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Prakruti_Patil
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Practice Questions Chat 06 Sep 2024, 04:14
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just have to stop being lazy and remember these fractions :)
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TheVDR
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Prakruti_Patil
Hi, I am struggling with approximation questions like these -- what’s the best way to round figure these numbers? the ans to this one is given 1/2
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286/556 is almost equal to 1/2

222 * 1.5 is 333. 222 * 2 is 444...375 is somewhere in between.
So, 222/375 is almost equal to 1/1.75

875/571 is almost equal to 1.5/1

Safe to assume : (1/2) * (1/1.75) * (1.5/1)

This should make life much simpler :)
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RahulKumarSinha
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Practice Questions Chat 06 Sep 2024, 05:16
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Prakruti_Patil
no, the ans choices are pretty close -- and I end up getting the wrong ones always -- is there a thumb rule to approximation -- Like I would have rounded off to 290, 560, 220, 380, 880, 580
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instead of 580, u should round off it as 570 and the approx answer will be 11/25 which is, nearer to 1/2, very less than 1/3
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Rod728
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Practice Questions Chat 06 Sep 2024, 10:25
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Prakruti_Patil
Hi, I am struggling with approximation questions like these -- what’s the best way to round figure these numbers? the ans to this one is given 1/2
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all of them are fractions: 0,286=(2/7) 0,556=(4/7) 0,222 =(2/9) 0,375 =3/8 0,875 = 7/8 0,571 = 4/7

is it easier to see when you know 1/7= 0,143, then I could see 0,286 its 2* (1/7) and the same for 0,556 and 0,571 ;)

edit: 0556 = 5/9 =(0,111*5)
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varuntodi30
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Practice Questions Chat 06 Sep 2024, 22:51
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hi i got 5 question wrong in my gmat focus edition but got 79 score in quant can i know how i can get so low cause when i give mocks for the same wrong questions i get 83
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Practice Questions Chat 06 Sep 2024, 22:53
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varuntodi30
hi i got 5 question wrong in my gmat focus edition but got 79 score in quant can i know how i can get so low cause when i give mocks for the same wrong questions i get 83
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Score also depends on the level of questions that you marked incorrect.
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Goflexshubz
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Hi I am preparing for BAT exam which is undertaken by ISB for PGP MFAB, the sample paper question are quite similar to GMAT but the format is difficult. I am lacking knowledge about how to start studying for the exam.
The exam is gonna be held in post Mid November, can someone guide me?
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Practice Questions Chat 07 Sep 2024, 01:43
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Can someone explain how to solve these problems
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RahulKumarSinha
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abhiprem
Can someone explain how to solve these problems
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is the answer to 20 - B ?
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shaurya_gmat
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I quickly glanced through 20- and the answer I think is b. I just substituted the values to get the answer.

As for 19 - underroot 5 is smaller than underroot 7 sovyou open the modulus with a - sign (- ve values require an additional minus sign as mods can only result in a positive value).. use this concept to open all moduluses.. while solving take care of the signs and i think you should be able to solve it
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abhiprem
Can someone explain how to solve these problems
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Both B

For 19th, check for options. Use 12th root of 2 as 2 ^ (1/12) and so on for all options and check for x....option B satisfies
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Understood. Thanks very much guys
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TheVDR
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For 20th, systematically open modulus, one at a time by seeing what will be bigger. For example, root3 - |root5 - root7| opens as root 3 - (root7 - root5) = root3 - root7 + root5....this way, taking care of signs while opening modulus and then the brackets will help

What is the source of these questions?
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abhiprem
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Practice Questions Chat 07 Sep 2024, 02:06
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It’s bunuel’s algebra ps diagnostic tests: ps

TheVDR
For 20th, systematically open modulus, one at a time by seeing what will be bigger. For example, root3 - |root5 - root7| opens as root 3 - (root7 - root5) = root3 - root7 + root5....this way, taking care of signs while opening modulus and then the brackets will help
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Got it!
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Bunuel
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Practice Questions Chat 07 Sep 2024, 02:07
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Solution for question 19 is here: bunuel-s-algebra-ps-diagnostic-test-399587.html#p3171774

Solution for question 19 is here: bunuel-s-algebra-ps-diagnostic-test-399587.html#p3171798

All question have solutions in "Bunuel’s Algebra PS Diagnostic Test", you just have to scroll:
bunuel-s-algebra-ps-diagnostic-test-399587.html
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Practice Questions Chat 07 Sep 2024, 02:10
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Question 19 - Official Solution:


What is the value of \(|\sqrt{3} - |\sqrt{5}-\sqrt{7}|| - ||\sqrt{3} - \sqrt{5}|-\sqrt{7}|\) ?


