Math, asked by ajomon5334, 10 months ago

Rakesh had some apples and he divided them into two lots A and B. He sold the first lot at the rate ₹ 2 for 3apples and the second lot at rate of ₹ 1 per apple and got a total of ₹ 400. If he had sold the first lot at the rate of ₹ 1 per apple and the second lot at the rate of ₹4 for 5 apples, his total collection would have been ₹460 . Find the total number of apples he had.

Answers

Answered by habishajahan93
0

Answer:

Let in lotA there are x apples

Let in lotB there are y apples

The above statement gives 2 equations

2/3*x+y=400

taking LCM in left side and solving the equation weget,

2x+3y=1200 - - - - - - - - - equation1

x+4/5*y=460

Similarly we can solve this equation also

We get

5x+4y=2300 - - - - - - - - - equation 2

Multiplying 5 in equation1,multiplying 2 in equation2 we get,

10x+15y=6000 - - - - equation3

(eq3-eq4) 10x+8y=4600 - - - - - equation4

(-) (-) (-)

___________

7y=1400

y=200

Sub y in equation1

2x+600=1200

2x=600

x=300

In lot A no. of apples=300

In lot B no. of apples =200

Answered by Anonymous
0

Step-by-step explanation:

Let the first lot =x and the second lot =y, both in Rs .

∴ total number of bananas =x+y

In the first case price of x bananas at the rate of Rs. 2 per 3 bananas =

3

2x

and price of y bananas at the rate of Rs. 1 per banana =y.

∴ by the given condition

3

2x

+y=400

⇒2x+3y=1200 ..............(i)

In the second case price of x bananas at the rate of Rs. 1 per banana =x and price of y bananas at the rate of Rs. 4 per 5 banana =

5

4

y

∴ by the given condition x+

5

4

y=460

⇒5x+4y=2300 ........(ii)

Multiplying (i) by 5 and (ii) by 2, we get

10x+15y=6000 ........(iii) and 10x+8y=4600 .........(iv)

Subtracting (iv) from (iii), we get

7y=1400

⇒y=200

Putting y=200 in (i), we get

2x+3×200=1200

⇒x=300

∴x+y=300+200=500

So, Vijay had 500 bananas.

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