A, B, C, D, E, F, G, and H are all integers, listed in order of increa

Question

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to \(5^{12}\) and \(5^{13}\), respectively, what is the value of A?

A. \(-24(5^{12})\)

B. \(-23(5^{12})\)

C. \(-24(5^6)\)

D. \(23(5^{12})\)

E. \(24(5^{12})\)

A.  -24(5^{12})

B.  -23(5^{12})

C.  -24(5^6)

D.  23(5^{12})

E.  24(5^{12})

PareshGmat · Accepted Answer

H = 5^{13}

G = 5^{12}

H - G = 5^{13} - 5^{12} = 4 * 5^{12}

A   ...... 1 .......   B   ........ 2 .......   C   ........ 3 .........   D   .......... 4 ...........   E   ...... 5 .......   F   ........ 6 .........   G   ...................   H

From point A,  4 * 5^{12}  has to be added 6 times to reach point  G = 5^{12}

A + 6* 4 * 5^{12} = 5^{12}

A = -23 * 5^{12}

Answer = B

BrentGMATPrepNow · Answer

H = 5^13
G = 5^12

Distance between H and G = 5^13 - 5^12
= 5^12(5^1 - 1)
= (5^12)(4)
=  (4)(5^12)

So, F = 5^12 -  (4)(5^12) 
E = 5^12 -  (4)(5^12)  -  (4)(5^12)  
D = 5^12 -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12) 
C = 5^12 -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12) 
B = 5^12 -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12) 
A = 5^12 -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)  -  (4)(5^12)

So, A = 5^12 -  (4)(5^12)  +  (4)(5^12)  +  (4)(5^12)  +  (4)(5^12)  +  (4)(5^12)  +  (4)(5^12) ] 
= 5^12 -  (24)(5^12) 
= (1)(5^12) -  (24)(5^12) 
= (-23)(5^12)
= Answer choice B

Cheers,
Brent

sreelakshmigs · Answer

Ans : B

Assume that the numbers appear as shown below on the number line

A-----B-----C-----D-----E-----F-----G-----H
 | |
 (5^12) (5^13)

As the values for G and H are given , we can calculate the difference between any two terms of the series .

Common Difference ,d = (5^13) - (5^12) 
 = (5^12) * 
 = (5^12)*(4)

Also F + d = G as the terms are in equidistant and in increasing order.

So F + (5^12)*(4) = (5^12).

That is , F = (5^12) - (5^12)*(4) 
 = (5^12) 
 = (5^12) (-3)

Similarly , E = F - d
 = (5^12) 
 = (5^12)*(-7)

You can see a -4 getting added to the non-exponent part of the values . That is , according to the pattern , D SHOULD BE (5^12)*(-7-4)= (5^12)*(-11)
Following this pattern , A = (5^12)*(-23)

Ans : B

JeffTargetTestPrep · Answer

We are given that A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size, and that the distance between any two consecutive numbers, which we can denote by the variable n, is constant.

Since G and H are equal to 5^12 and 5^13, respectively, then the distance between them is:

n = 5^13 - 5^12

n = 5^12(5 - 1)

n = 5^12(4)

Since A is 6 numbers ahead of G, A is 6n less than G, and therefore:

A = G - 6n

A = 5^12 - 6

A = 5^12

A = 5^12  = -23(5^12)

Answer: B

aashaybaindurgmat · Answer

How I solved it is:
A, B, C, D, E, F, G, H (Total of 8 numbers = n)
Since its a sequence and each term is a certain constant more than the other.
H - G = 5^13 - 5^12 = (5^12) (5-1) = 4 x (5^12) = Our common difference, d.

