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monirjewel
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
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Question Stats:
86% (01:05) correct
14%
(01:05)
wrong
based on 663
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What is the sum of the integers from -190 to 195 inclusive?
A) 0
B) 5
C) 375
D) 875
E) 965
A) 0
B) 5
C) 375
D) 875
E) 965
From -190 to +195 we know that all numbers from 1 to 190 will have both positive and negative terms. So they will just cancel each other out.
The numbers that remain will be 191, 192,193, 194 and 195 whose sum can be easily calculated. The answer is E.
The numbers that remain will be 191, 192,193, 194 and 195 whose sum can be easily calculated. The answer is E.
^^ Great explanation. My own approach was to use the formula (avg of 1st and last number) * total number . But I got stuck.
195 - 191/2 * 5 = 965
E
E
ulm
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-190+190=0
-189+189=0
...
we still have 191+192+193+194+195=965
(E)
-189+189=0
...
we still have 191+192+193+194+195=965
(E)
Posts: 17
What is the sum of the integers from -190 to 195, inclusive?
a) 0
b) 5
c) 375
d) 875
e) 965
a) 0
b) 5
c) 375
d) 875
e) 965
This can be viewed as a sum of consecutive terms problem
the sequence consists of integers between -191 and 196
the sequence starts at -190 and ends at +195
in an arithmetic progression,
the nth term is given by tn=a+(n-1)d
here tn=195, a=-190, d=1
hence, 195=-190+(n-1)
or n=386
Sum of n terms can be calculated by sn=n/2(a+l)
a=first term, l=last term, n=no. of terms
sn=386*(-190+195)/2
sn=193*5
sn=965
the sequence consists of integers between -191 and 196
the sequence starts at -190 and ends at +195
in an arithmetic progression,
the nth term is given by tn=a+(n-1)d
here tn=195, a=-190, d=1
hence, 195=-190+(n-1)
or n=386
Sum of n terms can be calculated by sn=n/2(a+l)
a=first term, l=last term, n=no. of terms
sn=386*(-190+195)/2
sn=193*5
sn=965
the series is in ap.
AP - sum of the series = (n/2)(2a+ (n-1)d)
n = 190+1+195 (190 negative terms, 1 zero and 195 positive terms)
= 386
a = -190
d = 1
=> sum of this series = (386/2)(-380+385) = 965
Answer is E.
AP - sum of the series = (n/2)(2a+ (n-1)d)
n = 190+1+195 (190 negative terms, 1 zero and 195 positive terms)
= 386
a = -190
d = 1
=> sum of this series = (386/2)(-380+385) = 965
Answer is E.
Sum of -190 to +190 will be zero. You just have to sum up 191 t0 195 which is 965.
Posts: 29
-190+(-189)+(-188+).....+(-4)+(-3)+(-2)+(-1) + 0+1+2+3+4+........+188+189+190+191+192+193+194+195
Everything will cancel out except the last 5 numbers(191+192+193+194+195)
So, the ans will be 965.
Everything will cancel out except the last 5 numbers(191+192+193+194+195)
So, the ans will be 965.
msvel2304
Joined: 21 Sep 2011
Last visit: 12 Oct 2014
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Concentration: Entrepreneurship, General Management
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GMAT 1: 530 Q42 V20
GMAT 2: 540 Q43 V28
GMAT 3: 680 Q48 V35
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Schools: Babson '15 (WL) Kelley '15 (D) Carlson '15 (A) Rotman '15 (A) ISB '14 (D) Wake Forest '15 (WL)
GMAT 3: 680 Q48 V35
Posts: 90
IMO E.
-190 to +190 everything will cancel out.
Sum of 191 to 195 = 965
-190 to +190 everything will cancel out.
Sum of 191 to 195 = 965
as the list is consecutive integers, we can use following formulas:
sum/n= average.
sum=(average)(n)
average=a+b/2=190+195/2=2.5
number of items(n)=B-A+1=195-(-190)+1=195+191=386.
sum=average*n=2.5*386=965.
answer is E
sum/n= average.
sum=(average)(n)
average=a+b/2=190+195/2=2.5
number of items(n)=B-A+1=195-(-190)+1=195+191=386.
sum=average*n=2.5*386=965.
answer is E
-190 to +190 will cancel out each other
Sum= 190*5 +(1+2+3+4+5)= 950+15= 965
E
Sum= 190*5 +(1+2+3+4+5)= 950+15= 965
E
What is the sum of integers from -190 to 195, inclusive?
A 0
B 5
C 375
D 875
E 965
A 0
B 5
C 375
D 875
E 965
yuerchang
integers from 1 to 190 gets cancelled from similar negative values
left only 191+192+193+194+195 = 965
Ans E
The series is: -190, -189, -188,........ -2, -1, 0, 1, 2, ........ 188, 189, 190, 191, 192, 193, 194, 195;
Hence, the -ve and +ve values from -190 to +190 cancel each other leaving only 191, 192, 193, 194, 195;
So, the sum of the series:
191 + 192 + 193 + 194 + 195 = 190*5 + (1+2+3+4+5) = 950 + 15 = 965; Answer E.
_________________
Hence, the -ve and +ve values from -190 to +190 cancel each other leaving only 191, 192, 193, 194, 195;
So, the sum of the series:
191 + 192 + 193 + 194 + 195 = 190*5 + (1+2+3+4+5) = 950 + 15 = 965; Answer E.
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BillyZ
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yuerchang
SUM = AVERAGE x NUMBER OF DATA POINTS
\(Average = \frac{(-190 + 195)}{2} = \frac{5}{2} = 2.5\)
Number of data point \(= 195 - (-190) + 1 = 386\)
\(Sum = 2.5 * 386 = 965\)
The answer is E.
yuerchang
Hi,
We know -190 to 190 will cancel out..
now what is important is how fast you can calculate 191+192+193+194+195
There are five terms which are close to 200, so ans will be close to 5*200=1000...
965 should be the answer..
If you want exact answer, it may be beneficial to calculate how far answer will be from 1000..
So they are 5+6+7+8+9 away, this is 35...
And 1000-35=965...
E
_________________
Abhishek009
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yuerchang
Integers from -190 to 190 is -
( -190 , -189 , -188...........-2 , -1 ,0 ) + ( 0 , 1 , 2 ............188 , 189 , 190) = 0
Sum of integers from -190 to 195 = Sum of Integers from -190 to 190 ( ie, 0 ) + 191 + 192 + 193 + 194 + 195 => 965
Hence, answer must be (E) 965
_________________
ScottTargetTestPrep
Expert
Target Test Prep Representative
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Expert reply
Posts: 22,738
yuerchang
In determining the sum of the integers from -190 to 195 inclusive, we should recognize that all of the negative integers will cancel out with their positive counterparts.
For instance, we have -190 and 190, -150 and 150, -10 and 10, etc.
Thus, the only positive numbers that won’t cancel out with their negative counterparts are 191, 192, 193, 194, and 195.
To determine the sum of these numbers, we can use this formula: sum = average x quantity.
Since we have an evenly spaced set, the average is equal to the median, which is 193, and the quantity is 5.
Thus, the sum = 193 x 5 = 965.
Answer: E
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