Bank Math

Question: If asinθ = 2 and acosθ = 2√3, then the value of √3tanθ - 1 is?

Solution:
asinθ = 2
acosθ = 2√3

Now,
asinθ/acosθ = 2/(2√3)
⇒ tanθ = 1/√3
⇒ √3tanθ = 1

∴ √3tanθ - 1 = 0

Let C.P. = Tk. 100.
Marked Price = 20% more than C.P.
∴ M.P. = Tk. 120
Discount = 30% on marked price
∴ S.P. = (100-30)% of M.P.
∴ S.P. = (70/100) × 120 = Tk. 84

Loss% = (100 - 84)/100 × 100
= 16%

Question: A can complete a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work the work together but A left 8 days before the completion of the work and B, 12 days before the completion of work. Only C worked up to the end. In how many days was the work completed?

Solution:
Let,
the work be completed in y days. So C works for y days
Therefore, A works for (y - 8) days B works for (y - 12) days.

ATQ,
{(y - 8)/36} + {(y - 12)/54} + (y/72) = 1
⇒ {6(y - 8) + 4(y - 12) + 3y}/216 = 1
⇒ 6y - 48 + 4y - 48 + 3y = 216
⇒ 13y = 216 + 96
⇒ 13y = 312
⇒ y = 312/13
∴ y = 24

Question: If log_m (1/√32) = - 5/2 what is the value of m?

সমাধান:
দেওয়া আছে,
log_m (1/√32) = - 5/2
⇒ m^{- 5/2} = 1/√32 [log_aM = x হলে, a^x = M হয়]
⇒ m^{- 5/2} = 1/(32^1/2)
⇒ m^{- 5/2} = 32^{- 1/2}
⇒ m^{- 5/2} = (2⁵)^{- 1/2}
⇒ m^{- 5/2} = 2^{- 5/2}
∴ m = 2

Question: A Water reservoir is (1/5)^th full and requires 20 liters more to make it (3/5)^th full. What is the capacity of the reservoir?

Solution: 
একটি চৌবাচ্চার ১/৫ অংশ পূর্ণ।
আরো ২০ লিটার যোগ করে ৩/৫ অংশ পূর্ণ হয়। 

অর্থাৎ, ২০ লিটার, (৩/৫) - (১/৫) = ২/৫ অংশের সমান 

চৌবাচ্চার ২/৫ অংশের সমান = ২০ লিটার 
∴ চৌবাচ্চায় পানি ধরে = ২০ × ৫/২ লিটার 
= ৫০ লিটার 

Question: In January, the stock price went up by 50%. It then dropped by 20% in February, increased again by 25% in March, and declined by 10% in April. If Tk. 200 was invested initially and sold after April, calculate the net percentage change in price.

Solution:
At the end of January,
The value of the stock is = Tk. 200 + 50% of (Tk. 200)
= Tk. 200 + Tk. 100 = Tk. 300.

At the end of February,
The value of the stock is = Tk. 300 - 20% of (Tk. 300)
= Tk. 300 - Tk. 60 = Tk. 240.

At the end of March,
The value of the stock is = Tk. 240 + 25% of (Tk. 240)
= Tk. 240 + Tk. 60 = Tk. 300.

At the end of April,
The value of the stock is = Tk. 300 - 10% of (Tk. 300)
= Tk. 300 - Tk. 30 = Tk. 270.

Now, the percentage change in price is,
= (Change in price/Original price) × 100%
= (270 - 200)/200 × 100%
= (70/200) × 100%
= 35%

Question: If 9^{x + y} = 1 and 9^{x - y} = 3, then what are the values of x and y respectively?

Solution:
Given,
9^x+y = 1
⇒ 9^{x + y} = 9⁰
⇒ x + y = 0 .......(1)|

Again,
9^{x - y} = 3
⇒ 9^{x - y} = 3¹
⇒ (3²)^{x - y} = 3¹
⇒ 3^{2(x - y)} = 3¹
⇒ 2(x - y) = 1
⇒ x - y = 1/2 .............(2)

Now, solving (1) and (2) we get,
x + y = 0
x - y = 1/2
⇒ 2x = 1/2
∴ x = 1/4

Now,
x + y = 0
⇒ 1/4 + y = 0
⇒ y = - 1/4

Adding the given equations:
12a + 3b + 7b - 2a = 10
Or, 10a + 10b = 10
Or, 10(a + b) = 10
Or, a + b = 1
So, average of a and b is 0.5

Question: Which of the following comes first in dictionary order?

