GMAT Club Forum - Forums > Quantitative Questions https://gmatclub.com:443/forum/index.php Join the largest GMAT and MBA community at GMAT Club. Get expert advice, free tests, questions and admission tips. Boost your GMAT score and MBA chances en-gb Copyright (c) 2002 GMAT Club Forum mail@gmatclub.com (GMAT Club Forum) mail@gmatclub.com (GMAT Club Forum) Sun, 28 Apr 2024 00:00:00 -0800 Sun, 28 Apr 2024 12:14:02 -0800 MSSTI RSS Feed 2.0 Version 1.0.9 - (C) 2008-2009 leviatan21 - http://www.mssti.com/ GMAT Club Forum - Forums > Quantitative Questions https://gmatclub.com:443/forum/tps://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/imageset/site_logo.gif https://gmatclub.com:443/forum/index.php 60 Problem Solving (PS) | Re: What is the sum of all possible solutions to the equation (2x^2-x-9)^ https://gmatclub.com:443/forum/viewtopic.php?p=3389848#p3389848
sb0541 wrote:

nive28 wrote:
Pritishd wrote:
I am assuming this question to be \(\sqrt{2x^2-x-9}=x+1\)

Squaring on both the sides:
\(2x^2-x-9=x^2+2x+1\)
Simplifying will yield:
\(x^2-3x-10=0\)
\((x-5)(x+2)\)
x = 5 or -2
Sum of all possible solutions is \(5 + (-2) = 3\)
Ans: C

Please confirm the OA.


OA: option d: 5


If we put back the solution x = -2 into the equation then it is not valid .
We get 1 = -(1) , which is not true .
However, for x = 5 then equation is valid giving the result 6 = 6 .

Hence only 5 is a valid solution for the equation .

­

Statistics : Posted by K-ja • on 11 Jul 2016, 01:37 • Replies 12 • Views 18569


]]>
sb0541 wrote:

nive28 wrote:
Pritishd wrote:
I am assuming this question to be \(\sqrt{2x^2-x-9}=x+1\)

Squaring on both the sides:
\(2x^2-x-9=x^2+2x+1\)
Simplifying will yield:
\(x^2-3x-10=0\)
\((x-5)(x+2)\)
x = 5 or -2
Sum of all possible solutions is \(5 + (-2) = 3\)
Ans: C

Please confirm the OA.


OA: option d: 5


If we put back the solution x = -2 into the equation then it is not valid .
We get 1 = -(1) , which is not true .
However, for x = 5 then equation is valid giving the result 6 = 6 .

Hence only 5 is a valid solution for the equation .

­]]>
(K-ja)Sun, 28 Apr 2024 12:10:20 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389848#p3389848
Problem Solving (PS) | Re: In how many ways can 4 people be divided into 2 groups of 2 people eac https://gmatclub.com:443/forum/viewtopic.php?p=3389847#p3389847
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Statistics : Posted by bumpbot • on 04 Nov 2021, 04:00 • Replies 2 • Views 4269


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(bumpbot)Sun, 28 Apr 2024 12:02:56 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389847#p3389847
Problem Solving (PS) | Re: In how many ways can we put 4 different balls in 3 different boxes whe https://gmatclub.com:443/forum/viewtopic.php?p=3389846#p3389846
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Statistics : Posted by bumpbot • on 28 Sep 2016, 10:40 • Replies 11 • Views 55878


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(bumpbot)Sun, 28 Apr 2024 12:01:05 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389846#p3389846
Problem Solving (PS) | Re: If x^2 = x + 1, which of the following is equal to x^3 ? https://gmatclub.com:443/forum/viewtopic.php?p=3389845#p3389845 Jake7Wimmer wrote:

Can someone show how to do this with plugging in numbers please.


The characteristics equation of x^2=x+1 follow the reoccurrence a_n=a_{n-1}+a_{n-2}. , commonly referred to as Fibonacci sequence.

This equation gives golden ratio and inverse of golden ratio as its roots

x^3 is equivalent to a_3 .
x^2 is a_2
x is a_1
1 is 1



You can find a_3 in terms of a_1
a_3=a_2+a_1=a_1+1+a_1=2a_1+1

This corresponds to 2x+1

Posted from my mobile device

Statistics : Posted by Oppenheimer1945 • on 12 Jan 2024, 06:57 • Replies 4 • Views 1626


]]>
Jake7Wimmer wrote:

Can someone show how to do this with plugging in numbers please.


The characteristics equation of x^2=x+1 follow the reoccurrence a_n=a_{n-1}+a_{n-2}. , commonly referred to as Fibonacci sequence.

