Problems 8.07.09
I could not post due to some engagements. Here are a bunch of problems to compensate 
Question 1)
A + B + C + D = D + E + F + G = G + H + I = 17 where each letter represent a number from 1 to 9. Find out number of ordered pairs (D,G) if letter A = 4.
a) 0 b) 1 c)2 d) 3 e) none of these
Question 2)
The sequence 1, 3, 4, 9, 10, 12….. includes all numbers that are a sum of one or more distinct powers of 3. Then the 50th term of the sequence is
a. 252 b. 283 c. 327 d. 360 e) none of these
Question 3)
Given that g(h(x)) = 2x² + 3x and h(g(x)) = x² + 4x − 4 for all
real x. WHich of the following could be the value of g(-4)?
a)1 b) -1 c) 2 d) -2 e) -3
Question 4)
If a, x, b and y are real numbers and ax+by = 4 and ax² +by² = 2 and
ax³ + by³= −3 then find (2x − 1)(2y − 1)
a)4 b) 3 c) 5 d) -3 e) cannot be determined.
Question 5)
K1,K2,K3…K30 are thirty toffees. A child places these toffees on a circle, such that there are exactly n ( n is a positive integer) toffees placed between Ki and Ki+1 and no two toffees overlap each other. Find n
a)4 b) 5 c) 9 d) 12 e) 13
Question 6)
For the n found in previous question, which of the two toffees are adjacently
placed on the circle? ( All other conditions remaining same)
a) K11 and K13 b) K6 and K23 c) K2 and K10 d) K11 and K18
e) K20 and K28
Question 5 is not complete… Just check it…
Shreenivas
July 9, 2009 at 8:30 am
My take on answers is
2)C
3)B
4)A
5)D
6)D
kindly let me know the correct answers!!!
abhishek
July 9, 2009 at 8:47 am
yeah, there was one line missing.
You have find n
I have fixed it…
Its a request that you post your method along with the answer..
That saves me from the effort to type the whole answers, and i can post more questions
good luck
Rahul
July 9, 2009 at 9:11 am
2) T1=1;T2=(10)3;T3=(11)3;T4=(100)3
hence T50= (110010)3=327
3)g(h(g(x)))=2g(x)^2 + 3g(x)
to calculate g(-4) we have to find out x such that h(g(x))=-4
i.e. x^2 + 4x=0 i.e. x=-4
hence we get g(-4)= 2g(-4)^2+ 3g(-4)
hence g(-4)=-1
abhishek
July 9, 2009 at 9:32 am
T1=1;T2=(10)3;T3=(11)3;T4=(100)3
hence T50= (110010)3=327.
Hey I don’t get that man . Can you please
explain that
Hari
July 16, 2009 at 11:43 am
as the problem says ‘includes all numbers that are a sum of one or more distinct powers of 3′ so (10)3=> 10 in base 3;(10)3= (3+0)10 ie 3 in base 10
(11)3=(3+1)10=(4)10
(100)3=(9)10
now (10)2=2,(11)2=3,(100)2=4 are the respective term numbers
so the 50th term would be (110010)3 as (110010)2=50
hope this makes it clear!!
abhishek
July 17, 2009 at 7:44 am
@ abhishek thanks man . I was trying to understand on similar lines . Its very clear now .
Hari
July 18, 2009 at 9:36 am
3) 2x-1=ax^3 + bxy^2 – 1; 2y-1= ayx^2 + by^3 – 1
(2x-1)(2y-1)=(ax^3 + byx^2 – 1)(ayx^2 + by^3 – 1)
after multiplying ,rearranging the terms and using the other conditions given in the problem we can easily get the ans as 4
4)if k1 is the first then k2 will be the (n+2)th toffee k3 will be (2n+3)th toffee. let us assume that in one complete circle there are k consecutive toffees, as there should be no overlap hence the (k+1)toffee shouldnot overlap onto the 1st toffee.
as there are 30 toffees each toffee will be placed at 12degrees
so (k+1)(n+1)12 shouldnt be equal to 360
ie (k+1)(n+1) =! 30
out of the options only for n=12 we cannot have any value of k satisfying (k+1)(n+1)=30
hence there should be 12 toffees in between!!!
abhishek
July 9, 2009 at 10:10 am
Answers
1) b
2) c
3) b
4) a
5) d
6) d
Rahul
July 16, 2009 at 10:49 am
The sequence 1, 3, 4, 9, 10, 12….. includes all numbers that are a sum of one or more distinct powers of 3. Then the 50th term of the sequence is
hi can any you explain the this sum clearly
like why abisheik took base all these things and how he changed the base
imman
July 17, 2009 at 5:37 pm