Number System: LCM and HCF Test-3

Since we are asked to find the largest number,
we find the HCF of the three numbers as that would be the only value that divides the three numbers.
In this case, we need to measure a certain amount of liquid.
How can we do this? We can only achieve the highest value of the cup volume if we take it as a common factor of the three volumes, thereby implying that the cup can measure the given quantities in a certain minimum number of times.
The HCF of the three numbers given here is 9.

Ideally we have to do such questions by options by the eliminating option one by one .
But given below is the actual approach for the question .
Number when divided by 9 gives 5 as remainder is in form = 9k +5
Number when divided by 11 gives 6 as remainder is in form = 11h +6
As the number is same hence
9k+5 = 11h + 6
9k – 1 = 11 h
So we will find the value for h by the equation: 9k - 1 = 11 h
Checking
9 x 1 -1 = 8 not divisible by 11
9 x 2 -1= 17 not divisible by 11
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9 x 5 =45-1=44 is divisible by 11So we have the smallest value for k is 5 and the smallest number is 50
,But the next number which will satisfy these criteria is LCM of 9 & 11 + 50
So the number will be in the form of 99p + 50.
Now we have to find the largest 3 digit number in the form of 99p+50 By substituting the value of p =10 approximately number is 1040 but which is greater than 1000
Hence for the value p=9
The number is = 99*9+50=941 , which is option C

The number which will give 7 as remainder when divided by both 17 and 19 is LCM of 17 & 19 + remainder i.e 7
Now l.c.m. of 17 and 19 is 323 so the number must be in the form of {P(323)+7}
So now we will find the value of p, now the largest 3 digit number in the form of 323p is 969 i.e. 323X3=969.
Therefore the number which when divided by 17 or 19 gives the remainder 7 is 976

We will solve the following problem with the following approach:
When the number N is divided by the numbers a, b, c and gives p, q ,r
when divided by N such that : (a-p)=(b-q)=(c-r)=Z,
then the number N must be in the form of {f(l.c.m. of a,b,c,)- D} in each case, where f is any positive integer.
Since the remainder are different so we first check the difference of remainder and the divisor i.e. 12-5 =13-6=7 so the difference is same.
Since the difference is same the number must be in the form of {f(l.c.m. of a,b,c,)- D}
So now the lcm of 12 and 13 is 156,so the number must be in the form of 156f – 7.
Now we have to find the value of f so the largest three digit number in the form of 156f
i.e 156 x 6=936
The number which gives the remainder 5 and 6 when divided by 12 and 13 is 929.

Numbers which are divisible by 2 from 1 to 900 are 450
Numbers which are divisible by 3 from 1 to 900 are 300
This means there are 750 numbers which are divisible by 2 and 3 between 1 to 900
but these 750 numbers also include the numbers which are also divisible by 2 and 3 i.e. divisible by  6.
There are 150 such numbers which are divisible by 6 between 1 to 900 and which we have already counter.
Therefore, the total numbers which neither divisible by 2 nor by 3 are
900-(450+300-150) = 900-600=300Hence answer is option B