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Exam Type: MCQ + Written
Post Name:
1. Trainee Assistant Officer (General)
2. Trainee Assistant Officer (Cash)
Exam Taker: BIBM
Post: Trainee Assistant Officer (Cash)
Selling 12 candies at a price of tk 10 yields a loss of x% and selling 12 candies at a price of tk 12 yields a profit of x%.What is the value of x ?
Solution:
Let,
12 candies cost price= Tk 100
x% loss selling price=(100-x)
Selling price(100-x) then cp 100 tk
sp 10. ” {(100*10)/(100-x)}
x% profit selling price=(100+x) tk
Selling price tk(100+x) then cp tk100
Sp. 12. ” {(100*12)/(100+x)}
According to the question,
{(100*10)/(100-x)}={(100*12)/(100+x)}
Or, x=9.09
So, The value of X is 9.09
Answer: 9.09
A merchant purchased a jacket for Tk. 60 and then determined a selling price that equaled the purchase price of the jacket plus a markup that was 25 percent of the selling price.During a sale, the merchant discounted the selling price by 20 percent and sold the jacket.What was the merchant’s gross profit on this sale?
Solution:
Given that,
Cost price=60 TK
Suppose,
Selling price=x TK
According to the question,
Selling price=cost price + selling price of 25%
Or, x =60+ x of 25%
Or, x=60+ x/4
Or, x-(x/4)=60
Or, 3x/4=60
Or, x =80
Hence, selling price= 80 TK
After 20% discount, actual selling price
=80-80 of 20%=64 TK
So, Gross profit on this sale
=64-60=4 TK
Answer: 4 TK
In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then estimate the quantity of water in litre to be further added in the mixture
Solution:
Given that:
Total milk & water mixture is 60 litres
Also give,
The ratio of milk & water is =2:1
Sum of Ratio=2+1=3
So, Initial mixture have milk
=(2/3)*60 =40 lit
And
Water be =(1/3)*60=20 lit
Let, only water added be x
According to the question,
Or,40:(20+x)=1:2
Or, 40*2=20+x
Or,x=80-20
Or, x=60
Answer: 60 litres
If x=3+22 then find the value of √x – 1/√x
Answer:24/5