Question

The prices of A1 and A2 are in the ratio 8:11.If the price of A2 is Rs.144 more than A1,then find the total price A1 and A2.

Answer 1

Step-by-step explanation:

a1+a2+a3+a4=16Let a1+a2=xand a3+a4=y⇒x+y=16⇒y=16-xNowProductP=xy⇒P=x16-x⇒P=16x-x2⇒dPdx=16-2xFor maixima or minima dPdx=0⇒16-2x=0⇒2x=16⇒x=8d2Pdx2=-2 ⇒P will be maximum when x=8y=16-x=16-8⇒y=8⇒Pmax=8×8=64⇒a1+a2a3+a4max=64

Answer 2

Given in the Question: We are provided that prices of $\sf{A_1}$ and $\sf{A_2}$ are in the ratio 8:11.

Need to Calculate: We are said to find the total price of $\sf{A_1}$ and $\sf{A_2}$.

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✰Let's say that,the price of both the objects are 8x and 11x respectively.

Therefore,

• The price of $\sf{A_2}$ is Rs.144 more than $\sf{A_1}$.

$\: \: \: \sf \implies \: 11x - 8x = 144 \\ \\ \: \: \: \sf \implies \: 3x = 144 \\ \\ \: \: \: \sf \implies \: x = \cancel\frac{144}{3} \\ \\ \: \: \: \sf \therefore \: x = 48$

Hence,

• Price of $\sf{A_1}$ = 8 × 48 = Rs.384.
• Price of $\sf{A_2}$ = 11 × 48 = Rs.528.

• We are also said that find the total price of $\sf{A_1}$ and $\sf{A_2}$.So let's find it!

$\: \: \sf \therefore \: total \: cost \: = 384 + 528 = 912 \: rupees$

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