Bank Math

Question: A rectangle has a diagonal length of 14 meters and a width of 12meters. What is the area of the rectangle in square meters?

Solution:
ধরি,
 


আয়তক্ষেত্র ABCD এর কর্ণের দৈর্ঘ্য AC = 14 মিটার এবং প্রস্থ AB = 12 মিটার
∴ দৈর্ঘ্য, BC = √(14² - 12²) মিটার
=√(196 - 144) মিটার
= √52 মিটার
= 2√13 মিটার

আয়তক্ষেত্রের ক্ষেত্রফল = (2√13 × 12) বর্গমিটার
= 24√13 বর্গমিটার

Question: The radius of a car wheel is 28 cm. How many revolutions will it make to cover a distance of 13.2 kilometers?

Solution:
আমরা জানি, 
চাকার পরিধি = 2πr
= 2 × (22/7) × 28
= 2 × 4 × 22
= 176 সে. মি.

মোট দূরত্ব = 13.2 কি. মি.
= 13.2 × 1000 × 100) সে. মি. [১ কি. মি. = 1000 মি., ১ মি = 100 সে. মি.] 
= 1320000 সে. মি.

∴ ঘূর্ণন সংখ্যা = 1320000/176
= 7500 টি

Question: If 2A = 3B = 4C, Than, A : B : C is.

Solution:
Let
2A = 3B = 4C = x
2A = x
A = x/2

3B =x
B = x/3

4C = x
C = x/4

A : B : C =  x/2 : x/3 : x/4
= (x/2) × 12 : (x/3) × 12 : (x/4) × 12
= 6 : 4 : 3

A Total Profit = Tk 3600
Profit ratio,
A : B = 5 : 4
B : C = 8 : 9
As B is common in both ratio, we make B equal in both ratio by multiplying One B in another.
A : B = 5 : 4 × 8
B : C = 8 × 4 : 9
So, ratio of
A : B : C = 40 : 32 : 36 = 10 : 8 : 9
Now,
C shares in profit = (3600×9)/27
= Tk 1200

Question: A, B, and C invest in a business in the ratio 2 : 1 : 3. If the total profit is Tk 30,000, how much more did C get than A?

Solution:
Let the invested shares ratio = 2x : x : 3x

Now
2x + x + 3x = 30000
Or, 6x = 30,000
∴ x = 5,000

∴ A’s share = 2 × 5,000 = 10,000
∴ C’s share = 3 × 5,000 = 15,000
∴ Difference = 15,000 - 10,000
= Tk 5,000

∴ C gets 5000 Tk more than A.

Question: What is the sum of the following sequence: 5, 12, 19, 26, ... , 54?

Solution:
এটি একটি সমান্তর ধারা (arithmetic series)।
প্রথম পদ, a = 5
সাধারণ অন্তর, d = 12 - 5 = 7
শেষ পদ= 54

আমরা জানি,
n তম পদ = a + (n - 1)d
⇒ 54 = 5 + (n - 1)7
⇒ 49 = 7(n - 1)
⇒ n - 1 = 7
⇒ n = 8

সমষ্টি, S_n = n/2{2a + (n - 1)d}
∴ S₈ = (8/2){2(5) + (8 - 1)7}
= 4{10 + (7 × 7)}
= 4{10 + 49}
= 4 × 59
= 236

অতএব, প্রদত্ত ধারাটির সমষ্টি হলো 236

Question: In how many ways can a group of 2 teachers and 5 students be formed from 5 teachers and 8 students?

Solution:
We have 5 teachers and 8 students.
We need to choose 2 teachers from 5 and 5 students from 8.

