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The next set of PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. 63 II. 126 III. 252

A. I only B. II only C. III only D. I and III only E. I, II and III

4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38

18! and 18!+1 are consecutive integers and so they do not have any common factor except 1. 15, 17, 33 (=3*11), and 39 (=3*13) are factors of 18! and none of those can be a factor of 18!+1. So, only 19 can be a factor of 18!+1

6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1 B. 6 C. 7 D. 30 E. 36

Numbers are: x, 2^6, and (2^5)*(3^5) LCM of the numbers = (2^6)*(3^6) As 3^6 is not part of second and third numbers, it must be part of x. So, lowest and highest values of x can be 3^6 and 6^6 {=(3^6)*(2^6)}. So the values that x can take are: (3^6)*(2^0), (3^6)*(2^1), (3^6)*(2^2), (3^6)*(2^3), (3^6)*(2^4), (3^6)*(2^5), and (3^6)*(2^6).

7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1 B. 2 C. 3 D. 4 E. 5

350 = 25 * 14 To have the GCD of two numbers to be 25, we need to split 14 into two co-prime numbers. Such pairs of numbers are: (1,13), (3,11), and (5,9).

1/3 + 1/9 + 1/27 + 1/37 =0.508508…………….. (recurring 508) --> 8 will be in every 3rd position --> 8 will be in 99th position --> 0 will be in 101st position

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

I. x=1 II. x=1 and y=0 III. x=1 or y=0

A. I only B. II only C. III only D. I and III only E. None

x^y=1 can be obtained by any of the following: (a) y = 0 and x = any real number (b) x = 1 and y = any real number (c) x = -1 and y = any even number

I. For x=2 and y=0, x^y=1. So, this condition is not necessarily true. II. For x=2 and y=0, x^y=1. For x=1 and y=5, x^y=1. So, this condition is not necessarily true. III. For x=-1 and y=2, x^y=1. So, this condition is not necessarily true.

Q4). The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range: Answer - C

Q7) The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

Answer is A

Since the GCM is 25, the 2 positive integers should have 5^2 as the only common factor. The 2 potential positive integers can be found by dividing 350 / 25 and finding unique prime factors that add up to the divisor.

Q1) The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. 63 II. 126 III. 252

Answer is E

S diagonal : R diagonal 1 : R diagonal 2 = 15:11:9

Let's assume x is the unknown multiplier and the ratio of diagonals are 15x:11x:9x

Area of S =\frac{15x}{\sqrt{2}}

Area of R = 1/2 * 11x * 9x

Area of S - Area of R = \frac{225x - 99x}{2} = 63x

And I, II, II are multiples of 63 when x = 1, 2, 3 _________________

1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. 63 II. 126 III. 252

A. I only B. II only C. III only D. I and III only E. I, II and III My answer is D Let the diagonals be 15x,11x and 9x. Area of a Rhombus is .5*d1*d2 .5*11x*9x=> 11*9*x2/2. = (99x^2)/2 Area of square with diagonal 15x is (225x^2)/2 Difference in areas is 63x^2. Difference can be 63, for X = 1 ; 126 for x=root2 and 252 for x=2. However when x= root2, the diagonals are not integers. Hence only 2 values are possible. 63 and 252

4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36 B. 30 < x < 37 C. 31 < x < 37 D. 31 < x < 38 E. 32 < x < 38 My answer C Squares less than 36 : 1,4,9,16,25 = 5 primes less than 36 : 2,3,5,7,11,13,17,19,23,29,31. =11 and sum =16. Hence the max value N has to be less than 37 as N =38 will increase the number of primes by 1 and squares by 1. And the sum will be 18 Minimum values of N has to be 32, as any value less than 32 will decrease the number of primes by 1.and the sum will be 15

A. 15 B. 17 C. 19 D. 33 E. 39 My answer is C. 15, 17,33 and 39 , perfectly divide into 18! Hence leave a reminder of 1, when they divide 18!+1. Hence can be eliminated , leaving 19 the answer.

6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1 B. 6 C. 7 D. 30 E. 36 My Answer C 66 = 26 *36 and 65 = 25 *35 . 4^3 = 26 Hence X should have 36 and It can have 20 till 26 that is 7 values from 20*36 till 26*36

7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1 B. 2 C. 3 D. 4 E. 5 My Answer C Let the numbers be 25a and 25b. for these numbers to have 25 as the GCD, a and b must be co-prime i.e have only 1 as the common factor. 25a+25b=350 => a+b =14 , a and b are co prime. Hence the pairs of a and b are => (1,13) , (3,11) ,(5,9).

8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10 B. 11 C. 55 D. 110 E. 330

My answer is D (337910*x)/3300 , reduced will become (12667 *x)/110 and 12667 is not divisible by any of the factors of 110( 2 or 5 or any combinations of both). Hence X has to be 110. At least.

9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

A. 0 B. 1 C. 5 D. 7 E. 8

My answer A 1/3 + 1/9 + 1/27 = 13/27. = .481481… 1/37=.027027… 13/27 + 1/27 = .508508.. Hence 101st digit is 0.(3*33 +2=> hence the second digit of the recurring decimal which is 0)

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

I. x=1 II. x=1 and y=0 III. x=1 or y=0

A. I only B. II only C. III only D. I and III only E. None My answer C. Since the question is what MUST be true I would go with C. as X can be any integer apart form 0 and Y can be 0. It’s not a must for X=1 and y=0.

1. The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. 63 II. 126 III. 252

A. I only B. II only C. III only D. I and III only E. I, II and III

Given that the ratio of the diagonal is d_s:d_1:d_2=15x:11x:9x, for some positive integer x (where d_s is the diagonal of square S and d_1 and d_2 are the diagonals of rhombus R).

area_{square}=\frac{d^2}{2} and area_{rhombus}=\frac{d_1*d_2}{2}.

The difference is area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2.

If x=1, then the difference is 63; If x=2, then the difference is 252; In order the difference to be 126 x should be \sqrt{2}, which is not possible.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...