This equation gives golden ratio and inverse of golden ratio as its roots

x^3 is equivalent to a_3 .
x^2 is a_2
x is a_1
1 is 1



You can find a_3 in terms of a_1
a_3=a_2+a_1=a_1+1+a_1=2a_1+1

This corresponds to 2x+1

Posted from my mobile device]]>
(Oppenheimer1945)Sun, 28 Apr 2024 11:56:23 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389845#p3389845
Problem Solving (PS) | Re: What is the remainder when 7^548 is divided by 10? https://gmatclub.com:443/forum/viewtopic.php?p=3389843#p3389843 Danou wrote:

Bunuel wrote:
guddo wrote:
What is the remainder when\(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E.9

Attachment:
2024-01-25_13-51-58.png



When dividing a positive integer by 10, the remainder is always the units digit of that integer. For instance, 123 divided by 10 yields the remainder of 3. Hence, essentially we need to find the units digit of\(7^{548}\).

    \( 7^{548} = (7^2)^{274} = 49^{274} = (50 -1)^{274}\).


When expanding\( (50 -1)^{274}\) , all terms but the last one will have 50 as their factors making them divisible by 10 and the last term will be\((-1)^{274}=1\) , which when divided by 10 yields the remainder of 1. Here, we could also note that when expanding we'll get all the terms with 50 and the last term\((-1)^{274}=1\) , hence the sum would be something with the units digit of 1, giving the remainder of 1 when divided by 10.

Alternatively, we can use the cyclicity of 7 in positive integer power, which is four, meaning that the units digit of 7 in positive integer power repeats in blocks of four {7,
...

Statistics : Posted by Bunuel • on 25 Jan 2024, 04:03 • Replies 4 • Views 1639


]]> Danou wrote:

Bunuel wrote:
guddo wrote:
What is the remainder when\(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E.9

Attachment:
2024-01-25_13-51-58.png



When dividing a positive integer by 10, the remainder is always the units digit of that integer. For instance, 123 divided by 10 yields the remainder of 3. Hence, essentially we need to find the units digit of\(7^{548}\).

    \( 7^{548} = (7^2)^{274} = 49^{274} = (50 -1)^{274}\).


When expanding\( (50 -1)^{274}\) , all terms but the last one will have 50 as their factors making them divisible by 10 and the last term will be\((-1)^{274}=1\) , which when divided by 10 yields the remainder of 1. Here, we could also note that when expanding we'll get all the terms with 50 and the last term\((-1)^{274}=1\) , hence the sum would be something with the units digit of 1, giving the remainder of 1 when divided by 10.

Alternatively, we can use the cyclicity of 7 in positive integer power, which is four, meaning that the units digit of 7 in positive integer power repeats in blocks of four {7,
...]]> (Bunuel)Sun, 28 Apr 2024 11:42:18 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389843#p3389843 Problem Solving (PS) | Re: What is the remainder when 1! + 2! + 3! 100! is divided by 18? https://gmatclub.com:443/forum/viewtopic.php?p=3389824#p3389824
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Statistics : Posted by bumpbot • on 22 Nov 2021, 03:15 • Replies 2 • Views 4012


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(bumpbot)Sun, 28 Apr 2024 10:14:32 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389824#p3389824
Problem Solving (PS) | Re: A box contains 10 pairs of shoes (20 shoes in total). If two https://gmatclub.com:443/forum/viewtopic.php?p=3389822#p3389822
Without restriction 20x19

Posted from my mobile device

Statistics : Posted by colfer • on 28 Oct 2009, 01:08 • Replies 18 • Views 69396


]]>

Without restriction 20x19

Posted from my mobile device]]>
(colfer)Sun, 28 Apr 2024 10:11:59 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389822#p3389822
Problem Solving (PS) | Re: (6^14*5^13) - (6^13*5^14) = https://gmatclub.com:443/forum/viewtopic.php?p=3389821#p3389821
Jake7Wimmer wrote:

Bunuel can you help with this question please

­\((6^{14}*5^{13})-(6^{13}*5^{14})=\)

A. \(0\)

B. \(1\)

C. \(30\)

D. \(11^{13}\)

E. \(30^{13}\)
____________________________________

­\(6^{14}5^{13} -6^{13}5^{14}=\)

­\(=6^{13}*5^{13}(6-5)=\)

­\(=6^{13}*5^{13}=\)

­\(=(6*5)^{13}=\)

­\(=30^{13}\)

Answer: E.­

Statistics : Posted by Bunuel • on 22 Oct 2023, 16:20 • Replies 3 • Views 1602


]]>
Jake7Wimmer wrote:

Bunuel can you help with this question please

­\((6^{14}*5^{13})-(6^{13}*5^{14})=\)

A. \(0\)

B. \(1\)

C. \(30\)

D. \(11^{13}\)

E. \(30^{13}\)
____________________________________

­\(6^{14}5^{13} -6^{13}5^{14}=\)

­\(=6^{13}*5^{13}(6-5)=\)

­\(=6^{13}*5^{13}=\)

­\(=(6*5)^{13}=\)

­\(=30^{13}\)

