Fractions everywhere

Relevant wiki: Fractions - Problem Solving

For both fractions I made their denominator equal, so:

$\frac{2}{3} = \frac{10}{15}$ and $\frac{4}{5} = \frac{12}{15}$ .

The smallest multiple they have on common is 60, so to get equality this means we need 6 times $\frac{2}{3}$ and 5 times $\frac{4}{5}$ . Hence the answer must be a multiple of 11, so the answer is 22.

Since we are given that $\begin{aligned} \dfrac23 + \dfrac23 + \cdots + \dfrac23 = \dfrac45 + \dfrac45 + \cdots + \dfrac45 \end{aligned}$ Let's say there are $n_1$ and $n_2$ identical fractions on respective sides . $\begin{aligned}& \dfrac{2n_1}{3} = \dfrac{4n_2}{5} \\& \dfrac{n_1}{3} = \dfrac{2n_2}{5} \\& 5n_1 = 6n_2 \\& \frac{n_1}{n_2} = \frac{6}{5} \implies n_1 + n_2 = 6+5 = 11 \\& \frac{2}{2}\left(\frac{n_1}{n_2}\right ) = \frac{12}{10} \implies 2(n_1+n_2) = 22\end{aligned}$

The smallest numbers of fractions $N=n_1+n_2 = 11$ also in general the total fractions will be in form of $11k$ for $k\in\mathbb N$ as the successive lowest common multiple of 5 and 6 in form of $30k$ .

Hence the required answer must be a multiple of 11 , among the options provided, only is $2(n_1+n_2) =2\times 11 = \boxed{22}$ is the answer.

A(2/3) = B(4/5) => 10A = 12B => A/B = 12/10 => A+B = 12 + 10 = 22

[ we didn't simplify it to 6/5 bcoz the sum has to be between 20 and 24... Or we can interpret 20 <= n*(6+5) <= 24 and n becomes 2]

A(2/3) = B(4/5) So, A:B = 6:5 , which means A+B would be a multiple of 6+5 = 11. Between 20 and 24, 22 is only multiple of 11.

Do not break your head with a lot of theory, since the function is an equality we must conceive that no matter how many parts exists for each side (A and B), the quantity of those inputs must respect the rule of keep the equality, so, the next step is only to get the arithmetic media and find the solution: 22.

... => A = 1.2B => A+B = 2.2B so out of 20,21,22,23,24 its gotta be 22:)

ATQ ,

(2/3)A =(4/5)B .....

ON Cross multiplying-

5A=6B -------- 🌟

...which means...that to equate 2/3 ND 4/5...we need at least 5terms of 2/3 ND 6 terms of 4/5. But the ques. Says..that the total no. Of fractions is - 19< A + B >25.

So to get the required no. Of fractions just multiply the 🌟 eq. By 2 to get the required fractions ---

10A=12B

10A +12B =22 fractions

A2/3=B4/5 A/B=12/10 12+10=22

A brute force Python solver:

import random

while True:
    A = random.randint(0, 50)
    B = random.randint(0, 50)

    if A * (2/3) == B * (4/5) and 20 <= A + B <= 24:
        print(A, B)
        break

I maybe thought a bit differently ... 2/3 = .666... 4/5= 0.8 .. I divided .666... into .8 and got 1.2 ... To me that meant clearly that if I had 10 4/5's I'd need 20% more 2/3's ... Therefore within the range specified 12 + 10 = 22 Just my thoughts :-) Ted

The first clue is that the response options are whole numbers.

The smallest whole number to be made of 2/3 is 2 (3 sets), and the smallest while from 4/5 is 4 (5 sets). The least common multiple of 2 and 4 is 4. That's 6 A = 5 B (aka 4=4) and 6+5=11. That means we need the next common multiple (8). 12 A = 10 B (aka 8=8) and 12+10=22, which satisfies the inequality.

$A \times \frac{2}{3} = B \times \frac{4}{5} \implies A = \frac{6}{5} \times B$

Use hit-and-trial to achieve the integral values of A for the integral values of B until you get values that satisfy the condition that $20 \le A+B \le 24$

Since $(A,B)=(12,10)$ satisfies $20 \le A+B \le 24$ , therefore the answer is $22$ .

In order to look at this problem in a neater way, you should convert both $\frac{2}{3}$ and $\frac{4}{5}$ into fractions with a common denominator. So, the equivalent fractions are $\frac{10}{15}$ and $\frac{12}{15}$ , respectively. In order for both sides of the expression to be equal, the numerator of the solution has to be a multiple of 10 and 12. Test the product of 10 and 12 as the numerator so the sum of both sides is $\frac{120}{15.}$ .

$\frac{120}{15}$ divided by $\frac{10}{15}$ is 12 or A.

$\frac{120}{15}$ divided by $\frac{12}{15}$ is 10 or B.

Therefore, A + B = 12 + 10 = 22.

A(2/3) and B(4/5) with A and B integers can be equal if and only if A(2/3) and B(4/5) are also integers. Therefore A must be a multiple of 3 and B a multiple of 5. Since 20 ≤ A + B ≤ 24, there is only one possible value for A + B: 22 with A = 12 and B = 10 to make 8 = 8.

i used x&y instead of a&b
2/3x=4/5y
10x=12y
answer is the sum of the coefficients for x&y. ie22
the test is 10 x 2/3=8=12 x 4/5

3x 2/3 = 2 . 5x 4/5 = 4 . 10x 4/5 = 8 = 12x 2/3 . 10+12=22 :)

Here is an easier way to reach the solution. From question it states;

20 <= A + B <= 24

From fractions we get

A * (2/3) = B * (4/5) 2A/3 = 4B/5 (divide both sides by 2) A/3 = 2B/5 (bring B to left side and 3 to right side) A/B = 2*3 / 5 = 6 / 5

Which means if you consider ratios you'd get A:B = 6:5. In other words A + B is a multiple of (5 + 6) = 11. So which multiple of 11 would lie between 20 & 24 ?

