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Senior Manager  V
Joined: 25 Dec 2018
Posts: 484
Location: India
Concentration: General Management, Finance
GMAT Date: 02-18-2019
GPA: 3.4
WE: Engineering (Consulting)
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 65% (02:39) correct 35% (02:48) wrong based on 127 sessions

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Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?

A) -1
B) −(10/3)
C) 12
D) 6
E) -2
Intern  B
Joined: 07 Apr 2018
Posts: 4
Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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a+1=a+b+c+d+5
b+2=a+b+c+d+5
c+3=a+b+c+d+5
d+4=a+b+c+d+5

Adding all the equations we get:
a+b+c+d+10=4(a+b+c+d)+20

Solving for (a+b+c+d) we get:

a+b+c+d=(-10/3)

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Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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1

Please Help needed why its should be B not D

Regards

K.y.Shrenik Bimbsar
Senior PS Moderator D
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 737
GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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Hi ShrenikBimbsar,

When so many variables are present the first order of business should be to convert everything into a single variable. This is easily possible in this case as multiple equations are given.

It comes out that,
b = a-1 , c= a-2 & d= a-3

On substituting the above and converting the everything into a it comes out that a = 2/3

And a+b+c+d = 4a - 6

Hence the answer is -10/3.

Hope it helps.

ShrenikBimbsar wrote:

Please Help needed why its should be B not D

Regards

K.y.Shrenik Bimbsar

Posted from my mobile device
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Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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Thanks

Regards
K.y.Shrenik Bimbsar
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Joined: 13 Sep 2018
Posts: 21
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Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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Hi ShrenikBimbsar,

When so many variables are present the first order of business should be to convert everything into a single variable. This is easily possible in this case as multiple equations are given.

It comes out that,
b = a-1 , c= a-2 & d= a-3

On substituting the above and converting the everything into a it comes out that a = 2/3

And a+b+c+d = 4a - 6

Hence the answer is -10/3.

Hope it helps.

ShrenikBimbsar wrote:

Please Help needed why its should be B not D

Regards

K.y.Shrenik Bimbsar

Posted from my mobile device

Thanks

Regards
K.y.Shrenik Bimbsar
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3536
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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mangamma wrote:
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?

A) -1
B) −(10/3)
C) 12
D) 6
E) -2

You do not have to solve this problem all the way.

(1) Express three of the variables in terms of one other.
Use $$a$$ or $$d$$ for the sake of visual simplicity; both are on one end of the first string of equalitites.

$$d+4=c+3$$
$$c=(d+1)$$

$$d+4=b+2$$
$$b=(d+2)$$

$$d+4=a+1$$
$$a=(d+3)$$

(2) Solve for $$d$$ using the final equality.
Substitute for $$a, b,$$ and $$c$$
$$(d+4)=a+b+c+d+5$$
$$d+4=d+3+d+2+d+1+d+5$$
$$d+4=4d+11$$
$$-7=3d$$
$$3d=-7$$
$$d=-\frac{7}{3}$$

Stop. The answer must be a fraction with a denominator of 3.
If we add an integer to a fraction, the answer is a fraction with the same denominator.

Only one option has a denominator of 3.

The answer is (B).

Why stop?

$$a, b,$$ and $$c$$ all have values that equal
$$d$$(a fraction) + an integer

Test: If we add $$1$$ to $$-\frac{7}{3}$$, then $$1$$ must be changed to $$\frac{3}{3}$$

$$c=(d+1)$$
$$c=(-\frac{7}{3})+(\frac{3}{3})$$
$$c=-\frac{4}{3}$$
$$c$$ = another fraction with a denominator of 3.
$$a$$ and $$b$$ will also be fractions with a denominator of 3.
-- The sum of all four variables will be a fraction with a denominator of 3.
-- Only one option has a denominator of 3.
The answer is (B).

