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Two Mobiles S1 and S2 Are Sold for Rs. 10,490 Making 4% Profit on S1 and 6% on S2. If the Two Mobiles Are Sold for Rs.10,510, a Profit of 6% is Made on S1 and 4% on S2. Find the Cost Price of Both - Mathematics

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Question

Two mobiles S1 and S2 are sold for Rs. 10,490 making 4% profit on S1 and 6% on S2. If the two mobiles are sold for Rs.10,510, a profit of 6% is made on S1 and 4% on S2. Find the cost price of both the mobiles.

Sum

Solution

Let the C.P. of S1 mobile = Rs. x
and the C.P. of S2 mobile = Rs. y
In 1st case :
S.P. odf S1 mobile
= Rs. x + 4% of Rs. x

= Rs.`(x + 4/100 x)`

= Rs.`(104)/(100)x`

= Rs.`(26x)/(25)`

S.P. of S2 mobile
= Rs. y + 6% of Rs. y

=  Rs.`(y + 6/100 y)`

= Rs.`(106)/(100) y`

= Rs. `(53y)/(50)`

∴ `(26x)/(25) + (53y)/(50)` = 104900
⇒ 52x + 53y = 524500    ....(i)
In 2nd case :
S.P. of S1 mobile
= Rs. x + 6% of Rs. x

= Rs.`(x + 6/100 x)`

= Rs.`(106)/(100)x`

= Rs.`(53x)/(50)`

S.P. of S2 mobile 
= Rs. y + 4% of Rs. y

= Rs.`(y + 4/100 y)`

= Rs. `(104)/(100)y`

= Rs. `(26y)/(25)`

∴ `(53x)/(50) + (26y)/(25)` = 10510
⇒ 53x + 52y = 525500      ....(ii)
Adding eqns. (i) and (ii), we get
105x + 105y = 1050000
⇒ x + y = 10000       ....(iii)
Subtracting eqn. (i) from eqn. (ii), we get
x - y = 1000       ....(iv)
Adding eqns. (iii) and (iv), we get
2x = 11000
⇒ x = 5500
⇒ 5500 - y = 1000
⇒ y = 4500
Thus, the cost price of S1 mobile is Rs.5500 and that of S2 mobile is Rs.4500.

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Methods of Solving Simultaneous Linear Equations by Elimination Method - Method of Elimination by Equating Coefficients
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Chapter 8: Simultaneous Linear Equations - Exercise 8.3

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Frank Mathematics [English] Class 9 ICSE
Chapter 8 Simultaneous Linear Equations
Exercise 8.3 | Q 28
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