Answer: E.­]]>
(Bunuel)Sun, 28 Apr 2024 09:58:55 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389821#p3389821
Problem Solving (PS) | Re: The retail price of a sweater is $32. If the store can sell the sweate https://gmatclub.com:443/forum/viewtopic.php?p=3389817#p3389817
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Statistics : Posted by bumpbot • on 18 Apr 2017, 05:05 • Replies 7 • Views 6163


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(bumpbot)Sun, 28 Apr 2024 09:38:35 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389817#p3389817
Problem Solving (PS) | Re: Ten strips of paper are numbered from 1 to 10 and placed in a bag. If https://gmatclub.com:443/forum/viewtopic.php?p=3389816#p3389816
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Statistics : Posted by bumpbot • on 23 Feb 2022, 08:45 • Replies 4 • Views 1673


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(bumpbot)Sun, 28 Apr 2024 09:37:48 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389816#p3389816
Problem Solving (PS) | Re: Set A consists of the values 1, 2, and 3. Set B consists of the values https://gmatclub.com:443/forum/viewtopic.php?p=3389815#p3389815
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Statistics : Posted by bumpbot • on 11 Oct 2018, 02:50 • Replies 6 • Views 3137


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(bumpbot)Sun, 28 Apr 2024 09:36:47 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389815#p3389815
Problem Solving (PS) | Re: A computer retailer sells only TFT and LCD monitors. In the graph abov https://gmatclub.com:443/forum/viewtopic.php?p=3389814#p3389814
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Statistics : Posted by bumpbot • on 26 Dec 2014, 08:37 • Replies 10 • Views 5422


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(bumpbot)Sun, 28 Apr 2024 09:35:09 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389814#p3389814
Problem Solving (PS) | Re: In how many ways can the letters of the word "COMPUTER" be arranged if https://gmatclub.com:443/forum/viewtopic.php?p=3389813#p3389813
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Statistics : Posted by bumpbot • on 27 Jul 2010, 20:50 • Replies 9 • Views 13897


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(bumpbot)Sun, 28 Apr 2024 09:33:53 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389813#p3389813
Problem Solving (PS) | Re: If n is a negative integer and q is a positive integer, then there is https://gmatclub.com:443/forum/viewtopic.php?p=3389812#p3389812
N=mq+r
N has to be -100
M is only one negative integer
q has to be 30
r has to be a integer that is positive -> that eliminates A) and B) already

New equation is: -100=(M)(30)+R
Literally just plug C,D,E into R and see if you can make equation work.

Only E) 20 works because M will equal -4

-100=(-4)(30) + 20

Statistics : Posted by Jake7Wimmer • on 22 Oct 2023, 16:10 • Replies 10 • Views 3508


]]>

N=mq+r
N has to be -100
M is only one negative integer
q has to be 30
r has to be a integer that is positive -> that eliminates A) and B) already

New equation is: -100=(M)(30)+R
Literally just plug C,D,E into R and see if you can make equation work.

Only E) 20 works because M will equal -4

-100=(-4)(30) + 20]]>
(Jake7Wimmer)Sun, 28 Apr 2024 09:32:33 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389812#p3389812
Quantitative Questions | Re: Quant Question of the Day Chat https://gmatclub.com:443/forum/viewtopic.php?p=3389809#p3389809 Statistics : Posted by HarshR9 • on 20 Oct 2022, 20:57 • Replies 4500 • Views 130741


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(HarshR9)Sun, 28 Apr 2024 09:22:01 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389809#p3389809
Problem Solving (PS) | Re: ­If b > 0, x = [square_root]b+9[/square_root], and y = [square_root]b https://gmatclub.com:443/forum/viewtopic.php?p=3389805#p3389805
andrewjohn8 wrote:

­If b > 0, \(x = \sqrt{b+9}\), and \(y = \sqrt{b}\), what is the value of the product of x - y and x + y ?

A. 3
B. 9
C. b
D. b + 3
E. b + 9­

­

\((x - y)(x + y) = \)

\(= x^2 - y^2 =\)

\(=(\sqrt{b+9})^2 - (\sqrt{b})^2 =\)

\( = (b+9) - b =\)

\(= 9 \)

Answer: B.­

Statistics : Posted by Bunuel • on 28 Apr 2024, 09:05 • Replies 1 • Views 76


]]>
andrewjohn8 wrote:

­If b > 0, \(x = \sqrt{b+9}\), and \(y = \sqrt{b}\), what is the value of the product of x - y and x + y ?

A. 3
B. 9
C. b
D. b + 3
E. b + 9­

­

\((x - y)(x + y) = \)

\(= x^2 - y^2 =\)

\(=(\sqrt{b+9})^2 - (\sqrt{b})^2 =\)

\( = (b+9) - b =\)

\(= 9 \)

Answer: B.­]]>
(Bunuel)Sun, 28 Apr 2024 09:10:03 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389805#p3389805
Problem Solving (PS) | Re: A certain company's yearly revenue was 20% less in 2005 than in 2004. https://gmatclub.com:443/forum/viewtopic.php?p=3389803#p3389803 '05 = 80
'06 = 88
'07 = 100

'06 -> '07 = 88 -> 100

Change to 8800 and 10,000

1% of 8800 is 88 -> Eliminate a) and B) and we will try D) First.

