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**Uttara Bank Ltd. Exam Questions Solution**

**NOTE : **

**Post: Probationary Officer**

**Written Exam Date: 27 April 2018**

**Written Exam Time: 3.00 PM to 4.30 PM**

**Full Question Solution Given Below :**

**Question-1**

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

**Question-2**

**Question-3**

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.

**Below of Uttara Bank Limited Written Question**

**UBL Exam Questions Solution**

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