(20 + 24) /2 = 22

• same solution as Peter van der Linden, but what nearly through me of was the 20 equal or smaller than A+B equal or smaller than 24. It is just not true with the given information of 20 to 24 fractions in total, because it states that A has to be a minimum of 20 Frictions, so in total we would get 44 fractions with that info. Pls change to 10 equal or smaller than A +B equal or smaller than 14!

A way to calculate this in Python script:

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for a in range(25):
    for b in range(25):
        left_value = a * 2 / 3
        right_value = b * 4 / 5

        if left_value == right_value:
             print('Fractions:', a + b, ' A=', a, 'B=', b) 

X(2/3)=Y(4/5) Therefore (2/3)*(5/4) = Y/X therefore 10/12 =Y/X therefore 10X=12Y For Y/X to be = to 1 X=10 and y=12 for a total of 22.

(2÷3)+(2÷3)+.....A times = (4÷5)+(4÷5)+.....B times => A(2/3)=B(4/5)

=>A/3=2B/5

=>A=6B/5. ....... equation (I)

Now 20≤A+B≤24

=> 20≤(6B/5)+B≤24. (From equation (l))

=>20≤(11B/5) ≤24

If 20=(11B/5)

=>11B=100

=>B=9.09

If (11B/5)=24

=>11B=120

=>B=10.9

Since B is a real number Also B must be between 9.09 & 10.9 =>B=10

Putting B=10 in equation (I) , we get A=(6x10)÷5 =>A=12

So A+B= 12+10=22

Let's first think what is A+B ?
1. Natural numbers are defined In mathematics, as numbers used for counting. They also are Whole numbers;
2.A and B are results from counting.

So A, B and A+B are natural numbers and
A+B is one of {20, 21, 22, 23,or 24},

2A/3=4B/5 -->10A=12B --> A=1.2B.

We need A to be a natural/whole number.
This happens when B = 5, 10, 15, ...
The corresponding values for A are 6, 12, 18, ...
Now A+B is either 11, 22, 33, ...
The answer is 22.

We can see that $\frac{2}{3}$ x A = $\frac{4}{5}$ x B so the ratio between A and B is 5:6. Since both of them are integers we know that their sum is a multiple of 11, which means that there is only one possibility that fits the inequality: 22.

A/B=6/5

Therefore A=6B/5

On substituting this value in the given inequations, we get

B={9, 10, 11}. A={11, 12, 13}

By hit and trial method, we get A=12 and B=10

Thus, A+B is 22

Well this one is quite simple, first you need to find common product of division(Take the smallest common product of division to make things easier) so in this case 2/3 x 6 = 4 and 4/5 x 5 = 4. From there you get the ratio of 2/3 : 4/5 which is 6:5 . From there just find the range of A:B that suits the 20 < A+B < 24 range. In this case it is obviously 22 because A:B = 12:10 , where A+B=22 .

$A\frac{2}{3}=B\frac{4}{5}$ , and by cross multiplication and simplification, $5A=6B$ . The only way that the equation becomes true and that A and B are both natural numbers is to find the LCM of 5 and 6, which is 30.

If $A=6$ and $B=5$ , then $30=5(6)=6(5)$ . But $A+B=11$ is not between 20 and 24. So we try the next common multiple, which is 60. Then $A=12$ and $B=10$ and $60=5(12)=6(10)$ . Now $A+B=22$ , which satisfies the inequality above. Therefore [boxed $A+B=22$ ]

2a/3=4b/5 a=6/5b Cross check a+b=20,21,22,23,24 If a+b=22 a, b are integar

The left hand side is equivalent to $A \times \frac{2}{3})$ while the right side of the equations is equivalent to $B \times \frac{4}{5})$

Cross multiplying yields $10 \times A$ = $12 \times B$

Dividing each side by two gives $5 \times A$ = $6 \times B$

Now we can just find A,B pairs that work and see if there sum is between 20 and 24.

Setting A to 6 and B to 5 makes the equation true but 6 + 5 = 11 But if we double both A and B the equation will still be true and there sum will be 22. Since 22 is between 20 and 24 it is the correct answer.

Let $A \cdot \frac{2}{3} = B \cdot \frac{4}{5}$ by rearanging the equations we get $5A = 6B$ . Since $5 | 5A \Rightarrow 5 | 6B \Rightarrow 5 | B$ , we can rewrite $B$ as $B = 5n$ , replacing this on our formula gives $A = 6n$ , which gives us a general solution where the special case we are interested on is when $n = 2$ so $A = 12$ , $B = 10$ and $A + B = 22$

We need 3 of \frac {2}{3} to make the value of 2. We need 5 of \frac {4}{5} to make the value of 4.

If we take \frac {4}{6} we need 6 of \frac {4}{6} to make value 4. If we take \frac {8}{10} we need 10 of \frac {8}{10} to make value 8.

So if we double the \frac {4}{6} fraction one more time our value should also end up as 8

So we end up with: \frac {8}{12} we need 12 of \frac {8}{12} to make value of 8. And we keep the same as above for \frac {8}{10}.

We now have both fractions equal 8 when we have 12 repeats of \frac {8}{12} (or \frac {2}{3}) and 10 repeats of \frac {8}{10} (or \frac {4}{5})

12 repeats + 10 repeats is 22