Official Quant thread for CAT 2013

MBA CAT 2024
11581
@freebull said:
If a.b.c=2^5.3^4*5^7. How many Natural No Solutions?
a=2^x1*3^y1*5^z1
b=2^x2*3^y2*5^z2
c=2^x3*3^y3*5^z3

x1+x2+x3=5=>7C2=21ways

y1+y2+y3=4=>6C2=15ways

z1+z2+z3=7=9C2=36ways

21*15*36
275*36
9900

11582
@sujamait said:
x, y and z are real numbers. If xyz = 1, x + y + z = 2 and x^2+ y^2+ z^2 = 16, then what is the value of1/(xy + 2z) +1/(yz + 2x)+1/(xz + 2y)=?a-10b- (-4/13)c- 0d (-1)e- 2/7
1/(xy + 2z) +1/(yz + 2x)+1/(xz + 2y)
= z/(xyz+2z^2) +x/(xyz + 2x^2)+y/(xyz + 2y^2)
= z/(1+2z^2) +x/(1 + 2x^2)+y/(1 + 2y^2)

Now, Numerator = z(1+ 2y^2)(1 + 2x^2)+x(1 + 2y^2)(1 + 2z^2)+y(1 + 2x^2)(1 + 2x^2)

Now, z(1+ 2y^2)(1 + 2x^2)
=z{ 1+ 2(x^2+y^2)+4x^2y^2)}
=z{ 1+ 2(16-z^2) + 4/z^2} (x^2+y^2+z^2=16, xyz=1)
= 33z-2z^3+4/z

similarly, x(1 + 2y^2)(1 + 2z^2) = 33x-2x^3+4/x
y(1 + 2x^2)(1 + 2z^2) = 33y-2y^3+4/y

=> summation = 33(x+y+z)-2(x^3+y^3+z^3)+4(1/x+1/y+1/z)

Now xy+yz+zx= {(x+y+z)^2-(x^2+y^2+z^2)}/2=-6

x^2y^2+y^2z^2+z^2x^2=(xy+yz+zx)^2-2*xy*yz*zx=36-2=34

x^3+y^3+z^3 -3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=> x^3+y^3+z^3-3=2*(16+6)=44
=> x^3+y^3+z^3=44+3=47

(1/x+1/y+1/z)
=(xy+yz+zx) ( xyz=1 )
=-6

Numeric value of summation = 33*2-2*47-4*6= -52

Denominator = (1+2z^2)(1+2x^2)(1+2y^2)
=(1+2z^2+2x^2+4x^2z^2)(1+2y^2)
=1+2z^2+2x^2+4x^2z^2 + 2y^2+ 4y^2z^2+4x^2y^2+8x^2y^2z^2
=1+2(x^2+y^2+z^2)+4(x^2y^2+y^2z^2+z^2x^2)+8x^2y^2z^2
= 1+ 2*16+4*34+8 (x^2y^2+y^2z^2+z^2x^2=34 as derived above)
=177

ˆ ´ value of expression = -52/177

P.S. @sujamait : Sir kahan se le aaye aisa question. 40 minutes se kr raha hun :banghead:
11583

Q) A faulty meter skips the digit 5 and moves to 6 directly from 4. Currently this meter is showing a reading of 3016. What is the actual reading ??

11584
@vijay_chandola said:
1/(xy + 2z) +1/(yz + 2x)+1/(xz + 2y)= z/(xyz+2z^2) +x/(xyz + 2x^2)+y/(xyz + 2y^2)= z/(1+2z^2) +x/(1 + 2x^2)+y/(1 + 2y^2) Now, Numerator = z(1+ 2y^2)(1 + 2x^2)+x(1 + 2y^2)(1 + 2z^2)+y(1 + 2x^2)(1 + 2x^2)Now, z(1+ 2y^2)(1 + 2x^2)=z{ 1+ 2(x^2+y^2)+4x^2y^2)}=z{ 1+ 2(16-z^2) + 4/z^2} (x^2+y^2+z^2=16, xyz=1)= 33z-2z^3+4/zsimilarly, x(1 + 2y^2)(1 + 2z^2) = 33x-2x^3+4/xy(1 + 2x^2)(1 + 2z^2) = 33y-2y^3+4/y=> summation = 33(x+y+z)-2(x^3+y^3+z^3)+4(1/x+1/y+1/z)Now xy+yz+zx= {(x+y+z)^2-(x^2+y^2+z^2)}/2=-6x^2y^2+y^2z^2+z^2x^2=(xy+yz+zx)^2-2*xy*yz*zx=36-2=34x^3+y^3+z^3 -3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=> x^3+y^3+z^3-3=2*(16+6)=44=> x^3+y^3+z^3=44+3=47(1/x+1/y+1/z)=(xy+yz+zx) ( xyz=1 )=-6Numeric value of summation = 33*2-2*47-4*6= -52Denominator = (1+2z^2)(1+2x^2)(1+2y^2)=(1+2z^2+2x^2+4x^2z^2)(1+2y^2)=1+2z^2+2x^2+4x^2z^2 + 2y^2+ 4y^2z^2+4x^2y^2+8x^2y^2z^2=1+2(x^2+y^2+z^2)+4(x^2y^2+y^2z^2+z^2x^2)+8x^2y^2z^2= 1+ 2*16+4*34+8 (x^2y^2+y^2z^2+z^2x^2=34 as derived above) =177 ˆ´value of expression = -52/177P.S. @sujamait : Sir kahan se le aaye aisa question. 40 minutes se kr raha hun
bha option match kar rha hai kya koi ? 😛
me khud koi shortcut dhoond rha tha..
11585
@ayushbhalotia

Since the Digit 5 is not operating, the base system involved here would be 9..

