BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Question of the Day - 27th August, 2009

Expert replies
by quant-master » Wed Aug 26, 2009 3:42 pm
Question 1:

How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?
a)1875
b)2000
c)2375
d)2500
e)3875

Question 2:

6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?
(A) 1/5
(B) 2/5
(C) 1/10
(D) 1/7
(E) 2/15

OA will be posted in 24 hrs time

Thanks,
Quant-Master
https://gmat-quants.blocked - My Blog Updated almost daily with new quant fundas. Find collection of quants question in my blog
Join the discussion
Source: — Problem Solving |

by PussInBoots » Wed Aug 26, 2009 4:59 pm
Q1:
1st digit: 4 possibilities
2nd digit: 5 possibilities (dont forget 0)
4th digit: 5 odd numbers, so 5 possibilities
3rd and 5th: 4 possibilities each

64 * 25 = 1600

Q2:
2/5. Could someone post explanation of the numeric solution, because I solved it my drawing table and fixing one of the choices?
Join the discussion

by capnx » Wed Aug 26, 2009 5:02 pm
1st Q:
One 4
1*4*5*5*5 = 500
3*1*5*5*5 = 375
No 4
3*4*5*5*5 = 1500
sum = 2375

2nd Q:
(2*4!)/5! = 2/5
Join the discussion

by tohellandback » Wed Aug 26, 2009 10:33 pm
Answer 1:
E-Even
O- ODD
Number is of the form EE OOO
no 4's
3*4*5*5*5=1500
One 4:
when 4 at first place:
4*5*5*5=500
when 4 at 2nd place:
3*5*5*5=375
total=2375
Answer C

Answer 2)
circular permutation:
total number ways of arranging 6 people in circular arrangement=5!
when two people are adjacent=4!*2
probability=4!*2/5!= 2/5
Answer B
Last edited by tohellandback on Thu Aug 27, 2009 2:16 am, edited 1 time in total.
The powers of two are bloody impolite!!
Join the discussion

My Answer

by enniguy » Thu Aug 27, 2009 1:31 am
Q 1. - E.
Expected format: EEOOO
With 4 as a possibility for first E (Then second E cannot be 4):-

4 * 4 * 5 * 5 * 5 = 16 * 125.

With 4 as a possibility for second E(Then first E cannot be 4):-

3 * 5 * 5 * 5 * 5 = 15 * 125.

Total = 31 * 125 = 1875.

Q 2. - B.
Number of ways of selecting 2 people out of 6:
6C2 = 15.

Imagine 6 people sitting on a round table. Then you will have 6 pairs of adjacent people sitting (Instead of 5 since this is a round table), Hence, total number of adjacent people "at a time" = 6

Probablity = 6 / 15 = 2/5.
Join the discussion