Bank Math

Question: A shop is offering a 15% discount on a television whose original price is Tk. 40,000. What will be the selling price of the television after applying the discount?

Solution:
The original price of the television = Tk. 40000
And the discount percentage = 15%

We know,
Discount Amount = (Discount Percentage ÷ 100) × Original Price
= (15 ÷ 100) × 40000
= 6000

∴ Selling Price = 40000 - 6000 = 34000

The selling price after the discount is Tk. 34000.

Distance covered = (108 + 112)
= 220 meter.
Time = 6 seconds.
Relative speed = 220/6 = 110/3 m/s.
= (110/3) × (18/5) km/hr
= 132 km/hr.
50 + Speed of second train = 132 km/hr.
Speed of second train = (132 - 50)
= 82 km/hr.

Question: Kobir walks 10 km towards North. From there he walks 6 km towards South. Then, he walks 3 km towards East. How far and in which direction is he with reference to his starting point?

Solution:

শুরুস্থান A এবং গন্তব্যস্থান D
AC = 10 - 6 = 4 কিমি

AD² = AC² + CD²
AD² = 4² + 3²
AD² = 16 +9
AD² = 25
AD² = 5²
AD = 5
 

Question: What is the H.C.F. of 8/15, 12/25 and 16/30?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M of denominators)

H.C.F of numerators:
H.C.F. of 8, 12 and 16 = 4

& L.C.M of denominators:
L.C.M. of 15, 25 and 30 = 150

∴ Required H.C.F. = 4/150
= 2/75

Amount of milk left after 3 operations
= [40 {1 - (4/40)}³] litres
= (40 × 9/10 × 9/10 × 9/10) litres
= 29.16 litres.

Question: If x⁴ ≤ 16 and y² ≤ 36, then the maximum possible value of (x - y) is:

Solution: 
Given that, 
x⁴ ≤ 16 
⇒ x⁴ ≤ 2⁴
⇒ x ≤ 2
∴ -2 ≤ x ≤ 2
And
y² ≤ 36
⇒ y² ≤ 6²
⇒ y ≤ 6
∴ - 6 ≤ y ≤ 6

Now, 
x could be positive 2 or negative 2. y could be positive 6 or negative 6 . The four possible values for (x - y) are as follows, 
1. 2 - 6 = - 4  ; [When x = 2 and y = 6]
2. 2 - (- 6) = 2 + 6 = 8  ; [When x = 2 and y = - 6]
3. - 2 - 6 = - 8  ; [When x = - 2 and y = 6]
4. - 2 - (- 6) ​= - 2 + 6 = 4  ; [When x = - 2 and y = - 6]

So the maximum value would be 8.

Question: If Tk. 945 is allocated into three portions according to the ratio (2/3) : (3/4) : (5/6), what is the amount of the first portion?

Solution:
The given ratio = 2/3 : 3/4 : 5/6

Take LCM of denominators 3, 4, 6 = 12

∴ The ratio = 8 : 9 : 10 (Multipy thr ratio with 12)

Sum of parts = 8 + 9 + 10 = 27

The first portion = 945 × (8/27) = 280 Tk

∴ First portion = 280 Tk.

Question: A, B, and C invest in a business in the ratio 3 : 4 : 5. If the total profit is Tk 24,000, how much more did C get than A?

Solution:
→ Profit shares = 3x : 4x : 5x

Now
3x + 4x + 5x = 24000
12x = 24,000
→ x = 2,000

→ A’s share = 3 × 2,000 = 6,000
→ C’s share = 5 × 2,000 = 10,000
→ Difference = 10,000 - 6,000 = Tk 4,000

Question: The number of boys and girls in a school is in the proportion 7 : 5. When 20 more girls are added and 14 boys leave, the ratio becomes 6 : 5. Determine the total boys.