Solution:
প্রদত্ত চারটি শব্দের মধ্যে Precedent, Precept এবং Precise -এর প্রথম চারটি অক্ষর "Prec" একই। অন্যদিকে, Preclude শব্দের প্রথম চারটি অক্ষরও "Prec" পর্যন্ত একই।

এখন পঞ্চম অক্ষরগুলো তুলনা করি:
Precedent: e
Precept: e
Precise: i
Preclude: l

ডিকশনারি ক্রম অনুযায়ী, 'e', 'i' এবং 'l' এর মধ্যে 'e' আগে আসে। সুতরাং, Precedent এবং Precept এই দুটি শব্দের মধ্যে প্রথম শব্দটি পাওয়া যাবে। বাকি শব্দ দুটি পরে আসবে।

এখন Precedent এবং Precept এর মধ্যে তুলনা করি। এই শব্দ দুটির প্রথম পাঁচটি অক্ষর "Prece" একই।
ষষ্ঠ অক্ষরগুলো হলো: 'd' (Precedent) এবং 'p' (Precept)।

ডিকশনারি ক্রম অনুযায়ী, 'd' এবং 'p' এর মধ্যে 'd' প্রথমে আসে।

সুতরাং, Precedent শব্দটি বাকি সব শব্দের আগে আসবে।

Question: Machine P prints x units in 20 minutes and machine Q prints 2x units in 10 minutes. In how many minutes will P and Q, working together, print 100x units?

Solution:
Machine P প্রতি মিনিটে প্রিন্ট করে = x/20 ইউনিট
Machine Q প্রতি মিনিটে প্রিন্ট করে = 2x/10 = x/5 ইউনিট

∴ P এবং Q একত্রে প্রতি মিনিটে প্রিন্ট করে = (x/20) + (x/5)
= (x + 4x)/20
= 5x/20
= x/4 ইউনিট

এখন,
x/4 ইউনিট প্রিন্ট করতে সময় লাগে 1 মিনিট।
∴ 100x ইউনিট প্রিন্ট করতে সময় লাগবে = (100x × 4)/x = 400 মিনিট
= 6 ঘণ্টা 40 মিনিট

Question: If log⁡_y81 = 4/2​, what is the value of y?

Solution: 
log⁡_y81 = 4/2
⇒ log⁡_y81 = 2
⇒ y² = 81  [log_ba = c ⇒ b^c = a]
⇒ y² = 9²
∴ y = 9

Question: In one hour, a boat goes 11 km/hr along the stream and 7 km/hr against the stream. The speed of the boat in still water (in km/hr) is-

Solution:
Speed in still water = (11 + 7)/2 kmph
= 18/2 kmph
= 9 kmph.

Let, distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
Time taken at 4 kmph : x/4 = 30 min earlier
Difference between time taken : 30 - (-20) = 50 mins = 50/60 hours.
x/3 - x/4 = 50/60
x/12 = 5/6
x = 10 km.

Question: A shopkeeper marks up his goods by 50% above the cost price. He then offers a discount of 20% on the marked price. What is the overall percentage profit?

Solution:
Let the cost price (CP) be Tk. 100

Marked Price = 50% more than cost price
= 100 + 50
= Tk. 150

Discount = 20% of 150
= (20/100) × 150
= Tk. 30

Selling Price (SP) = 150 - 30 = Tk. 120
∴ Profit = SP - CP = 120 - 100 = Tk. 20

∴ Overall percentage profit = (profit/cost price) × 100%
= (20/100) × 100%
= 20%

Question: A bag contains 7 red and 9 green marbles. One marble is drawn at random. What is the probability that the marble drawn is not red?