88x14 = (80+8)(10+4) = 800+320+80+32 = 1232 (I used foil just to multiply 88x14 so that I won't make a mistake)
8800+1232 = 10,032 which is aprox 10,000

Thus 14% and D) is answer

Statistics : Posted by Jake7Wimmer • on 24 Oct 2023, 20:31 • Replies 4 • Views 1731


]]>
'05 = 80
'06 = 88
'07 = 100

'06 -> '07 = 88 -> 100

Change to 8800 and 10,000

1% of 8800 is 88 -> Eliminate a) and B) and we will try D) First.

88x14 = (80+8)(10+4) = 800+320+80+32 = 1232 (I used foil just to multiply 88x14 so that I won't make a mistake)
8800+1232 = 10,032 which is aprox 10,000

Thus 14% and D) is answer]]>
(Jake7Wimmer)Sun, 28 Apr 2024 09:05:34 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389803#p3389803
Problem Solving (PS) | ­­When it operates at its own constant rate, hose A can fill an empty https://gmatclub.com:443/forum/viewtopic.php?p=3389802#p3389802
Rate of A: 1 tank/2 hours
Rate of B: 1 tank/4 hours

If it takes hose A 2 hours to fill a tank, how long will it take both to fill it?

t/2 + t/4 = 1

t = 4/3

which answer equals 4/3 when you plug in x, the time for hose A. That is x=2.

2(2)/3 = 4/3

Therefore Answer A­

Statistics : Posted by Fish181 • on 26 Apr 2024, 21:16 • Replies 6 • Views 261


]]>

Rate of A: 1 tank/2 hours
Rate of B: 1 tank/4 hours

If it takes hose A 2 hours to fill a tank, how long will it take both to fill it?

t/2 + t/4 = 1

t = 4/3

which answer equals 4/3 when you plug in x, the time for hose A. That is x=2.

2(2)/3 = 4/3

Therefore Answer A­]]>
(Fish181)Sun, 28 Apr 2024 09:05:18 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389802#p3389802
Problem Solving (PS) | Re: An Ameba is an organic life form that divides into two Amebas each rou https://gmatclub.com:443/forum/viewtopic.php?p=3389798#p3389798
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Statistics : Posted by bumpbot • on 05 Sep 2019, 04:19 • Replies 5 • Views 1685


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(bumpbot)Sun, 28 Apr 2024 08:56:11 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389798#p3389798
Problem Solving (PS) | Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each https://gmatclub.com:443/forum/viewtopic.php?p=3389789#p3389789 gmatophobia wrote:

Fish181 wrote:
­i missed this question because I got favorable cases = 2 and total cases = 6 for 1/3. Why are we multiplying theseresults

Fish181

F avorable cases =6

\(y\) can be filled in 2 ways and\(x\) can be filled in 3 ways

Hence, favourable cases = 2 * 3 = 6

Here are possible values ofxy

33
53
83
35
55
85

Total cases =9 ,because \(x\) can be filled in 3 ways and\(y\) can be filled in 3 ways

Here are all the possible values of xy

33
53
83
35
55
85
38
58
88

Hence required probability =\( \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{6}{9} =\frac{2}{3}\)

 ­

­Crystal clear thank you! I was not comprehending that 53x,y12 was the six digit number... my mind split it into "53x" , "y12"... no wonder this problem was so confusing to me. Definitely attainable if you understand divisibility rules and some combinatorics.
...

Statistics : Posted by Fish181 • on 01 Nov 2023, 11:43 • Replies 11 • Views 2800


]]> gmatophobia wrote:

Fish181 wrote:
­i missed this question because I got favorable cases = 2 and total cases = 6 for 1/3. Why are we multiplying theseresults

Fish181

F avorable cases =6

\(y\) can be filled in 2 ways and\(x\) can be filled in 3 ways

Hence, favourable cases = 2 * 3 = 6

Here are possible values ofxy

33
53
83
35
55
85

Total cases =9 ,because \(x\) can be filled in 3 ways and\(y\) can be filled in 3 ways

Here are all the possible values of xy

33
53
83
35
55
85
38
58
88

Hence required probability =\( \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{6}{9} =\frac{2}{3}\)

