|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
20 Jan 2022, 06:45
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:12) correct
30%
(02:05)
wrong
based on 149
sessions
History
Date
Time
Result
Not Attempted Yet
In a Math Olympiad, two representatives from a team A ,while solving quadratic equation committed the following mistakes:
(a) One of them made mistake in noting down the constant term and got the roots as 5,9
(b) The Second one committed an error in co-efficient term of x and got the roots as 12,4
But soon they realized their mistakes and managed to get it right jointly. What was the quadratic equation?
(A) \(x^2+4x+14\)
(B) \(2x^2 +7x-24\)
(C) \(3x^2 -17x+52\)
(D) \(x^2 -14x+48\)
(E) \(x^2+6x+60\)
(a) One of them made mistake in noting down the constant term and got the roots as 5,9
(b) The Second one committed an error in co-efficient term of x and got the roots as 12,4
But soon they realized their mistakes and managed to get it right jointly. What was the quadratic equation?
(A) \(x^2+4x+14\)
(B) \(2x^2 +7x-24\)
(C) \(3x^2 -17x+52\)
(D) \(x^2 -14x+48\)
(E) \(x^2+6x+60\)
20 Jan 2022, 07:52
Kudos
Bookmarks
(a) One of them made mistake in noting down the constant term and got the roots as 5,9
If the roots are 5 and 9, then the factors for the quadratic are \((x-5)(x-9)\). Expanding them:
\((x-5)(x-9)\)
\(x^2 - 14x + 45\)
In the quadratic formula \(ax^2 + bx + c\), c is your constant. We are told that the error is in the constant, therefore we can keep the rest and let the constant remain c.
\(x^2 - 14x + c\)
(b) The Second one committed an error in co-efficient term of x and got the roots as 12,4:
With the roots 12 and 4, the factors will be \((x-12)(x-4)\), which when expanded gives:
\(x^2 - 16x + 48\)
We are told that the error in this instance is in the co-efficient of x. The co-efficient of x will be \(bx\) in the form \(ax^2 + bx + c\).
which means we have \(x^2 + bx +48\)
Putting (a) and (b) together:
Putting it together we get: \(x^2 - 14x + 48\)
Answer D
If the roots are 5 and 9, then the factors for the quadratic are \((x-5)(x-9)\). Expanding them:
\((x-5)(x-9)\)
\(x^2 - 14x + 45\)
In the quadratic formula \(ax^2 + bx + c\), c is your constant. We are told that the error is in the constant, therefore we can keep the rest and let the constant remain c.
\(x^2 - 14x + c\)
(b) The Second one committed an error in co-efficient term of x and got the roots as 12,4:
With the roots 12 and 4, the factors will be \((x-12)(x-4)\), which when expanded gives:
\(x^2 - 16x + 48\)
We are told that the error in this instance is in the co-efficient of x. The co-efficient of x will be \(bx\) in the form \(ax^2 + bx + c\).
which means we have \(x^2 + bx +48\)
Putting (a) and (b) together:
Putting it together we get: \(x^2 - 14x + 48\)
Answer D
