Question

In a certain year the price of gasoline rose by 20% during January, fell by 20% during February, rose by 25% in March, and fell by x% in April. The price of gasoline at the end of April was the same as it had been at the beginning of January. Find the value of x.

Answer 1

x= 16 2/3

Explanation:

Price will be p

The price in January increases by 20%, it will be 6p/5. The price in February decreased by 20%, it becomes 24p/25. The price in March rose by 25%, it will be 6y/5.

April: 6y/5 times 100-x/100 = p 6(100-x)=500 100=6x x= 16 2/3

Answer 2

Answer: The value of x is
`(50)/(3)`

Explanation:

Let the original price of gasoline be y

Since we have given that

Price of gasoline rose by 20% during January fell by 20% during February rose by 25% in March and fell by x% in April.

So, According to question,

In January, Price became

`(100+20)/(100)* y=(120)/(100)* y=(6)/(5)* y=(6y)/(5)`

Similarly, In February, Price became

`(100-20)/(100)* (6y)/(5)=(80)/(100)* (6y)/(5)=(4)/(5)* (6y)/(5)=(4)/(5)* (6y)/(5)=(24y)/(25)`

Similarly, in March, Price became,

`(100+25)/(100)* (4* 6y)/(25)=(125)/(100)* (6* 4y)/(25)=(5)/(4)* (24y)/(25)=(5)/(4)* (24y)/(25)=(6y)/(5)`

similarly, in April , Price became,

`(100-x)/(100)* (6y)/(5)`

so, we have given that the price of gasoline at the end of April was the same as it had been at the beginning of January.

So, it becomes,

`(6y)/(5)* (100-x)/(100)=y\\\\6(100-x)=500\\\\600-6x=500\\\\600-500=6x\\\\100=6x\\\\x=(100)/(6)=(50)/(3)`

Hence, the value of x is
`(50)/(3)`