41. How many positive integers are there that are not larger than 1000 and are neither perfect squares nor perfect cubes?
We use the Inclusion-Exclusion Principle. There are 1000 integers from 1 to 1000;
among these numbers, 31 are perfect squares (indeed, 312 = 961 ≤ 1000, but 322 > 1000), 10
are perfect cubes (this is because 103 = 1000), and 3 are both squares and cubes (these three
numbers are 16 = 1, 26 = 64, and 36 = 729). Thus, by the Inclusion-exclusion principle, there
are
1000 − (31 + 10) + 3 = 962
numbers that are neither perfect squares nor perfect cubes.
42. There are 9 players including Mic and Jordan standing in a row. What is the probability of being 2 or less players between Mic and Jordan?
total no. of ways arranging 9 people is = 9factorial.
case-1--
both sit together
hence no. of ways of arranging them is 8fact*2fact.
case-2
1person sit between them
hence no. of way of arranging them is 7(2fact*7fact).
case--3
2 person sit between them
hence no. of ways arranging them is 7(2fact*6fact)
hence total no. of ways ==8!2!+7(2!7!)+6(2!7!)
divide this by 9!..
ans.. 7/12
43. If the decimal number 120 when expressed to the base a,b and c equals 60,80,100 respectively, then which of the following statement is true?
a) a,b,c are in geometric progression
b) a,b,c are in arithmetic progression
c) a,b,c are in harmonic progression
d) a-b-c=1
120 base 10= 60 base a i.e. a=(120/60)*10 =>a=20
120 base 10=80 base b i.e. b=(120/80)*10 =>b=15
120 base 10= 100 base c i.e. c=(120/100)*10 =>c=12
We can see clearly it is neither in A.P. nor in G.P. also not satisfying last option
Now for H.P. b=2ac/(a+c)
2*20*12/32=15
condition satisfied
Ans. c) a,b,c are in harmonic progression
44. If a=b*c then for any value of n, the equation (a-b)^n-(c-b)^n+c^n is always divisible by
a) bc
b) b but not c always
c)c but not b always
d)non of above
put n=1 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
put n=2 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
put n=3 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
for for any value of n the given expression is divisible by bc
option A.
45. A and B pick up a ball at random from a bag containing M red, N yellow and O green balls one after the other, replacing the ball every time till one of them gets a red ball.The first one to get the red ball is declared as the winner.If A begins the game and the odds of his winning the game are 3 to 2, then find the ration M:N.
odds in favour of winning the event = (prob that an event will occur):(prob. that an event will not occur)
if odds in favour of a winning the event is 3:2
it means 3 if for red and 2 is for yelow and green.
since the bag contains all three types so none can be zero.
that means yellow:green =1:1.
therefore M:N=3:1
