MathProblemSolver101

A box contains some red, blue and white marbles. There is a total of 65 red marbles and blue marbles. There is a total of 35 red marbles and white marbles. The number of blue marbles is thrice the number of white marbles. What fraction of the marbles is red?

I think there’s a typo - It should say what fraction of the marbles are red, but never mind that… haha

Let $r, b, w \in \text{red, blue, white}.$

$r+b = 65. (1)$

$r+w = 35. (2)$

$b=3w. (3)$

(2) & (3) $\Rightarrow r+3w = 65. (4)$

$\frac{(4) - (2)}{2} \Rightarrow \frac{2w}{2} = 30 \div 2 \Rightarrow w = 15 (5)$

(5) & (3) $\Rightarrow b = 45 (6)$

(6) & (1)$ \Rightarrow r = 20 (7)$

Steps to get desired answer: $\frac{(7)}{(7)+(6)+(5)} \Rightarrow \frac{20}{20+45+15} \Rightarrow \frac{20}{80} = \frac{1}{4}.$

Thus, the answer is $\boxed{\frac{1}{4}}$

Why? What happened?

Tom had some money, he spent $33 on books and 3/5 of the remainder on toys. After that, he had 1/4 as much money as he had at first. How much money did Tom have left?

Let $n$ be the initial temperature

We have $n - 33 - \frac{3}{5}(n-33) = \frac{1}{4}n$

Simplifying the LHS: $\frac{2}{5}(n-33) = \frac{1}{4}n.$

Multiplying both sides by $5 \cdot 4 = 20$: $8(n-33) = 5n.$

Divide both sides by $8$: $n-33=\frac{5}{8}n.$

Subtract both sides by $\frac{5}{8}n$: $\frac{3}{8}n-33=0.$

Add both sides by $33$: $\frac{3}{8}n = 33.$

Multiply both sides by $\frac{8}{3}$: $n = \frac{8}{3} \cdot 33 = 8 \cdot 11 = 88.$

$\frac{1}{4} \cdot 88 = \boxed{22 \text{dollars}.}$

John bought only cookies and Ben bought only tarts at first. John gave half of his cookies to Ben. Ben gave half of his tarts to John. John ate 14 tarts and Ben ate 16 cookies. The ratio of John's tarts to cookies became 1:7 and the ratio of Ben's tarts to cookies became 1:4.

(a) How many tarts did Ben buy?

(b) What is the ratio of the total number of tarts to the total number of cookies they had at first?

Let $c$ be the number of cookies John bought

Let $t$ be the number of tarts Ben bought

We have that $\frac{\frac{1}{2}t - 14}{\frac{1}{2}c} = \frac{1}{7} (1)$

We also have that $\frac{\frac{1}{2}c - 16}{\frac{1}{2}t} = \frac{1}{4} (2)$

$(1) \Rightarrow \frac{1}{2}t - 14 = \frac{1}{14}c \Rightarrow 7t - 196 = c (3)$

$(2) \Rightarrow \frac{1}{2}c - 6 = \frac{1}{8}t \Rightarrow 4c - 48 = t (4).$

Can you take it from here?

Sorry, I made a mistake.

2^30

4/9 + 4/81 + 4/729 + … = (4/9)/(1-1/9) = (4/9)/(8/9) = (4/9)(9/8) = 4/8 = 1/2

The area of a sector of a circle given the radius (r) and the angle (a) is $\frac{a}{360} \cdot r^2 \cdot \pi$

Here, the angle is $60$ and the radius is $6$ so we have $\frac{60}{360} \cdot 6^2 \pi= \frac{1}{6} \cdot 36 \pi = 6 \pi.$

The area of the equilateral triangle: We first must know the height and width, which is 6, but we do not know the height(I am assuming we do not know the formula for the area of an equilateral triangle.) Let the height be $h.$ We have $(\frac{6}{2})^2 + h^2 = 6^2 \Rightarrow 9 + h^2 = 36 \Rightarrow h^2 = 27 = 3 \sqrt{3}.$

The height is $3\sqrt{3},$ so the area is $\frac{1}{2} \cdot 3\sqrt{3} \cdot 6 = 9 \sqrt{3}.$

The answer is the sector - equilateral triangle = $6 \pi - 9 \sqrt{3}.$

There's no picture?

1 card from each suit: (1/4)^4

4 aces: (1/13)^4

(1/4)^4 + (1/13)^4