Question

51. Eighteen balls are kept in a box. Some balls are red and remaining balls are blue. Twice the number of
red balls exceeds thrice the number of blue balls by one. The number of red balls in the box is
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Answer 1

Answer:

red balls /18

Step-by-step explanation:

HOPE U UNDERSTAND

Answer 2

Step-by-step explanation:

Given :-

Eighteen balls are kept in a box. Some balls are red and remaining balls are blue. Twice the number of red balls exceeds thrice the number of blue balls by one.

To find :-

Find the number of red balls in the box ?

Solution :-

Total number of balls kept in the box = 18

There are red and blue balls

Red balls + Blue balls = 18

Let the number of blue balls be X

Thrice of the blue balls = 3X

Then the number of red balls = (18-X)

Twice of red balls = 2(18-X)

Given that

Twice the number of red balls exceeds thrice the number of blue balls by one.

=> 2(18-X) = 3X+1

=>36-2X = 3X+1

=> 36-1 = 2X+3X

=> 35 = 5X

=> X = 35/5

=> X = 7

Blue balls = 7

Red balls = 18-7 = 11

Answer:-.

Total number of red balls in the box = 11

Check:-

Blue balls = 7

Red balls = 11

total balls = 7+11 = 18

Twice the number of red balls exceeds thrice the number of blue balls by one.

=> 2(11) = 22 =21+1

=> 3(blue balls)+1

Verified the given relations in the given problem