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Directions (Q. 1 – 10): Each question contains Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.
Quantity II: The CI on a certain sum of money for 2 years at 6 % per annum is Rs. 3708. Then the principle is?
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
Quantity I: Value of 3x + y?
Quantity II: Value of 3y – x?
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
Quantity II: If 5y2 +11y =12, then the value of y is?
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
Quantity II: 5 years ago, the ratio of age of P and Q is 3: 4. P’s age after 6 years is equal to the present age of Q. Then find the present age of P?
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
Quantity II: A and B invested in the ratio of 2: 3. A invested the money for 9 months. The ratio of profit of A and B is 3: 4. Then, how many months B invested the money?
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
Quantity I: If 3 balls are drawn randomly, then the probability of getting at least one yellow balls?
Quantity II: If 2 balls are drawn randomly, then the probability of getting both the balls is either white or violet?
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
Quantity II: The perimeter of a rectangle is 212 m. The difference between the length and breadth of the rectangle is 18 m. Find the area of a rectangle
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
Quantity II: The boat travels 72 km downstream in 8 hours. The speed of the stream is 3 km/hr. Find the speed of the boat in still water?
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
Quantity II: The average age of 13 persons is 25. If the average age of first six persons is 24 and that of last six persons is 26, then find the age of remaining person?
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
Quantity II: 12 % of first number is equal to 25 % of second number. The difference of these two numbers is 78. Then find the largest number?
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:
1). Answer a
Quantity I:
SI = (Pnr)/100
4800 = (P*3*5)/100
P = (4800*100)/15 = Rs. 32000
Quantity II:
C.I = P[(1 + (r/100))2 – 1]
3708 = P[(1 + 6/100)2 – 1]
3708 = P[(106/100)2 – 1]
3708 = P[(53/50)2 – 1]
3708 = P[(2809/2500) – 1]
3708 = P[309/2500]
P = (3708*2500)/309 = Rs. 30000
Quantity I > Quantity II
2). Answer a
5x – 2y = 5 —> (1)
1 + (x/y) = 8/5
(y + x)/y = 8/5
5x + 5y = 8y
5x – 3y = 0 —> (2)
By solving the equation (1) and (2), we get,
X = 3, y = 5
Quantity I: 3x + y = 9 + 5 = 14
Quantity I: 3y – x = 15 – 3 = 12
Quantity I > Quantity II
3). Answer b
Quantity I:
5x2 -19x +12=0
5x2 -15x-4x +12=0
5x(x-3)-4 (x-3) =0
(5x-4) (x-3) =0
X= 4/5, 3
Quantity II:
5y2 +11y -12=0
5y2 +15y-4y -12=0
5y(y+3) -4(y+3) =0
(5y-4) (y+3) =0
Y=4/5, -3
Quantity I ≥ Quantity II
4). Answer c
Quantity I:
3 years ago, the ratio of age of A and B = 3: 4(3x, 4x)
According to the question,
3x + 4x + 10 = 45
7x = 35
X = 5
The present age of A = 3x + 3 = 18 years
Quantity II:
5 years ago, the ratio of age of P and Q = 3: 4(3x, 4x)
According to the question,
3x + 5 + 6 = 4x + 5
X = 6
The present age of P = 3x + 5 = 18 + 5 = 23 years
Quantity I < Quantity II
5). Answer c
Quantity I:
According to the question,
(2500*12)/(6000*x) = (5/7)
(25*12)/60x = (5/7)
2100 = 300x
X = 2100/300 = 7 months.
Quantity II:
According to the question,
(2*9)/(3*x) = (3/4)
18/3x = (3/4)
72 = 9x
X = (72/9) = 8 months.
Quantity I < Quantity II
6). Answer a
Quantity I:
Total probability n(S) = 15C3
Required probability n(E) = 1 – P(none is yellow)
Probability of getting none is yellow balls,
P(E) = n(E)/n(S) = 11C3/15C3
= > 33/91
Required probability = 1 – (33/91) = 58/91
Quantity II:
Total probability n(S) = 15C2
Required probability n(E) = 6C2 or 5C2
P(E) = n(E)/n(S) = 6C2 or 5C2 / 15C2
= > (15 + 10)/105 = 25/105
= > 5/21
Quantity I > Quantity II
7). Answer c
Quantity I:
Area of the square = a2 = 324
Side of the square (a) = 18m
Length of the rectangle = 54m
Breadth= 44m
Area of the rectangle = 54*44= 2376 Sq. m
Quantity II:
The perimeter of a rectangle = 212 m
2(l + b) = 212
l + b = 106 —> (1)
l – b = 18 —> (2)
By solving the equation (1) and (2), we get,
Length = 62 m, Breadth = 44 m
Area of the rectangle = 62*44 = 2728 Sq. m
Quantity I < Quantity II
8). Answer a
Quantity I:
Downstream speed= 64/8 = 8 km/hr
Upstream speed= 8*(3/4) = 6 km/hr
Speed of boat in still water = ½ *(8+6) = 7 km/hr
Quantity II:
Downstream speed= 72/8= 9 km/hr
Downstream speed = Speed of boat in still water + speed of stream
Let the speed of boat be x
=> x + 3 = 9
=> x= 6 km/hr
Speed of boat in still water = 6 km/hr
Quantity I > Quantity II
9). Answer a
Quantity I:
Total ages of A, B and C, after six years = 21*3 = 63 years
Total present ages of A, B and C = 63 – 18 = 45 years
Total present ages of A, B, C and D = 17*4= 68 years
The present age of D = 68 – 45 = 23 years
Age of D, after 7 years = 23 + 7 = 30 years
Quantity II:
The average age of 13 persons = 25
Total age of 13 persons = 13*25 = 325
The average age of first six persons = 24
Total age of first 6 persons = 24*6 = 144
The average age of last six persons = 26
Total age of last 6 persons = 26*6 = 156
The age of remaining person = 325 – (144 + 156) = 25 years
Quantity I > Quantity II
10. Answer c
Quantity I:
According to the question,
10x + 5x = 126 + 6x
15x – 6x = 126
9x = 126
X = 14
Largest number = 10x = 140
Quantity II:
(12/100)*x = (25/100)*y
(x/y) = (25/12)
x : y = 25 : 12
13’s = 78
1’s = 6
Largest number = 25x = 25*6 = 150
Quantity I < Quantity II
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This post was last modified on June 26, 2021 8:56 am