∴ Number of ways = ⁵C₂ × ⁸C₅
= {5!/(2!(5 - 2)!)} × {8!/(5!(8 - 5)!)}
= {5!/(2!×3!)} × {8!/(5!×3!)}
= {(5×4)/(2×1)} × {(8×7×6)/(3×2×1)}
= 10 × 56
= 560 ways

ধরি, আয়তক্ষেত্রটির দৈর্ঘ্য x একক এবং প্রস্থ y একক
সুতরাং ক্ষেত্রফল = xy বর্গ একক
৫০% বৃদ্ধিতে নতুন দৈর্ঘ্য = x + (xএর৫০%) = x + x/2 = ৩x/2
৫০% বৃদ্ধিতে নতুন প্রস্থ = y + (yএর ৫০%) = y + y/2 = ৩y/2
নতুন ক্ষেত্রফল = (৩x/2) × (৩y/2) = ৯xy/৪
ক্ষেত্রফল বৃদ্ধি পেয়েছে = ৯xy/৪ - xy = ৫xy/৪
∴ ক্ষেত্রফল শতকরা বৃদ্ধি পেয়েছে = (৫xy × ১০০)/৪xy = ১২৫% 

শর্টকাটঃ ক্ষেত্রফল বৃদ্ধি পাবে = [50 + 50 + (50×50 / 100)] % = 125%

Question: X can complete a work in 12 days, and Y alone can do it in 18 days. They work together for 6 days, and Z completes the remaining work in 3 days. If the total payment for the work is Tk. 600, how much should Z get?

সমাধান:
X-এর একদিনের কাজ = 1/12
Y-এর একদিনের কাজ = 1/18

X ও Y একসাথে ৬ দিন কাজ করে:
= 6 × (1/12 + 1/18)
= 6 × {(3 + 2)/36} = 6 × (5/36) = 30/36 = 5/6

∴ বাকি কাজ = 1 − 5/6 = 1/6

Z এই 1/6 কাজ ৩ দিনে করেছে, অর্থাৎ Z-এর কাজ = 1/6

X-এর কাজ = 6 × 1/12 = 1/2
Y-এর কাজ = 6 × 1/18 = 1/3
Z-এর কাজ = 1/6

তাহলে অনুপাত = 1/2 : 1/3 : 1/6
= 3 : 2 : 1

মোট টাকা = 600
Z-এর অংশ = 1/(3+2+1) = 1/6
∴ Z পাবে = 600 × (1/6) = 100 টাকা

Given amount = Tk. 1348.32
Principle = Tk. 1200
And time = 2 years

According to the law,
A = P(1 + R/100)ⁿ
1348.32 = 1200(1 + R/100)²
1348.32/1200 = (1 + R/100)²
11236/10000 = (1 + R/100)²
(106/100)² = (1 + R/100)²
(1 + 6/100)² = (1 + R/100)²
R = 6% per annum.

Let, A³ = 27 = 3³
So, A = 3
and, A² = 9

Question: The sides of a triangular field are 13 m, 14 m, and 15 m respectively. What is its area?

Solution:
Let the sides of the triangle are,
a = 13 m, b = 14 m, c = 15 m

We know,
Semi-perimeter, s = (a + b + c)/2
= (13 + 14 + 15)/2
= 42/2
= 21 m

We know the formula for the area of a triangle, 
Area = √[s(s - a)(s - b)(s - c)]
= √[21(21 - 13)(21 - 14)(21 - 15)]
= √[21 × 8 × 7 × 6]
= √[21 × 8 × 42]
= √7056
= 84

Therefore, the area of the triangular field is 84 square meters.

Question: A worker union contract specifies a 6% salary increase plus a Tk. 450 bonus for each worker. For a worker, this is equivalent to an 8% salary increase. What was this worker's salary before the new contract?

Solution:
ধরি, কর্মীর পূর্বের বেতন = x টাকা।

6% বৃদ্ধিতে বেতন = x + x এর 6%
= x + (6x/100) = 106x/100

বোনাস হিসেবে 450 টাকা যোগ করলে মোট বেতন = (106x/100) + 450

8% বৃদ্ধিতে বেতন = x + x এর 8%
= x + (8x/100) = (108x/100)

প্রশ্নমতে,
(106x/100) + 450 = (108x/100)
⇒ 450 = (108x/100) - (106x/100)
⇒ 450 = (2x/100)
⇒ x = (450 × 100)/2
∴ x = 22500

অর্থাৎ, কর্মীর পূর্ববর্তী বেতন ছিল 22500 টাকা।

Question: If the volume of a sphere is 288π cm³, what is the surface area of the sphere?

