Right side:
(1/27)^(2x+10)
=(3^-3)^(2x+10)
= 3^(-6x - 30)
Re-write both sides:
3^(4x-5) = 3^(-6x - 30)
from here, you can solve x
4x - 5 = -6x - 30
4x + 6x = -30 + 5
10x = -25
x = -2.5
Answer:
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Step-by-step explanation:
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.
