Q4

f(n) = # perfect squares < n
g(n)= # primes < n

we need f(n) + g(n) = 16

lets try with the options.
A. x between 30 and 36
f(n) = 5. No of squares = 5 (1,4,9,16 and 25)
g(n) = 11 (Primes are 2,3,5,7,11,13,17,19,23,29,31)
f(n) + g(n) = 16

hence A

Q6.

LCM of x, 2^6 and (2*3)^5 = (2*3)^6
therefore x must at least have a 3^6 in it. Plus it can have other powers of 2 ranging from 2^0 to 2^6
that makes 7 possible values for X
And C

Q7,
gcf = 25 = 5^2
Sum of 2 integers = 350

25 , 325 -> GCF = 25, SUM = 350 (OK)
50 , 300 -> GCF = 50, SUM = 350 (NOT OK)
75 , 275 -> GCF = 25, SUM = 350 (OK)
100 , 250 -> GCF = 50, SUM = 350 (NOT OK)
125 , 225 -> GCF = 25, SUM = 350 (OK)
150 , 200 -> GCF = 50, SUM = 350 (NOT OK)
175 , 175 -> GCF = 175, SUM = 350 (NOT OK)

Hence there will be 3 such pairs of integers.
Ans C

Q8

x * 2 * (3^2) * 5 * 13 * 17 * 19 / (2^2 * 3 * 5^2 * 11)
after simplification we are left with:

x * 3 * 13 * 17 * 19 / 2 * 5 * 11

therefore x must at least be divisible by 110 (2*5*11)

Ans D

Q9.
reducing the equation makes it 508/999
when any number is divided by 999, the remainder after the decimal keeps repeating itself infinetely.

there the above fraction will be 0.508508508 ... and so on
every 2, 5, 8th ... didgit will be 0 and
every 3, 6, 9, 12th ... didgit will be 8
since 101 = 3*x + 2
101st digit in decimal will be 0
ans A

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If x is not equal to 0 and x^y=1, then which of the following must be true?

The Q itself says that x is not equal to zero. Therefore, x can be +ve or -ve both okay.
now, option 1 says...... x=1 it is false becoz it can't be must be true as x can be one but it can be some other number as well. for ex. this also satisfies the condition given. 2^0=1. therefore it must not be true.

Option second says ...... II. x=1 and y=0 , similar to the above one the given condition can be true but it is not
must to be true as 2^0=1 again satisfies the condition given. Therefore False for me again.
Option -III. it says either x=1 or y=0 now this option fulfills the criteria as if x =1 then no matter the power raise to it it will alwz remains 1 or if y will be equal to zero then no matter the base but the exp. remains in every case. Hence,
III must be true ...... Therefore, C.

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2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

Now in this Q, the Q for at most 3 letters in a subset that means it can have 0 or 1 or 2 or 3.
Therefore, when the set contains 0, it can be represented as 7c0 which yields us 1.
Second.... when the set contains 1, it can be represented as 7c1 which yields us 7.
Third, .........when the set contains 2, it can be represented as 7c2 which yields us 21.
Forth, .........when the set contains 3, it can be represented as 7c3 which yields us 35.
Now as we know that these four conditions are joined together with the help of "or"
so we must add them.... Therefore, 1+7+21+35=64. Hence, 64 answer....... E.

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First of all, before solving this problem, we must remember that two consecutive integers have a GCF of 1. That means 2 consecutive integers only share one factor that is 1. They never share any other factor. therefore, the factors of 18! can never be the factors of 18!+1. Okay!!!

Now, lets take a look at the answer choices.
1. 15, it is a factor of 18! as 18! contains both 5 & 3. Therefore, Eliminated.
2. 17, it is a factor of 18! as 18! contains 17. Therefore, Eliminated.
3. 19, it is not a factor of 18! as 18! can't contain 19 in it. Therefore, Chosen... Correct.
4. 33, it is a factor of 18! as 18! contains both 11 & 3. Therefore, Eliminated.
5. 39, it is a factor of 18! as 18! contains both 13 & 3. Therefore, Eliminated.

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In this Q. we have to find the least value of X ????

Now for this first of all we have prime factorize 377,910 i.e.;

377,910 = 4199*3^3*2*5
3300 = 2*3*5*11*2*5.

