GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jun 2019, 12:30 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

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.  D01-36

Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags 00:00

Difficulty:   5% (low)

Question Stats: 90% (01:19) correct 10% (01:58) wrong based on 188 sessions

HideShow timer Statistics

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

1
1
Official Solution:

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Total # of balls in the jar is $$B + (6B + 10) + (2B + 5) = 9B + 15$$;

The probability of picking a blue or green ball is $$\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}$$.

_________________
Intern  B
Joined: 15 Mar 2015
Posts: 13
Concentration: Strategy, Finance
GMAT 1: 700 Q47 V39 GMAT 2: 710 Q47 V41 GMAT 3: 710 Q48 V39 GPA: 3.3
WE: Information Technology (Real Estate)

Show Tags

Bunuel wrote:
Official Solution:

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Total # of balls in the jar is $$B + (6B + 10) + (2B + 5) = 9B + 15$$;

The probability of picking a blue or green ball is $$\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}$$.

Hey Bunuel,

Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity? Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

bigchuck wrote:
Bunuel wrote:
Official Solution:

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Total # of balls in the jar is $$B + (6B + 10) + (2B + 5) = 9B + 15$$;

The probability of picking a blue or green ball is $$\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}$$.

Hey Bunuel,

Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity? $$\frac{B+(2B + 5)}{9B + 15}=\frac{3B+5}{3*(3B+5)}=\frac{1}{3}$$.
_________________
Board of Directors P
Joined: 17 Jul 2014
Posts: 2543
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)

Show Tags

1
I doubt this is a 700 lvl question...Even I, not a lover or probability questions, solved it under 1 minute:
B+2b+5 / 9B+15
3B+5 / 3(3B+5) => simplify by 3B+5 and get 1/3
Manager  B
Joined: 27 Aug 2014
Posts: 71

Show Tags

Hi

Is it because only one ball is being selected that we are not applying the intersection part where both are possible??

Bunuel wrote:
Official Solution:

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Total # of balls in the jar is $$B + (6B + 10) + (2B + 5) = 9B + 15$$;

The probability of picking a blue or green ball is $$\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}$$.

Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

sinhap07 wrote:
Hi

Is it because only one ball is being selected that we are not applying the intersection part where both are possible??

Bunuel wrote:
Official Solution:

A jar contains only $$B$$ blue balls, $$6B + 10$$ yellow balls and $$2B+5$$ green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{3}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

Total # of balls in the jar is $$B + (6B + 10) + (2B + 5) = 9B + 15$$;

The probability of picking a blue or green ball is $$\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}$$.

Yes, we are selecting only one ball and we want it to be either blue or green.
_________________
Current Student Joined: 21 Jul 2013
Posts: 106
WE: Securities Sales and Trading (Commercial Banking)

Show Tags

hi EMPOWERgmatRichC ,

can you show how to apply the 'Test It' method to this question? thanks a lot
Intern  Joined: 23 Apr 2016
Posts: 14

Show Tags

I think this is a high-quality question and I agree with explanation. I still do not understand how we get from 3B+5 / 9B+15 to 1/3 ..

Even if I divide by 3B+5 then I end up with
1/3B+3
Intern  Joined: 23 Apr 2016
Posts: 14

Show Tags

I think this is a high-quality question and I agree with explanation. Someone explained it to me.....EASY....

factoring out is key.

3B+5/9B+15 = 3B+5/ 3(3B+5) --- divide both now by 3B+5 and you are left with 1/3
Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

expertesp wrote:
I think this is a high-quality question and I agree with explanation. I still do not understand how we get from 3B+5 / 9B+15 to 1/3 ..

Even if I divide by 3B+5 then I end up with
1/3B+3

This is algebra 101:

$$\frac{B+(2B + 5)}{9B + 15}=\frac{3B + 5}{3*(3B + 5)}=\frac{1}{3}$$.
_________________
Intern  Joined: 23 Apr 2016
Posts: 14

Show Tags

Yes indeed. I find it easy to forget about the basics when you see something visually confusing.

Again I appreciate that some people are really into math and left brained but put yourselves in the shoes of those who are right brained. It's like someone who speaks with someone they like and suddenly forget how to speak because they are nervous ! Posted from my mobile device
SVP  V
Joined: 26 Mar 2013
Posts: 2237

Show Tags

2
whitehalo wrote:
hi EMPOWERgmatRichC ,

can you show how to apply the 'Test It' method to this question? thanks a lot

Hi,

As Probability is ratio between two things so we can plug-in number to make it easy.
Put B=1

We have 1 blue, 16 Yellow, 7 green, total balls= 24

We need prob. of choosing blue or green

Prob of choosing blue = 1/24

Prob of choosing green = 7/24

Prob of choosing Blue ORGreen= 1/24 +7/24 = 8/24 =1/3

I hope it helps
Intern  Joined: 23 Apr 2016
Posts: 14

Show Tags

And also worth pointing out is the way the question is phrased:

Either - OR, which means we need to "add" the two individual probabilities.

If the question had said: AND I suppose I would have to use both individual probabilities and multiply them instead.
Intern  B
Joined: 10 May 2017
Posts: 27

Show Tags

I liked this question. It is a clear example of using "OR" condition.

Blue balls = B/(B+6B+10+2B+5)
= B/ (9B+15)

Green balls = (2B+5)/(9B+15)

Total= (B+2B+15)/(9B+15) due to common denominator.

Solve and you will be left with 1/3.

Long process? Yes. But one thing I learnt from the members in GMAT club is don't rely on formulas. Step it out and understand.
Intern  B
Joined: 08 Jun 2017
Posts: 2

Show Tags

We need to find the probability of getting a blue or a green ball.
So 1st we could pick up green then blue or 1st we could pick up blue then green. so shouldn't 1/3 be multiplied by 2.
If we were given that 1st green and second is blue then the answer 1/3 would be correct. Looking forward to hearing from you.

Regards
Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

Shikh27 wrote:
We need to find the probability of getting a blue or a green ball.
So 1st we could pick up green then blue or 1st we could pick up blue then green. so shouldn't 1/3 be multiplied by 2.
If we were given that 1st green and second is blue then the answer 1/3 would be correct. Looking forward to hearing from you.

Regards

We are selecting only one ball and we want it to be either blue or green.
_________________
Manager  B
Joined: 02 Jan 2016
Posts: 121

Show Tags

Studying for more than 2 years and I am unable to understand this question:
What does A jar contains only B blue balls, 6B+10 yellow balls and 2B+5 green balls means ?

what is 6B + 10 yellow balls ? does it mean , also why is it stated contains only B blue balls ? this is so confusing.
Math Expert V
Joined: 02 Sep 2009
Posts: 55732

Show Tags

hero_with_1000_faces wrote:
Studying for more than 2 years and I am unable to understand this question:
What does A jar contains only B blue balls, 6B+10 yellow balls and 2B+5 green balls means ?

what is 6B + 10 yellow balls ? does it mean , also why is it stated contains only B blue balls ? this is so confusing.

B is some number. For example, if B = 1, then the jar contains only 1 blue ball, 16 yellow balls and 7 green balls. "Only" there indicates that there are no other color balls in the jat.
_________________
Intern  B
Joined: 21 Jul 2016
Posts: 1

Show Tags

If B=3, then

3 Blue balls
28 Yellow balls
11 Green balls

So, Prob(A) ou (B) = Prob (A) + Prob (b) then (3 +11) / 42 = 14/42 = 7/21 = 1/3 D01-36   [#permalink] 07 Feb 2019, 10:48

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by

D01-36

Moderators: chetan2u, Bunuel  