 ­

­Crystal clear thank you! I was not comprehending that 53x,y12 was the six digit number... my mind split it into "53x" , "y12"... no wonder this problem was so confusing to me. Definitely attainable if you understand divisibility rules and some combinatorics.
...]]> (Fish181)Sun, 28 Apr 2024 08:45:34 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389789#p3389789 Problem Solving (PS) | Re: A committee of 10 members consists of men and women. If two members of https://gmatclub.com:443/forum/viewtopic.php?p=3389787#p3389787
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Problem Solving (PS) | Re: An electrical appliances store sold this month 400% more than the aver https://gmatclub.com:443/forum/viewtopic.php?p=3389786#p3389786
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(bumpbot)Sun, 28 Apr 2024 08:35:38 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389786#p3389786
Problem Solving (PS) | Re: A closed aluminum rectangular box has inner dimensions x centimeters https://gmatclub.com:443/forum/viewtopic.php?p=3389784#p3389784
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(bumpbot)Sun, 28 Apr 2024 08:29:59 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389784#p3389784
Problem Solving (PS) | Re: Three surgeons - Anna, Claire, and Diane - are asked to perform an eme https://gmatclub.com:443/forum/viewtopic.php?p=3389781#p3389781
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(bumpbot)Sun, 28 Apr 2024 08:24:54 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389781#p3389781
Problem Solving (PS) | Re: In how many ways can 6 different letters be posted in 3 letterboxes su https://gmatclub.com:443/forum/viewtopic.php?p=3389778#p3389778
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(bumpbot)Sun, 28 Apr 2024 08:18:46 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389778#p3389778
Problem Solving (PS) | Re: If the tens digit of the positive integer x and the tens digit of the https://gmatclub.com:443/forum/viewtopic.php?p=3389767#p3389767
DmitryFarber wrote:

­We don't know the number of digits, but it doesn't matter. When we add, no digit affects the value of those to its right, only those to its left. 
For instance, if we add 2169 + 53,869, the answer still ends in 38, just as when we add 69 + 69. Similarly, when we double the value, those additional places to the left don't change the fact that 38 doubles to 76. 

So in short, we can focus only on the relevant digits and not worry about any other potential digits. This is common on place value/digit problems.

­Understood, thank you. I think you meant this, correct?  "When we add, no digit affects the value of those to its left, only those to its right."­

Statistics : Posted by Engineer1 • on 03 Jan 2024, 06:46 • Replies 8 • Views 2288


]]>
DmitryFarber wrote:

­We don't know the number of digits, but it doesn't matter. When we add, no digit affects the value of those to its right, only those to its left. 
For instance, if we add 2169 + 53,869, the answer still ends in 38, just as when we add 69 + 69. Similarly, when we double the value, those additional places to the left don't change the fact that 38 doubles to 76. 

So in short, we can focus only on the relevant digits and not worry about any other potential digits. This is common on place value/digit problems.

­Understood, thank you. I think you meant this, correct?  "When we add, no digit affects the value of those to its left, only those to its right."­]]>
(Engineer1)Sun, 28 Apr 2024 07:36:48 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389767#p3389767
Quantitative Questions | Re: How do you memorize formulas? https://gmatclub.com:443/forum/viewtopic.php?p=3389765#p3389765
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Statistics : Posted by bumpbot • on 25 Oct 2017, 00:56 • Replies 3 • Views 3590


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Problem Solving (PS) | Re: If P is a prime number greater than 5, what is the remainder when P^2 https://gmatclub.com:443/forum/viewtopic.php?p=3389761#p3389761
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(bumpbot)Sun, 28 Apr 2024 07:15:48 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389761#p3389761
Problem Solving (PS) | If k, (k+200), (k+350), and 15*K are all multiples of P, the https://gmatclub.com:443/forum/viewtopic.php?p=3389760#p3389760 P is factor of K+200 means (K+200)/P should be an integer . We know K/P is an integer since K is a multiple of P. Therefore 200/P must be an integer. Similarly 350/P must be an integer. Only 25 meets both the criteria from the given numbers

Posted from my mobile device

Statistics : Posted by colfer • on 20 Jun 2014, 00:39 • Replies 9 • Views 8297


]]>
P is factor of K+200 means (K+200)/P should be an integer . We know K/P is an integer since K is a multiple of P. Therefore 200/P must be an integer. Similarly 350/P must be an integer. Only 25 meets both the criteria from the given numbers

Posted from my mobile device]]>
(colfer)Sun, 28 Apr 2024 07:04:06 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389760#p3389760
GMAT Club Tests | Re: M30-02 https://gmatclub.com:443/forum/viewtopic.php?p=3389748#p3389748 Rebaz wrote:
Bunuel wrote:
OfficialSolution:

In how many ways can nine family members, consisting of four grandparents and five grandchildren (three brothers and two sisters), be arranged around a circular table so that the two sisters sit next to each other and are positioned immediately between two of the brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640


Consider treating two brothers and two sisters as a single unit, represented as {BSSB}.

With this grouping, we have a total of 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}, where {G} represents a grandparent and {B} represents a brother outside the unit.

These 6 units can be arranged around a circular table in\((6-1)!=5!\) ways.

Next, we'll analyze the {BSSB} unit in more detail:

We can choose 2 brothers out of the 3 for the unit in\(C^2_3=3\) ways.

Within the unit, these selected brothers can be arranged in 2! ways, such as\( \{B_1, S, S,B_2\}\) or\( \{B_2, S, S,B_1\}\) .

