Bank Math

Question: A student got twice as many sums wrong as he got right. If he attempted 42 sums in all, how many did he solve correctly?
Solution:
Let he has solved correctly X no. of sums.
Therefore incorrect no. of sums = 2X
Now,
X + 2X = 42
⇒ 3X = 42
⇒ X = 14

∴ 14 sums he has done correctly.

Question: If the area of a right triangle is 6 square meters and the hypotenuse is 5 meters, what is the perimeter of the triangle? 

Solution: 
যেহেতু ত্রিভুজটির অতিভুজ = 5 মিটার
∴ ত্রিভুজটি সমকোণী ত্রিভুজ।

ধরি, ত্রিভুজের দুই বাহু যথাক্রমে x ও y
∴ x² + y² = 5² = 25

আবার, ক্ষেত্রফল, (1/2)xy = 6
⇒ xy = 12

আমরা জানি, (x + y)² = x² + y² + 2xy
= 25 + (2 × 12)
= 25 + 24 
= 49
x + y = 7

পরিসীমা = (x + y + 5)
= 7 + 5
= 12 মিটার

Hand of minute moves 360° in 60 minutes
In 20 minutes it will move = (360° × 20) / 60 = 120°

We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Simple interest is the same as the rate of interest.
Hence,
Rate of interest = R% and Time = R years

S.I. = (P × R × R)/100
⇒ 60 = (1500 × R²)/100
⇒ 15R²= 540
⇒ R² = 36
⇒ R=6 %

Rate of Interest = 6%.

Question: If a regular polygon has each of its interior angle equal to 175°, then find the number of sides of the polygon.

Solution:
Given that,
Each interior angle = 175°

We know,
Sum of interior angles = (n - 2) × 180
Each interior angle = Sum of interior angles ÷ n

Now,
175 = [(n - 2) × 180] ÷ n
⇒ 175 × n = (n - 2) × 180
⇒ 175n = 180n - 360
⇒ 180n - 175n = 360
⇒ 5n = 360
⇒ n = 360/5
∴ n = 72

So, the polygon has 72 sides.

Let the products actual price be x
ATQ, 76% of x = 912
Or, 76x/100 = 912
Or, x = (912 × 100) / 76
= 1200

Question: Two pipes X and Y can fill a cistern in 10 and 15 hours respectively. Both pipes are opened together. After how many hours should pipe X be turned off so that the cistern is filled in 9 hours?

সমাধান:
ধরি, মোট সময় 9 ঘন্টা পর চৌবাচ্চাটি পূর্ণ হয়। এই সম্পূর্ণ সময়ে কেবল নল Y খোলা ছিল।

নল Y, 15 ঘন্টায় চৌবাচ্চাটি পূর্ণ করতে পারে।
∴ 1 ঘন্টায় Y পূর্ণ করে 1/15 অংশ।
∴ 9 ঘন্টায় Y পূর্ণ করে = 9/15 অংশ
= 3/5 অংশ।

অবশিষ্ট অংশ যা X পূর্ণ করেছিল = 1 - 3/5 অংশ
= 2/5 অংশ।

নল X, 10 ঘন্টায় পূর্ণ করে 1 অংশ।
∴ 1 অংশ পূর্ণ করে 10 ঘন্টায়।
∴ 2/5 অংশ পূর্ণ করে = (10 × 2/5) ঘন্টা
= 4 ঘন্টা।

∴ নল X কে 4 ঘন্টা পর বন্ধ করতে হবে।

Cost of a CD = 200/10 = 20
And, cost of a DVD = 20 + 20 = 40
∴ Number of DVD = 200/40 = 5

Question: A man completes (1/3) of a job in 13 days. At this rate, how many more days will it take him to finish the job?

Solution:
 

Work done = 1/3

Balance work =(1−1/3) = 2/3

Less work, Less days ( Direct proportion)
Let the required number of days be x
Then,
⇔ (1/3) : (2/3) : : 13 : X
⇔ (1/3)/(2/3) = 13/X
⇔ 1/2 = 13/X
⇔ X = 13 × 2 = 26 days

Question: At what rate of compound interest per annum will a sum of Tk. 4000 becomes Tk. 4840 in 2 years?

​Solution:
Principal, P = Tk. 4000
Compound Amount, C = Tk. 4840
Time, n = 2 years
Rate, r = ?