Solution:
Number of red marbles = 7
Number of green marbles = 9
Total number of marbles = 7 + 9 = 16

P (red marble) = 7/16

∴ P (not red marble) = 1 - (7/16)
= 9/16

If Rashed is walking 5/6 of his usual speed that means he is taking 6/5 of using time.

According to the question,
6/5 of usual time - usual time = 10 mins
1/5 of usual time = 10 mins
Usual time = 50 mins.

Question: The difference between the compound interest and the simple interest on a sum at 10% per annum for 2 years is Tk. 800. Find the principal.

Solution:
Given That,
r = 10%

We know,
Difference between C.I and S.I for 2 years
= p(r/100)²
= P(10/100)²

ATQ,
P(10/100)² = 800
⇒ P(1/10)² = 800
⇒ P(1/100) = 800
⇒ P = 800 × 100
∴ P = 80,000

Hence, Principal = Tk. 80,000

Length of the triangle = √(10² - 8²) = 6
∴ Area of the triangle = 1/2 × 8 × 6 = 24 cm²

Alternative approach,
ত্রিভুজটির লম্ব, AB হলে
AB²  = AC² - BC² 
        = (10)² - (8)² 
        = 100 - 64 
        = 36 
∴ AB = 6 
∴ ত্রিভুজটির ক্ষেত্রফল = 1/2 × 8 × 6 = 24 cm²

Let the numbers be 3x, 2x and 5x.
Then,
9x² + 4x² + 25x²=1862
=> 38x² = 1862
=> x² = 49
so, here x = 7.
So, middle number = 2x = 14

Let's, 4A = 6B = 10C = x
Now, A = x/4; B = x/6; C = x/10
So, A : B : C = x/4 : x/6 : x/10 
                    = x/4 × 60 : x/6 × 60 : x/10 × 60
                    = 15 : 10 : 6
 ∴ Profit of C = 4650 × 6/(15+10+6) = 900Tk.

Distance between two parallel tangents = 18 cm
That means, diameter = 18 cm
Therefore, radius of the circle = 18/2 = 9 cm

Question: If today is Wednesday. After 71 days, it will be:

Solution:
We know that each day of the week is repeated after 7 days.

71 ÷ 7 = 10 (remainder 1)

So, after (7 × 10) = 70 days it will be Wednesday.
∴ after 71 days it will be (Wednesday + 1 day) = Thursday.

Question: If 0.24 ÷ q² = 6, then q equals:

Solution:
0.24 ÷ q² = 6
⇒ q² = 0.24/6
⇒ q² = 0.04
⇒ q = √0.04
⇒ q = 0.2

∴ The value of q is 0.2.

Length of bridge = Distance travelled by the person in 4 min
= speed X time
Speed = 54 Km/hr = 54×5/18 = 3 × 5 = 15 m/s.
Time =4 min = 4 × 60 = 240 s
Required length = 15 × 240 = 3600 m.

Question: In a rectangle, the diagonal length is 15 and the width is 9. What is the perimeter of the rectangle?

Solution:
Given:
Diagonal of rectangle, d = 15
Width, w = 9
Length = l

We know, by Pythagoras theorem,
l² + w² = d²
⇒ l² + 9² = 15²
⇒ l² + 81 = 225
⇒ l² = 225 - 81
⇒ l² = 144
∴ l = 12
∴ Length of the rectangle, l = 12

∴ Perimeter of a rectangle = 2(l + w)
= 2(12 + 9)
= 2 × 21
= 42

So the perimeter of the rectangle is 42.

We know, I = pnr
= 505 × 5/100 × 4
= 101

We know that the curved surface area of a cylinder is 2πrh
Given that, r = 2R, h= H/2
Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH
Therefore, the answer is “Same”.

Question: What percentage of numbers from 1 to 50 has 2 or 9 in the unit digits?

Solution:
Numbers from 1 to 50 has 2 or 9 in the unit digits
= 2, 9, 12, 19, 22, 29, 32, 39, 42, 49
= 10 Numbers

∴ Amount in percentage
= 10/50 × 100%
= 20%