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Answer :
Let the number be x.
And the other number be y
As given y= 5x and also,
Condition I: x + 21 and 5x + 21
Let the number y be twice of x
Condition II: 5x + 21 = 2 (x + 21)
\(\Rightarrow\) 5x + 21 = 2x + 42 (Solving the bracket)
\(\Rightarrow\) 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)
\(\Rightarrow\) 3x = 21
\(\Rightarrow\) x = 21 ÷ 3 = 7 (Transposing 3 to RHS)
Hence, the required numbers are 7 and 7 × 5 = 35.