Solution:
Given that, 
the ratio of the boys and girls is 7 : 5
Let the number of the boys and girls be , 
boys = 7x
girls = 5x

After changing the number of the students, 
Number of boys = 7x - 14
Number of girls = 5x + 20

Now the ratio becomes, 
⇒ (7x - 14)/(5x + 20) = 6/5
⇒ 35x - 70 = 30x + 120
⇒ 35x - 30x = 120 + 70
⇒ 5x = 190
∴ x = 38

So the number of the boys = 7x = 7 × 38 = 266

Question: The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 5% per annum is Tk. 3. The sum is-

Solution:
Let, Sum = x
Here, r = 5% = 5/100 and n = 2 

Now, S.I. = (x × 5 × 2)/100
= x/10

And, C.I. = [x(1 + (5/100))2 - x]
= (441x - 400x)/400
= 41x/400

ATQ,
(41x/400) - (x/10) = 3
⇒ (41x - 40x)/400 = 3
⇒ x/400 = 3
∴ x = 1200

Question: In one hour, a boat goes 17 km/hr along the stream and 9 km/hr against the stream. The speed of the boat in still water (in km/h) is-

Solution:
Let x be the boat speed.
And, y be the stream speed.

Down stream speed  = x + y = 17 km/h
Upper stream speed = x - y = 9 km/h

Now, x + y + x - y = 17 + 9
⇒ 2x = 26
⇒ x = 13

∴ The speed of the boat in still water is 13 km/h

Question: Find the H.C.F. of p(x) = x³ - 9x and q(x) = x² + 2x - 15.

Solution:
Given that, p(x) = x³ - 9x and q(x) = x² + 2x - 15

Now, the factors of p(x) = x³ - 9x
⇒ x(x² - 9)
⇒ x(x + 3)(x - 3)

And, the factors of q(x) = x² + 2x - 15
⇒ x² + 5x - 3x - 15
⇒ x(x + 5) - 3(x + 5)
⇒ (x - 3)(x + 5)

∴ The required H.C.F. is (x - 3)

Question: If the price of an article is increased by 25% and then decreased by 20%, the net change in the price is -

Solution: 
Let,
The price of an article is 100 Tk.

If the price increased by 25%,
So, the new price will be after increase = 100 + {100 × (25/100)} Tk.
= 100 + 25 Tk.
= 125 Tk.

Then the new price decreased by 20%,
So, the new price will be after decrease = 125 - {125 × (20/100)} Tk.
= 125 - 25 Tk.
= 100 Tk.

∴ The net effect on the price of the article is = (100 - 100) 
= 0 Taka

So, after both the increase and the decrease, the price remains the same.

Question: What is the least number which when doubled is exactly divisible by 8, 12, 16, and 20?

Solution:
Let the number be x.
Doubled the number is 2x.

8 = 2 × 2 × 2
12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2
20 = 2 × 2 × 5

∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240
∴ x = 240/2 = 120

Question: A rectangular field will be fenced on three sides, leaving one side of 20 feet uncovered. If the area of the field is 600 square feet, how many feet of fencing is required?

Solution:
আয়তাকার মাঠের ক্ষেত্রফল = 600 বর্গ ফুট
যে পাশে বেড়া দেওয়া হবে না তার দৈর্ঘ্য = 20 ফুট
অতএব, আয়তাকার মাঠের অন্য পাশের দৈর্ঘ্য = ক্ষেত্রফল/একপাশের দৈর্ঘ্য
= 600 / 20 = 30 ফুট

চতুর্দিকে বেড়া দেওয়ার প্রয়োজন নেই, কারণ একপাশ উন্মুক্ত থাকবে।
যে তিনটি পাশে বেড়া দিতে হবে, তাদের দৈর্ঘ্য হবে (30 + 20 + 30) ফুট।
সুতরাং, প্রয়োজনীয় বেড়ার মোট দৈর্ঘ্য = 30 + 20 + 30 = 80 ফুট।

Question: The least perfect square number divisible by 3, 4, 5, 6 and 8 is:

Solution:
LCM 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
= 4 × 4 × 9 × 25
= 3600

Question: In a trapezoid, the lengths of the two parallel bases are 12 and 20 units, respectively. If the height of the trapezoid is 5, find the area of the trapezoid.

Solution:
Given that,
Trapezoid with bases a = 12 and b = 20
Height, h = 5

We know,
Area of trapezoid = (1/2) × (sum of bases) × height
= (1/2) × (a + b) × h
= (1/2) × (12 + 20) × 5
= (1/2) × 32 × 5
= 80

So, the area of the trapezoid is 80 square units.

