Mrs.thapa bought 150 kg of oranges.18 kg of oranges were rotten and not fit for selling.If she sold the remaining quantity of or
2 answers:
Answer:
Rs64.00/kg
Step-by-step explanation:
1. Calculate the amount of good oranges
Total amount of oranges = 150 kg
Amount of rotten oranges = 18 kg
Amount of good oranges = 132 kg
2. Calculate the income from the good oranges
3. Calculate the net unit rate of income
The income from the good oranges was Rs10 560.
Based on 150 kg of oranges, the unit rate of income was
4. Calculate the original unit rate of cost
Let x = original unit cost. Then
0.10x = 10 % of original unit cost and
1.10x = 10 % over original unit cost
1.10x = RS70.40/kg
x = (RS70.40/kg)/1.10 = Rs64.00/kg
The original unit rate of cost was Rs64.00/kg.
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Work the information to set inequalities that represent each condition or restriction.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
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</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
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4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
<span />
4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.