vnf

I'm a "sis", Jay :p

The unit digit of the total remaining cards is 6 => 180 - 176 = 4 (the value of each card is not greater than 9)

4(1 + 2 + ... + 9) = 180 (1) => The removed card must be 4 => suff (2) suff Hence, (D)

(E) is the answer, Jay. Just solve the equalities and represent the values on the horizontal axis, you can see it clearly. Only values of x from (E) equality ( -2/3

(E) it is

For (B)

© it is J1 = Jason's salary in 1995 J2 = Jason's salary in 1998 K1 = Karen's salary in 1995 K2 = Karen's salary in 1998 (J2 - J1)/J1 = (K2 - K1)/K1 (1) K1 = 2,000 + J1 => insuff (2) K2 = 2,440 + J2 => insuff (1) & (2) => (J2 - J1)/J1 = (2,440 + J2 - 2,000 - J1)/(2,000 + J1) => (J2 - J1)/J1 = (440 + J2 - J1)/(2,000 + J1) => 2,000(J2 - J1) + J1(J2 - J1) = 440J1 + J1(J2 - J1) => (J2 - J1)/J1 = p = 440/2,000 => suff

4a = 2*pi*r => a = pi*r/2 Area S/Area C = (a^2)/(pi*r^2) = (pi^2)*(r^2)/4*(pi*r^2) = pi/4 = 3/4 (approx) Hence, (B)

For (B)

-2x > 3y Is x negative? (1) y > 0 => -2x > 3y > 0 => x must be -ve => suff (2) 2x + 5y - 20 = 0 => - 2x = 5y - 20 > 3y => 2y > 20 => y > 10 => -2x > 3y > 30 > 0 => y must be -ve => suff Hence, (D)

x = a/2 + b/2^3 + c/a^4 = (8a + 2b + c)/16 If a, b, c are equal to 0 or 1, x can be equal to all the choices except ©

(B) it is (1) Clearly insuff (2) (s1 + s2 + s3)/3 = 78 (1) (s1 + s2 + s3 + s4)/4 = 90 (2) (1) & (2) => can firgure out s4 => suff

(x+1)x/2 = 1 + (10+x+1)(10-x)/2 => x^2 + x - 56 = 0 => x = 7

(D) it is 4^17 - 2^28 = 2^34 - 2^28 = 2^28(2^6 - 1) = 2^28 * 7 * 3 * 3 => The greatest prime factor is 7

Total switches = x => Defective switches = (1/5)x => Nondefective swithches = (4/5)x Rejected switches = 4/5(x/5) + 1/20(4x/5) = x/5 => Sold swithches = x - x/5 = 4x/5 Defective sold switches = (1/5)*(x/5) = x/25 => The percent of the switches sold by the factory are defective = (x/25)*100/(4x/5) = 5% Hence, (B) =>