A. \(2\sqrt{5}-2\sqrt{7}-2\sqrt{3}\)
B. \(2\sqrt{5}-2\sqrt{7}\)
C. \(2\sqrt{7}-2\sqrt{5}\)
D. \(2\sqrt{5}-2\sqrt{7}+2\sqrt{3}\)
E. \(2\sqrt{7}+2\sqrt{5}\)


To answer this question we should recall the property of the absolute value:

\(|x| = x\), when \(x \geq 0\);

\(|x| = -x\), when \(x < 0\).
So, we should evaluate the expressions in the modulus to see whether they are positive or negative.

STEP 1:

Since \(\sqrt{5}-\sqrt{7} < 0\), then \(|\sqrt{5}-\sqrt{7}|=-(\sqrt{5}-\sqrt{7})=\sqrt{7}-\sqrt{5}\);

Since \(\sqrt{3} - \sqrt{5} < 0\), then \(|\sqrt{3} - \sqrt{5}|=-(\sqrt{3} - \sqrt{5})= \sqrt{5} -\sqrt{3}\).
Thus, \(|\sqrt{3} - |\sqrt{5}-\sqrt{7}|| - ||\sqrt{3} - \sqrt{5}|-\sqrt{7}|\) will become:

\(|\sqrt{3} -(\sqrt{7}-\sqrt{5})| - |(\sqrt{5} -\sqrt{3})-\sqrt{7}|=|\sqrt{3} +\sqrt{5}-\sqrt{7}| - |\sqrt{5} -\sqrt{3}-\sqrt{7}|\).
STEP 2:

\(\sqrt{3} +\sqrt{5}-\sqrt{7}\) must be positive because \(\sqrt{3} +\sqrt{5}=1.something + 2.something=3.something\), while \(\sqrt{7} < 3.something\). So, \(|\sqrt{3} +\sqrt{5}-\sqrt{7}| =\sqrt{3} +\sqrt{5}-\sqrt{7}\);

\(\sqrt{5} -\sqrt{3}-\sqrt{7}\) is obviously negative. So, \(\sqrt{5} -\sqrt{3}-\sqrt{7}=-(\sqrt{5} -\sqrt{3}-\sqrt{7})=\sqrt{7}+\sqrt{3}-\sqrt{5}\).
Thus, \(|\sqrt{3} +\sqrt{5}-\sqrt{7}| - |\sqrt{5} -\sqrt{3}-\sqrt{7}|\) will become:

\((\sqrt{3} +\sqrt{5}-\sqrt{7})-(\sqrt{7}+\sqrt{3}-\sqrt{5})=2\sqrt{5}-2\sqrt{7}\).

Answer: B­
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Question 20 - Official Solution:

If \(x > 0\) and \(x^{(3*x^{12})}=4\), what is the value of \(x\) ?

A. \(\sqrt[12]{2}\)
B. \(\sqrt[6]{2}\)
C. \(\sqrt[3]{2}\)
D. \(\sqrt{3}\)
E. \(\sqrt{2}\)


\(x^{(3*x^{12})}=4\);

Take to the fourth power:

\((x^{(3*x^{12})})^4=4^4\);

\(x^{(12*x^{12})}=4^4\);

\((x^{12})^{(x^{12})} =4^4\);

\(x^{12}=4=2^2\);

\(x=(2^2)^{(\frac{1}{12})}\);

\(x=2^{(\frac{1}{6})}\);

\(x=\sqrt[6]{2}\)


Answer: B­
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Practice Questions Chat 07 Sep 2024, 03:28
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Out of 16 cricket players,5 are good bowlers,3 are wicket keepers and the left are normal bowlers but not wicket keepers.Out of all, a team of 11 players is to be formed where at least 4 are good bowlers and 2 are wicket keepers?

Pls solve it..
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Practice Questions Chat 07 Sep 2024, 03:49
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Tarek143
Out of 16 cricket players,5 are good bowlers,3 are wicket keepers and the left are normal bowlers but not wicket keepers.Out of all, a team of 11 players is to be formed where at least 4 are good bowlers and 2 are wicket keepers?
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Kinda confused by what this question wants. Is the answer 4 good bowlers, 2 wicket keepers and 5 regular bowlers? or does this question asks for all the cases like 4 Good Bowlers and 3 Wicketkeepers and 4 Regular bowers, or 5 good bowlers, 3 wicket keepers and 3 regular bowlers?
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Seems like questions says, "In a pool of 16 cricketers, 5 are good bowlers, 3 are wicket keepers, and the rest 8 are normal bowlers" and we need to make a team of 11 players with with 2 conditions - at least 4 good bowlers and at least 2 wicket keepers.

If that is the case, we’ll have to make cases - (4,2,5) + (5,2,4) + (4,3,4) + (5,3,3) where (X,Y,Z) represents number of Good Bowlers, Wicket Keepers and Normal Bowlers respectively.......Basically, a (5C4 * 3C2 * 8C5) + (5C5 * 3C2 * 8C4) +....and so on
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