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find the numbers if their sum is 60 and the ratio between them is 5:7.

find the numbers if their sum is 60 and the ratio between them is 5:7?

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find two numbers in the ratio of 2 to 5 that have a sum of 119.


The difference between the two numbers is 10. Find their sum ... the ratio 2:7.if sum of the numbers is 99 , find ... numbers have a sum of 60. what is the ...
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How to Calculate Ratios: 9 Steps (with Pictures) - wikiHow


How to Calculate Ratios. ... When comparing two numbers in a ratio, ... If we convert these ratios to their fraction forms, we get 2/5 and x/20.
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What are two numbers in the ratio 7:3 that have the ...


The ratio between two numbers is 4:5 and their sum is 540. ... and they have difference of 20 between them, x ... The sum of two numbers is 20 and their difference is 25.
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The ratio between two numbers is 4:5 and their sum is 540 ...


Let the numbers be 4x and 5x Sum of the numbers = 540i.e. 9x=540x=60 ... The ratio between two numbers is 4:5 and their ... of them, they will be in the ratio 5:7.
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The ratio of two numbers is 7:9. The difference between the ...


... .Without assigned those numbers find difference between them ... between the 2 numbers is 5 and their sum is ... ratio 2:7.if sum of the numbers is 99 , find ...
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Help with Ratios - WebMath


A ratio is a statement of how two numbers compare. It is a comparison of the size of one number to the size of another number. All of the lines below are different ...
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The ratio of two numbers is 5:3.Their sum is 64.What is the ...


The ratio of two numbers is 5:3.Their sum is 64.What ... there are 60 million students ... there are cards each with one of the numbers {1, 2, 3, 4, 5} written on them.
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I Love Maths -A complete, Indian site on Maths


Ratio and Proportion. Compounded Ratio of ... 1·2 kg and 60 gm. Find the compounded ratio of (i) 5 : 7 ... 20 and 43 to make them proportional? Find two numbers such ...
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Ratio of natural numbers. Evolution of the real numbers.


THE RATIO OF TWO NATURAL NUMBERS. The definition. ... The ratio of two natural numbers is their relationship with respect to relative size, which we can always name.
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algebra precalculus - Ratio between two numbers is 6:7 and ...


... 7 and the difference between them is 10. ... Maintain relationship between two numbers, when their sum equals 1. 0. ... Find ratio / division between two numbers. 1.
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Suggested Questions And Answer :


Sum of two numbers is 300 and the two numbers have a ratio of 20:10. Find the larger number

Sum of two numbers is 300 and the two numbers have a ratio of 20:10. Find the larger number A ration of 20 to 10 means that one number is twice as large as the other. So, we have x and we have 2x. x + 2x = 300 3x = 300 x = 100 The other number is 2x; 2 * 100 = 200    < The answer 100 + 200 = 300
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find the sum of all positive numbers from 5 to 1,555 that are divisible by 5

find the sum of all positive numbers from 5 to 1,555 that are divisible by 5 The numbers from 5 to 1555 that are divisible by 5 are: 5, 10, 15, 20, ... 1545, 1550, 1555 which is five times 1, 2, 3, 4, ..., 309, 310, 311. So, the sum of all positive numbers from 5 to 1,555 that are divisible by 5 is 5 times the sum of all positive numbers from 1 to 311. The sum of all positive numbers from 1 to n is: (1/2)n(n+1) The sum of all positive numbers from 1 to 311 is: (1/2)*311*(311 + 1) = 311*156 = 48,516 Answer: 242, 580
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Plz help I need it now

Question: The mean of 5 numbers is 11.the numbers are in ratio of 1:2:3:4:5. Find the smallest number. Let the samllest number be n. The numbers are in the ratio 1:2:3:4:5. So the 5 numbers are: n, 2n, 3n, 4n, 5n. Sum of the 5 numbers is n + 2n + 3n + 4n + 5n = 15n So mean of these 5 numbers is 15n/5 = 3n. But mean = 11 (given)Therefore 3n = 11. i.e. n = 11/3 Answer: smallest number is 3 2/3
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The number of degrees in an angle of a triangle is to the number of grades in the second is to the number of radians in the third as 288:280:pi. Find all the angles in degrees.

The number of degrees in an angle of a triangle is to the number of grads in the second is to the number of radians in the third as 288:280:pi. Find all the angles in degrees. We have the following definitions,   360 degrees = 2.pi radians 400 grads = 2.pi radians 400 grads = 360 degrees   So pi rad = 180 degrees And 280 grads = (280/400)*360 = 252 degrees So, the three angles are: 288, 252, 180 degrees   OK, I undersdtand your point. I originally thought there must be 3 separate triangles. I have the three angles as 288, 252 and 180, which are in the ratios given. As you said, the sum of the angles in a triangle add up to 180. These three angles I got add up to 288 + 252 + 180 = 540 + 180 = 720 And 720 is 4 times 180. So just divide all the three angles by 4, to give 72, 63, 45, which are all in the proper ratio and add up to 180 degreres.   Sorry about the mixup.
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find the numbers if their sum is 60 and the ratio between them is 5:7.

????????? 2 numbers ????????? sum =60 num2=(7/5)num1 num1+num2=(12/5)num1=60 num1=(5/12)*60=25 num2=35
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two numbers are in the ratio 4:7.

If the ratio is 4:7 thew the answer is 4/7 then the number is 4 and 7 you have de possibility to calculate like this 4/7=X/Y 3x+2y=59 and you gona have x=0.59 and y=1,0325 if you transforme 0,59 like a 4  then 0,59-------4 and y -------1.0325 then y=1,0325*4/0.59<=> y=7
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geometric sequence

Carlos who weights 104kg plans to lose 8kg during the first 3months of a diet,6kg during the next 3months,4.5kg during the following 3months and so on. this pattern of weight loss forms a geometric sequence. find the persons weight after a large number of years, assuming their diet is a successfu? Weight = 104 kg. a0 = 8 kg : weight loss in 1st quarter a1 = 6 kg:       “         “    “  2nd     “ a2 = 4.5kg:     “        “    “   3rd     “ By observation, the common ratio, r = ¾ i.e. a_(n+1) = a_n * r,  or a_n = a_0*r^n For a geometric sequence, the sum to infinity is S = a0/(1 – r), where a0 = 8 and r = ¾ Then, S = 8 / (1 – ¾) = 8 / (1/4) = 32 Therefore total weight loss over a large number of years is 32 kg. Weight after this time would be 104 – 32 = 72 kg
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magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
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how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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The result of a number multiplied by a positive integer is always larger than the original number

The sum of two positive integer is 105 nd the ratio is 2:3 find the number by linear exp
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