12/720 has even numbers in num. and denom. It should be reduced.
The most efficient way of reducing fractions is to find GCD.
How is GCD to be found is shown in ry1. After you get experienced, you may use fractions' utilities. In this case it is 12.
The second step is to get equal denominators in all fractions taking part in addition/subtraction. To this end LCM is needed.
How is LCM to be found is shown in ry2.
Just place the numbers into their correct places .
We want you to get acquainted with the matrices via this simple operation .
Neglect the upper half and divide the result you obtained by 2 .
Sum of positive terms: 22 2/5 .
It is not posssible to reduce 2/945; try if 1680 = 700 + 700 + 280 is divisible by 7 .
Finding the GCD of 7 and 1680 is in ry1 and finding LCM is in ry2. Just as an exmple .
Take the simplest case of 80 seats. In Theatre 2 and 3 there should be 53 seats. 53/2 = 26 1/2 .
There should be 15 more in T-2 which means 26 1/2 - 7 1/2 = 19 seats in T-3. And of course 34 in T-2 .
Before food he must have $300. Let the price of PSP be P, (1+1/2)xP should be $200
A-36 should be divisible by A. 1 is an integer which means 36/A should be an integer. 36 is divisible by A if A=1, ...
Each element of the numerator is twice the corresponding element in the denominator .
1/2 - 3/20 + 3/28 - 9/110 = (110-33-18)/220 + 3/28 = 59/220 + 3/28...
Informal solution: Posssbile weights are 3 and 4 , 6 and 8 , 9 and 12 , 12 and 16 ... Third is O.K.
For the formal solution have a look at ry1.
3/5 - 1/2 = 1/10 .
Let Janet's allowance be A_{J} and Janet's A_{T}. A_{J}/A_{T} = 5/6. What is true for a year should also be valid for a
month since the rate is the same over the year. In a month Janet saves A_{T}/12 which should be equal to (5xA_{T})/6xC . This says 1/12 = 5/6xC ...
These type of problems are more easily solved by trial and error. Just try the answers .
Let the amount of mixture be B at the beginning and E at the end. We have to find 2 equations:
1. E = B - 20 .
2. (3/10)xE + 20 = B/2.
Putting 1 in 2 we have (3/10)x(B-20) + 20 = B/2 which means B/2 - 3xB/10 = 14
In a second the machine does 1/16 ths of the whole lifting. After 15 seconds 1/16 ths of the work (w.r.t. the machine) remains.
The man does the work in 240 seconds. What the machine does in a second, he does in 240/16 = ... seconds . See ry1 also.
The last two are not very likely .
The first two are 7 and 8 and 43/5 is 8 3/5
5/(A+5) ≥ 1 1/2 . The first is certainly less than 1 as long as A is natural number.
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