Uttara Bank PO-2018 Written Math Solution

#Uttara Bank PO-2018 Written Math
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Question:From a number of apples a man sells half the number of existing apple plus 1 to the first customer, sells 1/3 of the remaining apple plus 1 to the second consumer, and sells 1/5 of the remaining apple plus 1 to the third consumer. He then finds that he has 3 apples left. How many apples did he have originally?
Solution:
Suppose,
The number of apples =x
He sells apples to the first customer
=(x/2+1)=(x+2)/2
Remaining Apples
=x-(x+2)/2
=(x-2)/2
He sells apples to the second customer
=1/3{(x-2)/2}+1
=(x+4)/6
Remaining apples
={(x-2)/2}-{(x+4)/6}
=(2x-10)/6
=(x-5)/3
He sells third customer
=1/5{(x-5)/3}+1
=(x+10)/15
Remaining apples
={(x-5)/3}-{(x+10)/15}
=(4x-35)/15
According to the question,
(4x-35)/15=3
Or, x=20
Answer:20

#Alternative:
Let,number of apples = x
He sells apples to the first customer = x/2+1
He sells apples to the second customer =1/3{x-( x/2+1)}+1
=(x + 4)/6
He sells apples to the third customer=1/5{x-[ (x/2)+1)+(x+4)/ 6]+1
=(x+10) /15
Finally, he has 3 apples left then
So,
x=(x+2)/2+(x+4)/6+(x+10)/15+3
Or,x ={15(x+2)+5(x + 4)+2(x+10)+ 90}/30
Or,30x=22x+160
Or,8x=160
Or,x=20

The number of apples = 20
Answer:20

Question-2:.A farmer sold a cow & Ox for Tk 80,000 and got a profit on 20% on the cow & 25%on the ox.If he sells the cow the cow and the ox for tk.82,000 and got a profit 25% on cow and 20% on the ox. Find the individual cost price of both?[Uttara Bank _PO_2018]
Solution:
Let
Cost price of cow be x tk
Cost price of Ox be y tk
According to the first condition,
120%x + 125% Y=80,000
Or, 120x +125y=80,000*100
Or, 24x+25y=16,00,000..........(i)
Again,Second condition
125%x + 120%y= 82,000
Or, 125x + 120y=82,000*100
Or, 25x +24y=16,40,000----------(ii)
(i)*25-(ii)*24=»
600x + 625y=4,00,00,000
600x + 576y=3,93,60,000
-----------------------------------------
Or, 49y=640000
Or, y=13061.22
Putting the value of y=13061.22 equation (i)=»
24x=1600000-25*13061.22
Or, 24x =1600000-326530.61
Or, 24x=1273469.4
Or,x=53061.22
Answer:53061.22 tk & 13061.22 tk.

Question-3:A train starts from station a with some passengers. At station b, 10% of the passengers get down 100 passengers get in. At station c 50% get down and 25 get in. At station d 50% get down and 50 get in making the total number of passengers 200. How many passengers did board the train at station a?
Solution:
Let, x passengers board the train at station A
At Station B,passenger get down
=x of 10%=x/10
Remaining passenger
=(x-x/10)=9x/10
Passenger get in=100
Now total passenger
=(9x/10)+100=(9x+1000)/10

At Station C,
Passenger get down
={(9x/10)+100} of 50%
=(9x+1000)/20
Remaining
={(9x+1000)/10}-{(9x+1000)/20}
=(9x+1000)/20
Passenger get in=25
Now total passenger,
{(9x+1000)/20}+25
=(9x+1500)/20
At Station D:
Passenger get down
=(9x+1500)/20 of 50%
=(9x+1500)/40
Remaining passenger
={(9x+1500)/20}-{(9x+1500)/40}
=(9x+1500)/40
Now total passenger
{(9x+1500)/40}+50
=(9x+3500)/40
According to the question,
(9x+3500)/40=200
Or, 9x=8000-3500
Or, x =500
Answer:500