Question

8.Eight times the first of three consecutive odd integers is 4 less than thrice the sum of second and third. The second integer is a)7 b)9 c)13 d)11 e)15

Answer 1

Answer:

11

Step-by-step explanation:

Answer 2

${\huge{\sf {\blue{\underline{Answer}}}}}$

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${\orange{\boxed{ (b) \: 9 }}}$

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${\huge{\sf {\blue{\underline{Explanation : }}}}}$

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Let, the three consecutive odd integers be (x+1) , (x+3) , (x+5)

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So,

Eight times the first number = 8(x+1)

Sum of second and third number = (x+3) + (x+5)

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According to Sum,

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8 ( x + 1 ) + 4 = 3 [( x + 3 ) + ( x + 5 )]

↬8x + 8 + 4 = 3 [ x + 3 + x + 5 ]

↬ 8x + 12 = 3 ( 2x + 8)

↬ 8x + 12 = 6x + 24

↬ 8x - 6x = 24 - 12

↬ 2x = 12

↬ x = 6

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Second number =

⇢ ( x + 3 )

⇢ ( 6 + 3 )

⇢ 9

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${\huge{\sf {\blue{\underline{Verification : }}}}}$

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First number = ( x + 1 ) = 7

Second number = 9

Third number = 11

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RHS,

8 ( x + 1 ) + 4

⇢ 8 [ 7 ] + 4

⇢ 56 + 4

⇢ 60

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LHS,

3 [( x + 3 ) + ( x + 5 )]

⇢ 3 [ 9 + 11 ]

⇢ 3 ( 20 )

⇢ 60

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So, LHS = RHS