Topic - Quant Mixed Bag

Solved ? - Yes

Source - CAT Preparation With IIMpossible Forum. ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - CAT Preparation With IIMpossible Forum. ]]>

a) 4

b) 5

c) 6

d) 7 ]]>

Product of numbers and LCM= HA x HB x HAB = H( A+B+AB) =89

Therefore H=1 and ( A+B+AB)= 89. OR (A+1)(B+1)= 90. Therefore

i) A= 8, B=9

ii) A=44, B=1

iii) A=29, B=2

iv) A=4, B= 17

v) A= 5, B=14

Hence (b) ]]>

a) 3

b) 4

c) 5

d) 6 ]]>

Hab-H= 27 OR

H(ab-1)=27

Possibilities

H ab-1 ab a,b Numbers

1 27 28 28,1 or 7,4 28,1 or 7,4

3 9 10 10,1 or 5,2 30,3 or 15,6

9 3 4 4,1 36,9

27 1 2 2,1 54,27

Hence (d) ]]>

a) 6

b) 4

c) 7

d) 8 ]]>

H+Hab= 30 OR

H(1+ab)=30

a and b are co prime to each other

Possibilities

H 1+ab ab a,b Numbers

1 30 29 29, 1 29,1

2 15 14 14,1 or 7,2 28,2 or 14,4

3 10 9 9,1 27,3

5 6 5 5,1 25,5

6 5 4 4,1 24,6

10 3 2 2,1 20,10

15 2 1 15,15 15,15

30 1 0 Not possible Not possible

Hence (d)

N=9k+6

and N=21l+12

∴ 9k+6=21l+12

⟹ 9k-21l=6

or (3k-7l)=6

or 3k=7l+2 or k=(7l+2)/3

So put the min. possible value of l such that the value of k an integer or in other words numerator (i.e.,7l+2) will be divisible by 3.

Thus at l=1, we get k=3 (an integer). So the least possible number N=9×3+6=21×1+12=33.

Now the higher possible values can be obtained by adding 33 in the multiples of LCM of 9 and 21 i.e., The general form of the number is 63m+33. So the other number in the given range including 33 are 96, 159, 222, 285, 348, …, 1104. Hence there are total 18 numbers which satisfy the given condition. ]]>

N=13k+3 ….(1)

and N=5l+2 ….(2)

where k and l are the quotients belong to the set of integers.

Thus 5l+2=13k+3

⟹ 5l-13k=1

⟹ 5l=13k+1

⟹ l=(13k+1)/5

Now we put the value of k such that numerator will be divisible by 5 or l must be integer so considering k=1,2,3,… we find that k=3, l become 8. So the number N=5×8+2=42

Thus the least possible number = 42

To get the higher numbers which satisfy the given conditions in the above problem we just add the multiples of the given divisors (i.e., 13 and 5) to the least possible number (i.e., 42)

NOTE The next higher number =(Multiple of LCM of 13 and 5)+42

=65m+42

In the above problem what is the greatest possible number of 4 digits ?

Since the general form of this number is 65m+42 so by putting m=153 we get (9945+42=)9987, which is required number.

HINT To get the value of 9945 we can simply divide 9999 (which is the greatest 4 digit number) by 65 and then subtract the remainder from 9999. ]]>

= 24m-1

Since, the least 5 digit number is 10000. So the required number must be atleast 10000. So putting the value of m=417, we get (10008-1)=10007, which is the required number. ]]>

Hence the least possible number

=(LCM of 18,35 and 42)-16

=630-16 [∴(18-2)=(35-19)=(42-26)=16]

=614

What is the least possible number which when divided by 2, 3, 4, 5, 6 it leaves the remainders 1, 2, 3, 4, 5 respectively ?

Since the difference is same as

(2-1)=(3-2)=(4-3)=(5-4)(6-5)=1

Hence the required number =(LCM of 2,3,4,5,6)-1

=60-1=59

In the above problem what is the least possible 3 digit number which is divisible by 11 ?

Since the form of the number is 60m-1, where m=1,2,3,…but the number (60m-1) should also be divisible by 11 hence at m=9 the number becomes 539 which is also divisible by 11. Thus the required number = 539. ]]>

=(LCM of 21,25,27 and 35)m + 2

=4725m+2

=4727 (for the least possible value we take m=1) ]]>

=LCM of 24,32,42)+(5)=672+5=677

Hence such a least possible number is 677.

In the above question how may numbers are possible between 666 and 8888 ?

Since the form of such a number is 672m +5, where m=1,2,3,…So, the first no.=672×1+5=677 and the highest possible number in the given range

=672×13+5

=8736+5=8741

Thus the total numbers between 666 and 8888 are 13. ]]>

15 = 3 x 5

So 15 has 4 factors

So 4 possible values of N ]]>

So 100 has 9 factors

However N cannot be 1

So N can take 8 values ]]>

a) 22

b) 21

c) 18

d) 23 ]]>

so (ORDERED - 1 )/2 is ur answer ]]>