If ages in the numerical are mentioned in the ratio A : B, then A: B will be Ax and Bx
1) At present : Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.
Rohan's age 4 years ago = 5x – 4
Rahul's age after 4 years = 3x + 4

2)Ratio of Rohan’s age 4 years ago and Rahul's age after 4 years is 1: 1
Therefore,
(5x - 4)/(3x + 4) = 1/1
⇒ 5x - 4 = 3x + 4
⇒ 5x - 3x = 4 + 4
⇒ 2x = 8
⇒ x = 4
Solving, we get x = 4

3)We are asked to find the ratio between Rohan's age 4 years hence and Rahul's age 4 years ago.
Rohan's age : (5x + 4)
Rahul's age: (3x – 4)
Ratio of Rohan's age and Rahul’s age
(5x + 4)/(3x - 4) = 24/8
(5x + 4)/(3x - 4) = 3/1
(5x + 4) : (3x - 4) = 3 : 1.

Question: One pipe can fill a tank in 20 minutes and the other in 30 minutes. If both pipes are opened at the same time, how many minutes will it take to fill two-thirds of the tank?

Solution:

প্রথম নল দ্বারা,
20 মিনিটে পূর্ণ হয় = 1 অংশ
∴ 1 মিনিটে পূর্ণ হয় = 1/20 অংশ

দ্বিতীয় নল দ্বারা,
30 মিনিটে পূর্ণ হয় = 1 অংশ
∴ 1 মিনিটে পূর্ণ হয় = 1/30 অংশ

দুইটি নল একসঙ্গে খুলে দিলে 1 মিনিটে পূর্ণ হয় = (1/20) + (1/30)
= (3 + 2)/60
= 5/60
= 1/12 অংশ

এখন,
1/12 অংশ পূর্ণ হয় = 1 মিনিটে
∴ 1 অংশ পূর্ণ হয় = 12 মিনিটে 
∴ দুই-তৃতীয়াংশ বা 2/3 অংশ পূর্ণ হয় = 12 × (2/3) = 8 মিনিটে

Question: If the volume of a sphere is 2304π cm³, what is the surface area of the sphere?

Solution:
দেওয়া আছে,
গোলকের আয়তন, V = 2304π cm³

আমরা জানি,
গোলকের আয়তন, V = (4/3)πr³
⇒ (4/3)πr³ = 2304π
⇒ r³ = 2304 × (3/4)
⇒ r³ = 576 × 3
⇒ r³ = 1728
⇒ r = 12 সেমি

এখন,
গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল, A = 4πr²
⇒ A = 4π(12)²
⇒ A = 4π × 144
⇒ A = 576π cm²

অতএব, গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল হলো 576π cm²।

Number of shares purchased = 5508/102
= 54.

Income from each share = 4% of Tk. 100
= Tk. 4

∴ Original income = Tk. (54 × 4) = Tk. 216

Money incurred from sale of share = Tk. (105 × 54)
= Tk. 5670

Number of new shares purchased = 5670/126 = 45

New income = Tk. (45 × 5)
= Tk. 225

∴ Change in income = Tk. (225 - 216)
= Tk. 9.

Question: Find the next number in the series: 8, 27, 64, 125, 216, ?

Solution:
The series represents the cubes of consecutive natural numbers:
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343

So, the next number will be 343.

Question: A boy of height 1.5 m is walking away from the base of a lamp post at a speed of 0.8 m/sec. Find the height of the lamp post from the ground, if the shadow of the boy is 2.0 m after walking for 4 sec.

Solution:

Given that,
Height of the boy = 1.5 m
Speed of the boy = 0.8 m/s
Distance travelled by boy in 4 sec = 0.8 × 4 = 3.2 m
Total distance of shadow of boy and distance from base of lamp post = 2.0 + 3.2 = 5.2 m

Let the height of lamp post be 'h' m
According to question,
⇒ 1.5/2.0 = h/5.2
⇒ h = (5.2 × 1.5)/2.0
⇒ h = 3.9 m

So, The height of the lamp post is 3.9 meters.

