Question

The cost of 20 pens, 22 erasers and 25 sharpeners is rs.196. The cost of 23 pens, 27 erasers and 30 sharpeners is rs.233. The cost of 109 pens, 125 erasers and 140 sharpeners is rs.K. What can be the value of k?

Answer 1

1pen = x    1eraser =y      1sharpner = z

R₁  ⇒     20x + 22y + 25z = 196

R₂  ⇒     23x + 27y + 30z = 233

R₃  ⇒     109x + 125y + 140 = k

2*R₁ = R₄  ⇒     40x + 44y + 50z    =   392

3*R₂ = R₅  ⇒    69x  + 81y  + 90z   =  699

R₄ + R₅      ⇒     109x + 125y + 140z =  1091                = R₃

k = 1091

Answer 2

Answer:

k=1091

Step-by-step explanation:

• Because 20 pens, 22 erasers, and 25 sharpeners cost Rs.196
• 40 pens, 44 erasers, and 50 sharpeners cost Rs.392, respectively.

Since 23 pens, 27 erasers, and 30 sharpeners cost Rs.233,implies,

The total cost is Rs.699 for 69 pens, 81 erasers, and 90 sharpeners.

As a result, the total cost of,

(40+69=109) pens,

(44+81=125) erasers,

and,

(50+90=140) sharpeners

= Rs.(392+699=1091).

For pens : ((20+23)*3)-20) = 109

Like in the same way for erasers and sharpners,

((22+27)*3)-22 = 125 and ((25+30)*3)-25)

implies,

k = ((196 + 233)*3)-196 = 1091

#SPJ3