(3) if you are not sure, do all of the arithmetic

Find each value

$$a=(d+3)$$
$$a=(-\frac{7}{3})+(\frac{9}{3})$$
$$a=\frac{2}{3}$$

$$b=(d+2)$$
$$b=(-\frac{7}{3})+(\frac{6}{3})$$
$$b=-\frac{1}{3}$$

From above, $$c=-\frac{4}{3}$$

$$a+b+c+d=$$

$$\frac{2}{3}+((-\frac{1}{3})+(-\frac{4}{3})+(-\frac{7}{3}))=$$

$$\frac{2}{3}+(-\frac{12}{3})=(\frac{2}{3}-\frac{12}{3})=-\frac{10}{3}$$

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Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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mangamma wrote:
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?

A) -1
B) −(10/3)
C) 12
D) 6
E) -2

So this is what I observed:

a + 1 = b + 2 = c + 3 = d + 4
This means the values of the variables are reducing by 1 as we move to the right. So b is 1 less than a. c is 1 less than b and so on.
So a, b, c and d could be something like 4, 3, 2, 1 respectively.

d + 4 = a + b + c + d + 5
a + b + c + d = d - 1

So (a + b + c + d) is further 1 less than d so taking the above example, it will go down one step further and be 0.
Hence, a, b, c, d, a+b+c+d are consecutive numbers in decreasing order.

Now this is not possible if a, b, c and d are all positive numbers since their sum will be greater than individual numbers.
So a+b+c+d cannot be 6 or 12.

If a+b+c+d = -1, then d = 0, c = 1, b = 2 and a = 3
Does not satisfy.

If a+b+c+d = -2, then d = -1, c = 0, b = 1 and a = 2
Does not satisfy.

If a+b+c+d = -(10/3), then d = -7/3, c = -4/3, b = -1/3 and a = 2/3
Satisfies.

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Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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a+1 = b+2
=> b = a-1 ——— (1)
a+1 = c+3
=> c = a-2 ——— (2)
a+1 = d+4
=> d = a-3 ——— (3)

a+1 = a+b+c+d+5
=> b+c+d = -4 ——— (4)

(1)+(2)+(3)
=> b+c+d = 3a-6 ——— (5)

From (4) and (5)
3a-6 = -4
=> a = 2/3 ——— (6)

From (4) and (6)
a+b+c+d = 2/3 - 4
=> -10/3

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Joined: 25 Feb 2019
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Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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mangamma wrote:
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?

A) -1
B) −(10/3)
C) 12
D) 6
E) -2

I have attached the solution.

All are equal, let us assume that all are equal to K (any constant)

we can find the value of K by solving equation in the picture

amd then find the value of required expression

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File comment: Solution IMG_20190312_220112.jpg [ 1.23 MiB | Viewed 965 times ]

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Posts: 17
Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is  [#permalink]

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mangamma wrote:
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?

A) -1
B) −(10/3)
C) 12
D) 6
E) -2

So this is what I observed:

a + 1 = b + 2 = c + 3 = d + 4
This means the values of the variables are reducing by 1 as we move to the right. So b is 1 less than a. c is 1 less than b and so on.
So a, b, c and d could be something like 4, 3, 2, 1 respectively.

d + 4 = a + b + c + d + 5
a + b + c + d = d - 1

So (a + b + c + d) is further 1 less than d so taking the above example, it will go down one step further and be 0.
Hence, a, b, c, d, a+b+c+d are consecutive numbers in decreasing order.

Now this is not possible if a, b, c and d are all positive numbers since their sum will be greater than individual numbers.
So a+b+c+d cannot be 6 or 12.

If a+b+c+d = -1, then d = 0, c = 1, b = 2 and a = 3
Does not satisfy.

If a+b+c+d = -2, then d = -1, c = 0, b = 1 and a = 2
Does not satisfy.

If a+b+c+d = -(10/3), then d = -7/3, c = -4/3, b = -1/3 and a = 2/3
Satisfies.

I have taken another approach, what will be your suggestion?

Attachments 20190322_170735.jpg [ 2.11 MiB | Viewed 677 times ] Re: Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is   [#permalink] 22 Mar 2019, 09:14
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