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problem solving q's

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by hareesh860 » Sun Jan 22, 2012 2:37 am
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9) T= 1/2-1/2^2+1/2^3-...-1/2^10 = 1/4+1/4^2+1/4^3+1/4^4+1/4^5
Notice that 1/4^2+1/4^3+1/4^4+1/4^5 < 1/4, we can say that 1/4<T<1/2.
Answer is D
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by hareesh860 » Sun Jan 22, 2012 2:39 am
10)If each term in the sum a1+a2+a3...+an is either 7 or 77 and the sum is 350, which of the following could be n?

a)38
b)39
c)40
d)41
e)42
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by hareesh860 » Sun Jan 22, 2012 2:39 am
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10)We know that the unit's digit in each term in the sum is 7. We also know that the sum ends in 0. The only way to get a bunch of 7s to add up to a 0 is if the number of terms is a multiple of 10.
Only (c) is a multiple of 10
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by hareesh860 » Sun Jan 22, 2012 2:40 am
11)Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6 <= y <= 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100
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by hareesh860 » Sun Jan 22, 2012 2:42 am
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11)No. of possible value for x=10 and y=11
1) information given is right angle at P, PR(ll) to x axis information inferred y coordinates of p and r is same.
2) PQ (llel) y therefore x coordinates of P and Q is same.
For P x can be selected in 10 ways and y in 11 ways = 10*11
For R x in 9 ways and y in 1 way (as same of P) =9*1
For q x in 1 way and y in 10 ways (one already selected for P) =10*1
Total ways=10*11*9*10=9900
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by hareesh860 » Sun Jan 22, 2012 2:42 am
12) A password of a computer used five digits where they are from 0 and 9. What is the probability that the password solely consists of prime numbers and zero?
A 1/32
B 1/16
C 1/8
D 2/5
E ½
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by hareesh860 » Sun Jan 22, 2012 2:43 am
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12) There are 10 possible options (0,1,2,3,4,5,6,7,8,9) for each digit.
5 of the options (0,2,3,5,7) are zero or prime.
So, P(a given digit is zero or prime) = 5/10 = 1/2
A quick way is to look at this as an AND probability.
P(all five digits are zero or prime) = P(1st digit is zero or prime AND 2nd digit is zero or prime AND 3rd digit
is zero or prime AND 4th digit is zero or prime AND 5th digit is zero or prime)
This is equal to P(1st digit is zero or prime) x P(2nd digit is zero or prime) x P(3rd digit is zero or prime) x
P(4th digit is zero or prime) x P(5th digit is zero or prime)
So, we get 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32
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