50 POINTS AND BRAINLIEST! There are 111 balls of four different colors in a box. Suppose that if 100 balls are taken then at least one ball of each color appears among the taken balls. What is the minimum number of balls that must be taken so that at least three different colors appear among the taken balls?

88

Step-by-step explanation:

If there are 111 balls of 4 colors, if 100 are taken, and each at least 1 of each color appears, there has to be at least 12 of each color (if you take 100 marbles of 3 colors, all 11 of 1 color could still be in the box, so there has to be at least 12 of each). If there are 12 marbles of 3 colors, there would be 111-(12×3)=111-36-75 marbles of the last color. To get at least 1 marble in 3 colors add 75+12+1=88. If there are 12 marbles of 2 colors, there would be a combined total of 87 marbles for the other 2 colors. 87+1=88. If there is 12 marbles of 1 color, there would be a combined total of 99 marbles for the other 3 colors. If there was a divide of 97+1+1, the 2 colors with only 1 marble could have still been left in the box when 100 balls were taken. So there has to be at least 12 of the colors. So we get 111-12-12-12=75, which is the maximum number of marbles that 1 color can have. 75+12+1=88, our previous answer.

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step-by-step explanation:

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

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