Also unrelated, 
 G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12) 
As n=7, because its the 7th term in the sequence

Therefore,
A = (5^12) - 24 (5^12)
 = (5^12) (1-24) Taking 5^12 common  
 = - 23 (5^12)
 = Answer choice B

GMATinsight · Answer

Values of G and H are 5^12 and 5^13, respectively, and there is only one gap between them

so each gap between any two adjacent numbers = 5^13 - 5^12 = 5^12 (5-1) = 4* 5^12

order of numbers is A___B___C___D___E___F___G___H

A is 6 gaps smaller than G

i.e. A = 5^12 - 6*(4* 5^12) = -23* 5^12

Answer: option B

Izzyjolly · Answer

Let distance between any two consecutive numbers is d.
Distance between G and H is  d= 5^{13} - 5^{12} = 4*5^{12}

Sequence {A, B, C, D, E, F, G, H}
A is the first term of sequence, G is the 7th term of sequence, thus:  A +6d = G 
--> A + 6*4*5^{12} = 5^{12}  -->  A + 24*5^{12} = 5^{12} --> A = (-23)*5^{12}

Answer: B.

pandeyashwin · Answer

Diff between last and second last term = 5^13 - 5^12 = 4 * 5^12
Since difference is constant , we can use decreasing AP. (changing this order from ascending to descending)
So 1st term is 5^13 and difference = - 4 * 5^12
= a + (n-1) d
= 5^13 + 7 * (- 4 * 5^12)
= 5^13 - 28* 5^12
= 5^12 * -23

GMATGuruNY · Answer

A-B-C-D-E-F-G-H

Since the values above are equally spaced:
H-A = 7(H-G)
-A = 7H - 7G - H
-A = 6H - 7G
A = 7G - 6H =  7*5^{12} - 6*5^{13} = 5^{12}  = 5^{12}(-23)

B

GmatPoint · Answer

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to \(5^{12}\) and \(5^{13}\), respectively, what is the value of A?

A. \(-24(5^{12})\)

B. \(-23(5^{12})\)

C. \(-24(5^6)\)

D. \(23(5^{12})\)

E. \(24(5^{12})\)

﻿Given the numbers are listed in the increasing order with an equal difference.
﻿Hence the difference between the two consecutive numbers remains the same.
﻿If H =  5^{13}  and G =  5^{12} . The difference is  5^{13}-5^{12}\ =\ 5^{12}\left(4\right) 
﻿Hence F =  5^{12}-4\cdot5^{12}=\ -3\cdot5^{12} 
﻿E =  ﻿\ -3\cdot5^{12}-4\cdot5^{12}=\ -7\cdot5^{12} 
﻿D =  \ -11\cdot5^{12} 
﻿C =  \ -15\cdot5^{12} 
﻿B =  \ -19\cdot5^{12} 
﻿A =  \ -23\cdot5^{12}

dipanjandas792 · Answer

Why everybody here is taking G why not H and solve?

ThatDudeKnows · Answer

dipanjandas792

Either way is fine!

A is 6 "steps" below G.
A is 7 "steps" below H.
So we can either start with G and subtract 6 "steps" or start with H and subtract 7 "steps."
Each "step" is  5^{13}-5^{12}

Since the other replies in the thread have started from G and you're asking about starting from H, I'll show the latter.

H =  5^{13} , so A =  5^{13}-7(5^{13}-5^{12}) 
A =  5^{13}-7(5^{12})(5-1) 
A =  5^{13}-28(5^{12}) 
A =  5(5^{12})-28(5^{12}) 
A =  -23(5^{12})

Shhh103 · Answer

the easier approach should be finding the difference = 5^13-5^12 = 4 * 5^12

and subtracting 4* 5^12 from 5^12 6 times

ans = -23 * 5^12

jerinpeter · Answer

The distance from G to H is 5^13 - 5^12.

The distance between two consecutive points is constant, so the distance from A to G will be 6 times the distance from G to H or 6(5^13 – 5^12).

The value of A, therefore, will be equal to the value of G minus the distance from A to G: 
5^12 – 6(5^13 – 5^12) ---> 5^12 – 6  ---> 5^12 – 6(5^12)(4) ---> 5^12(1 – 24) ---> (-23)5^12.

The correct answer is B.

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