B's 3 day's work = (1/21) × 3
= 1/7.
Remaining work = 1 - (1/7)
= 6/7.
(A + B)'s 1 day's work
= (1/14) + (1/21)
= 5/42.
Now, 5/42 work is done by A and B in 1 day.
∴ 6/7 work is done by A and B in (42/5) × (6/7)
= 36/5 days.
Hence, total time taken = 3 + (36/5) days
= 10(1/5) days.

Upstream 120 km in 8 hours = 15 km/hr
⇒ 5x = 15
∴ x = 3
Downstream speed = 9x = 27 km/hr
In 3 hrs. Distance is = 27 × 3 = 81 km

Question: Find the HCF of 4/9, 8/15, 12/25 is-

Solution:
We know,
HCF = (HCF of all numerators)/(LCM of all denominators)

Now, numerators: 4, 8, 12
4 = 2 × 2
8 = 2 × 2 × 2
12 = 2 × 2 × 3
∴ HCF = 2 × 2 = 4

And, denominators: 9, 15, 25
9 = 3 × 3
15 = 3 × 5
25 = 5 × 5
∴ LCM = 3 × 3 × 5 × 5 = 9 × 25 = 225

∴ HCF = (HCF of all numerators)/(LCM of all denominators) 
= 4/225

Question: In the list S = {17, 9, 24, X, 14, 17, 21} the mean, median and mode are all equal to one another. What is the value of X?

Solution:
Let the list be: 9, 14, 17, 17, 21, 24, X (sorted, except X).
Since mean = median = mode, and mode = 17 (appears twice, others once),
⇒ mean = median = 17.

We know,
Mean = (sum of all numbers)/7 = 17
17 = (17 + 9 + 24 + X + 14 + 17 + 21)/7
⇒ 17 = (102 + X)/7
⇒ 102 + X = 119
⇒ X = 119 - 102
∴ X = 17 

 So the value of X is 17.

By investing Tk. 1552, income = Tk. 128
By investing Tk. 97, income = (128 × 97)/1552
= 8
Hence, the dividend is 8%.

Question: Three unbiased coins are tossed. What is the probability of getting exactly two heads?

Solution:
তিনটি মুদ্রার জন্য মোট ফলাফল = 2³ = 8

মোট নমুনা বিন্দু = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
= 8 টি

ঠিক দুইটি হেড আসার অনুকূল ফলাফল = (HHT, HTH, THH)
∴ অনুকূল ঘটনার সংখ্যা = 3

∴ সম্ভাবনা = অনুকূল ঘটনা/মোট সম্ভাব্য ঘটনা
= 3/8

Let A’s age 10 years ago be = x years
Then, B’s age 10 years ago = 2x years
ATQ, (x + 10) / (2x + 10) = 3/4
Or, 6x + 30 = 4x + 40
Or, x = 5
So, the total of their present ages = (x + 10 + 2x + 10) = (5 + 10 + 2.5 + 10) = 35 years.

Question: If x = 1 + √2 and y = 1 - √2, find the value of (x² + y²)².

Solution: 
Given that, x = 1 + √2 and y = 1 - √2
∴ x + y = 1 + √2 + 1 - √2
= 2

And, xy = (1 + √2)(1 - √2)
= 1² - (√2)²
= 1 - 2
= - 1

Now,
x² + y² = (x + y)² - 2xy
= (2)² - 2(- 1)
= 4 + 2
= 6

∴ (x² + y²)² = 6²
= 36

The solution of this question is based on the rule,
The HCF of (a^m - 1) and (aⁿ - 1) is given by (a^{HCF of m, n} - 1)
Thus for this question the answer is (3⁵ - 1)
Since, 5 is the HCF of 35 and 125

Profit earned by manufacturer = 20%
Profit earned by wholesaler = 25%
Profit earned by retailer = 30%
S.P. of shoes = Tk. 50

Therefore, 130% of 125% of 120% of x = 50.50
⇒ 120/100 × 125/100 × 130/100 × x = 5050/100
⇒ (195/100) x = 5050/100
⇒ x = (5050 × 100)/(195 × 100)
⇒ x = 25.89
Cost price of shoes = Tk. 25.89