We know,
C = P × (1 + r/100)ⁿ
⇒ 4840 = 4000 × (1 + r/100)²
⇒ (1 + r/100)² = 4840/4000 
⇒ (1 + r/100)² = 484/400
⇒ 1 + r/100 = 22/20 [উভয়পাশে বর্গমূল করে]
⇒ r/100 = (11/10) - 1
⇒ r/100 = (11 - 10)/10
⇒ r/100 = 1/10
⇒ r = (1 × 100)/10
∴ r = 10

∴ Interest Rate = 10%

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(E) = 27
∴ P(E) = n(E)/n(S) = 27/36 = 3/4.

Question: The ratio of length and breadth of a rectangular park is 7 : 5. A man runs along its boundary at 8 km/hr and takes 9 minutes for one round. Find its area in sq. meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the man covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 8 × (9/60)
= 1.2 km
= 1200 meters

Let,
Length = 7x and breadth = 5x
So, Perimeter,
2(7x + 5x) = 1200
⇒ 24x = 1200
∴ x =1200/24 = 50 meters

So, Length = 7 × 50 = 350 meters
And, Breadth = 5 × 50 = 250 meters

Area = Length × Breadth
= 350 × 250
= 87500 sq. m.

Question: The average of 3, 5, 7, and x is 6, and the average of 6, 2, x, and y is 7. What are the values of x and y?

Solution: 
Given that,
The average of 3, 5, 7, and x is 6

Therefore,
6 = (3 + 5 + 7 + x​)/4
⇒ 24 = 15 + x
⇒ x = 24 - 15
∴ x = 9

Therefore,
7 = (6 + 2 + x + y​)/4
⇒ 7 = (6 + 2 + 9 + y​)/4
⇒ 28 = 17 + y
⇒ y = 28 - 17
∴ y = 11

The sum of all prime numbers from 1 to 20
= (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)
= 77

Question: The perimeter of a rectangular field is 110 meters. If the length of the field is 5 meters less than three times the width, what is the area of that field in square meters?

Solution:
ধরি, আয়তাকার ক্ষেত্রটির প্রস্থ = x মিটার
সুতরাং, ক্ষেত্রটির দৈর্ঘ্য = 3x - 5 মিটার

আয়তক্ষেত্রের পরিসীমা = 2(দৈর্ঘ্য + প্রস্থ)
প্রশ্নমতে,
2((3x - 5) + x) = 110
⇒ 2(4x - 5) = 110
⇒ 4x - 5 = 55
⇒ 4x = 60
⇒ x = 15 মিটার

সুতরাং, প্রস্থ = 15 মিটার।
দৈর্ঘ্য = 3x - 5 = (3 × 15) - 5 
= 45 - 5 = 40 মিটার।

আয়তক্ষেত্রের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
ক্ষেত্রফল = 40 × 15 = 600 বর্গ মিটার।

সুতরাং, ক্ষেত্রটির ক্ষেত্রফল হলো 600 বর্গ মিটার।

মনে করি, AB = h
খুঁটিটি ভূমির সাথে 60° কোণ তৈরি করে BC = √3 মিটার ছায়া তৈরি করে
তাহলে খুঁটির উচ্চতা h = ?
প্রশ্নমতে, tan60° = AB / BC
⇒ √3 = h/√3
∴ h = √3.√3 = 3 মিটার 

Question: Which trigonometric ratio is undefined at 90°?

Solution:
sin90° = 1
cos90° = 0
cot90°= 0

∴ tan90° = sin90°/cos90°
= 1 / 0
= ∞ (Undefined)

ক) 8/(3⁴ × 7³) = 8/27,783 = 8/27,783
খ) 27/(3⁵ × 7³) = 1/3,087 = 8/24,696
গ) 12/(3³ × 7³) = 4/3,087 = 8/6,174
ঘ) 2/(3³ × 7²) = 2/1,323 = 8/5,292

লব একই হলে যে ভগ্নাংশের হর বড় সে ভগ্নাংশটি ছোট
 সঠিক উত্তর: অপশন ক

Question: An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of-

Solution: 
Given that,
Speed = 240 kmph 
Time = 5 hours 

We know,
 Distance = Speed × Time 
= 240 × 5
∴ Distance = 1200 km

Again,
Given: 
Distance = 1200 km 
New Time = 5/3 hours 
 
 New Speed = Distance/New Time
= 1200/(5/3)
= (1200 × 3)/5
= 720 kmph
 
 ∴ The aeroplane must travel at a speed of 720 kmph.

