Posted on by bellcatblog

Quant 16th August

  1. Answer:15
    Solution:<ECD=<DCB+<BCE=90+60=150
    <CDE=<DEC
    Hence <DEC=15

  2. Answer:B
    Solution: first check for division for 8
    so can be 8 or 0
    Now check for division by 9 for 9 the sum of all the numbers should be divisible by 9
    hence 8 and 0

  3. Answer:4 values
    Solution:

Posted on by bellcatblog

Quant 2nd August

  1. Answer:b 3/2
    solution:complete the figure by replicating it.
    semicircle will be a circle inscribed in ∆ of side 5,5,6
    A=r*s A=1/2(6)(4)=12 S=a+b+c/2=8 r=A/S=12/8=3/2
  2. Answer:
    Solution:
  4. 3sina + 5cosa=5
    5sina – 3cosa=…Answer=+-3Solution=3sina + 5cosa=5
    5sina – 3cosa=x
    9 sin^2 a + 25 cos^2a + 25 sin^2a + 9cos^2a= 25+x^2
    ==> x^2 = 9
    x= +/-3
  5. AB & CD are two parallel chords of a circle such that AB=6cm , CD=12cm . If the distance between two chords is 3cm, find radius of the circleSolution:
  6. What is the difference in the total number of factors and the number of prime factors in 60^3 – 32^3 – 28^3 ?Answer:116
    Solution:the concept is  a+b+c = 0 then
    a^3 + b^3 + c^3 = 3abc3*60*32*28 = 2^9*3^2*5^1*7^1
    (10)(3)(2)(2) = 120

    total factor =120
    prime factors =4

  7. Solution

  8. Answer :8

    Solution:

  9. Find the last three digits of 3^1994Answer:369
    Solution:3^1994 mod 8
    9^… mod 8 = 1
    3^1994 mod 125 =
    e(125)= 100
    3^2000 mod 125 =1
    3^6 * 3^1994 mod 125 = 126
    27*3^1994 mod 125 = 88
    3*3^1994 mod 125 = 107.
    so 3^1994 nod 125 = 119
    125a +119 = 8b +1
    so 125+119+125= 369

  10. Answer:15
    Solution:x cannot be equal to 0,1,9,13,17

  11. Answer:65/3
    Solution:a+b+c=25
    ab+bc+ca=75
    now a^2+b^2+c^2=25^2-2*75=625-150= 475
    now (b*1+c*1)^2<=(b^2+c^2)*2
    (25-a)^2<=(475-a^2)*2
    625+a^2-50a<=950-2a^2
    3a^2-50a-325<=0
    3a^2-65a+15a-325<=0
    a(3a-65)+5(3a-65)<=0
    (a+5)(a-65/3)<=0
    so max =65/3 and min=-3

  12. Capture.PNGAnswer:7
    Solution:
Posted on by bellcatblog

Foreign Words

  1. sui generis Unique
  2. Quel dommage.–>what a shame

  3. boondoggle  a wasteful or impractical project or activity often involving graft

  4. Diddly Small Amount
  5. Whodunnit :mysterious murder story ,we are not able to find murderer till the end.
  6. hermeneutics:the branch of knowledge that deals with interpretation, especially of the Bible or literary texts.
  7. honoris causa:as a mark of honour
  8. Ceteris paribus:other things equal
  9. Gallicize:to become french
  10. shibboleths:a custom, principle, or belief distinguishing a particular class or group of people, especially a long-standing one regarded as outmoded or no longer important.
  11. Fealty:a feudal tenant’s or vassal’s sworn loyalty to a lord.

 

Posted on by bellcatblog

Quant 29th July

  1. Q1. Find the remainder when 50^51^52 is divided by 11.
    A) 6 B) 4 C) 7 D) 3Answer:6
    Solution:50,mod11 so 6
  2. If the roots of the equation (x-a)(x-b)-k=0 are c & d , then prove that roots of
    (x-c)(x-d)+k =0 are a & bSolution:x^2-(a+b)x+ab-k and x^2-(c+d)x+cd+k=0…..

    a+b=c+d and vice versa,
    cd=ab-k,
    ab=cd+k……

    Simple inspection.

  3. Suppose you have a currency named Miso, in three denominations: 1 miso,10 misos, and 50 Misos…..in how many way can you pay a bill of 107 misos….??Answer:18

    Solution:
    1x +10y +50z =107
    z=0 , x +10y = 107 ,
    y= 0,1 , 2, 3 to 10 = 11 ways
    z=1 , 1+10y = 57,
    y= 0, 1, to 5 = 6 ways
    z= 2 , 1+10y = 7 ,
    y= 0 , 1 way
    total = 11+6+1 = 18

  4. If (1-p) is a root of quadratic equation
    x^2+px+(1-p)=0, then find its rootsAnswer:0,1

    Solution: Put 1-p in the equation u get p =1

    Then put that in the equation u get roots 0,-1

  5. Find the number of triples (a, b, c) of positive integers such that (a+ab+abc) = 11

    Answer:3
    Solution:a(1+b+bc)=11
    so a must be 11 or 1
    a=11
    Then 1+b+bc=1 then b=0 so not acceptable
    If a=1 then (1+b+bc)=11
    b+bc=10
    b(1+c)=10
    b=1,2,5,10
    b=1 then c=9
    b=2 then c=4
    b=5 then c= 1
    b=10 then c=0 (not acceptable)

    So (1,1,9)(1,2,4)&(1,5,1)

  6.  Find the remainder for (73*79*81)/11

    Answer:1
    Solution:-4*2*4/11–>-(-1)=1

Posted on by bellcatblog

LOVE

http://www.filtercopy.com/posts/28-romantic-phrases-from-literature-that-ll-probably-give-you-goosebumps-if-you-re-in-love

 

“You’re beautiful, but you’re empty…One couldn’t die for you. Of course, an ordinary passerby would think my rose looked just like you. But my rose, all on her own, is more important than all of you together, since she’s the one I’ve watered. Since she’s the one I put under glass, since she’s the one I sheltered behind the screen. Since she’s the one for whom I killed the caterpillars (except the two or three butterflies). Since she’s the one I listened to when she complained, or when she boasted, or even sometimes when she said nothing at all. Since she’s my rose.”