Last visit was: 23 Apr 2026, 18:09 It is currently 23 Apr 2026, 18:09
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,872
 [11]
Given Kudos: 105,853
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,872
 [11]
A man arranges to pay off a debt of $3600 by 40 annual installments wh 03 Dec 2020, 07:00
1
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

61% (02:55) correct 39% (03:02) wrong based on 138 sessions

History

Date
Time
Result
Not Attempted Yet
A man arranges to pay off a debt of $3600 by 40 annual installments which are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). When 30 of the installments are paid he dies leaving one-third of the debt unpaid. What was the value of the 8th installment ?

A. $35

B. $50

C. $65

D. $70

E. $75
ShowHide Answer
Official Answer
C
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 22 Apr 2026
Posts: 4,846
Own Kudos:
9,181
 [1]
Given Kudos: 226
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,846
Kudos: 9,181
 [1]
A man arranges to pay off a debt of $3600 by 40 annual installments wh 04 Dec 2020, 02:42
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
A man arranges to pay off a debt of $3600 by 40 annual installments which are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). When 30 of the installments are paid he dies leaving one-third of the debt unpaid. What was the value of the 8th installment ?

A. $35

B. $50

C. $65

D. $70

E. $75
Show more


The sum of an AP = \(\frac{n}{2}[2a + (n - 1)d]\)

When n = 40, Sum = 3600 = \(\frac{40}{2}[2a + (40 - 1)d] = 20*(2a + 39d)\)

So 2a + 39d = 180 .. (1)

When he died, 10 installments amounting to 1/3rd of his debt are remaining = 1/3 * 3600 = 1200

So the sum of the 1st 30 installments = 3600 - 1200 = 2400 = \(\frac{30}{2}[2a + (30 - 1)d] = 15*(2a + 29d)\)

2a + 29d = 160 .. (2)

Equation (1) - (2) gives, 10 d = 20. Therefore d = 2

Putting d = 2 in (1), 2a = 180 - 78 = 102

a = 51

The 8th installment = a + (n - 1)d = a + 7d = 51 + (7*2) = 51 + 14 = $65


Option C

Arun Kumar
avatar
studybee5
avatar
Joined: 22 Sep 2016
Last visit: 09 Jan 2022
Posts: 2
Given Kudos: 200
Posts: 2
Kudos: 0
A man arranges to pay off a debt of $3600 by 40 annual installments wh 06 Dec 2020, 18:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
2/3*3600=30(a1+a30)/2 >>>> a1+a30=160

3600=40*(a1+a40)/2, a1+a40=180

160-a30=180-a40 >>> a40-a30=20 >>>d=2

a1+(a1+d39)=180
2a+39d=180
2a+39*2=180 >> a1=51

a8=51+7*2=65
User avatar
TestPrepUnlimited
User avatar
Joined: 17 Sep 2014
Last visit: 30 Jun 2022
Posts: 1,223
Own Kudos:
1,138
 [1]
Given Kudos: 6
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Expert
Expert reply
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Posts: 1,223
Kudos: 1,138
 [1]
A man arranges to pay off a debt of $3600 by 40 annual installments wh 06 Dec 2020, 20:02
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A man arranges to pay off a debt of $3600 by 40 annual installments which are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). When 30 of the installments are paid he dies leaving one-third of the debt unpaid. What was the value of the 8th installment ?

A. $35

B. $50

C. $65

D. $70

E. $75
Show more

An arithmetic sequence has the property median = mean. The first 40 terms has a median of \(\frac{a(1) + a(40)}{2}\), the first 30 terms has a median of \(\frac{a(1) + a(30)}{2}\). We can use these to replace the mean, hence build the sum using the median * number of terms.

First 40 installments: \(\frac{a(1) + a(40)}{2} * 40 = 3600\), \(a(1) + a(40) = 180\).
First 30 installments: \(\frac{a(1) + a(30)}{2} * 30 = 2400\), \(a(1) + a(30) = 160\).

Then we can eliminate a(1) to get: \(a(40) - a(30) = 20\). \(a(40) - a(30)\) is also the difference of 10 terms, so the differerence between consecutive terms is 20/10 = 2.

Finaly \(a(1) + a(30) = a(1) + a(1) + 29*2 = 160\) and \(a(1) +29 = 80\), \(a(1) = 51\). Add 7 differences which is 14 to get 65.

Ans: C
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 23 Apr 2026
Posts: 1,809
Own Kudos:
2,090
Given Kudos: 679
Posts: 1,809
Kudos: 2,090
A man arranges to pay off a debt of $3600 by 40 annual installments wh 10 Dec 2020, 10:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A man arranges to pay off a debt of $3600 by 40 annual installments which are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). When 30 of the installments are paid he dies leaving one-third of the debt unpaid. What was the value of the 8th installment ?

