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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Gmat prep quants doubts!

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Thu Dec 23, 2010 5:16 am
Question Number 1

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.

The correct answer is B.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Thu Dec 23, 2010 5:20 am
Question Number 2:

Remember that √(x²) is always a positive quantity. Thus for positive x, its value is x but for negative x, its value is -x i.e. in other words it is equal to |x|.

Thus, √(x²)/x = |x|/x

The correct answer is E.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Thu Dec 23, 2010 5:23 am
nitesh_2110 wrote:
Image
As per the 4th one:

Maximum 4 letter code can be generated by using the 26 alphabets = 26^4
Maximum 5 letter code can be generated by using the 26 alphabets = 26^5

Thus, Total maximum = (Maximum 4 letter code can be generated) + (Maximum 5 letter code can be generated)
= 26^4 + 26^5
=26^4(1+26)
= 27(26^4)

IMO C
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Thu Dec 23, 2010 5:30 am
Question Number 3:

Let's analyze each of the options individually:
  • 1. (a + b)² ≠ (a² + b²) => f(a + b) ≠ f(a) + f(b)
    2. (a + b + 1) ≠ (a + 1) + (b + 1) = (a + b + 2) => f(a + b) ≠ f(a) + f(b)
    3. √(a + b) ≠ (√a + √b) => f(a + b) ≠ f(a) + f(b)
    4. 2/(a + b) ≠ (2/a) + (2/b) = 2(a + b)/ab => f(a + b) ≠ f(a) + f(b)
    5. (-3(a + b)) = (-3a) + (-3b) => f(a + b) = f(a) + f(b)
The correct answer is E.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Thu Dec 23, 2010 5:35 am
Question Number 4:

Maximum number of codes that can be generated = (Maximum number of 4-letter code) + (Maximum number of 5-letter code)

As there is no restrictions regarding repetition of letters,
  • Maximum number of 4-letter code = (26)^4
    Maximum number of 5-letter code = (26)^5
Maximum number of codes that can be generated = (26^4) + (26^5) = (26^4)*(1 + 26) = 27*(26^5)

The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Sun Jul 25, 2010 2:51 am

by nitesh_2110 » Thu Dec 23, 2010 9:48 am
thank you so much! Mr anurag!..:)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1