Solution: 
Given that the volume, V = 288π cm³
or, (4/3)πr³ = 288π
or, r³ = 216 
∴ r = 6 cm

Surface area of a sphere, A = 4πr²
= 4π(6)²
= 144π cm²

Question: In a class of 100 students, 55 are taking Biology, 35 are taking Chemistry and 10 are taking both courses. How many students are not enrolled in either course?

Solution:
দেওয়া আছে,
মোট শিক্ষার্থীর সংখ্যা, n(U) = 100
জীববিজ্ঞান নেওয়া শিক্ষার্থীর সংখ্যা, n(B) = 55
রসায়ন নেওয়া শিক্ষার্থীর সংখ্যা, n(C) = 35
উভয় বিষয় নেওয়া শিক্ষার্থীর সংখ্যা, n(B ∩ C) = 10

কমপক্ষে একটি বিষয় নেওয়া শিক্ষার্থীর সংখ্যা, n(B ∪ C) = n(B) + n(C) - n(B ∩ C)
⇒ n(B ∪ C) = 55 + 35 - 10
⇒ n(B ∪ C) = 90 - 10
⇒ n(B ∪ C) = 80

কোনোটিই নেয়নি এমন শিক্ষার্থীর সংখ্যা = n(U) - n(B ∪ C)
= 100 - 80
= 20

সুতরাং, 20 জন শিক্ষার্থী কোনো কোর্সই গ্রহণ করেনি।

So the maximum value of a² b² c² = (1/3 × 1/3 × 1/3) = 1/27
( When the sum of the three positive quantities is fixed, the product will be maximum when the quantities are equal)
Hence, the maximum value of ABC = 1/√27
= 1/(3√3)

log_x125 = 3
Or, x³ = 125 = 5³
Or, x = 5

Question: A square field is surrounded by a path of uniform width 4 meters. If the area of the path is 192 square meters, find the side length of the field.

Solution:
Let the side of the field = x meters.
Then, the side of the field including the path = x + (2 × 4)
= x + 8 meters.

Area of path = Area of field with path - Area of field
⇒ 192 = (x + 8)² - x²
⇒ 192 = x² + 16x + 64 - x²
⇒ 192 = 16x + 64
⇒ 16x = 192 - 64
⇒ 16x = 128
⇒ x = 128/16
⇒ x = 8 meters

∴ Therefore, the side length of the field is 8 meters.

Question: A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

Solution:
Let man's rate upstream be x kmph.
Then, his rate downstream = 2x kmph.

∴ (Speed in still water) : (Speed of stream) = {(2x + x)/2} : {(2x - x)/2}
= (3x/2) : (x/2)
= 3 : 1

Let the total value be Tk. X.
The value of 3/4 th = Tk. 3X/4 then the value of 1/4 th = Tk. X/4
Since he made a profit of 10% on 3X/4 and loss 2% on X/4
Selling price = Tk. [110% of 3X/4] + [98% of X/4]
= Tk. 330X/400 + 98X/400
= Tk. 428X/400
Since the total cost price is Tk. X and the total profit is Tk. 1500 then
Tk. {(428X /400) - X} = Tk. 1500
⇒ {(428X - 400X)/400} = 1500
⇒ 28X = 1500 x 400
⇒ 28X = Tk. 6,00,000
⇒ X = 6,00,000/28 = Tk. 21428.57
= Tk. 21428.57
Hence the answer is Tk. 21428.57

Question: The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

Solution:
The number leaves a remainder 8 when divided by 12, 15, 20 and 54.
So the required number = LCM(12, 15, 20, 54) + 8

Now, 
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3

∴ LCM(12, 15, 20, 54) = 540

∴ Required Number = 540 + 8 = 548 

Question: When 40% of the first number is added to the second number, the second number becomes 7/5 times the first number. What is the ratio of the first number to the second number?

