = 18/2
= 9 hours

Let, - 2x – 8 < 9
Or, - 2x < 17
Or, x > -8.5
So, the required value is -8

Question: If sec²θ + tan²θ = 7, then the value of θ when 0° ≤ θ ≤ 90° is?
 
solution: 
Given that, 
sec²θ + tan²θ = 7
⇒ 1 + tan²θ + tan²θ = 7.  ; [sec²θ = 1 + tan²θ]
⇒ 2tan²θ = 6
⇒ tan²θ = 3
⇒ tanθ = √3
⇒ θ = 60∘

Question: Rahim is currently four times as old as Kabir. Five years ago, Rahim was seven times as old as Kabir. What will be the sum of their ages after 12 years?

Solution:
Let Kabir's current age = x years
Rahim's current age = 4x years

5 years ago:
Rahim's age = 4x - 5
Kabir's age = x - 5

According to the question,
4x - 5 = 7(x - 5)
⇒ 4x - 5 = 7x - 35
⇒ - 5 + 35 = 7x - 4x
⇒ 30 = 3x
∴ x = 10

So, Kabir's age = 10 years, Rahim's age = 4 × 10 = 40 years

After 12 years:
Kabir's age: 10 + 12 = 22
Rahim's age: 40 + 12 = 52

∴ Sum: 22 + 52 = 74 

Students like all the three subjects = 60 - (20 + 25 + 30 - 5 - 7 - 8 + 3) = 2

Let the length of the train be x metres and its speed be y m/sec.
Then,
x/y = 1.75
⇒ x = 1.75y

Since the train takes less time to pass a moving object than a stationary object,
it means that the cyclist is moving in a direction opposite to that of the train.

∴ x/(y + 10) = 1.5
⇒ x = 1.5 y + 15
⇒ 1.75 y = 1.5 y + 15
⇒ 0.25 y = 15
⇒ y = 15/0.25
⇒ y = 60.

Length of the train = 1.75 y = (1.75 × 60) m
= 105 m.

Question: In a football team of 12 players, the captain is 30 years old and the goalkeeper is 4 years older than the captain. If these two players are excluded, the average age of the remaining players becomes 2 years less than the team’s overall average age. Find the overall average age of the team?

Solution:
Let the average age of the whole team be x years.

Captain’s age = 30 years
Goalkeeper’s age = 30 + 4 = 34 years

Total age of all 12 players = 12x

After excluding the captain and goalkeeper,
Total age of remaining players = 12x - (30 + 34)

Given that the new average age = (x - 2) years,
Total age of remaining players = 10(x - 2)

So,
12x - 64 = 10(x - 2)
⇒ 12x − 64 = 10x - 20
⇒ 12x - 10x = 64 - 20
⇒ 2x = 44
⇒ x = 22

∴ The average age of the team is 22 years.

Question: In what ratio must a grocer mix two varieties of pulses costing Tk. 15 and Tk. 20 per kg respectively so as to get a mixture worth Tk. 16.50 per kg?

Solution:
Let the ratio be x : y

Cost of first variety = Tk. 15 per kg
Cost of second variety = Tk. 20 per kg 
Cost of mixture = Tk. 16.50 per kg

For x kg of first variety and y kg of second variety:
Total cost of mixture = (x × 15) + (y × 20) = 15x + 20y
Total quantity of mixture = (x + y) kg

According to the question,
(15x + 20y)/(x + y) = 16.50
⇒ 15x + 20y = 16.50(x + y) 
⇒ 15x + 20y = 16.50x + 16.50y 
⇒ 20y - 16.50y = 16.50x - 15x 
⇒ 3.50y = 1.50x 
⇒ y/x = 1.50/3.50 
⇒ y/x = 150/350 
⇒ y/x = 3/7

Therefore, x : y = 7 : 3

বিকল্প সমাধান:
By the rule of alligation,

 ∴ Required rate = 3.50 : 1.50 = 7 : 3

Present age of A + B = 50
A + 4 = C
So, A = C - 4
B = C + 2

Now,
C + 2 + C - 4 = 50
Or, 2C = 52
Or, C = 26

So,
B = 26 + 2
= 28
8 years Ago B age = (28 - 8) = 20 years

Question: In a certain code language, 'BACK' is coded as '89', 'LEGS' is coded as '63'. How will 'CHEST' be coded in that code language?

Solution:
 
Logic: Sum of reverse alphabetical values - 2 = Number.

Now, BACK
B = 25, A = 26, C = 24, K = 16
∴ BACK = 25 + 26 + 24 + 16
= 91
= 91 - 2
= 89 

LEGS
L = 15, E = 22, G = 20, S = 8
∴ LEGS = 15 + 22 + 20 + 8
= 65
= 65 - 2
= 63 

Now for CHEST
C = 24, H = 19, E = 22, S = 8, T = 7
∴ CHEST = 24 + 19 + 22 + 8 + 7
= 80
= 80 - 2
= 78

So the code for CHEST is 78. 

Question: The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in 2.5 hours?

Solution: 
Speed of car = 950/19 
= 50 km/h

Speed of train = 220% × speed of car
= (220/100) × 50
= 110 km/h

Distance covered by the train = 110 × 2.5
= 110 × (25/10)
= 275 km 

Question: Three numbers are in the ratio 2 : 3 : 5, and the sum of their squares is 608. Find the smallest number.

Solution:
Let the numbers be 2x, 3x, 5x

ATQ,
(2x)² + (3x)² + (5x)² = 608
⇒ 4x² + 9x² + 25x² = 608
⇒ 38x² = 608  
⇒ x² = 608/38
⇒ x² = 16 = 4²
∴ x = 4

∴ Smallest number = 2x = 2 × 4 = 8

প্রশ্ন: একটি দড়িকে বৃত্তাকারে বাঁকালে তার ব্যাসার্ধ হয় ৭ সেমি। যদি দড়িটিকে একটি বর্গের আকারে বাঁকানো হয়, তবে এর ক্ষেত্রফল কত হবে?

