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Question2: I. 2b= 2c ==> 2(b-c) = 0 ==> b = c ----Sufficient II. c^2 = b^2 ==>c^2 - b^2 = 0 ==>(c+b)(c-b) = 0 --Insuff Ans. (A)

2) Since (-a,b) and (-b,a) are in same quadrant. ==> both a & b are either positive or negative. I. xy > 0 ==> both x & y are either positive or negative. Therefore, (-x,y) can be either in 2nd quad or in 3rd quad. --Insuff II. ax > 0 ==> both a & x are either positive or negative. --Insuff Combined, a,b,x and y all are either positive or negative. --Suff Ans ©

1) mv = 6 ==> (m,v) = (2,3), (3,2), (6,1), (1,6), (-2,-3), (-3,-2) --Insuff 2) (m+v)^2 = 25 ==> (m+v) = +5, -5 --Insuff Combined, Insuff Ans (E)

Total no divisible by 8 = 12 Total no which is less by 1 from the no divisible by 8 = 12 Total no which is less by 2 from the no divisible by 8 = 12 Required Probability = (12 + 12 + 12)/96 = 3/8

Suppose no of students in first, 2nd, 3rd and 4th grade is G1, G2, G3 & G4 respectively. G1/G3 = (G1/G2)*(G2/G4)*(G4/G3) = (3/4)*(8/5)*(2/3) = 4/5 Ans (E)

Question 1. Total sets of 3 condidates = 7C1*10C2 = 315 Question 2. f(n) = a^n f(1) = a (I) f(2) = a^2 = 100 ==>a^2 = 100 ==> a = +-10 -------Insufficient (II) f(3) = a^3 = -1000 ==> a^3 = -1000 ==> a = -10 --------Sufficient Ans (B) Question 3. P = 1*2*3*..............*29*30 = (3*6*9*12*......*27*30)(1*2*4*7*10*11*.........*28*29) = (3*(3*2)*(3*3)*(3*4)*(3*5)*(3*3*2)*(3*7)*(3*8)*(3*3*3)*3*10) (1*2*4*7*10*11*.........*28*29) = 3^14 (1*2*4*5*2*7*8*10)(1*2*4*7*10*11*.........*28*29) Ans © Question 4. 10x/(x+y) + 20y/(x+y) = K ==> 10 + 10y/(x+y) = K Put k = 15 in eqn (1), x = y --- But we have, x Put k = 18 in eqn (1), y = 4x --- this is the correct choice as it satifies given condition

(1) e (2) c

Problem 1. (1) n = 6, 9 i.e. n can be odd or even. Therefore, (1) is insufficient. (2) n = 9, divisor = 1,3,9 2n = 18, divisor = 1,2,3,6,9,18 n = 15, divisor = 1,3,5,15 2n = 30, divisor = 1,2,3,5,6,10,15,30 i.e. n should be odd. Therefore, (2) is sufficient. Answer should be (2). Problem 3. Distance travelled by cyclist in 5 mins = 20*5/60 = 5/3 miles Time taken by hiker to tranvel 5/3 miles = (60/4)*(5/3) = 25 mins Total wait time for cyclist = 25 - 5 = 20 mins Answer should be ©.

x = 3x -7 ==> y = x/3 + 7/3 Slope of this eqn = 1/3 Slope of the euation passing thru two points (a,b) and (a+3,b+k) = (+k-b)(+3-a)= k/3 Therefore, k/3 = 1/3 ==>k=1

x1 = no of teachers who teach 1 class x2 = no of teachers who teach 2 class x3 = no of teachers who teach 3 class x1 + x2 + x3 = 37 x1 + 2x2 + 3x3 = 64 From above two equations, we have, x2 + 2x3 = 27 We can conclude that, Minimum x3 = 0 and Maximum x3 = 13

a(n+1) = 1 + 1/an n = 1, a2 = 1 + 1/1 = 2 n = 2, a3 = 1 + 1/2 = 3/2 n = 3, a4 = 1 + 1/(3/2) = 5/3 n = 4, a5 = 1 + 1/(5/3) = 8/5

Initial selling price = $44, profit = 10% ==> Cost of the game = $40 New selling price = 46, profit = 46 - 40 = $6 % profit = 6*100/40 = 15%

Shiva, below is an explaination: It will be (total ways) - (total ways with 2 of the 3 group members being a married couple) Total Ways = 8C3 = 56 Total Ways w/couple = Total way 1 couple can be selected out of four people * total way one person can be selected from remaining 6 people = 4C1 * 6C1 = 4 * 6 = 24 Total way to select group members = 56-24 = 32.

Put k = 1, 2, 3, 4 .........10 in the given expression T = (1/2) + (-1/4)+(1/8)+()+(-1/16)+...................+(1/2^10) This is a series in geometric progression. Hence, T = [(1/2){1-(-1/2)^10}]/{1-1/2} = 1-1/2^10, which is less than 1 and greater than 1/2 Correct answer should be c.

Let us assume that son has done 1/x part of work. therefore wayne did 3/x part of work. Now, 1/x+3/x = 1 ==> x = 4 It means son does 1/4 part of work in 3 hrs. therefore, son will complete total work in 3*4 = 12 hrs.

(1) stmt, n(n+2) = 15, ==> n=3, -5 (2)stmt, (n+2)^n = 125, This stmt is satisfied by n = 3 only. By looking into both stmts only, it can be deduced that n = 3, none of the stmt alone gives one value of n. Therefore, ans should be C.

.25P = .10R R = 2.5P Lets pick up any no, say p = 100, then R = 250 (1) stmt satisfies above eqn (2) stmt also satiesfies above eqn. Therefore, correct ans should be D.