ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

According to the question,
(x + y)/(x - y) = 4
⇒ x + y = 4x - 4y
⇒ 4x - x = 4y + y
⇒ 3x = 5y
⇒ x/y = 5/3
⇒ x²/y² = 25/9
Now,
⇒ x² + y²/x² - y²
= {(x² /y²) + 1}/{(x²/y²) - 1}
= {(25/9) +1}/{(25/9 - 1}
(34/9) × (9/16)
= 17/8

A : B = 4 : 5;

B : C = 7 : 9
= (7 × 5/7) : (9 × 5/7)
= 5 : 45/7

C : D = 3 : 4
= (3 × 15/7) : (4 × 15/7)
= 45/7 : 60/7.

A : B : C : D
= 4 : 5 : 45/7 : 60/7
= 28 : 35 : 45 : 60.

Let the shares of A, B, C, D be 28x, 35x, 45x, 60x respectively.

Then,
28x = 1680
⇒ x = 1680/28
= 60

∴ D's Share = 60x = Tk. (60 × 60)
= Tk. 3600.

Let,
income in the first year be Tk. x and expenses in the second year be Tk. y
Then, x/45000 = 2/3 and 25000/y = 5/9
⇒ x = (2 × 45000)/3
= 30000
and y (25000 × 9)/5
= 45000.
∴ Total savings for 2 years
= Tk. [(30000 - 25000) + (45000 - 45000)]
= Tk. 5000.

Let,
the cost of the table and chair be Tk. 5x and Tk. 7x respectively.
New cost of chair = 120% of Tk. 7x = Tk (6/5 × 7x)
= Tk. 42x/5.
New cost of table = 110% of Tk. 5x = Tk.(11/10 × 5x)
= Tk. 55x/10.
∴ New ratio = 55x/10 : 42x/5
= 55 : 84

এছাড়াও,
5 : 7
নতুন অনুপাতঃ
5 এর 110% : 7 এর 120%
= 5.5 : 8.4

Let,
the full marks for each paper be x.
Let the marks obtained in the five papers be 6y, 7y, 8y, 9y and 10y respectively.
Then, (6y + 7y + 8y + 9y + 10y)/5x = 60/100
⇒ 40y/5x = 3/5
⇒ 40y = 3x
⇒ x = (40/3)y
50% of x = (50/100) × (40/3)y
= 20y/3
= 6(2/3)y.
Clearly, the student got more than 50% marks in each of the last 4 papers.

Let,
the daily wage of a man, a woman and a boy be Tk. 4x, Tk. 3x and Tk. 2x respectively.
Then, 15 × 4x + 18 × 3x + 12 × 2x = 2070
⇒ 60x + 54x + 24x = 2070
⇒ 138x = 2070
⇒ x = 15
∴ daily wages of 1 man, 2 women and 3 boys
= Tk. (4x + 2 × 3x + 3 × 2x)
= Tk. (4x + 6x + 6x)
= Tk. 16x
= Tk. (16 × 15)
= Tk. 240.

Let third proportional be x
⇒ 25 : 30 : : 30 : x
⇒ 25 × x = 30 x 30
⇒ x = (30 x 30)/25
= 36.

a² + b² + c² - ab - bc - ca = 0 .....(i)
Multiple equation (i) by 2 we get
⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
⇒ (a² + b² - 2ab) + (b² + c² - 2bc) + (c² + a² - 2ca) = 0 [ (a + b)² = a² + b² + 2ab]
⇒ (a - b)² + ((b - c)² + (c - a)² = 0 [if x² + y² + z² = 0 then x = 0, y = 0, z = 0]
∴ a - b = 0
⇒ a = b
b - c = 0
⇒ b = c
c - a = 0
⇒ c = a
∴ a : b : c = 1 : 1 : 1

Zinc in first allow = 2/3 units;
Zinc in second alloy = 4/5 units.
copper in first alloy = 1/3 units;
copper in second alloy = 1/5 units.
Let the first and second alloys be mixed in the ratio 1 : y.
Then, {(2/3) + (4y/5)}/{(1/3 + (y/5)) = 3/1
⇒ 10 + 12y = 3 (5 + 3y)
⇒ 10 + 12y = 15 + 9y
⇒ 3y = 5
⇒ y = 5/3.
∴ Required ratio = 1 : 5/3
= 3:5

Ratio of capitals = 7/2 : 4/3 : 6/5 = (7/2 × 30) : (4/3 × 30) : (6/5 × 30)
= 105 : 40 : 36.
Let the initial capitals of A, B and C be Tk. 105x, 40x, 36x respectively.
Then, ratio of profit = [105x × 4 + (150% of 105x) × 8] : (40x × 12 ) : (36x × 12)
1680 : 480 : 432
= 35 : 10 : 9.
∴ A's share = Tk. (2430 × 35/54) = Tk. 1575; B's share = Tk (2430 × 10/54) = Tk. 450
C's share = Tk. (2430 × 9/54) = Tk. 405.

Sima : Rajon = (50000 × 36) : (80000 × 30)
1800000 : 2400000
= 3 : 4
∴ Sima's share = Tk. (24500 × 3/7)
= Tk. 10500.

Let the total investment be Tk. z
Then, 20% of z = 98000
⇒ z = 98000 × 100)/20
= 490000.
Let the capitals of P, Q and R be Tk. 5x, Tk. 6x and Tk. 6x respectively.
Then, (5x × 12) + (6x 12) + (6x × 6) = 490000 × 12
⇔ 168x = 490000 × 12 ⇔ x = (490000 × 12)/168 = 35000.
∴ R's investment = 6x = Tk. (6 × 35000)
= Tk. 210000.

Let the total capital be Tk. x.
Then, Rahul's share = Tk. x/2
Amin's share = Tk. x/3
Akash's share = [x - {(x/2) + (x/3)}]
= Tk. x/6
∴ Required ratio = x/2 : x/3 : x/6 = 1/2 : 1/3 : 1/6
= 3 : 2 : 1

Let
amount of water be 8x.
So, the amount of sugar is 3x.
According to question,
8x/(3x + 2) = 2/1
Solving this equation, we get, x = 2
Therefore, the amount of sugar in the original solution = 3 × 2 = 6 kg.

salt = 16 × 1/4 = 4 Kg
water = 16 - 4
= 12 kg.
এখন water/salt = 4
∴ water = 4 × 4 = 16 Kg
∴ (16 - 12) = 4 Kg water add করতে হবে।

Let y be the amount of flavour.
(0.2 × 20) + 1 × y = 0.25 (20 + y)
Or, 4 + y = 5 + 0.25y
Or, 0.75y = 1
So, y = 1.33

Let C.P of 1 litre milk be Tk. 1.
Then, S.P of 1 litre of mixture = Tk. 1, Gain = 25%
C.P. of 1 litre mixture = Tk. (100/125) × 1
= Tk. 4/5
By the rule of alligation, we have:

∴ Ratio of the milk to water = 4/5 : 1/5
= 4 : 1.
Hence, percentage of water in the mixture
= {(1/5) × 100}%
= 20%

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