Question: A man invested Tk. 36,000 when he bought Tk. 100 shares at Tk. 144. If 16% dividend is declared, find his annual income.

Solution:
For Tk. 144 he gets Tk. 16
∴ For Tk. 1 he gets Tk. 16/144
∴ For Tk. 36,000 he gets Tk. (16 × 36,000)/144
= Tk. (1/9) × 36,000
= Tk. 4,000

∴ The man's annual income is Tk. 4000.

Let there be x pupils in the class.
Total increase in marks = x . 1/2
= x/2
∴ x/2 = 83 - 63
=> x/2 = 20
=> x = 40

Question: If log_x16 = 4/3, what is the value of x?

Solution:
log_x16 = 4/3
⇒ x^4/3 = 16 [log_ba = c ⇒ b^c = a]
⇒ (x^4/3)^3/4 = 16^3/4
⇒ x = 16^3/4
⇒ x = (2⁴)^3/4
⇒ x = 2³
∴ x = 8

Correct sum = 36 × 100 + 90 - 40
= 3650
Correct average = 3650/100 = 36.5
Error = (36.5 - 36) = 0.5
∴ Error% = {(0.5/36.5) × 100}% = (100/73)%
= 1.36%

Question: Find the equation of the line with x-intercept = 5 and y-intercept = 2.

Solution:
Given,
x-intercept = 5, the line passes through (5, 0).
y-intercept = 2, the line passes through (0, 2).

We know,
The intercept form of a line is:
(x/a) + (y/b) = 1, where a = x-intercept, b = y-intercept.
⇒ (x/5) + (y/2) = 1
⇒ (2x + 5y)/10 = 1
⇒ 2x + 5y = 10
⇒ 2x + 5y - 10 = 0

∴ The equation of the line is 2x + 5y - 10 = 0

Some of five numbers = 5 × 27
After excluding one number, the sum of the remaining four numbers = 4 × 25
Excluded number = (5 × 27) - (4 × 25)
= 135 - 100
= 35.

Question: What is the highest power of 5 in the prime factorization of 625?

Solution:
Given that, Highest power of 5 in the prime factorization of 625
Now, The prime factorization of 625 = 5 × 5 × 5 × 5 = 5⁴
Highest power of 5 = 4

∴ The highest power of 5 is 4.

এখানে,
(2x² + 3x - 5)(x + 2) = 2x³ + 4x² + 3x² + 6x - 5x - 10 = ax³+bx²+cx+d
⇒ 2x³ + 7x² + x - 10 = ax³ + bx² + cx + d
উভয় দিকে তুলনা করে পাই
a = 2, b = 7, c = 1, d = -10  
 অতএব, ac - bd = 2×1 -7(-10)
                    = 2 + 70
                    = 72

Let total boys is 4 and participated in sports is 1
Total girls is 8 and participated in sports 3
So, total student 4 + 8 = 12,
and participant = 1 + 3 = 4
Therefore, the proportion = 4/12

Question: HCF and LCM of two numbers are 6 and 180 respectively. If the numbers are between 30 and 80, what is the sum of the numbers? 

Solution:
Given that,
HCF and LCM of two numbers are 6 and 180.

Let the numbers be 6x and 6y (where x and y are co-prime)
∴ LCM = 6xy
∴ 6xy = 180
⇒ xy = 180/6 = 30

Possible co-prime factor pairs of 30: (1, 30), (2, 15), (3, 10), (5, 6)

The numbers must lie between 30 and 80:
6 × 5 = 30 and 6 × 6 = 36, both between 30 and 80

Other pairs give at least one number ≤ 18 or ≥ 90, not valid
∴ Required numbers are 30 and 36
∴ Sum of the numbers = 30 + 36 = 66

So, the sum of the numbers is 66.

Question: A 42 liter mixture contains milk and water in the ratio 3 : 4. How many liters of milk must be added to the mixture so that the ratio of milk to water becomes 1 : 1?

Solution:
The ratio of milk to water is 3 : 4
Total portion = 3 + 4 = 7

Quantity of milk = 42 × (3/7) = 18 liters.
Quantity of water = 42 × (4/7) = 24 liters.

Let,
Quantity of milk to be added = x liters

According to the question,
(18 + x) : 24 = 1 : 1
⇒ (18 + x)/24 = 1/1
⇒ 18 + x = 24
⇒ x = 24 - 18
⇒ x = 6

∴ Quantity of milk to be added = 6 liters