Math, asked by Charisma22331, 18 days ago

Which scenario reflects an annual inflation rate of 3%? A. In year 1, the price of a computer is $325.00. In year 2, the same computer costs $328.00. B. In year 1, the price of a soda is $0.75. In year 2, the same soda costs $0.78. C. In year 1, the price of a board game is $10.00. In year 2, the same board game costs $10.03. D. In year 1, the price of a sofa is $500.00. In year 2, the same sofa costs $515.00.Which scenario reflects an annual inflation rate of 3%? A. In year 1, the price of a computer is $325.00. In year 2, the same computer costs $328.00. B. In year 1, the price of a soda is $0.75. In year 2, the same soda costs $0.78. C. In year 1, the price of a board game is $10.00. In year 2, the same board game costs $10.03. D. In year 1, the price of a sofa is $500.00. In year 2, the same sofa costs $515.00.

Answers

Answered by y2krahulsingh
0

Answer:

First Case (Computer),

Price in year 1 = ₹325

Price in year 2 = ₹328

increased price = ₹328 - ₹325

= ₹ 3

Therefore, increased percentage = (3/325 × 100) ℅

= 300/325℅ (divide numerator & denominator both by 25)

= 12/13℅

Second Case (Soda),

Price in year 1 = ₹0.75

Price in year 2 = ₹0.78

increased price = ₹0.78- ₹0.75

= ₹ 0.03

Therefore, increased percentage = (0.03/0.75× 100) ℅

= (3/75 × 100)℅

= (1/25 × 100)℅

= 100/25 ℅ (divide numerator & denominator both by 25)

= 4℅

Third Case (Board game),

Price in year 1 = ₹10

Price in year 2 = ₹10.03

increased price = ₹10.03- ₹10

= ₹ 0.03

Therefore, increased percentage = (0.03/10 × 100) ℅

= 0.003 × 100℅

= 0.3℅

Forth Case (Sofa),

Price in year 1 = ₹500

Price in year 2 = ₹515

increased price = ₹515 - ₹500

= ₹ 15

Therefore, increased percentage = (15/500 × 100) ℅

= (3/100 × 100)℅

= 300/100℅

= 3℅

Answer :- Question no. D reflects an annuall inflation rate of 3℅

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