ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage =(11628/20400)x 100%
= 57%.

Total number of fruits = (14 + 23) = 37
Let x oranges be removed.
Then,
70% of (37 - x) = 14
⇒ 7 (37 - x) = 140
⇒ 37 - x = 20
⇒ x = 17

Let the maximum marks be x.
Then,
(40−4)% of x =261
⇒ 36% of x =261
⇒ 36x/100 =261
⇒ x = (261×100/36)
⇒ x = 725

Let total quantity of original milk = 1000 gm
Milk after first operation = 80% of 1000 = 800 gm
Milk after second operation = 80% of 800 = 640 gm
Milk after third operation = 80% of 640 = 512 gm
∴ Strength of final mixture = 51.2%

Let the 3rd number is 100
According to the question,
1st :2nd: 3rd= 20:50:100
Required % = (20/50) × 100 = 40%

Winner gets 65% 0f valid votes and loser gets 35% of votes.
Difference between this two = 2748
(65-35)% = 2748
30% = 2748
Total number of voters, 100%
= (2748×100)/30
= 9160

1000000 × 0.75 × 0.80 × 0.85=  5,10,000

According to question,        
XY/100 + YX/100
= 2XY/100
= 2% o fXY

Let the number be x
The problem could be 40 is subtracted from 60% of a number, the result is 50.
60% of x = 60/100×x= 60x/100
60x/100 - 40 = 50
(60x - 4000) /100 = 50
Cross multiplication
60x - 4000 = 50 ×100
60x -4000 = 5000
60x = 5000 + 4000
60x = 9000
x = 9000/60
x = 150
Hence, the number is 150

CP× (76/100) = 912
CP = 12 × 100
=> CP = 1200

Total cost price = Tk (40 x 7/2) = Tk 140
Cost per litre = Tk n2,
Total quantity = 140/2 = 70 Litres.
Water to be added =(70-40) =30 Litres.

Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]n × T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(90 - 9)/90]2 (90) =  72.9 litres.

 Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352
=> T = 44

Apply Allegation Method and first calculate the ratio in which they have to be mixed.
= 8 : 12 = 2 : 3
Thus, the two varieties of oil should be mixed in the ratio 2 : 3. So, if 240 liters of the second variety are taken, then 160 liters of the first variety should be taken.

Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 × 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 × (150 + P)
120 + 4P = 150 + P
=> P = 10 liters.

Students passed in English = 80%
Students passed in Math's = 85%
Students passed in both subjects = 73%
Then, number of students passed in at least one subject
= (80+85)-73
= 92%.
Thus, students failed in both subjects = 100-92 = 8%

Let the initial expenses on Sugar was Tk. 100.
Now, Price of Sugar rises 25%. So, to buy same amount of Sugar, they need to expense,
= (100 + 25% of 100) = Tk. 125.
But, They want to keep expenses on Sugar, so they have to cut Tk. 25 in the expenses to keep it to Tk. 100.
Now, % decrease in Consumption,
(25/125)×100=20%

Quantity of alcohol in vessel P = 62.5/100 × 2 = 5/4 litres
Quantity of alcohol in vessel Q = 87.5/100 × 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed
 = 4.75 : 1.25 = 19 : 5.

Pass mark = (17 + 10) = 27
Let maximum marks be x
Then 36% of x = 27
Or, 36x/100 = 27
Or, 36x = 2700
Hence, x = 75

Water content in 40 litres of mixture.
= 40×(10/100) =4 litres
Milk in the mixture.
= 40−4=36 litres
Let x litres of water is mixed
⇒(4+x)/(40+x)=20/100
⇒x=5 litres