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Find four consecutive even integers whose sum is 42.

Find three consecutive integers such that twice the smallest is 12 more than the largest.

2006-11-11 05:42:02 · 7 answers · asked by Rani 1 in Science & Mathematics Mathematics

7 answers

Answer 1:
Let the four integers be 'x', 'x+1', 'x+2' and 'x+3'
x + x + 1 + x + 2 + x + 3 = 42
4x + 6 = 42
4x = 42 - 6
4x = 36
x = 36/4
x = 9
The integers are 'x', 'x+1', 'x+2' and 'x+3'
The integers are 9, 10, 11 and 12

Answer 2:
Let the integers be 'x', 'x+1' and 'x+2'
2x = (x + 2) + 12
2x = x + 14
2x - x = 14
x = 14
The integers are 'x', 'x+1' and 'x+2'
The integers are 14,15 and 16

2006-11-11 16:09:32 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0 1

Answer-1
well, there are no consecutive 4 even integers whose sum is 42.


Let the first integer be 2n, then
2n + (2n+2) + (2n+4) + (2n+6) = 42
=> 4*2n= 42-12
=> 8n= 30
=> n= 30/8 which is not an integer.

Answer-2

Let the first integer be n, then the following integers are n+1, n+2.

SInce, twice the smallest is 12 more than the largest
2n-(n+2)=12
=>2n-n-2=12
=>2n-n=12+2=14
=>n=14
Thus, the required three consecutive integers are 14, 15,16.
Observe that 2*14=28 and 28-16=12.

I hope this help you.

All the best.

2006-11-11 06:02:33 · answer #2 · answered by Paritosh Vasava 3 · 1 0

42/ 4 = about 10 so try numbers around 10
9+10+11+12 = 42

2n -12 = n+2 {some number,n, 2n-12 is equal to n+2 }
n =14 2*14=28
28-12 =16

14, 15, 16 works
not sure why you need the middle number

2006-11-11 05:56:45 · answer #3 · answered by metaraison 4 · 0 0

The formula is

n + n + 1 + n + 2 + n + 3 = 42

4n + 6 = 42

4n + 6 - 6 - 42 - 6

4n = 36

4n/4 = 36/4

n = 9 The answer is n = 9

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The numbers are 9 10 11 and 12

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2006-11-11 06:46:26 · answer #4 · answered by SAMUEL D 7 · 0 0

let the even cosecutive integers be
n,n+2,n+4,n+6
adding and equating to42
4n+12=42
adding -12
4n=30
n=7.5
sum of four consecutive even integers cannot be 42

2.let the three consecutive integers be
n-1,n,n+1
2(n-1)=12+n+1
2n-2=n+13
ading 2
2n=n+15
adding -n
n=15
the integers are 14,15 1nd 16

2006-11-11 05:57:14 · answer #5 · answered by raj 7 · 0 0

A)4.00x10^2 B)6.0000x10^4 C)7.50000x10^5 D)5x10^-3 E)3.4x10-3 F)6.457x10-2 All you're able to desire to do is pass the decimal till there's a single digit. Then, you multiply by ability of 10 to the nth capability. in case you're able to desire to pass forward the exponent would be beneficial and in case you progression back the exponent would be detrimental.

2016-10-21 22:06:36 · answer #6 · answered by ? 4 · 0 0

9,10,11,12
14,15,16

2006-11-11 05:52:19 · answer #7 · answered by mjbayunl 2 · 0 0

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