CHSL & FCI PREPARATION

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@lalit6600 said:
@surajpandey ..............bt still example bhi to satisfy hona chahiye na.......
hum aapne example liya 41.

lets check.
according to your solution.
answer is to be 11.
but it is not even in the option bro.

there could be infinite of number that can satisfy the value... but only one method..so better to be go with that method.
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CHSL-2011 Questions

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@custom said:
thank u suraj jee aapne kya xplaination banai thi
mujhe yeh question nahi aata tha bhai..par ab aa gaya.. ;)
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@gautam7821 said:
q6 . ans x= 6
2^x/2 + 2*2^x = 320
2^x = 320 /5
2^x = 64
x = 6
bhai their is a slight error. when you will take
2^x common then inside bracket [{1/2} +2] i.e 5/2

toh finally solution wil be 2^x= 320*2/5= 2^7 or 128
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guys some questions have been posted by anticipator bhai..so dont go..try to solve them.. 😃

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@surajpandey said:
bhai their is a slight error. when you will take 2^x common then inside bracket [{1/2} +2] i.e 5/2toh finally solution wil be 2^x= 320*2/5= 2^7 or 128
Already spotted my error .. So deleted the post .. anyway thanks for super fast reply .. :)
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@anticipator.lad said:
CHSL-2011 Questions
102.) its simple
square all the numbers
3rt2=18
3rt7=63
6rt5=180
4rt5=80

as you can see 6rt5 is biggest number among all.
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@gautam7821 said:
Already spotted my error .. So deleted the post .. anyway thanks for super fast reply ..
you could have edited than deleting the post bhai.
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@anticipator.lad said:
CHSL-2011t 2 Questions
ques 102 option (c)
try to use elimination as much as possbile b/w 3rt2 3rt7 = 3rt7 is bigger
now 6rt5 is bigger than 3rt7 as for 2rt20 it is 4rt5 so 6rt5 is the biggest among these

ques 104 option (D)
you have to actually solve this one as i tried for a shortcut and was confused btw 9 and 8
multiply 7*98 subtract 47 and divide by 71 u get 9

ques 105 option (d)
again no is divisble by 7 so 18000 is eliminated leaves a remainder of 4 when divided by 25 so one ending in 2 is eliminated check for one if it divides its ur answer if it does nt other one is
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@anticipator.lad said:
CHSL-2011 Questions
in reply to your Q103

x/(x+3)

new eq.

(x+7)/(x+3-2)=2
solve this and get x=5
so numerator is 5 and denominator is 8.
so there sum will be 5+8=13
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Q104. answer is 3 cos 3 is divisible of 9.

solt.
structure of question: (71*x + 47) is divisible by 7. (eq 1)

the new number is 98 *7=686
as 47 is added in it . why dont we remove it? ;)
686-47=639

now 639 is divisible of 71 = 9 times. value of x=9 incase you hv doubt check eq 1

now we need to find divisible of x i.e 3

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@anticipator.lad said:
CHSL-2011 Questions
103 B 13

+

simple p/q = x/x+3

new equation x+7/x+3 = 2




so x = 5




p= 5 & q = 8

so p+ q = 13
where p = num & q = den
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@anticipator.lad 105 D 18004
as the answer need to be divisible of 7 and leaves reaminder when divided by 25.
18002, 18000 are eliminated.

now we are rest with.
17004
18004

only 18004 is a multiple of 7 and thats the answer.


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@anticipator.lad bhai ho gaye saare ab answer batado sahi hai ya nahi...
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CHSL 2010 questions


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@anticipator.lad said:
CHSL 2010 questions
D 597


exp

99*6 + 3


as 99 is added by 6 times plus 21/7
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@anticipator.lad said:
CHSL 2010 questions
ques 108 => option 4
just check remainders 28+ 21 = 49 divided by 33 gives remainder 16

ques 109 => option 2
sum = 84
hcf = 12
divide sum by hcf u get sum = 7 and hcf =1 find such pairs (1,6) (2,5) (3,4)
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@anticipator.lad said:
CHSL 2010 questions
108 4. 16 sum of two numbers is 2*33 x+ 49 , when divided by 33 remainder is 16.



109. 2. 3 pairs 1&6, 2&5, 3&4
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@anticipator.lad said:
CHSL 2010 questions
Q107 answer is 597
99 *6 and add the rest [1/7 +2/7+--------6/7]
594 + [21/7]
594+3=597


Q108 answer is 16
solt.
1st no. 33k +21 2nd no. 33k +28

add them both i.e 66k +49
when you will divide 49/33
remainder is 16

Q109 answer is c i.e 3
12x+12y = 84 [ as 12 is hcf so it will include in both the numbers]
now x+y=7
1 and 6
2and 5
3 and 4

so these 3 pairs could be the possiblity.
remember the relationship between number should be as like co-primes.
id any pair is 4 and 6 .. it wonbt work and wilnot be considered.

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waise 2010 ke asaan the 2011 ke thode time consuming so on that trend 2012 main ......