Gradutate Record Exam

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Anonymous said...

NOV 5th

1)Ans:294
if a number is divisible by bothb2 & 3 then it is divisible by 6
so range of numbers divisible by 6 between 400 to 700

400/6=66.66 So 67*6=402
700/6=116.66 So 116*6=696

range is 696-402=294

2)Ans: A < B
so stupid question to give Explanation

3)Ans: A < B
Col A:P1=1
P2=24*1+8=32 remainder of 32/6=2
P3=24*32+8=776 remainder of 776/6=2
P4=24*776+8=18632 remainder of 18632/6=2
......................................
....................
Remainder of P66/6=2
Col B:4
A < B

4)Ans: A > B
Col A:
pen=x stapler=y
1pen &1stapler=10$
x+y=10 ----------------Eq(1)
3pens&2staplers=25$
3x+2y=25 -----------------Eq(2)
Eq(2)-2*Eq(1)
3x+2y=25
-2x-2y=20
---------------------
x=5
cost of 1pen=5$
Col B:4.20$


5)Data insufficient
we cannot determine the radius only from centre
it can be any number in range of 0 < r < infinity

6) Ans: A = B
Col A:
Given (a+b)/2=10 So a+b=20
then (a+b+c)/3=10
a+b+c=30 but a+b=20 So c=30-20=10
Col B:10

7)Ans:0.8x
Since given x is -ve number between 0 and -1
sqrt(x) is an imaginary number
so 0.8x has greater value

8)Ans: A = B
length of side of triangle with sides a&b be say x
since it is a right angled triangle
x=sqrt(a^2+b^2)
Since the other triangle is also a right angled triangle
f=sqrt(x^2+c^2)=sqrt(a^2+b^2+c^2)
Col A:(a^2+b^2+c^2)
Col B:f^2=(a^2+b^2+c^2)

***************************************************************************************************************************
FAQS NOVEMBER

1)Ans:A=B
series is 1,-1,2,-2,3,-3,4,-4,.................
Col A: Sum of the first 67 terms
since the successive terms are negations of each other they cancel each other when calculating the sum
but since sum is only for the 67 terms last terms negation will not be considered.
So the value of the sum be 67th terms
the 67th term=66/2+1=34
Sum=34
Col B:34
So A=B


2)Ans:58
for the value to be minimum,
t must be minimum r must be maximum
So minimum value of t=156
maximum value of r=98
so minimum value of t-r=156-98=58


3)Ans:3/17
Here 30% of 1 is 0.3=3/10
Of given options value less than 3/10 is 3/17


4)Ans: A < B
Col A:P1=1
P2=24*1+8=32 remainder of 32/6=2
P3=24*32+8=776 remainder of 776/6=2
P4=24*776+8=18632 remainder of 18632/6=2
......................................
....................
Remainder of P66/6=2
Col B:4
A < B
ANY MORE EASY APPROACHES

5)Ans:3/4
since the no.of even numbers=no.of odd numbers in the range of 1 to 10
this is similar to probalility of atleast one head when two unbiased coins are tossed
so the probability is 3/4

Nov 11, 2008, 7:16:00 AM  

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