# Quantitative Aptitude Questions (Inequality) for SBI Clerk/PO Mains 2018 Day-111

Dear Readers, SBI is conducting Online Examination for the recruitment of Clerical Cadre and  Probationary Officers. Examination of SBI Clerk/PO Mains was scheduled from August 2018. To enrich your preparation here we have providing new series of Inequality- Quantitative Aptitude Questions. Candidates those who are appearing in SBI Clerk/PO Mains Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.

[WpProQuiz 2484]

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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.

1. Quantity I: The SI on a certain sum of money for 3 years at 5 % per annum is Rs. 4800. Then the principle is?

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

1. 5x – 2y = 5 and 1 + (x/y) = 8/5.

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

1. Quantity I: If 5x2 = 19x -12, then the value of x is?

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

1. Quantity I: 3 years ago, the ratio of age of A and B is 3: 4. After 2 years, the sum of their ages is 45. Then find the present age of A?

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

1. Quantity I: P and Q started a business by Rs. 2500 and Rs. 6000 respectively. P invested for whole year and they divided their shares after a year. The ratio of profit of P and Q is 5: 7. Then, how many months Q invested the money?

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

1. A box contains 6 white balls, 5 violet toys and 4 yellow toys.

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

1. Quantity I: The length of the rectangle is thrice the side of a square whose area is 324 Sq. m. The breadth of the rectangle is 10 m less than the length of the rectangle. Find the area of rectangle?

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

1. Quantity I: The ratio between the downstream and upstream speed of the boat is 4:3. The boat travels 64 km downstream in 8 hours. Find the speed of the boat in still water?

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

1. Quantity I: The average age of three persons A, B and C, after six years is 21 years while the average ages of A, B, C and D at present is 17 years then find the age of D, after 7 years?

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

1. Quantity I: There are three numbers in the ratio 5: 6: 10. The sum of the largest and the smallest numbers is 126 more than the other number. Find the largest number?

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

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

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

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

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

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

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

Quantity I:

Area of the square = a2 = 324

Side of the square (a) = 18m

Length of the rectangle = 54m

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

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

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

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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