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Combined 8 Bank Senior Officer Written Exam Question Solution:
Exam Type: Written
Post Name: Senior Officer
Written Exam Date: 22 September 2018
Written Exam Time: 3.00 PM to 5.00 PM
Exam Taker: Ahsanullah University of Science and Technology (AUST)
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See Combined 8 Bank Senior Officer Written Exam Question Solution in below:
Q-1:Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, 5/6 hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit?
Answer: 20 km/hr
Full Solution:
Let second rabbit takes x hr with speed s2
First rabbit takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2= 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x= 5/2 in s2 –> 20 km/hr
Shortcut:
Let, speed of faster & slower rabbit be x & y respectively
Atq, x+y=50
And, 50/y-50/x=5/6
From the above equations, y=20 km/h (ans)
2. Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hours, Pipe C can fill Tank in 6 Hours. The Tank is already filled up to 1/6 of its capacity. Now Pipe A is opened in the First Hour alone, Pipe B is opened in the Second Hour alone and Pipe C is opened in the Third Hour alone. This cycle is repeated until the Tank gets filled. Then in How many Hours does the rest of Tank gets filled?
Full Solution:
In First Hour Tank filled = 1/6+1/18
Second Hour = 1/6+1/18-1/12
Third Hour = 1/6+1/18-1/12+1/6 = 11/36 is filled
25/36 is left
From then 3 hours work = 1/18-1/12+1/6 = 5/36
5*3 Hours = 5*5/36 = 25/36
Total = 5*3+3 = 18 Hours
Shortcut:
Remaining part=1-1/6=5/6
In 3hrs 3pipes can fill=1/18-1/12+1/6=5/36
So,
To fill 5/6part, it requires=3*36/5*5/6=18 hrs (ans)
Ans: 18 Hours
3. A Jar contains ‘x’ liters of Milk, a seller withdraws 25 liter of it and sells it at tk. 20 per liter. He then replaces it water. He repeated the process total three times. Every time while selling he reduces selling price by tk. 2. After this process Milk left in the mixture is only 108 liters so he decided to sell the entire Mixture at tk. 15 per liter. Then how much profit did he earned if bought Milk at tk. 20 per liter?
Ans: 70 Tk. profit
Shortcut:
Seller sells Milk at Tk. 20,18 and 16 respectively for three times
= 25*(20+18+16) = 1350
108 = x(1-25/100) 3
x =256 liter
He sold entire 256 at Tk.15 =256*15 = 3840
Cost price = 256*20 = 5120
profit = 5190-5120 = 70 Tk. (ans)
Ans: 70 Tk. profit
4. A container contains milk and water in the ratio of 3:1. How much mixture should be taken out and replaced with milk so that the container contains milk and water in the ratio of 15:4. What portion of original mixture had been replaced by milk.
Full Solution:
Let total original quantity = x litres, Let y litres replaced.
After y litres of mixture drawn out,
Milk = [3/(3+1)] * x – [3/(3+1)] * y
Water = [1/(3+1)] * x – [1/(3+1)] * y
Now y litres of milk poured in can. Milk becomes (3/4)*x – (3/4)*y +y = (3/4)*x +(1/4)*y
Now [(3/4)*x +(1/4)*y] / [(1/4)*x – (1/4)*y] = 15/4
Solve, y = (3/19)* x
So 3/19 of original mixture removed.
Ans: 3/19 portion had been replaced by milk
5. A man rows to a place 40 km distant and back in a total of 18 hours. He finds that he can row 5 km with the stream in the same time as 4 km against the stream. What is the speed of boat in still water?
Full Solution:
Suppose he moves 5km downstream in x hours
Then, downstream speed a= 5/x km/hr
Speed upstream speed b = 4/x km/hr
40 / (5 /x) + 40 / (4/x) = 18
8x + 10x = 18
x = 1
a = 5 km/hr, b = 4 km/hr
speed of boat = ½ (5 + 4 ) = 9/2 km/hr= 4.5 Km/hr
Shortcut:
Speed in downstream=5/t
Speed in upstream=4/t
ATQ, 40/(5/t)+40/(4/t)=18
=>t=1
So,
Speed in downstream=5
Speed in upstream=4
Now speed of boat in steel water=(5+4)/2=4.5 km/hr (ans)
Ans: Speed of the boat in still water 4.5 km/hr
6. Probability Solution A Bag contains some White and Black Balls. The probability of picking two white balls one after other without replacement from that bag is 14/33. Then what will be the probability of picking two Black balls from that Bag if bag can hold maximum 15 balls only?
Full Solution:
Wc2/ (B+W)c2 = 14/33
W (W-1)/(W+B)*(B+W-1) = 14/33
Now expressing 14/33 in the above format by multiplying 4 in numerator and denominator
W (W-1)/(W+B)*(B+W-1) = 8*7/12*11 (note = balls <15)
W =8
W+B =12 B =4
Probability = 4c2/12c2 = 1/11
Shortcut:
4/33=(8/12)*(7/11)
So we can write,
Total ball=12
White ball=8
Black ball=12-8=4
So, Required probability=4c2/12c2= 1/11(ans)
Ans: 1//11
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Combined 8 Bank Senior Officer Written Exam Question Solution:
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