Question

✨Guys my new Question ✌✍

i)▶A confectioner had some boxes for burger and he bought 50 more boxes. After two days half of these boxes were used and only 40 boxes were left. Find the number of boxes he had in the beginning.

ii)▶ solve the equation....
x/2 + x/3 -x/4 = 7..

✨✨the person who answered the first will be marked brainlist☺⚽⚽

Answer 1

Answer:

(i) 30.

(ii) 12

Step-by-step explanation:

(i)

Let the number of boxes he had be 'x'.

According to the given condition,

⇒ (x + 50)/2 = 40

⇒ x + 50 = 80

⇒ x = 30.

Therefore, In the beginning he had 30 boxes.

(ii)

Given: (x/2) + (x/3) - (x/4) = 7

LCM of 2,3,4 = 12.

⇒ (x/2) * 12 + (x/3) * 12 - (x/4) * 12 = 7 * 12

⇒ 6x + 4x - 3x = 84

⇒ 7x = 84

⇒ x = 12.

Therefore, the value of x = 12.

Hope it helps!

Answer 2

______Heyy Buddy ❤________

______Here's your Answer.._______

1.☆☆☆☆☆☆☆☆☆

Let the Number of box be x.

New No. of box = x + 50.

Half boxes are used =( x + 50)/2

A.T.Q.

=> (x + 50) - (x+50)/2 = 40

=> 2x + 100 - x - 50)/2 = 40

=> x + 50 = 40× 2

=> x = 80 - 50

=> x = 30.

No. of initial boxes = 30 boxes.
✔✔✔


2.☆☆☆☆☆☆☆☆☆☆

x/2 + x/3 -x/4 = 7

=>( 6x + 4x - 3x)/12 = 7

=> 7x = 12 × 7

=> 7x = 84

=> x = 84/7

=> x = 12.
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