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Quant Boosters - Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta - Set 1

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badalravi last edited by @Guest

@soumil_jain 4
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zabeer last edited by @Guest

@badalravi correct!
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zabeer last edited by zabeer @Guest

0 for n = even and n - 2 for n = odd.
CBD
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badalravi last edited by @Guest

@soumil_jain 1
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badalravi last edited by @Guest

@soumil_jain ordered 180 unordered 124
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badalravi last edited by @Guest

@soumil_jain divisibility rule of 32 as last 4 digit to be divisible by 32
Thus (7475/32) = 19
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badalravi last edited by @Guest

@soumil_jain 4
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badalravi last edited by @Guest

@soumil_jain 13/64
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badalravi last edited by @Guest

@soumil_jain 102!/100 = 1
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badalravi last edited by @Guest

@soumil_jain 25 ways
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quant - mixed bag 73 question bank 47

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cat questions86 quant - mixed bag73 question bank47 theory of equations14 permutation & combination11 time speed & distance10 cat questions lr10 set theory7

Question Bank - Number theory - Sibanand Pattnaik
Quant BBQ - Best of Best Questions • question bank • • sibanand_pattnaik

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Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 4
Quant - Boosters • quant - mixed bag • • shashank_prabhu

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@Naman-Jain-0 ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___There are a total of 8 steps represented by (___) and let the dash ('|') represent if he is stopping at the previous step. There are 7 such dashes. Each dash can take a 0 or 1. 0 indicates, he is stopping at the step immediately before it. As each dash can take a 0 or 1, the number of ways is 128.And as the maximum steps u can take is 6, the cases where he takes 7 steps at a time - given by (1,0,0,0,0,0,0) and (0,0,0,0,0,0,1)- and 8 steps at a time - given by (0,0,0,0,0,0,0)- are eliminated. Therefore it is 125.
Quant Boosters - Kamal Lohia - Set 4
Quant - Boosters • quant - mixed bag • • kamal_lohia

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@kamal_lohia thank you @zabeer also , has anyone posted memory based LRDI questions from 2015 and 2016 ?
Quant Boosters - Ravi Handa - Set 2
Quant - Boosters • quant - mixed bag • • handakafunda

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@handakafunda Sir is answer 20c10-18c8 is correct for this question??
Quant Boosters - Maneesh - Set 2
Quant - Boosters • quant - mixed bag • • maneesh_p

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Angle = 30x - 11y/2 = 30 * 7 – 11 * 30/2 = 210 – 165 = 45 degree.
Quant Boosters - Vikas Saini - Set 2
Quant - Boosters • quant - mixed bag • • vikas_saini

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2x^2 + 6 + 2x^2 + x + 6 = 4x^2 + 2x + 4 = > x + 12 = 2x + 4. = > x = 8.
Quant Boosters - Sibanand Pattnaik - Set 2
Quant - Boosters • quant - mixed bag • • sibanand_pattnaik

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1000 = 2^3 * 5^ When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases .. Case -1 ; ( 1, 2^3, 5^3) ..now here ordered arrangements = 3^2*3! = 54 Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)now total arrangement of these cases is 3!/2! ( 15) = 45 case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1 So finally 54 + 45 + 1 = 100
Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 5
Quant - Boosters • quant - mixed bag • • shashank_prabhu

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@Vikrant-Garg Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?) This is possible in 2 ways - if we get DD or if we get GG as the result of first 2 tests. In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases. So total number of ways = 4!/(2! x 2!) = 6 ways. Favourable cases = 2 Probability = 2/6 = 1/3 Is it clear ?

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