সমাধান:
এখানে,
বৃত্তের ব্যাসার্ধ, r = ৭ সেমি
∴ বৃত্তের পরিধি = ২πr
= ২ × (২২/৭) × ৭
= ৪৪ সেমি

ধরি,
বর্গের প্রতিটি বাহুর দৈর্ঘ্য = ক সেমি
∴ বর্গের পরিসীমা = ৪ক

এখন,
বৃত্তের পরিধি = বর্গের পরিসীমা
⇒ ৪৪ = ৪ক
⇒ ক = ৪৪/৪ = ১১ সেমি

∴ বর্গের ক্ষেত্রফল = ক² = ১১² = ১২১ বর্গ সেমি

Question: If sinθ = cosθ then what is the value of θ?

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

The age before 10 years increased by 25 % in 10 years.
In other words, the age before 10 years would have increased by 100 % in 40 years.
That is, the age before 10 years is 40 years. Therefore, the present age is 50.

Question: The average mark of 40 students is 62. One mark wrongly entered as 72 instead of 47. Corrected average? 

Solution:
Original total marks
= 62 × 40
= 2480
One student’s mark was recorded as 72, but the correct mark was 47.
So the error = 72 - 47 = 25

Since 25 extra marks were added, subtract them from the total.
= 2480 - 25
= 2455

Corrected average
= 2455/40
= 61.3

Question: Two pipes, Pipe A and Pipe B, can fill a tank in 15 hours and 20 hours, respectively. If both pipes are opened together, after how many hours should Pipe B be closed so that the tank is completely filled in 12 hours?

Solution:
ধরি, ট্যাঙ্কটির ধারণক্ষমতা হলো LCM (15, 20) = 60 ইউনিট
পাইপ A-এর কর্মদক্ষতা = 60/15 = 4 ইউনিট/ঘন্টা
পাইপ B-এর কর্মদক্ষতা = 60/20 = 3 ইউনিট/ঘন্টা
পাইপ A এবং B-এর মিলিত কর্মদক্ষতা = 4 + 3 = 7 ইউনিট/ঘন্টা

ধরি, পাইপ A এবং পাইপ B একত্রে চলে n ঘন্টা।
∴ পাইপ B বন্ধ করার পর পাইপ A একা (12 - n) ঘন্টা চলে।

প্রশ্নানুসারে,
7n + 4(12 - n) = 60
⇒ 7n + 48 - 4n = 60
⇒ 3n + 48 = 60
⇒ 3n = 60 - 48
⇒ 3n = 12
⇒ n = 12/3
⇒ n = 4

সুতরাং, পাইপ B-কে 4 ঘন্টা পর বন্ধ করা উচিত।

Total Distance = 300 + 150 = 450 m
Speed = 60 kmph = 60×(5/18)=(50/3) m/sec
Distance = Speed×Time
450 =(50/3)×Time
Time = 27 seconds

Let,
A's speed = x km/hr.
Then,
B's speed = (7 - x) km/hr.
So,
24/x + 24/(7 - x) = 14
⇒ {24(7 - x) + 24x} = 14x(7 - x)
⇒ 168 - 24x + 24x = 98x - 14x²
⇒ 14x² - 98x + 168 = 0
⇒ x² - 7x + 12 = 0
⇒ x² - 4x - 3x + 12 = 0
⇒ x(x - 4) - 3(x - 4) = 0
⇒ (x - 3) (x - 4) = 0
⇒ x = 3 or x = 4
Since A is faster than B, so A's speed = 4 km/hr and B's speed = 3 km/hr.

Question: What will be the difference between simple and compound interest at 6% on a sum of Tk. 15000 after 2 years?

Solution:
দেওয়া আছে, Principal, P = 15000 টাকা
Rate of interest, r = 6%
Time, n = 2 বছর

Simple Interest (SI):
SI = (P × r × n) / 100
= (15000 × 6 × 2) / 100
= 180000 / 100
= 1800 টাকা

Compound Interest (CI):
CI = P × (1 + r/100)ⁿ - P
= 15000 × (1 + 6/100)² - 15000
= 15000 × (1.06)² - 15000
= 15000 × 1.1236 - 15000
= 16854 - 15000
= 1854 টাকা

∴ Difference between CI and SI:
= 1854 - 1800
= 54 টাকা

Question: A sum of Tk. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is-

Solution:
Let,
C′s share = Tk. x
Then,
B′s share = Tk. x/4,
A′s share = Tk.(2/3) × (x/4) = Tk. x/6

∴ x/6 + x/4 + x = 1360
⇒ 17x/12 = 1360
⇒ x = (1360 × 12)/17
∴ x = Tk. 960

∴ B′s share = Tk.(960/4) = Tk. 240

There are two possible values of k 7 and 14.
So, their summation is 21 and 4 will be the remainder when 21 is divided by 17.

i.e. Average speed = Total distance / Time
Distance =Time x Speed

Total distance covered by Mithila = Distance covered in first 4 hours + distance covered in next 6 hours
= (80 x 4) + (30 x 6)
= 500 miles / hr

Total time taken to complete the journey = 4 + 6 = 10 hrs

Therefore,
Average speed = Total Distance/Time
= 500 / 10
= 50 miles/hr

Let,
the original price be Tk. 100
New final price = 120% of (Tk. 125% of 100)
= Tk. {(120/100) × (125/100) × 100)}
= Tk. 150
∴ Increase = (150 - 100)%
= 50%.