Now as given in the Question that, The product of a positive integer x and 377,910 is divisible by 3,300.
Therefore, 4199*3^3*2*5*x / 2*3*5*11*2*5.

After cancelling the required. we get X= 2*5*11 =110.
Therefore 110........................ D.

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Here, it is given that the LCM of x, 4^3 & 6^5 is 6^6.
we can write is like this way. LCM (x,2^6, 2^5*3^5) is 2^6*3^6.
Tha means that the LCM has 2^6*3^6 & we have 2^6, 2^5*3^5 therefore x must contain 3^6. alongwith 0,1,2,3,4,5,6. Therefore, x can be 7 different values. Therefore, 7. ........................C.

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Yeah, Here if we add 1/3 + 1/9 + 1/27 + 1/37 we will get the value as ::::: 0.50850850850850850850850850850851. it will keep on repeating itself in groups of three & @ 101th place it has a value. 0. Therefore, ......A.

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It is given that F(n) = # of perfect Squares < n &
G(n) = # of primes less than n. & it is given that F(n)+G(n) = 16 so, the values in the two functions can be anything,
so to keep the options limited, lets look at the answer choices.
A. 30 < x < 36 .........
Perfect Squares < 30 = 1 4 9 16 25 i.e. 5.
Prime Numbers < 36 = 2 3 5 7 11 13 17 19 23 19 31 i.e. 11.
Therefore, this satisfies the condition given. Therefore,............. A.

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The Q asks, How many of the subsets contains the five elements leaving behind 0.
so we can calculate the # of subsets as 2^5 = 32. Therefore, 32 . .........D.

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Here it is given that the GCF of two integers is 25 or 5^2. That means that they both share 5^2 only.
Now we can depict the intergers as 25a & 25b sharing only 25 & a & b don't share anything b/w them. okay !!
Now, as given in the question that sum of the integers is 350. therefore, 25a+25b=350 or 25(a+b)=350
or a+b =14. Now we look for those values of a & b those don't share anything b/w them. So,
a+b= 3+11 or 1+13 or 5+9.........so Three values. Therefore, 3. C..........................

Question 1
Diagonals: 15x, 11x, 9x
Area of S: 225.x^2/2
Area of R: 99.x^2/2

Difference: 63.x^2

For I to be possible: x has to be equal to 1 (ok!)
For II to be possible: x has to be equal to sqrt(2) (NOT ok, since diagonals must be integers)
For III to be possible: x has to be equal to 2 (ok!)

Answer: D

Question 2:

Subset with 0 letters: 1
Subset with 1 letter: 7
Subset with 2 letters: C(7,2) = 21
Subset with 3 letters: C(7,3) = 7.6.5/6 = 35

Total: 1 + 7 + 21 + 35 = 64

Answer: E

Question 3:

Total number of subsets: 2^6 = 64
Number of subsets that include 0: 2^5 = 32

Total of subsets that do not contain 0: 64 - 32 = 32

Answer:D

Question 4:
Easier if you test numbers (considering that f(x) + g(x) is always crescent).

x = 30 -> f(30) = 5, g(30) = 10 -> f+g = 15
x = 31 -> f(31) = 5, g(31) = 10 -> f+g = 15
x = 32 -> f(32) = 5, g(32) = 11 -> f+g = 16
x = 33 -> f(33) = 5, g(33) = 11 -> f+g = 16
x = 34 -> f(34) = 5, g(34) = 11 -> f+g = 16
x = 35 -> f(35) = 5, g(35) = 11 -> f+g = 16
x = 36 -> f(36) = 5, g(36) = 11 -> f+g = 16
x = 37 -> f(37) = 6, g(37) = 11 -> f+g = 17

Therefore: 30 < x < 37

Answer: B

Question 5:
Best solution is through elimination:
18! is divisible by 15, and 17. Therefore, 18! + 1 is not divisible by neither of these.
18! is also divisible by 33 as it has the factors 11 and 3. Therefore, 18! + 1 is not divisible by 33
18! is also divisible by 39 as it has the factors 13 and 3. Therefore, 18! + 1 is not divisible by 39

The only remaining option is: 19

Answer: C

Question 6
LCM (x,2^6,3^5.2^5) = (2^6).(3^6)

x = 2^a. 3^b
b must be equal to 6 (since 3^6 is a factor of the LCM)
a can be equal to 0,1,2,3,4,5,6

Therefore: 7 possibilities

Answer: C