Question

At a wholesale market, Ali bought 8 apples and 7 pears for $4.10, while Ahmed bought 4 apples and 9 pears for $3.70. Find the cost of 2 apples and 2 pears.​

Answer 1

Answer:

$1.1

Step-by-step explanation:

Let,

1 apple price = x$, and

1 pear price = y$

A/Q,

8x+7y = $4.10 (i)

4x+9y = $3.70 (ii)

doing (i)-2×(ii)

(8x+7y)-2(4x+9y) = $4.10 - 2×$3.70

8x+7y-8x-18y = $4.10 - $7.4

-11y = -$3.3

y = $0.3

putting value in (i)

8x+7(0.3) = $4.10

8x+2.1 = $4.10

8x = $2.0

x = $0.25

Now,

cost of 2 apples and 2 pears = 2x+2y

= 2($0.25)+2($0.3)

= $0.5+$0.6

= $1.1

Answer 2

Given: Ali bought 8 apples and 7 pears for $4.10, while

Ahmed brought 4 apples and 9 pears for $3.70.

To find: We have to find the cost of 2 apples and 2 pears

Solution:

Let the cost price of an apple be x and pear is y.

Ali bought 8 apples and 7 pears for $4.10. So, we can write

$8x + 7y = 4.10$

Ahmed bought 4 apples and 9 pears for $3.70.

So, we can write-

$4x + 9y = 3.7$

Multiplying the above equation by 2 we get-

$8x + 18y = 7.4$

Subtracting the two equations we get-

$8x + 7y - 8x - 18y = 4.1 - 7.4 \\ - 11y = 3.3 \\ y = 0.3$

Putting the value of y in equation two we get-

$4 \times x + 9 \times 0.3= 3.7 \\ 4x = 3.7 - 2.7 \\ x = 0.25$

Thus the cost of two apples is 0.25×2=0.5 and the cost of two pears is 0.3×2=0.6.

Thus the cost of two apples is $0.5 and the cost of two pears is $0.6.