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

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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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, \(c=-\frac{4}{3}\)

\(a+b+c+d=\)

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

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

Answer B
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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.

Answer (B)
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chetan2u, Bunuel, VeritasKarishma, Gladiator59, generis

Please Help needed why its should be B not D


Regards

K.y.Shrenik Bimbsar

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.



Posted from my mobile device
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Thanks

Regards
K.y.Shrenik Bimbsar

Thanks

Regards
K.y.Shrenik Bimbsar

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

Answer (B)

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

Posted from my mobile device

Hi "VeritasKarishma"

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

Thanks in advance.

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

Adding the first 4 terms = 4 times the last term

(a + 1) + (b + 2) + (c + 3) + (d + 4) = 4 * (a + b + c + d + 5)

(a + b + c + d) + 10 = 4 * (a + b + c + d) + 20

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

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

Option B

Arun Kumar
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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) --> correct
C) 12
D) 6
E) -2

Solution:
a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5 = k i.e let say x = a + b + c + d, x+5 =k ---(i)
=>k + k + k + k + k = 5k
(a + 1)+(b + 2)+(c + 3)+(d + 4)+(a + b + c + d + 5) = 5k = 5*(x+5) from (i)
=> x + 1+2+3+4+ x + 5 = 5x+25
=> 2x+15 = 5x+25
=> 3x = -10
=> x = -10/3

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

=> 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, we get

a + b + c + d + 10 = 4 (a + b + c + d + 5)

=> a + b + c + d = 4((a + b + c + d ) + 20 - 10

=>a + b + c + d = 4((a + b + c + d ) + 20 - 10

=>a + b + c + d = 4((a + b + c + d ) + 10

=> -10 = 4 (a + b+ c+ d) - (a + b+ c+ d)

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

=> a + b + c + d = \(\frac{-10}{3}\)

Answer B

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