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[WpProQuiz 6851]
Directions (1 – 5): Study the following graph carefully and answer the given questions
The bar graph shows the no. of Red, Blue and Pink colour toys in a box A, B, C, D and E
The line graph shows the probability of a green colour toy taken from each boxes
1) If 25% of red toys taken from box A and added to box E, then find the probability of taken two toys both are pink colour in box E
a) 98/1378
b) 105/1353
c) 105/1378
d) 102/1378
e) 115/1378
2) If the total no of red colour toys and pink colour toys from all the boxes (A, B, C, D and E) are taken out and filled in a new box P, then 2 toys taken out from box P find the probability of different colours
a) 3854/7875
b) 3869/7875
c) 3824/7875
d) 3857/7875
e) 3814/7875
3) Four toys taken from box C, then find the probability of different colour
a) 116/2303
b) 226/2303
c) 256/2303
d) 216/2303
e) 96/2303
4) Total number of balls in box B and D together is approximately what percentage of the total number of balls in box C and E together?
a) 60
b) 50
c) 40
d) 80
e) None of these
5) Two balls taken randomly from box B. What is the probability of at least one pink colour ball?
a) 42/65
b) 41/65
c) 36/65
d) 38/65
e) 48/65
Directions (6 – 10): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.
6) Quantity I: 5, 8, 21, 80, 395, ?
Quantity II: 15, 40, 100, 230, 500, ?
a) Quantity I > Quantity II
b) Quantity I ≥ Quantity II
c) Quantity II > Quantity I
d) Quantity II ≥ Quantity I
e) Quantity I = Quantity II or Relation cannot be established
7) Quantity I: 6x2+11x-35=0
Quantity II: 2x2 – 40x + 200= 0
a) Quantity I > Quantity II
b) Quantity I ≥ Quantity II
c) Quantity II > Quantity I
d) Quantity II ≥ Quantity I
e) Quantity I = Quantity II or Relation cannot be established
8) Quantity I: Time taken by P to complete 1/5th of work if Q takes 6 days to complete 3/5th of work and together they take 5 days to complete 3/4th of work
Quantity II:Â Time taken by a train to cross a platform of length 60 km given that it crosses a pole in two and a half hours running at 60 km/hr
a) Quantity I > Quantity II
b) Quantity I ≥ Quantity II
c) Quantity II > Quantity I
d) Quantity II ≥ Quantity I
e) Quantity I = Quantity II or Relation cannot be established
9) Quantity I: Age of M five years ago if 6 years hence ratio of age of M to N will be 14: 11, and 1 year ago the ratio was 7: 5.
Quantity II:Â Average age of 2 students included in a group of 5 students with average age 18 years given that addition of these 2 students in group makes average increase by 1.
a) Quantity I > Quantity II
b) Quantity I ≥ Quantity II
c) Quantity II > Quantity I
d) Quantity II ≥ Quantity I
e) Quantity I = Quantity II or Relation cannot be established
10) A shopkeeper gets a discount of 20% on an article and sells it at 10% profit.
Quantity I: Labeled price as a percent of cost price.
Quantity II:Â Profit percent if discount given is 60%
a) Quantity I > Quantity II
b) Quantity I ≥ Quantity II
c) Quantity II > Quantity I
d) Quantity II ≥ Quantity I
e) Quantity I = Quantity II or Relation cannot be established
Answers :
Directions (1 – 5):
Box A:
Let us take the no. of green colour toys be x
Probability = xc1/(32+x)c1
9/25 = x/(32+x)
32*9 +9x = 25x
16x = 32*9=> x= 18 green colour toys
Box B:
Let us take the no. of green colour toys be x
Probability = xc1/(34+x)c1
3/20 = x/(34+x)
3*34+ 3x = 20x
=>3*34= 17x
X= 6 green colour toys
Box C:
Let us take the no. of green colour toys be x
Probability = xc1/(40+x)c1
1/5 = x/(40+x)
=>40+x= 5x
=>4x=40 =>x= 10 green colour toys
Box D:
Let us take the no. of green colour toys be x
Probability = xc1/(26+x)c1
2/15 = x/(26+x)
52+2x= 15x
=>13x= 52
=> x= 4 green colour toys
Box E:
Let us take the no. of green colour toys be x
Probability = xc1/(40+x)c1
1/5 = x/(40+x)
=>40+x= 5x
=>4x=40 =>x= 10 green colour toys
1) Answer: c)
Total number of balls in box E = [(12*25/100)+15] +10+15+10
= 18+10+15+10
= 53
Required probability = 15c2/53c2 = (15*14)/(53*52)
= 105/1378
2) Answer: b)
Total number of red colour boys in box P = (12+8+10+8+15) = 53
Total number of pink colour balls in box P = (12+16+18+12+15) = 73
Required probability = (53*73)/(53+73)c2
= 53*73/(126*125/1*2) = 3869/7875
3) Answer: d)
Required probability = (10*12*18*10)/50c4
= (10*12*18*10)/(50*49*48*47/1*2*3*4)
= (24*9)/(49*47)
= 216/2303
4) Answer: e)
Required percentage
= [(8+10+16+6+8+6+12+4)/(10+12+18+10+15+10+15+10)]*100
= 70/100 * 100 = 70%
5) Answer: a)
Required probability = 1- 24c2/40c2
= 1 – (24*23/40*39)
= 1- 23/65
= 42/65
Direction (6-10) :
6) Answer: a)
Quantity I:
5*2-2 = 8
8*3-3 = 21
21*4-4 = 80
80*5-5 = 395
395*6-6 = 2364
Quantity II:
15*2+10 = 40
40*2+20 = 100
100*2+30 =230
230*2+40 =500
500*2+50 =1050
Quantity I > Quantity II
7) Answer: c)
Quantity I: 6x2 + 11x – 35 = 0
6x2 – 21x + 10x – 35 = 0
3x (2x – 7) + 5(2x – 7) = 0
(2x – 7) + (3x + 5) = 0
X = 7/2, -5/3
Quantity II: 2x2 – 40x + 200= 0
x2– 20x + 100 = 0
x (x – 10) – 10 (x – 10) = 0
(x – 10) (x – 10) = 0
X = 10, 10
Quantity I < Quantity II
8) Answer: a)
Quantity I: P and Q completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days
Q complete 3/5 work in 6 days, so complete work in 5/3 * 6 = 10 days
So in 1 day P completes = 3/20 – 1/10 = 1/20
So to complete 1/5th work = 1/5 * 20 = 4 days
Quantity II: length of train = 2.5 * 60 = 150 km
So time taken to cross platform of length 60 km with speed 60 km/hr
= (150+60)/60 = 3.5hrs
Hence, Quantity I >Quantity II
9) Answer: c)
Quantity I: (M+6)/(N+6) = 14/11
And (M-1)/(N-1) = 7/5
Solve the equations, M = 22, so 5 years ago = 22 – 5 = 17 years
Quantity II: Total age of 2 students added to group = 18*2 + 7*1 = 43
So their average age = 43/2 = 21.5 years
Hence Quantity I < Quantity II
10) Answer: c)
Quantity I: MP = (100+10)/(100-20) * CP = 11/8 of CP
So required % = MP/CP * 100 = (11CP/8)/CP * 100 = 137.5%
Quantity II: % profit = (100+10) [(100-10)/(100-60)] – 100 = 590/4 = 147.5%
Hence Quantity I < Quantity II