Similarly, the sisters within the unit can be arranged in 2! ways, either[m] \{B,
...

Statistics : Posted by Bunuel • on 16 Sep 2014, 01:45 • Replies 12 • Views 17959


]]> Rebaz wrote:
Bunuel wrote:
OfficialSolution:

In how many ways can nine family members, consisting of four grandparents and five grandchildren (three brothers and two sisters), be arranged around a circular table so that the two sisters sit next to each other and are positioned immediately between two of the brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640


Consider treating two brothers and two sisters as a single unit, represented as {BSSB}.

With this grouping, we have a total of 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}, where {G} represents a grandparent and {B} represents a brother outside the unit.

These 6 units can be arranged around a circular table in\((6-1)!=5!\) ways.

Next, we'll analyze the {BSSB} unit in more detail:

We can choose 2 brothers out of the 3 for the unit in\(C^2_3=3\) ways.

Within the unit, these selected brothers can be arranged in 2! ways, such as\( \{B_1, S, S,B_2\}\) or\( \{B_2, S, S,B_1\}\) .

Similarly, the sisters within the unit can be arranged in 2! ways, either[m] \{B,
...]]> (Bunuel)Sun, 28 Apr 2024 06:42:43 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389748#p3389748 Problem Solving (PS) | Re: K = 10^26 + 2^26, and K is a multiple of 2^n but NOT a multiple of 2 https://gmatclub.com:443/forum/viewtopic.php?p=3389745#p3389745 Bunuel wrote:

vraju87 wrote:
­Could anyone please explain how we can say with certainity that(\(5^{26}\) +1) will not have a power of 2 greater than 1.Thanks!­

Good question.
­
First ofall:


Divisibility rule for\(2^n\):

An integer is divisible by 2 if its last digit is divisible by 2.

An integer is divisible by 4 if its last two digits is divisible by 4.

An integer is divisible by 8 if its last three digits is divisible by 8.

...

Generally, an integer is divisible by\(2^n\) if its last\(n\) digits is divisible by\(2^n\).

Next, 5 raised to any positive integer power greater than 1 ends with 25. Thus, 5^26 + 1 = ...25 + 1 = ...26. Since 26 is not divisible by 4, then 5^26 + 1 is not divisible by any power of 2 greater than 1.

Hope it'sclear.­

Thank you for the clear explanation!
...

Statistics : Posted by vraju87 • on 12 Aug 2020, 09:32 • Replies 7 • Views 3797


]]> Bunuel wrote:

vraju87 wrote:
­Could anyone please explain how we can say with certainity that(\(5^{26}\) +1) will not have a power of 2 greater than 1.Thanks!­

Good question.
­
First ofall:


Divisibility rule for\(2^n\):

An integer is divisible by 2 if its last digit is divisible by 2.

An integer is divisible by 4 if its last two digits is divisible by 4.

An integer is divisible by 8 if its last three digits is divisible by 8.

...

Generally, an integer is divisible by\(2^n\) if its last\(n\) digits is divisible by\(2^n\).

Next, 5 raised to any positive integer power greater than 1 ends with 25. Thus, 5^26 + 1 = ...25 + 1 = ...26. Since 26 is not divisible by 4, then 5^26 + 1 is not divisible by any power of 2 greater than 1.

Hope it'sclear.­

Thank you for the clear explanation!
...]]> (vraju87)Sun, 28 Apr 2024 06:27:35 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389745#p3389745 Problem Solving (PS) | Re: If one of two parallel lines has 7 points on it and the other has 8 https://gmatclub.com:443/forum/viewtopic.php?p=3389734#p3389734 waddap wrote:

I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missingsomething

Hope below wouldhelp.
­
If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 455
B. 364
C. 196
D. 168
E.56


Approach#1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points:\(C^2_8*C^1_7=28*7=196\) ;

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points:\(C^2_7*C^1_8=21*8=168\) ;

Total number of triangles:\(196+168=364\).

Approach#2:

Any three distinct points chosen from the total of\(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as\(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\) . Here,[m]
...

Statistics : Posted by Bunuel • on 25 Oct 2022, 01:22 • Replies 3 • Views 781


]]>
waddap wrote:

I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missingsomething

Hope below wouldhelp.
­
If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 455
B. 364
C. 196
D. 168
E.56


Approach#1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points:\(C^2_8*C^1_7=28*7=196\) ;

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points:\(C^2_7*C^1_8=21*8=168\) ;

Total number of triangles:\(196+168=364\).

Approach#2:

Any three distinct points chosen from the total of\(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as\(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\) . Here,[m]
...]]>
(Bunuel)Sun, 28 Apr 2024 05:55:13 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389734#p3389734
Problem Solving (PS) | A number n can be written as the product of four prime numbers https://gmatclub.com:443/forum/viewtopic.php?p=3389729#p3389729
delphsan wrote:

­A number n can be written as the product of four primer numbers, exactly two of which are the same. How many different positive divisors does n have, including 1 and n?

