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Quant Boosters - Hemant Malhotra - Set 14

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ramsri last edited by @hemant_malhotra

@Rakesh-Achanta Steps please ......................
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Dhruv Poonia last edited by @hemant_malhotra

@akash achanta use venn diagram
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Tanuj79 last edited by @hemant_malhotra

@hemant_malhotra is it 4(2 root2 +root5)?
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Tanuj79 last edited by @hemant_malhotra

@hemant_malhotra
a = 1, b = 0, c = 16.
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balaji.k last edited by balaji.k @hemant_malhotra

This post is deleted!
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anandk last edited by @hemant_malhotra

@hemant_malhotra Is ans 65?
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Yash H Chaudhari last edited by @hemant_malhotra

@hemant_malhotra THE ANS IS WRONG...IT SHOULD BE IN FIBONACCI SERIES
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aneeeshp last edited by @hemant_malhotra

@Yash-H-Chaudhari Could you please provide your solution?
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Jyoti Singh last edited by Jyoti Singh @hemant_malhotra

@hemant_malhotra said in Quant Boosters - Hemant Malhotra - Set 14:

The sum of the squares of three numbers, x, y and z is 138, while the sum of their products taken two at a time is 131. Their sum is

(a+b+C)^2 = a^2 + b^2 + c^2 + 2(ab +bc +ca) = 20
Hence
None of the above
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Jyoti Singh last edited by @hemant_malhotra

@hemant_malhotra 8..............................................................
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quant - mixed bag 73

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cat questions86 quant - mixed bag73 question bank47 theory of equations15 permutation & combination11 time speed & distance10 cat questions lr10 set theory7

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 - Kamal Lohia - Set 2
Quant - Boosters • quant - mixed bag • • kamal_lohia

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Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square. Only possible case is 26² = 676 i.e. O = 7.
Quant Boosters - Rajesh Balasubramanian - Set 5
Quant - Boosters • question bank theory of equations • • rajesh_balasubramanian

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The roots are a and b: a + b = p and ab = 12 (a + b)^2 = p^2 (a - b)^2 = (a + b)^2 - 4ab => (a - b)^2 = p^2 - 12 * 4 = p^2 - 48 If |a - b| ≥ 12 { Difference between the roots is at least 12} then, (a - b)^2 ≥ 144 p^2 - 48 ≥ 144 p^2 ≥ 192 P ≥ 8√3 or P ≤ -8√3
Quant Boosters - Sagar Gupta - Set 3
Quant - Boosters • quant - mixed bag • • zabeer

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14 to 28 : 15 students 30, 31, 32 and 1 to 12 : 15 students so total 32
Quant Boosters - Gaurav Sharma - Set 2
Quant - Boosters • quant - mixed bag • • gaurav_sharma

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S = 1 + 4x + 7x^2 + 10x^3 + … Sx = x + 4x^2 + 7x^3 + 10x^4 + … S (1 – x) = 1 + 3x + 3x^2 + 3x^3 + … S (1 – x) = 1 + 3x/( 1 – x) 35/16(1 – x) = (1 – x + 3x) / ( 1 – x) = ( 1 + 2x ) / ( 1 – x) 35 (1 – x) ^2 = 16 (1 + 2x ) 35x^2 – 102x + 19 = 0 (7x – 19) (5x – 1) = 0 x = 19/7, 1/5 But x # 19/7 (because 19/7 > 1) So x = 1/5
Quant Boosters - Vikas Saini - Set 4
Quant - Boosters • quant - mixed bag • • vikas_saini

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Here given E x E = E and A x A =A. So E = 1 or 5, A = 5 or 1. Let’s put E = 1 and A = 5. 5B1 = 1 x B1 Not possible. Let’s put E = 5 and A = 1. 1B5 = 5 x B5 So B = 2. BBA + CCD = EEE. 221 + CCD = 555. 1 + D = 5. D = 4. 2 + C = 5. C = 3. AB x CE = 12 x 35 = 420.
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Quant Boosters - Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta - Set 1
Quant - Boosters • quant - mixed bag question bank • • A Former User

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@soumil_jain 25 ways
Quant Boosters - Hemant Malhotra - Set 6
Quant - Boosters • quant - mixed bag • • hemant_malhotra

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a + 5b + 10c + 25d = 50 , Put d = 0, a + 5b + 10c = 50 now solve this Put d = 1 then d = 2

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