Question

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​

Answer 1

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 2

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