Thus, While the Meter Reading moves from 1-1000, the actual reading is 9^3= 729..

This is true for 1001-2000 and 2001-3000..Thus a total of 729*3= 2187..

Now, For 3001-3016, the Number 15 is expressed in base 9..We have to convert it in Base 10..
=>1*9 + 5=14..

Thus, the correct reading is 14 + 2187= 2201..
11586
@ayushbhalotia said:
Q) A faulty meter skips the digit 5 and moves to 6 directly from 4. Currently this meter is showing a reading of 3016. What is the actual reading ??
3*729+9+5=2201


@sujamait said:
bha option match kar rha hai kya koi ? me khud koi shortcut dhoond rha tha..
Sir rechecked more than 4-5 times :embarrassed:
Koi option match ni kr ra :angry:
11587
@ayushbhalotia said:
Q) A faulty meter skips the digit 5 and moves to 6 directly from 4. Currently this meter is showing a reading of 3016. What is the actual reading ??


iss no ko base 9 mein convert kardo bus..
2202 - 1 ?
11588
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?

The last number in the product series contains 999 number of 9s.
OPTIONS

1) 89
2) 109
3) 129
4) None of these
11589
@ayushbhalotia said:
Q) A faulty meter skips the digit 5 and moves to 6 directly from 4. Currently this meter is showing a reading of 3016. What is the actual reading ??
(3016 = (3015)base 9..
converting to base 10....3*9^3 + 1*9 + 5 = 2201
11590
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?

The last number in the product series contains 999 number of 9s.
OPTIONS

1) 89
2) 109
3) 129
4) None of these
11591
There were 320 students in the premises of an ongoing college fest. A student can take part in any of the
three activities €” Dance, Debate and Drama. Ten students took part in all the three activities and
201 students took part in Debate. Number of students who took part in Dance and Drama but not in Debate
is one-third the number of students who took part only in Drama. Number of students who took part in
Dance and Debate but not in Drama is equal to the number of students who took part in Debate and Drama
but not in Dance and is 1 more than the number of students who took part only in Debate. Twenty-nine
students just came to enjoy the performances. Number of students who took part in Dance is 27 more than
the number of students who took part in Drama. Find the number of students who took part in Dance.
A. 54
B. 45
C. 137
D. 154
E. 68
11592
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
109?
11593
@vijay_chandola said:
3*729+9+5=2201 Sir rechecked more than 4-5 times Koi option match ni kr ra
-4/13 hai OA

11594
@ankita14 said:
109?
method ?
11595
@sujamait

Should be 109..

Here, 9*99*999=(10-1)*(100-1)*(1000-1) =>1000*x + 109..

The Next terms in the series, 9999*99999=(10000-1)*(100000-1) =>1000x + 1..

This, will be true for all the remaining pairs..

Thus, Rem[109*(1*1*1* ......*448 times)/1000]= 109*1 = 109..
11596
@sujamait said:
method ?
9*99= 891. Rest of the terms leave -1 as remainder. Final remainder is -891=109
11597
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
rem= 9*99*(0-1)*(0-1)....../1000=rem(-891/1000)= 109
11598
R=? 61^67^71^101/103
11599
@sujamait said:
What is the remainder when 9 × 99 × 999 × … × 99… 9 is divided by 1000?The last number in the product series contains 999 number of 9s.OPTIONS1) 89 2) 109 3) 129 4) None of these
109??
-9*99=-891/1000=109
11600
@vijay_chandola said:
1/(xy + 2z) +1/(yz + 2x)+1/(xz + 2y)= z/(xyz+2z^2) +x/(xyz + 2x^2)+y/(xyz + 2y^2)= z/(1+2z^2) +x/(1 + 2x^2)+y/(1 + 2y^2) Now, Numerator = z(1+ 2y^2)(1 + 2x^2)+x(1 + 2y^2)(1 + 2z^2)+y(1 + 2x^2)(1 + 2x^2)Now, z(1+ 2y^2)(1 + 2x^2)=z{ 1+ 2(x^2+y^2)+4x^2y^2)}=z{ 1+ 2(16-z^2) + 4/z^2} (x^2+y^2+z^2=16, xyz=1)= 33z-2z^3+4/zsimilarly, x(1 + 2y^2)(1 + 2z^2) = 33x-2x^3+4/xy(1 + 2x^2)(1 + 2z^2) = 33y-2y^3+4/y=> summation = 33(x+y+z)-2(x^3+y^3+z^3)+4(1/x+1/y+1/z)Now xy+yz+zx= {(x+y+z)^2-(x^2+y^2+z^2)}/2=-6x^2y^2+y^2z^2+z^2x^2=(xy+yz+zx)^2-2*xy*yz*zx=36-2=34x^3+y^3+z^3 -3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=> x^3+y^3+z^3-3=2*(16+6)=44=> x^3+y^3+z^3=44+3=47(1/x+1/y+1/z)=(xy+yz+zx) ( xyz=1 )=-6Numeric value of summation = 33*2-2*47-4*6= -52Denominator = (1+2z^2)(1+2x^2)(1+2y^2)=(1+2z^2+2x^2+4x^2z^2)(1+2y^2)=1+2z^2+2x^2+4x^2z^2 + 2y^2+ 4y^2z^2+4x^2y^2+8x^2y^2z^2=1+2(x^2+y^2+z^2)+4(x^2y^2+y^2z^2+z^2x^2)+8x^2y^2z^2= 1+ 2*16+4*34+8 (x^2y^2+y^2z^2+z^2x^2=34 as derived above) =177 ˆ´value of expression = -52/177P.S. @sujamait : Sir kahan se le aaye aisa question. 40 minutes se kr raha hun
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