Algebra

please note the question - Must be less. SO we just need to find the case(s) when it's possible for the following expressions to be NOT less than 1
a. r/s where 0<r<1 and 1<s<2, r=1/2 and s=3/2, 1/2 : 3/2= 2/6 Always less than 1-Correct
b. rs, 3/4 * 4/3 = 1 Is not less than 1-Incorrect
c. s-r, 1.99(s)-0.9(r)=1.9 Is not less than 1-Incorrect

Tagne wrote:1. If 0<r<1<s<2, which of the following must be less than 1?
a. r/s
b. rs
c. s – r

We note that each number in sequence is decreasing by 3. Let's take -4 and look for -4+(3*X)=23, 3X=27, X=9
Let's take 9 and and add 1 to it - we have n=10

IOM C

Tagne wrote: 2. In the arithmetic sequence t1, t2, t3, ……..tn,….t1 = 23 and tn = tn-1 - 3 for n > 1. What is the value of n when tn = -4.
a. -1
b. 7
c. 10
d. 14
e. 20

n (n is integer) >0, {1* ... *n}/990, find n lowest? In other words the problem is asking to n! which has divisor 990.
let's perform prime factorization of 990
990 (2) 495
495 (3) 165
165 (3) 55
55 (5) 11
11 (11) 1 --> 2*3^2 *5*11

lets list all possible factors from 1 to n until we get our required 2*3^2 *5*11

1*2*3 {we have 2*3^1} *4*5 {we have 5} *6 {we have the other 3^1} {and continue till 11, as 11 is prime} *....*11, hence our number must be 11! OR n=11

IOM B

Tagne wrote: 3. If n is a positive integer and the product of all the integers from 1 to n, inclusive is a multiple of 990, what is the least possible value of n?
a. 10
b. 11
c. 12
d. 13
e. 14

simply substitute (1-x) instead of x
a. f(x)=1-x, f(1-x)=1-(1-x) Not Correct, as 1-x=1-(1-x) Is not true, 1-x=x
b. 1-x^2=1-(1-x)^2 ---> {(1-x)^2=1-2x+x^2} Not Correct
c. x^2-(1-x)^2= (1-x)^2-(1-(1-x))^2 ---> {1-2x+x^2 -x^2} Not Correct
d. x^2(1-x)^2= (1-x)^2 *(1-(1-x))^2 ---> {(1-2x+x^2)*(x^2) Correct, as Left-hand side is F(x)= x^2(1-x)^2 OR F(x)= x^2(1-2x+x^2)

e. testing this answer choice just for assurance
e. x/(1-x) =(x-1)/(1-(1-x) Not Correct

IOM D

Tagne wrote: 4. For which of the following functions f is f(x) = f(1-x) for all x?
a. F(x) = 1-x
b. F(x) = 1-x^2
c. F(x) = x^2 – (1-x)^2
d. F(x)= x^2(1-x)^2
e. F(x) = x/(1-x)

Tagne wrote:1. If 0<r<1<s<2, which of the following must be less than 1?
a. r/s
b. rs
c. s – r

2. In the arithmetic sequence t1, t2, t3, ……..tn,….t1 = 23 and tn = tn-1 - 3 for n > 1. What is the value of n when tn = -4.
a. -1
b. 7
c. 10
d. 14
e. 20
3. If n is a positive integer and the product of all the integers from 1 to n, inclusive is a multiple of 990, what is the least possible value of n?
a. 10
b. 11
c. 12
d. 13
e. 14

4. For which of the following functions f is f(x) = f(1-x) for all x?
a. F(x) = 1-x
b. F(x) = 1-x^2
c. F(x) = x^2 – (1-x)^2
d. F(x)= x^2(1-x)^2
e. F(x) = x/(1-x)

1) since r<s; therefore r/s will always yields an answer that is going to be less than 1; hence a

2) as tn= tn-1 - 3;
therefore t2= t1-3; t2= 23-3; t2=20; i.e. an ap with a common difference of 3;

therefore by using formula of nth term of ap we have -4=23 - 3(n-1); n-1=9 , n=10 hence c

3)1*2*3*4*--------n =n! and as per question n!=990K;

for n to be smallest we must have 11 as the biggest prime no. since the biggest prime factor of 990 is 11;

therefore n must be 11; hence b

4) simply plug the options to the given equation only option d satisfy the required condition hence answer is D