1. The number of units of a certain item sold by a company in 2010 was 20% higher than the number of units it sold in 2009. If the number of units of the item sold in 2009 was 3,000 less than that in 2010, how many units did the company sell in 2009?
A. 12,000
B. 15,000
C. 16,000
D. 18,000
E. 24,000
Answers
Answer:
Given:
\tt{\implies Items\: sold\:in\:2010=20\%\:higher\:than\:in\:2009}⟹Itemssoldin2010=20%higherthanin2009
\tt{\implies Items\:sold\:in\:2009\:less\:3000\:than\:2010}⟹Itemssoldin2009less3000than2010
\sf\large\underline{To\: Find:}ToFind:
\tt{\implies Items\:sold\:in\:2010=?}⟹Itemssoldin2010=?
\sf\large\underline{Solution:}Solution:
\tt{\implies The\: items\:be\:x\:sold\:in\:2009}⟹Theitemsbexsoldin2009
\sf\small\underline{Given\:that:}Giventhat:
The number of units of a certain item sold by a company in 2010 was 20% higher than the number of units it sold in 2009. If the number of units of the item sold in 2009 was 3,000less than that in 2010]
\rm{\implies x\times\:120\%-x=3000}⟹x×120%−x=3000
\tt{\implies \dfrac{120x}{100}-x=3000}⟹100120x−x=3000
\tt{\implies \dfrac{120x-100x}{100}=3000}⟹100120x−100x=3000
\tt{\implies \dfrac{20x}{100}=3000}⟹10020x=3000
\tt{\implies \dfrac{x}{5}=3000}⟹5x=3000
\tt{\implies x=3000\times\:5}⟹x=3000×5
\tt{\implies x=15000}⟹x=15000
Now calculate the item sold in 2010]
\tt{\implies Items\: sold\:_{(in\:2010)}=No\:of\: items\:in\:2009+3000}⟹Itemssold(in2010)=Noofitemsin2009+3000
\tt{\implies Items\: sold\:_{(in\:2010)}=15000+3000}⟹Itemssold(in2010)=15000+3000
\tt{\implies Items\: sold\:_{(in\:2010)}=18000}⟹Itemssold(in2010)=18000
\bf\large{Hence,}Hence,
\tt\orange{\implies Items\: sold\:_{(in\:2010)}=18000}⟹Itemssold(in2010)=18000