Answer:
50 of D and 10 of C
Step-by-step explanation:
First of all, put this data into 2 equations.
You sold 60 items, so C+D=60
C is $5 so we can represent it by 5C
D is $7 so we can represent it by 7D
You made $400 total from C and D, so 5C+7D=400
We can use simultaneous equations to solve this.
To eliminate one of these variables, we'll multiply the first one by 5 to make it 5C like the other.
5(C+D=60) (make sure you multiply both sides.)
so 5C+5D=300
5C+7D=400
Now we solve it:
5C-5C+7D-5D=400-300
7D-5D=100
2D=100
D=50
Now we can substitute this with one of the equations to find C
C+50=60
C=60-50
C=10
so, (10x5)+(50x7)=400
Answer:
Rs 120.
Step-by-step explanation:
10=0.85SP-CP; CP+10=0.85SP; SP=[CP+10]/0.85 Eq 1. Let SP= Selling Price and CP= Cost Price
-2 =0.75SP-CP; 0.75SP=C-2; SP=[CP-2]/0.75 Eq 2
[CP+10]/0.85=[CP-2]/0.75 : SP of Eq 1=SP of Eq 2
0.75[CP+10]=0.85[CP-2]
0.75CP+7.5=0.85CP-1.7
0.85CP-0.75CP=-1.7–7.5=9.2
0.10CP=9.2; CP=9.2/0.10
CP=Rs 92 Cost Price of pen
10=0.85SP-92; 0.85SP=92+10=102; SP=102/0.85=Rs 120 Marked Price of pen (answer)
From Eq2: -2=0.75SP-CP; 0.75SP=CP-2=92–2=90; SP=90/0.75=Rs120; -2=0.75(120)-CP; CP-2=0.75(120); CP-2=90; CP=90+2=Rs 92
Set CP of Eq 1=CP of Eq 2:
CP=0.85SP-10 from Eq 1; CP=0.75SP+2 from Eq 2;
0.85SP-10=0.75SP+2; 0.85SP-0.75SP=10+2=12
0.10SP=12; SP=12/0.10=Rs120 is the Marked Price(answer)
Normally, the Selling Price is the marked price. The seller will not disclose the Cost Price because it is the price when the item was acquired or procured, otherwise the buyer will ask for more discounts and based his buying price from the Cost Price if it is known. The calculated SP and CP satisfy both Eq 1 and Eq 2. Both Eq 1 and Eq 2 satisfy the given conditions of the problem above.
Answer:A
Step-by-step explanation:
You use the equation:
m= (y2-y1) ÷ ( x2-x1)
m= (12-9) ÷ (2-1)
m= 3÷1
m= 3
What is the math problem? I’ll try to answer it
Answer:
41
Step-by-step explanation:
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