Please solve all in a copy
Answers
Answer:
If a/b = c/d = e/f = g/h = k then k=a+e+c+gb+d+f+h
If a1 / b1 , a2/b2, a3/b3... an/bn are unequal fractions then we have a1+a2+a3+..+anb1+b2+b3+..+bn lies between the lowest and the highest of these fractions.
If we have two equations containing three unknowns as a1x + b1y + c1z = 0 and a2x + b2y + c2z = 0 then we can find the proportion x : y : z. This will be given by b1∗c2−b2∗c1:a2∗c1−c2∗a1:a1∗b2−a2∗b1
If the ratio ab>1 (called a ratio of greater inequality) and if k is a positive number: a+kb+k<ab and a−kb−k>ab Similarly if ab<1 then a+kb+k>ab and a−kb−k<ab
Maintenance of equality when numbers are added in both the numerator and the denominators. This if best illustrated through an example: 2030=20+230+3 i.e. ab=a+cb+d if and only if cd=ab
Consequently, if cd>ab then a+cb+d>ab and if cd<ab then a+cb+d<ab
To get the consolidated ratio A:B:C:D:E when individual ratios A:B, B:C, C:D, D:E are given
A:B = 2:3, B:C = 4:5, C:D = 6:11, D:E = 12:17
In order to create one consolidated ratio for this situation using the LCM process becomes too long. The short cut goes as follows:
A will correspond to the product of all numerators (2 × 4 × 6 × 12) while B will take the first denominator and the last 3 numerators (3 × 4 × 6 × 12). C on the other hand takes the first two denominators and the last 2 numerators (3 × 5 × 6 × 12), D takes the first 3 denominators and the last numerator (3 × 5 × 11 × 12) and E takes all the four denominators (3 × 5 × 11 × 17).
In mathematical terms this can be written as:
ab=N1D1,bc=N2D2,cd=N3D3,de=N4D4 then a:b:c:d:e=N1N2N3N4:D1N2N3N4:D1D2N3N4:D1D2D3N4:D1D2D3D4