# Quant Questions With Solutions- 5

41. How many positive integers are there that are not larger than 1000 and are neither perfect squaresnor 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.

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

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

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.

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 isdeclared as the winner.If A begins the game and the odds of his winning the game are 3 to 2, then findthe 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

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

46. There is a circular table and 60 people can sit in that. If there are N number of people sitting and amonster comes and want to occupy a seat such that he has someone on his side. What is the minimum value of N?

a) 15

b) 20

c) 29

d) 30

a) 15

b) 20

c) 29

d) 30

Say, one person is sitting in seat no. 2 then d monster sits in seat no. 1 or 3 (monster will find a person beside it)

Again there is a person sitting in seat no. 5, then d monster can sit in seat no 4 and 6...

again there is a person sitting in seat no. 8

Thus d pattern of ppl sitting is-

sit no. 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59

thus min no. of n is 20

Again there is a person sitting in seat no. 5, then d monster can sit in seat no 4 and 6...

again there is a person sitting in seat no. 8

Thus d pattern of ppl sitting is-

sit no. 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59

thus min no. of n is 20

47. How many 3 digits numbers less than 1000 are there in which, if there is a 3 it is followed by 7? Given that no two digit of the number should be same.

suppose d 3 digit no. starts with 1

10__(This blank space can b filled in 7 ways 2,4,5,6,7,8,9)

in d above case we cannot take 3 since it should be followed by 7 which is not possible here.

now we take-

11__ this space can also be filled in 7 ways

12__ 7 ways

13__ this space can be filled only by 1 way and dat is 7.

14__ again 7 ways

15__ 7 ways

16__ 7 ways

17__ 7 ways

18__ 7 ways

19__ 7 ways

that gives us 57 ways in total

We will follow d same procedure for 3 digits starting with 2,4,5,6,8,9(but not in case of 3 and 7)

thus we get 57*7= 399 ways

now lets check condition for 3. 3 is always followed by 7 and nothing else

thus

37__ this space can be filled in 8 ways(0,1,2,4,5,6,8,9)

no other condition is possible in case of 3.

that makes our total to be 399+8=407

now lets check condition for 7

70__ this can be filled in 7 ways

71__ 7 ways

72__ 7 ways

73__ this condition is not possible since 3 should be followed by 7 but we cannot repeat 7.

74__ 7 ways

75__ 7 ways

76__ 7 ways

78__ 7 ways

79__ 7 ways

this gives us 56 ways

therefore our total becomes 407+56=463 and this is the answer.

10__(This blank space can b filled in 7 ways 2,4,5,6,7,8,9)

in d above case we cannot take 3 since it should be followed by 7 which is not possible here.

now we take-

11__ this space can also be filled in 7 ways

12__ 7 ways

13__ this space can be filled only by 1 way and dat is 7.

14__ again 7 ways

15__ 7 ways

16__ 7 ways

17__ 7 ways

18__ 7 ways

19__ 7 ways

that gives us 57 ways in total

We will follow d same procedure for 3 digits starting with 2,4,5,6,8,9(but not in case of 3 and 7)

thus we get 57*7= 399 ways

now lets check condition for 3. 3 is always followed by 7 and nothing else

thus

37__ this space can be filled in 8 ways(0,1,2,4,5,6,8,9)

no other condition is possible in case of 3.

that makes our total to be 399+8=407

now lets check condition for 7

70__ this can be filled in 7 ways

71__ 7 ways

72__ 7 ways

73__ this condition is not possible since 3 should be followed by 7 but we cannot repeat 7.

74__ 7 ways

75__ 7 ways

76__ 7 ways

78__ 7 ways

79__ 7 ways

this gives us 56 ways

therefore our total becomes 407+56=463 and this is the answer.

48. How many numbers are there between 1100 - 1300 which are divisible by and that all 4-digits of number should be odd (ex:1331)

1300-1100=200 now divide it by 2 we have 100 even number hence 100 odd no's

now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there

hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197

hence 9 no's

now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there

hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197

hence 9 no's

49. Three dice are rolled simultaneously. Find the probability of getting at least one six.

a) 1/6*5/6*5/6

b) 5/6*5/6*5/6

c) 3*1/6*5/6*5/6

d) none of the above

condition for no sixes is-

(5/6)*(5/6)*(5/6)

i.e. 125/216

therefore for atleast 1 six-

=1-(125/216)

=91/216

therefore option d)

(5/6)*(5/6)*(5/6)

i.e. 125/216

therefore for atleast 1 six-

=1-(125/216)

=91/216

therefore option d)

50. If there are n numbers of square cubes and you are given two colors, black and white. How many unique square cubes can u color?

Let B= Black color W=white color

If all 6 face colured with B and none with W -> 1 unique cube

5 B and 1 W->1

4 B and 2 W->2

3 B and 3 W->2

2 B and 4 W->2

1 B and 5 W->1

0 B and 6 W->1

---------- Answer=10

If all 6 face colured with B and none with W -> 1 unique cube

5 B and 1 W->1

4 B and 2 W->2

3 B and 3 W->2

2 B and 4 W->2

1 B and 5 W->1

0 B and 6 W->1

---------- Answer=10

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