two numbers are in the ratio of 3:5 and if 10 be subtracted from each of them,the remainders are in the ratio of 1:5,find the numbers
Answers
Let the two numbers be 3x and 4x.
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25 ⟹21x−20x=−25+35
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25 ⟹21x−20x=−25+35 ⟹x=10
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25 ⟹21x−20x=−25+35 ⟹x=10 The required numbers are
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25 ⟹21x−20x=−25+35 ⟹x=10 The required numbers are3x = 3×10 = 30 and
Let the two numbers be 3x and 4x.Subtract 5 from each of these numbers. We get 3x-5 and 4x-5. From the given problem, the ratio of these new numbers is 5:7. Then,3x−54x−5=57 Cross-mutiplying,(3x−5)×7=(4x−5)×5 Multiplying 7 inside 3x-5 and 5 inside 4x-5,21x−35=20x−25 ⟹21x−20x=−25+35 ⟹x=10 The required numbers are3x = 3×10 = 30 and4x = 4×10 = 40.
Answer:
12 and 20
Step-by-step explanation:
let the numbers = 3x and 5x
According to question
3x-10/5x-10=1/5
by cross multiplication
15x-50=5x-10
=>10x=40
=>x=4
So the numbers are
3x=3×4=12
5x=5×4=20