A. $35

B. $50

C. $65

D. $70

E. $75
Show more


So amt. paid in the 30 first installments is 2400
Hence \(\frac{30}{2}\)\((x+x+29(d)) = 2400\)
\(2x+29(d)= 160\) (i)

Also amt. paid in the last 10 installments is 1200 this can be expressed as
The sum from \(31^{th}\) to \(40^{th}\) term is 1200
so\( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)
after simplification we get :
\(2x+69d=240\) (ii)


Solving (i) an (ii)
We get d= 2 and x= 51

The value of the eight term is \(x + 7d\) where \(x\) is the first term and d is the common difference of the A.P.
Putting the values of x and d we get : 51+7(2)= 65

Ans-C

Hope it's clear.
User avatar
lucasd14
User avatar
Current Student
Joined: 24 Nov 2021
Last visit: 22 Aug 2024
Posts: 37
Own Kudos:
24
Given Kudos: 92
Location: Argentina
Schools: ESADE (A)
Schools: ESADE (A)
Posts: 37
Kudos: 24
A man arranges to pay off a debt of $3600 by 40 annual installments wh 13 Apr 2022, 09:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stne

So amt. paid in the 30 first installments is 2400
Hence \(\frac{30}{2}\)\((x+x+29(d)) = 2400\)
\(2x+29(d)= 160\) (i)

Also amt. paid in the last 10 installments is 1200 this can be expressed as
The sum from \(31^{th}\) to \(40^{th}\) term is 1200
so\( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)
after simplification we get :
\(2x+69d=240\) (ii)


Solving (i) an (ii)
We get d= 2 and x= 51

The value of the eight term is \(x + 7d\) where \(x\) is the first term and d is the common difference of the A.P.
Putting the values of x and d we get : 51+7(2)= 65

Ans-C

Hope it's clear.
Show more

Hi stne, could you please explain how you obtained the LHS of the formulas \(\frac{30}{2}\)\((x+x+29(d)) = 2400\) & \( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)? Thanks!
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 23 Apr 2026
Posts: 1,809
Own Kudos:
2,090
Given Kudos: 679
Posts: 1,809
Kudos: 2,090
A man arranges to pay off a debt of $3600 by 40 annual installments wh 14 Apr 2022, 08:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lucasd14
stne

So amt. paid in the 30 first installments is 2400
Hence \(\frac{30}{2}\)\((x+x+29(d)) = 2400\)
\(2x+29(d)= 160\) (i)

Also amt. paid in the last 10 installments is 1200 this can be expressed as
The sum from \(31^{th}\) to \(40^{th}\) term is 1200
so\( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)
after simplification we get :
\(2x+69d=240\) (ii)


Solving (i) an (ii)
We get d= 2 and x= 51

The value of the eight term is \(x + 7d\) where \(x\) is the first term and d is the common difference of the A.P.
Putting the values of x and d we get : 51+7(2)= 65

Ans-C

Hope it's clear.

Hi stne, could you please explain how you obtained the LHS of the formulas \(\frac{30}{2}\)\((x+x+29(d)) = 2400\) & \( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)? Thanks!
Show more

Hi lucasd14,

You need to know TWO basic properties of terms that are in A.P:

(1) The sum of n terms in an A.P. is given by \(\frac{n}{2}\){ 1st term + last term}...(i)

Also any given term in an AP can be found if we know the first term and common diff.

(2) nth term= a + (n-1)d ...(ii)

Lets say we need the \(5\)th term
Lets say \(a \) is the first term
lets say \(d \) is the diff. between consecutive terms

Hence the \(5\)th term:
\(= a + (5-1)d\)
\(= a+4d \)

Coming back to the question :

\(\frac{1}{3}*3600\) was unpaid so \(1200 \) was unpaid .

Hence he paid remaining \(2400\) in \(30\) installements which were in A.P.

Now that you know the sum of n terms that are in A.P is given by \( \frac{n}{2}\){ 1st term + last term} you need to plug in the values

\(n = 30\)
let \(x= 1\)st term
let the diff. between two terms be \(d\)
Last term \(( 30\)th term ) \(= x+ (30-1)d\)... using (ii)
Now we know the sum of the \(30\) installements \(= 2400\)

Hence \( \frac{30}{2}\)\(({ x+ x+ (30-1)d }) = 2400 \)... using (i)

\( \frac{30}{2}\) \(({ x+ x+ (29)d }) =2400\)

Now in the same way you should be able to get the second part of your query:

In this part we are finding the sum of \(10\) terms (that are in A.P.)and the sum equals \(1200\)
\(1\)st term is the \(30\)th term
Last term is the \(40\)th term
\(1\)st term= \(30\)th term = \(x+ 29\)d ...( using ii)
Last term =\(40\)th term\(= x + 39\)d ...(using ii)

Hence finally \(\frac{10}{2}(x+ 29d + x + 39d) = 1200\)...using (i)

Should be clear, if anything is still unclear, feel free to ask. All the best.
Moderators:
Bunuel
Math Expert
109785 posts
Krunaal
Tuck School Moderator
853 posts