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Heix
Joined: 21 Feb 2024
Last visit: 13 Apr 2026
Posts: 460
Own Kudos:
Given Kudos: 63
Location: India
Concentration: Finance, Entrepreneurship
Schools: Ross MM "26 NUS Columbia '27
GMAT Focus 1: 485 Q76 V74 DI77 
GPA: 3.4
WE:Accounting (Finance)
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11 Jul 2025, 09:18
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We need to find the range of values for the expression: lala - 2blbl + 3lcic - 4didi + 5labcdlabcd
First, let's simplify each term by recognizing that for any non-zero number x:
IxIx = x*sign(x)
So our expression becomes:
a? sign(a) - 2b*sign(b) + 3c?sign(c) - 4d/sign(d) + 5(abcd)? sign(abcd)
Since we're looking for the range and a, b,c, and d are non-zero but can be any values, we can simplify by setting lal = IbI = Icl = Idl =1. This doesn't affect the range because we can
always scale these values to achieve any specific result within the range.
With this simplification, our expression becomes:
sign(a) - 2 sign(b) + 3-sign(c) - 4-sign(d) + 5 sign(abcd)
Let's denote:
sign(a) = A (either 1 or -1)
sign(b) = B (either 1 or -1)
sign(c) = C (either 1 or -1)
sign(d) = D (either 1 or -1)
Then sign(abcd) = A-B-C-D
Our expression is now:
A - +3C- 4D+ 5(A-B.C-D)
A - 2B + 3C - 4D + 5(A-B-C-D)
To find the maximum value, we need to maximize each term:
For A: choose A = 1 to get +1
For -2B: choose B = -1 to get +2
For 3C: choose C = 1 to get +3
For -4D: choose D = -1 to get+4
For 5(A-B.C-D): with A=1, B--1, C=1, D=1, we get 5(1-11-1) = 5(1) = +5
Maximum value = 1 + 2+3+4+5 = 15
To find the minimum value, we need to minimize each term:
For A: choose A = -1 to get -1
For -2B: choose B= 1 to get get-2
For 3C: choose C=-1 to get -3
For -4D: choose D= 1 to get 4
For 5(A-B-C-D): with A=-1, B=1, C--1, D=1, we get 5(-1-1-1-1) = 5(-1) = -5
Minimum value = -1 - 2-3-4 - 5=-15
Therefore, the range of the expression is [-15, 15], which means the range is 15 - (-15) = 30.
The answer is E) 30.
First, let's simplify each term by recognizing that for any non-zero number x:
IxIx = x*sign(x)
So our expression becomes:
a? sign(a) - 2b*sign(b) + 3c?sign(c) - 4d/sign(d) + 5(abcd)? sign(abcd)
Since we're looking for the range and a, b,c, and d are non-zero but can be any values, we can simplify by setting lal = IbI = Icl = Idl =1. This doesn't affect the range because we can
always scale these values to achieve any specific result within the range.
With this simplification, our expression becomes:
sign(a) - 2 sign(b) + 3-sign(c) - 4-sign(d) + 5 sign(abcd)
Let's denote:
sign(a) = A (either 1 or -1)
sign(b) = B (either 1 or -1)
sign(c) = C (either 1 or -1)
sign(d) = D (either 1 or -1)
Then sign(abcd) = A-B-C-D
Our expression is now:
A - +3C- 4D+ 5(A-B.C-D)
A - 2B + 3C - 4D + 5(A-B-C-D)
To find the maximum value, we need to maximize each term:
For A: choose A = 1 to get +1
For -2B: choose B = -1 to get +2
For 3C: choose C = 1 to get +3
For -4D: choose D = -1 to get+4
For 5(A-B.C-D): with A=1, B--1, C=1, D=1, we get 5(1-11-1) = 5(1) = +5
Maximum value = 1 + 2+3+4+5 = 15
To find the minimum value, we need to minimize each term:
For A: choose A = -1 to get -1
For -2B: choose B= 1 to get get-2
For 3C: choose C=-1 to get -3
For -4D: choose D= 1 to get 4
For 5(A-B-C-D): with A=-1, B=1, C--1, D=1, we get 5(-1-1-1-1) = 5(-1) = -5
Minimum value = -1 - 2-3-4 - 5=-15
Therefore, the range of the expression is [-15, 15], which means the range is 15 - (-15) = 30.
The answer is E) 30.
11 Jul 2025, 09:21
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for range we have to find max. value and min value
Let's find max value and to find it we need sum of positive integers to get the maximum number, lets consider |a|>0, |b|<0 because it has -ve sign associate so we can take -ve value to make it positive, |c|>0, |d|<0 same logic applies as |b|, |abcd|>0 because two negatives *two positives is positive
Now the max value will be 1+2+3+4+1=15
Let's find the min value and for the min. value we find max number in negative integers
if we take |a|<0, |b|>0, |c|<0. |d|>0 then |abcd|>0 because two negatives *two positives is positive
then it would be -1-2-3-4+5=-5
from above number we can still see it can be reduced further if we make 5 as -5 and for this we need |abcd|<0 so we need to make one variable +ve and 3 -ve that way we can get |abcd|<0 and we should make lower absolute value as +ve to get the min number so we take |a|>0 as it has the lowest absolute multiple value associate with it
Now, +1-2-3-4-5=-13
So range is max-min=15-(-13)=28 so answer is D
Let's find max value and to find it we need sum of positive integers to get the maximum number, lets consider |a|>0, |b|<0 because it has -ve sign associate so we can take -ve value to make it positive, |c|>0, |d|<0 same logic applies as |b|, |abcd|>0 because two negatives *two positives is positive
Now the max value will be 1+2+3+4+1=15
Let's find the min value and for the min. value we find max number in negative integers
if we take |a|<0, |b|>0, |c|<0. |d|>0 then |abcd|>0 because two negatives *two positives is positive
then it would be -1-2-3-4+5=-5
from above number we can still see it can be reduced further if we make 5 as -5 and for this we need |abcd|<0 so we need to make one variable +ve and 3 -ve that way we can get |abcd|<0 and we should make lower absolute value as +ve to get the min number so we take |a|>0 as it has the lowest absolute multiple value associate with it
Now, +1-2-3-4-5=-13
So range is max-min=15-(-13)=28 so answer is D
Bunuel
11 Jul 2025, 09:28
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Range means we have to find the difference between the lowest and highest possible values of this exp.
Now the lowest value will be when the maximum terms and highest value terms are negative
for lowest value we see that we can either have a or c <0
(we can't have both as it will have the last term as positive. we will just take one negative and it will be c as it has a multiply of 3 before it
so we will get 1-2-3-4-5=-13
for maximum value we will take b , d >0
then we will have
+ (-)(-) + (-)(-) + (-)(-) [just shown the sign for better understanding]
1++2+3+4+5=15
range of value 15-(-13)=28 option D
Now the lowest value will be when the maximum terms and highest value terms are negative
for lowest value we see that we can either have a or c <0
(we can't have both as it will have the last term as positive. we will just take one negative and it will be c as it has a multiply of 3 before it
so we will get 1-2-3-4-5=-13
for maximum value we will take b , d >0
then we will have
+ (-)(-) + (-)(-) + (-)(-) [just shown the sign for better understanding]
1++2+3+4+5=15
range of value 15-(-13)=28 option D
Bunuel