A. 5
B. 8
C. 9
D. 12
E.16­

 

\( ­n =a^2*b*c\) where a, b and c are distinct primenumber

CONCEPT : If\( N =a^p*b^q*c^r...\)
where a, b, c... are distinct primes
Number of factors of\( N =(p+1)*(q+1)*(r+1)*...\)


i.e. Factors of \( ­n =a^2*b*c\)  will be (2+1)*(1+1)*(1+1) = 12

Answer: OptionD

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---
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Statistics : Posted by GMATinsight • on 25 Apr 2024, 16:18 • Replies 2 • Views 179


]]>
delphsan wrote:

­A number n can be written as the product of four primer numbers, exactly two of which are the same. How many different positive divisors does n have, including 1 and n?

A. 5
B. 8
C. 9
D. 12
E.16­

 

\( ­n =a^2*b*c\) where a, b and c are distinct primenumber

CONCEPT : If\( N =a^p*b^q*c^r...\)
where a, b, c... are distinct primes
Number of factors of\( N =(p+1)*(q+1)*(r+1)*...\)


i.e. Factors of \( ­n =a^2*b*c\)  will be (2+1)*(1+1)*(1+1) = 12

Answer: OptionD

[spoiler=]

Get the VIDEO solutions of ALL QUANT problems of "GMAT Official Advanced Questions"here.


---
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...]]>
(GMATinsight)Sun, 28 Apr 2024 05:45:09 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389729#p3389729
GMAT Club Tests | M20-32 https://gmatclub.com:443/forum/viewtopic.php?p=3389727#p3389727 Smriti5645 wrote:

well, please explain -
10*1/4 - x*1/5 = 1
why this equation is equals to 1,
thanks.

Posted from my mobile device

­1 there represents the job done. Specifically, 1 pool is filled while water equal to 10*1/4 = 2.5 of the pool's capacity is poured in, and the amount of water equal to x*1/5 of the pool's capacity is drained out.

Statistics : Posted by Bunuel • on 16 Sep 2014, 01:09 • Replies 6 • Views 28483


]]>
Smriti5645 wrote:

well, please explain -
10*1/4 - x*1/5 = 1
why this equation is equals to 1,
thanks.

Posted from my mobile device

­1 there represents the job done. Specifically, 1 pool is filled while water equal to 10*1/4 = 2.5 of the pool's capacity is poured in, and the amount of water equal to x*1/5 of the pool's capacity is drained out.]]>
(Bunuel)Sun, 28 Apr 2024 05:42:16 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389727#p3389727
Problem Solving (PS) | Last Year a Company Gave Bonuses to a number of employees https://gmatclub.com:443/forum/viewtopic.php?p=3389725#p3389725
Fish181 wrote:

Last Year a company gave bonuses to a number of employees, but only in the three amounts of $750, $1500, and $7,350. If the total amount of the bonuses was $64800 and each of the three amounts was given to at least one employee, what is the fewest number of bonuses that the company could have given to employees last year?­

A 10
B 11
C 12
D 13
E14­

­Three amounts of Bonus are $750, $1500, and $7,350

Total Bonus = $64800

i.e $750a + $1500b + $7,350c= $64800

Now we need to find\( (a+b+c)_{Min} =?\)

i.e $750a + $1500b + $7,350c = $64800
i.e. $75a + $150b + $7,35c = $6480

i.e. 5a + 10b + 49c =432

IMPORTANT: Now to minimize a+b+c, the value of c MUST be Maximized so that values of a and b are theleast

Also, 5a + 10b = 432- 49c

IMPORTANT: also, 5a+10b must be multiple of 5 so we need to fund a multiple of 49 (i.e. 49c) which when subtracted from 432 leaves a result that is divisible by5

IMPORTANT: so the unit digit of 49c must be either 2 or7

for max value of x=8,
...

Statistics : Posted by GMATinsight • on 27 Apr 2024, 12:57 • Replies 2 • Views 105


]]>
Fish181 wrote:

Last Year a company gave bonuses to a number of employees, but only in the three amounts of $750, $1500, and $7,350. If the total amount of the bonuses was $64800 and each of the three amounts was given to at least one employee, what is the fewest number of bonuses that the company could have given to employees last year?­

A 10
B 11
C 12
D 13
E14­

­Three amounts of Bonus are $750, $1500, and $7,350

Total Bonus = $64800

i.e $750a + $1500b + $7,350c= $64800

Now we need to find\( (a+b+c)_{Min} =?\)

i.e $750a + $1500b + $7,350c = $64800
i.e. $75a + $150b + $7,35c = $6480

i.e. 5a + 10b + 49c =432

IMPORTANT: Now to minimize a+b+c, the value of c MUST be Maximized so that values of a and b are theleast

Also, 5a + 10b = 432- 49c

IMPORTANT: also, 5a+10b must be multiple of 5 so we need to fund a multiple of 49 (i.e. 49c) which when subtracted from 432 leaves a result that is divisible by5

IMPORTANT: so the unit digit of 49c must be either 2 or7

for max value of x=8,
...]]>
(GMATinsight)Sun, 28 Apr 2024 05:40:49 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389725#p3389725
GMAT Club Tests | Re: M15-28 https://gmatclub.com:443/forum/viewtopic.php?p=3389724#p3389724 thanhtra01011 wrote:

I think the question is wrong. How can Sep, Oct and Nov be in autumn?­


In the Northern Hemisphere, ­the autumn months are September, October, and November.

