Let total no. of switches is 100 Defective: 20 (1/5 of the light switches are defective.) Rejected: 16 (Four-fifths of the defective switches are rejected)
defective yet not rejected = 4
non defective = 100 - 20 = 80 non defective yet rejected by error = 4 (1/20 of the nondefective switches are rejected by mistake) non defective and not rejected = 80 - 4 = 76
Six mobsters have arrived at the theater for the premiere of the film Goodbuddies. One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankies requirement is satisfied?
Six mobsters have arrived at the theater for the premiere of the film Goodbuddies. One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankies requirement is satisfied?
1) 6 2) 24 3) 120 4) 360 5) 720
Is the answer D - i.e 360?
positions from front view: 1 - 2 - 3 -4 - 5- 6 When joey is 1, there are 5 places for frankie and 4! other 4 can arrange When joey is 2, there are 4 places for frankie and 4! other 4 can arrange When joey is 3, there are 3 places for frankie and 4! other 4 can arrange When joey is 4, there are 2 places for frankie and 4! other 4 can arrange When joey is 5, there are 1 places for frankie and 4! other 4 can arrange
Six mobsters have arrived at the theater for the premiere of the film Goodbuddies. One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankies requirement is satisfied?
1) 6 2) 24 3) 120 4) 360 5) 720
I think its D
lets the 6 friends be 1,2,3,4,J,F In every arrangement J should be infront of F.
Case 1:F at 6th position J can have any of the 5 positions infront of F and the rest of the friends can be arranged in 4! ways.So total no: of ways = 5*4! Case 2:F at 5th position J can have any of the 4 positions infront of F and the rest of the friends can be arranged in 4! ways.So total no: of ways = 4*4! . . . . . Case 5:F at 2th position J can have only 1position infront of F and the rest of the friends can be arranged in 4! ways.So total no: of ways = 1*4! So total no: of ways=(1+2+3+4+5)*4!=360
Let total no. of switches is 100 Defective: 20 (1/5 of the light switches are defective.) Rejected: 16 (Four-fifths of the defective switches are rejected)
defective yet not rejected = 4
non defective = 100 - 20 = 80 non defective yet rejected by error = 4 (1/20 of the nondefective switches are rejected by mistake) non defective and not rejected = 80 - 4 = 76
total non rejected = 76 + 4 = 80
defective but not rejected = 20 -16 = 4
solving => 80 * (x/100) = 4 => x = 5%
Yes 5% is the answer......i feel more comfortable solving in terms of x....sometimes going by "100" method confuses me.
Yes 5% is the answer......i feel more comfortable solving in terms of x....sometimes going by "100" method confuses me.
Just a thought Srikant267 - Be sure you differentiate between an 'x' and 'x'(multiplication) on the scratch pad... From what i hear its pretty tough job.
A fruit salad contains 2/3 blueberries and the rest raspberries. Chen loves raspberries, so she added 12 quarts of raspberries to the salad. If the mixture is now 5/7 raspberries, how many quarts of fruit salad were there to begin with? (A) 9 (B) 12 (C) 15 (D) 21 (E) 25
A fruit salad contains 2/3 blueberries and the rest raspberries. Chen loves raspberries, so she added 12 quarts of raspberries to the salad. If the mixture is now 5/7 raspberries, how many quarts of fruit salad were there to begin with? (A) 9 (B) 12 (C) 15 (D) 21 (E) 25
Is the answer A - 9?
Let fruits salad is 'a' Quarts to start with.. Blue Berries = (a * (2/3)) Rasp Berries = (a/3)
dear alchemist-mba, i agree with answer fot q26...i just ignored tht the eqn of circle wud yield 2 values of x and y
for q30, rx and ry are not the absolute rates of X and Y nstead rx is the ratio of rates of x and Y and ry is the ratio of rates of Y and Z.
thanks
Yep. But r-x = robot x / robot y r-y = robot y/ robot z (1) => r-x => ( robot x/ robot y) => robot x incsufficent we donot get any info on Z (2) r-y => (robot y/ robot z) => robot y insufiicent we dont get any info on x
dear jishnu, attached file contains answers. i cud nt solve the first ques on DS as P&C; s really giving me a tough time.
wat r the OAs?
guys any alternate solutions??
thanks
Here goes the leftover qn...
since 3 m can be 6 or 7
if its 6 => we cannot distribute 14 evenly if its 7 => we can distribute 14 evenly => so not sufficinet
(2) 13 n - cosnider least possible value for n - i.e 14 13n = 182. so m can be 7.
Sufficient.
This also hold good because since the factor of 13 n - i.e 13 donot share a common factor with any of the values b/w 3 - 13. hence n should contain a factor between 3 to 13. hence.. it is sufficinet..
Wow the O.A is really correct :2gunfire:. This should tell us to look into the question more carefully. The qn asked is length of segment PQ.
(i) draw a circle... now when x = -30, y = -40 or + 40. but the distance b/w (50,0) and (-30,40) or (50,0) and (-30,-40) is the same. the line x=-30 is parallel to line x = 50. darn.... so sufficient
(ii) draw a circle... now when y = -40, x = -30 or + 30. but the distance b/w (50,0) and (-30,-40) is different from (50,0) and (30,-40). so not suffiencet..
Let me know if you cant get it.. i will draw a diagram and will send you.