please solve this question.. no irrelevant answer please… the sum of three numbers, whose ratios are 3 \times \frac{1}{3} : 4 \times \frac{1}{5} : 6 \times \frac{1}{8} is 4917. find the numbers. About the author Mia
Answer: let the three numbers be x,y,z sum of the three numbers x+y+z = 4917 their ratios x:y:z = 3⅓ : 4⅕ : 6⅛ converting into improper fractions x:y:z = 10/3 : 21/5 : 49/8 they all should have a common factor. let it be k x = 10/3 k y = 21/5 k z = 49/8 k substituting 10/3 k + 21/5 k + 49/8 k = 4917 400k + 504k + 735k / 120 = 4917 1639k / 120 = 4917 k = 3×120 k = 360 applying the values x = 10/3 × 360 x = 1200 y = 21/5 × 360 y = 1512 z = 49/8 × 360 z = 2205 so the numbers are 1200, 1512, 2205 Reply
Answer:
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Answer:
let the three numbers be x,y,z
sum of the three numbers
x+y+z = 4917
their ratios x:y:z = 3⅓ : 4⅕ : 6⅛
converting into improper fractions
x:y:z = 10/3 : 21/5 : 49/8
they all should have a common factor. let it be k
x = 10/3 k
y = 21/5 k
z = 49/8 k
substituting
10/3 k + 21/5 k + 49/8 k = 4917
400k + 504k + 735k / 120 = 4917
1639k / 120 = 4917
k = 3×120
k = 360
applying the values
x = 10/3 × 360
x = 1200
y = 21/5 × 360
y = 1512
z = 49/8 × 360
z = 2205
so the numbers are 1200, 1512, 2205