Statistics : Posted by Bunuel • on 16 Sep 2014, 00:56 • Replies 5 • Views 17131


]]>
thanhtra01011 wrote:

I think the question is wrong. How can Sep, Oct and Nov be in autumn?­


In the Northern Hemisphere, ­the autumn months are September, October, and November.]]>
(Bunuel)Sun, 28 Apr 2024 05:31:36 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389724#p3389724
Problem Solving (PS) | If x = 4^3^(1/2) and y = 3*5^(1/2) , which of the following is the https://gmatclub.com:443/forum/viewtopic.php?p=3389719#p3389719
Danou wrote:

­If\( x =4\sqrt{ 3}\) and\(y=3\sqrt{5}\) ­, which of the following is the value of\(\frac{ x^4 - y^4 }{ x^2 -y^2}\) ?

A. -93
B. -3
C. 3
D. 27
E.93

Attachment:
Capture question.JPG

 ­

­\( x =4\sqrt{ 3}\) and\(y=3\sqrt{5}\)­

 \(\frac{ x^4 - y^4 }{ x^2 -y^2} =\frac{ (x^2 - y^2)*(x^2 + y^2) }{ x^2 -y^2} = (x^2 +y^2)\)

 \( (x^2 + y^2) = (4√3)^2 + (3√5)^2) = 16*3 + 9*5 =93\)

Answer: OptionE
---
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Statistics : Posted by GMATinsight • on 27 Apr 2024, 17:40 • Replies 4 • Views 160


]]>
Danou wrote:

­If\( x =4\sqrt{ 3}\) and\(y=3\sqrt{5}\) ­, which of the following is the value of\(\frac{ x^4 - y^4 }{ x^2 -y^2}\) ?

A. -93
B. -3
C. 3
D. 27
E.93

Attachment:
Capture question.JPG

 ­

­\( x =4\sqrt{ 3}\) and\(y=3\sqrt{5}\)­

 \(\frac{ x^4 - y^4 }{ x^2 -y^2} =\frac{ (x^2 - y^2)*(x^2 + y^2) }{ x^2 -y^2} = (x^2 +y^2)\)

 \( (x^2 + y^2) = (4√3)^2 + (3√5)^2) = 16*3 + 9*5 =93\)

Answer: OptionE
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(GMATinsight)Sun, 28 Apr 2024 05:25:55 -0800https://gmatclub.com:443/forum/viewtopic.php?p=3389719#p3389719
Problem Solving (PS) | ­Which of the following is equal to (1+2+3)^2(2+3)^2 https://gmatclub.com:443/forum/viewtopic.php?p=3389718#p3389718
Danou wrote:

­Which of the following is equal to\((1+√2+√3)^2−(√2+√3)^2\) ?

A. 1
B. 2
C. -1 + 2√2
D. 1 + 2√2+2√3
E. 2+ 2√2+2√3


Attachment:
Capture question 2.JPG
­

­\((1+√2+√3)^2−(√2+√3)^2\)

Knowing that √2 = 1.4 and √3 = 1.7, I LOVE to approximate such values which are smaller and easiertoo

i.e. \( (1+√2+√3)^2−(√2+√3)^2 = (1+1.4+1.7)^2−(1.4+1.7)^2 = (4.1)^2−(3.1)^2 ≈ 16-9 ≈7\)

A. 1NO
B.2 NO
C. -1 + 2√2 = -1+2.8 =1.8 NO
D. 1 + 2√2+2√3 = 1+2.8+3.4 = 7.2YES
E. 2 + 2√2+2√3 = 2+2.8+3.4= 8.2 NO

Answer: OptionE
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­\((1+√2+√3)^2−(√2+√3)^2\)

Knowing that √2 = 1.4 and √3 = 1.7, I LOVE to approximate such values which are smaller and easiertoo

i.e. \( (1+√2+√3)^2−(√2+√3)^2 = (1+1.4+1.7)^2−(1.4+1.7)^2 = (4.1)^2−(3.1)^2 ≈ 16-9 ≈7\)

A. 1NO
B.2 NO
C. -1 + 2√2 = -1+2.8 =1.8 NO
D. 1 + 2√2+2√3 = 1+2.8+3.4 = 7.2YES
E. 2 + 2√2+2√3 = 2+2.8+3.4= 8.2 NO

Answer: OptionE
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