If the sum of the last 3 integers in a set of 6 consecutive integers

Question

If the sum of the last 3 integers in a set of 6 consecutive integers is 624, what is the sum of the first 3 integers of the set?

A. 600
B. 605
C. 610
D. 615
E. 620

Bunuel · Answer

Say the 6 consecutive integers are x-3, x-2, x-1, x, x+1, x+2.

The sum of the last 3 integers = x + (x+1) + (x+2) = 3x + 3 = 624. 
The sum of the first 3 integers = (x-3) + (x-2) + (x-1) = 3x -6 = (3x + 3) - 9 = 624 - 9 = 615.

StormedBrain · Answer

Let numbers be x, x+1, x+2,x+3,x+4, x+5.

sum of last three = 3x+12 = 624 => 3x= 612.

Sum of first three = 3x+3 = 612+3 = 615.

Nunuboy1994 · Answer

To represent a consecutive series (constantly increasing by 1) let the series be

+   +   + x +   +

x +   +   = 624

3x + 3 =624
 3x = 621
 x= 207

3x -6 = 
 3(207) -6 = 615

Thus

615

amanvermagmat · Answer

Look at the 6 consecutive integers: x, x+1, x+2, x+3, x+4, x+5.

First three: x, x+1, x+2
Last three: x+3, x+4, x+5

As we can see, every number in second series is 3 more than every corresponding number in first series. Overall difference = 3*3 = 9 (since there are three numbers in each series)

So sum of first three numbers = 624 - 9 = 615.

NDND · Answer

the middle number in the last 3 numbers = 624/3 = 208 ==> the last 3 numbers are:207, 208, 209
the first 3 numbers are: 204, 205, 206
the sum of the first 3 integers: 204+205+206= 615

avsiqueiros · Answer

DON'T CASE THIS REPLY

just because I didn't read well the word "consecutive", I went to another conclusion that also could make sense but is totally wrong.

middle number of the last three numbers: 208. The sum of the 4th integer + last number is: 416.

first prime number of 416 is 13 and the sum of 195, 208, 221 = 624

my own trouble

ScottTargetTestPrep · Answer

Solution:

We can create the following equation:

x + 3 + x + 4 + x + 5 = 624

3x = 612

x = 204

Thus, the sum of the first 3 integers is 204 + 205 + 206 = 615.

Alternate solution:

We can let x, x + 1, x + 2, x + 3, x + 4, x + 5 be the six consecutive integers. We see that each of the first 3 integers is 3 less than each of the 3 corresponding last integers. Therefore, the sum of the first 3 integers should be 3 * 3 = 9 less than the sum of the last 3 integers. Since the sum of the latter 3 is 624, the sum of the first 2 is 624 - 9 = 615.

Answer: D

Kinshook · Answer

Asked: If the sum of the last 3 integers in a set of 6 consecutive integers is 624, what is the sum of the first 3 integers of the set?

Let the set of integers be {a-2,a-1,a,a+1,a+2,a+3}

The sum of the last 3 integers in a set of 6 consecutive integers is 624
a + 1 + a + 2 + a + 3 = 3a + 6 = 624

The sum of the first 3 integers of the set = a -2 + a - 1 + a = 3a - 3 = 3a + 6 - 9 = 624 - 9 = 615

IMO D

dimri10 · Answer

X-1+X+ X+1 is 624 X is 208 so
207, 208, 209
206, 205, 204 are the First numbers.
Minus 1 , minus 3, minus 5
So 624-9 is 615

Posted from my mobile device

TheNightKing · Answer

You can just look at the difference.
x,x+1,x+2,x+3,x+4,x+5. So the sum of first 3 is 3x+3 and Sum of last 3 is 3x+12. The difference is 9.

So answer will be 624-9=615

Abhishek009 · Answer

Let the nos be :  ( a - 3 ) , ( a - 2 ) , ( a - 1 ) , a , ( a + 1 ) , ( a + 2)

Sum of the last 3 integers is 624

So,  3a + 3 = 624

Thus,  a = 207

So, the first 3 integers are 204 , 205 & 206, their sum will be 615, Answer must be (D)

If the sum of the last 3 integers in a set of 6 consecutive integers

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If the sum of the last 3 integers in a set of 6 consecutive integers

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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