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

In the solution given their, first the base is calculated i.e. a,b,c in this case. Then we check for the option which satisfies the calculated values of a,b,c...

please elaborate solution to question no 43

ReplyDeleteIn the solution given their, first the base is calculated i.e. a,b,c in this case. Then we check for the option which satisfies the calculated values of a,b,c...

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