Each chef at "Sushi Emperor" prepares 151515 regular rolls and 202020 vegetarian rolls daily. On Tuesday, each customer ate 222 regular rolls and 333 vegetarian rolls. By the end of the day, 444 regular rolls and 111 vegetarian roll remained uneaten.

Question

04-02-2020
Mathematics

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Each chef at "Sushi Emperor" prepares 151515 regular rolls and 202020 vegetarian rolls daily. On Tuesday, each customer ate 222 regular rolls and 333 vegetarian rolls. By the end of the day, 444 regular rolls and 111 vegetarian roll remained uneaten.

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iwillanswer iwillanswer · Answer 1 · 2020-02-12T07:01:34+01:00

Question:

Each chef at "sushi emperor" prepares 15 regular rolls and 20 vegetarian rolls daily. On tuesday, each customer ate 2 regular rolls & 3 vegetarian rolls. by the end of the day, 4 regular rolls & 1 vegetarian roll remained uneating. how many chefs were on tuesday ? and how many customers were they ?

Answer:

There were 2 chefs and 13 customers on tuesday

Solution:

Let x be the number of chefs at Sushi Emperor and y be the number of customers on Tuesday.

From given,

Each chef prepares 15 regular rolls and 20 vegetarian rolls daily

If each chef prepares 15 regular rolls, then x chefs prepare 15x regular rolls

If each customer ate 2 regular rolls, then y customers ate 2y regular rolls

By the end of the day, 4 regular roll remained un eating

Therefore,

15x - 2y = 4 --------- eqn 1

If each chef prepares 20 vegetarian rolls, then x chefs prepare 20x vegetarian rolls

If each customer ate 3 vegetarian rolls, then y customers ate 3y vegetarian rolls

By the end of the day, 1 vegetarian roll remained uneating

Therefore,

20x - 3y = 1 ---------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 3

45x - 6y = 12 ------- eqn 3

Multiply eqn 2 by 2

40x - 6y = 2 ------- eqn 4

Subtract eqn 4 from eqn 3

45x - 6y = 12

40x - 6y = 2

( - ) --------------

5x = 10

x = 2

Substitute x = 2 in eqn 1

20(2) - 3y = 1

40 - 3y = 1

3y = 39

y = 13

Thus there were 2 chefs and 13 customers