1) If |2-3/x|<=1/3 and |3-x/y|<=1/6 and ,x,y are positive reals. Which of the follwoing is a possible value of
x^2/(2x-y) ?
A) sqrt(3)-1 B) sqrt(2) C) sqrt(1/2) D) sqrt(3)/2 E) sqrt(3/2)
2)There are two numbers A and B. A can be expressed as a product of 13 and a two-digit
prime number and B can be expressed as a product of 17 and a two-digit prime number. If
the unit’s digit of the product of A and B is 7, then how many distinct products of A and B
are possible?
A . 48 B. 55 C. 80 D. 110 E. 120
3) 10 beads and two similar diamond pendants are required to form a diamond necklace. If the beads are same, how many different diamond necklaces can be formed?
A) 1 B 5 C) 6 D) 8 E 10
4) Assume the given sum of the series, 7.5 + 15.5 + 10.5 + 13.0 + 13.5 + 10.5....+ x – y + z is
20634, (x, y, z > 0) then what is the value of the expression (2x + 3y + 5z)?
A) 1084 B) 1284 C) 1464 D) 1684 E) None of These.
5) Let P be the product of all natural numbers between 45 and 293 that have an odd number of
factors. Find the highest power of 12 in P.
A) 6 B) 8 C) 10 D) 12 E) none of these
6)Assume r, s and t be three distinct integers between 0 and 10. if 1/r+1/s-3/t=2/(5r)
then find the
number of distinct values of (r + s – t).
A) 1 B) 2 C) 4 D) 3 E) none of these
7 How many integers less than 300 are relatively prime to either 10 or 18?
A) 140 B) 141 C) 142 D) 139 E) 138
Source: OLD mocks( 2007 and before)
pehla tho i made a mistake and got root 2 as answer
ReplyDeleteagle ka 55 .. 6 ,5 prime each so 11*5 ..
beads walla 0,1,2,3,4,5 .. so 6 is the answer
4 th walla tho tumne hi hint diya .. hai ..
5th ka 8 ..
3.6 possibilities..When there is no gap b/w the diamonds to there are 5 beads in b/w the diamonds
ReplyDelete2.55 5 primes with digits ending 1, and 6primes with 7,5primes with 9 and 6primes with 3.Now 13*17 ends in 1. Now for A*B to end with 7, x*y should end with 7 so units digit of x and y can be 9*3 or 1*7 so total possibilities are 6*5+5*5
ReplyDelete5.The numbers with odd number of factors are all perfect squares. So from 7^2 to 17^2..The 3 has the greatest power 8.2 easily has power>16.So max power of 12 is 8
ReplyDeletePost lunch:
ReplyDelete7. I don't get the flaw in my reasoning.
10 and 18 have 2,3,5 as factors. So up to 299 whatever numbers are multiples of these numbers should be removed.
So factors of 2, 149 factors of 3 without 2,50(99-49) and number of factors of 5 alone excluding 2 and 3 are 20(59-29-19+9) so total number of numbers that are not relatively prime are 149+50+20=219. So number of numbers below 300 that are relatively prime to 10 or 18 are 299-219 = 80. Implex, can you throw some light on this?
6.Rearranging, we get 1/t = (1/5r)+(1/3s)..Now r,s,t are +ve integers<=10..So the only two possibilities for the sum of two fractions yielding yet another fraction of the same form are (1/2+1/2 and 1/3+1/6), Now the only possible form is 1/3+1/6 when r=6,s=3 and t=10 or r=3,s=10 and t=10. So there can be two possible values for r+s-t
ReplyDeletequestion no 7 is giving me issues
ReplyDeleteI had got 80 myself, but CL solution says 140
which i am sure is wrong. but it is an old mock , can't help
No problem, that's fine!
ReplyDelete6)Assume r, s and t be three distinct integers between 0 and 10. if 1/r+1/s-3/t=2/(5r)
ReplyDeletethen find the
number of distinct values of (r + s – t).
i tried like this 3/5r +1/s = 3/t
now 1,2,3,4,5,6,7,8,9 are the only values
now for r =1 s needs to 5 but t is not integer
for r=2 s = 5 and t =6
so r+s-t = 1 ,
now r =3 1/5 +1/s = 3/t no..
1. solving the eqn. 1>=x>= 1/3, 19/6>=y>=17/6, so x/2-y/x lies between (17/28, 19/93) . only sqrt(1/2) lies in this interval. so 3 is the answer.
ReplyDelete2. let p, q be the two prime numbers. so 3*p*7*q = 7
not (p,q) can be( 3,9 ) and the total pairs can be ({13, 23, 43,53,73,83} * {19,29,59,79,89}) total 30 values.
(p,q) can also be (1,7) and the total pairs for that can be ({11,31,41,61,71} *{17,37,47,67,97}) so total 25 values..
thus total of 55 values..
3. {(0,10) (1,9) .. (5,5) } so 6 different pandents can be made.
can u pls post me the method and answer for 4.
ReplyDeleteAre there no official solutions to the problems
ReplyDeleteall the problems have been solved well
ReplyDeleteHi @implex..
ReplyDeletequestion no:7) How many integers less than 300 are relatively prime to either 10 or 18?
A) 140 B) 141 C) 142 D) 139 E) 138
Numbers less than N (x^a y^b)and prime to it is N(1-1/x)(1-1/y)
so numbers prime to 10 less than 300 is = 120
numbers prime to 18 less than 300 is 100
numbers prime to 180 less than 300 is = 220-80 equal to 140..
i dont know where i am wrong or how am i right but i am getting this answer as 140..could you see into it ??
~~~~~~~A little change ~~~~
ReplyDeleteHi @implex..
question no:7) How many integers less than 300 are relatively prime to either 10 or 18?
A) 140 B) 141 C) 142 D) 139 E) 138
Numbers less than N (x^a y^b)and prime to it is N(1-1/x)(1-1/y)
so numbers prime to 10 less than 300 is = 120
numbers prime to 18 less than 300 is 100
numbers prime to 180 less than 300 is = 80
so number prime to 10 or 18 is 220-80= 140..
i dont know where i am wrong or how am i right but i am getting this answer as 140..could you see into it ??
see all even numbers, all multiples of 5 and all multiples of 3
ReplyDeletewill add up to more than 160 so answer is much less than 140
it should be 80
hmm good logic..but using the formula i am getting it as expected..dont know how ?? why is it so..!! i am confused in the ease of formula :-0..could you find out??
ReplyDelete