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Sadhvi09
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Practice Questions Chat 29 Dec 2023, 02:51
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my bad ..(PB+BC+CP) denotes exactly 2 groups ...but the formula is still crct T-N = P+B+C-(Excatly 2 groups PB,BC,CA)- 2(All three)
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sunngupta
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Practice Questions Chat 29 Dec 2023, 03:31
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Please explain this
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Malar95
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Practice Questions Chat 29 Dec 2023, 03:34
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sunngupta
Please explain this
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On my phone rn but I am convinced that this Q is present in FORUM. Pretty standard. And given that there’s modulus, the expert response there will be useful.

sunngupta
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Is the answer confirm A? If yes, then I can try my shot at explaining.

inequality-and-absolute-value-questions-from-my-collection-86939.html#p652806
Similar concept the Q 2 here and this is a fun and tricky set from Bunuel himself.
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Bunuel
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Practice Questions Chat 29 Dec 2023, 04:25
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Data Sufficiency Butler: December 2023
December 29DS 1DS 2
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Bunuel
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Practice Questions Chat 29 Dec 2023, 04:26
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Problem Solving Butler: October 2023
December 29PS 1PS 2
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Practice Questions Chat 29 Dec 2023, 09:39
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This is random but I was hanging around Bunuels post dazed again so back to the who is Bunuel Q?
- Well I am guessing he is India-Indian or there’s someone in the group who is India-Indian. Definitely a dude too.
When you tried to mind your business for so long and then suddenly spot something.
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imeanup
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Practice Questions Chat 29 Dec 2023, 11:38
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@bb @Bunuel why don’t the normal latex syntax doesn’t work in the post solution section? I would like to get this feature added. And keyboard shortcut to add LaTex or Tex syntax. The current layout is time consuming and confusing.
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Ayisa
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Practice Questions Chat 29 Dec 2023, 12:57
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Please Scholars, someone should help me solve this One.


Every year a tractor loses 20% of the value it had at the beginning of the year . If the tractor now has a value of $120000, what will it value be in five years ?
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SamAn12213
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Practice Questions Chat 29 Dec 2023, 13:09
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120000-(120000*20/100)= 24000
Value at the end of 1st year
1,20,000-24000=96000

Next year depreciation
96000*20/100= 19200
Value at the end of second year 96000-19200= 76800

Third year dep= 76800*20/100= 15360
Value at the end of third year
76800-15360= 61440

Dep for the 4th year = 61440*20/100
12288
Value at the end of 4th year
61440-12288=49152
Value at the Beggining of 5 th year =49152

If you want u can calculate the 5th year Dep. Too... And the balance amt ....at the end of the year...

Maybe there’s a direct method too...

Yeah the direct formula I thought of jus now is

Original Cost * ((100-%)/100)^n

Where n = number of years given in the question

Learnt it somewhere I think

But it would definitely work
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Practice Questions Chat 29 Dec 2023, 13:16
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Ayisa
Please Scholars, someone should help me solve this One. Every year a tractor loses 20% of the value it had at the beginning of the year . If the tractor now has a value of $120000, what will it value be in five years ?
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It is a GP with common ratio (r) = 0.8

Value after 5 yrs would be given by the 6th term in this GP.

a6 = a*r^5 = 120000*(0.8)^5 =$39321.6
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Thank you all for your responses
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Ayisa
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Please help me put on this one too;

A carpenter wishes to construct a ladder with nine rungs whose lengths decrease uniformly from 24 inches at the base to 18 inches at the top. Determine the lengths of the seven intermediate rungs.
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Valx
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Practice Questions Chat 29 Dec 2023, 23:45
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Good morning scholars
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gauravbansal8
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Ayisa
Please help me put on this one too; A carpenter wishes to construct a ladder with nine rungs whose lengths decrease uniformly from 24 inches at the base to 18 inches at the top. Determine the lengths of the seven intermediate rungs.
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My approach:

Consider it as Arithmetic Progression with first term (a) = 24

Last term is the 9th term (a9) = 18

n’th term of AP is given by an = a +(n-1)*d where ’d’ is the common difference of AP.

a9=18=24+(9-1)*d
Therefore, d=-3/4

Length of 7 intermediate rungs = Total length of all 9 rungs -24-18

= (9/2)*[2*24 + 8*(-3/4)] -24-18

=147 inches
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Practice Questions Chat 31 Dec 2023, 15:00
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gauravbansal8
Ayisa
Please help me put on this one too; A carpenter wishes to construct a ladder with nine rungs whose lengths decrease uniformly from 24 inches at the base to 18 inches at the top. Determine the lengths of the seven intermediate rungs.
My approach:

Consider it as Arithmetic Progression with first term (a) = 24

Last term is the 9th term (a9) = 18

n’th term of AP is given by an = a +(n-1)*d where ’d’ is the common difference of AP.

a9=18=24+(9-1)*d
Therefore, d=-3/4

Length of 7 intermediate rungs = Total length of all 9 rungs -24-18

= (9/2)*[2*24 + 8*(-3/4)] -24-18

=147 inches
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Didn't think to do it this way, great solution!
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Practice Questions Chat 31 Dec 2023, 15:31
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why am not getting the source tab on gmat varc-rc forum?
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Data Sufficiency Butler: January 2024
January 1DS 1DS 2
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Practice Questions Chat 01 Jan 2024, 06:27
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Problem Solving Butler: January 2024
January 1PS 1PS 2
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If someone is targetting 700+ in GMAT focus then ideally what is the maximum number of questions one can afford to answer incorrectly?
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Malar95
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horrorslive
If someone is targetting 700+ in GMAT focus then ideally what is the maximum number of questions one can afford to answer incorrectly?
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Thanks for asking this Q.
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