Solution:
Let the first number = x
and the second number = y.

According to the question,
y + 40% of x = (7/5)x
⇒ y + (40/100) x = (7/5)x
⇒ y + (4/10)x = (7/5)x
⇒ y = (7/5)x - (4/10)x
⇒ y = (14 - 4)x/10
⇒ y = 10x/10
⇒ y = x

Therefore, x : y = x : x = 1 : 1

Working 5 hours a day, A can complete the work in 8 days i.e.
= 5 × 8 = 40 hours
Working 6 hours a day, B can complete the work in 10 days i.e.
= 6 × 10 = 60 hours
(A + B)'s 1 hour's work,
= 1/40 + 1/60 = (3+2)/ 120
= 5/120
= 1/24
Hence, A and B can complete the work in 24 hours i.e. they require 3 days to complete the work.

Question: In a selection process, the ratio of applicants who were selected to those who were rejected is 10 : 4. If 420 applicants were rejected, what was the total number of applicants?

Solution:
Given ratio,
Selected : Rejected = 10 : 4
This means for every 10 selected applicants, 4 are rejected.

Let, Number of selected applicants = 10k
Number of rejected applicants = 4k

According to the question,
Rejected applicants = 420
So, 4k = 420
⇒ k = 420/4
∴ k = 105

∴ Selected applicants = 10k = 10 × 105 = 1050
∴ Rejected applicants = 4k = 420 (given)

∴ Total number of applicants = Selected + Rejected
= 1050 + 420
= 1470

So the total number of applicants was 1470.

0.001/(0.1×0.1)
= 0.001/0.01
= 100/1000
= 0.1

Question: The average weight of three friends is 35 kg. None of the friends weighs less than 33 kg. What can be the maximum weight of any of the three friends?

Solution:
Here,
The average weight of three friends is 35 kg

∴ Total weight of three friends = (35 × 3) kg 
= 105 kg

Minimum weight of two friends (33 × 2) kg
= 66 kg

∴ The maximum weight of any three friends is (105 - 66) kg
= 39 kg

Let the speed of first car = 7x
2nd car = 8x.

According to the question
8x = 200/5 = 40 [using speed = distance time]
x = 40/8 = 5

Speed of first car = 7x = 7 X 5 = 35 Km/hr

Let, Length = a, width = b
So, area, ab = 20

ATQ,
diagonal = √(a² + b²) = √41
⇒ a² + b²= 41
⇒ (a + b)² – 2ab = 41
⇒ (a + b)² – 2×20 = 41
⇒ (a + b)² = 41 + 40 = 81
⇒ a + b = 9
∴ Perimeter = 2(a + b) = 2×9 = 18 cm

Question: what is the angle between the hour and minute hands of a clock when it is 3 : 20?

Solution:
3 টা 20 মিনিট = 3 + (20/60) ঘণ্টা
= 3 + 1/3 = 10/3 ঘণ্টা

আমরা জানি,
ঘণ্টার কাঁটা 12  ঘণ্টায় 360° ঘোরে
ঘণ্টার কাঁটা 1  ঘণ্টায় = 360°/12 = 30°
ঘণ্টার কাঁটা 10/3 ঘণ্টায় = 30° × 10/3 = 100° ঘোরে

আবার,
মিনিটের কাঁটা 60 মিনিটে ঘোরে 360°
মিনিটের কাঁটা 1 মিনিটে ঘোরে 360°/60 = 6°
মিনিটের কাঁটা 20 মিনিটে ঘোরে (6 × 20) = 120° ঘোরে

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = |120° - 100°| = 20°

আমরা জানি,
রম্বসের ক্ষেত্রফল = 1/2 × কর্ণদ্বয়ের গুণফল ।
অর্থ্যাৎ 1/2 × 16 × অপর কর্ণ = 96
⇒ অপর কর্ণ = (96 × 2 )/16 c.m.
⇒ অপর কর্ণ = 12 c.m.
Answer: অপর কর্ণের দৈর্ঘ্য 12 c.m.