Quantitative Aptitude Questions (Inequality) for SBI Clerk 2018 Day- 87

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Practice Aptitude Questions (Inquality) Day-87

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Directions (Q. 1 – 10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship between them. Mark your answer accordingly.

  1. Quantity I: Amit and Arnav can do a piece of work in 24 days. Arnav takes 6/5 times of Amit can do the work. In how many days can Amit alone finish the work?

Quantity II: Sourav, Rohit and Sumit can do a piece of work in 20, 24 and 30 days respectively. They started the work together but after 4 days Sourav left the work and Rohit and Sumit continued the work with 3/4 of their usual efficiency then find the number of days required to complete the whole work?

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

2. Quantity I: The present age of Ankit is 12 years more than the present age of Soumen. After 10 years the ratio of ages of Soumen and Ankit will be 2:3 then find the age of Ankit after 4 years?

Quantity II: The present age of Madhu is twice the present age of Priyanka. The ratio between the present ages of Ankit and Priyanka is 4:5 and the age of Ankit after 5 years will be 21years then find the age of Madhu?

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: In an election between three candidates A, B and C. A got 30%, B got 48% vote and C got rest of the votes polled. If B got 720 more votes than A then find how many votes were polled?

Quantity II: In a village panchayat election between two candidates A and B, 10% of the votes were invalid. The ratio of votes received by A and B is 7:5. If the number of votes polled in favor of A was 630, then find the number of votes polled?

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 a trader sells an article at one-fourth less than the actual selling price, he gains 20% then find the percentage gain if he sells at 20% more than the actual selling price?

Quantity II: A trader purchased goods worth Rs. 28000. He sold half of the goods at 20% profit, 35% of the remaining at 5/4 of the cost price and remaining at a profit of 10% then find his overall profit percentage in the transaction?

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 of the two positive numbers A and B is 4:5. A is increased by 50% and 5 is added to it while B is increased by 100% and 4 is added to it. The new ratio between A and B becomes 2:3. What was the original value of A?

Quantity II: A vessel contains 63 liters of a mixture of milk and water in the ratio of 4:3. How much amount of water should be added to the mixture so that the ratio of milk and water becomes 2:3?

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. 12 men and 20 women can complete a piece of work in 20 days while 6 men and 15 women can complete the same piece of work in 30 days. If 8 men 12 women started the work and worked for ‘x’ days while 2 men and 6 women left the work after ‘x’ days and the remaining work got completed in 25 days more.

Quantity I: the value of x

Quantity II: In how many days 24 men and 15 women will complete the whole work?

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. Difference between CI and SI on a sum for 2 years at 10% is Rs 80.

Quantity I: The amount of simple interest on the same sum of money after 2 years at 12% interest.

Quantity II: Amount of compound interest that will be earned if 10% of the sum is retained and remaining is invested for 2 years at 8% interest rate.

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 area of a rectangular field is twice the area of a square field whose measure of a side 16 m. If the length of the rectangle is 2 times the breadth of the rectangle, then find the cost of fencing the rectangular field at the rate of Rs. 8 per meter?

Quantity II: The length and breadth of a rectangular field are in the ratio 5: 4.  Rs. 7200 is spent on flooring the whole field with stones at the rate of Rs. 10 per meter then find the cost of fencing the field at the rate of Rs. 12 per meter?

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. There is a certain number of red balls and 4 green balls in a bag. The probability of getting a red ball is 1/3 when 1 ball is picked at random.

Quantity I: Number of red balls.

Quantity II: Number of green balls.

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 of time taken by Rohit to travel upstream and downstream is 5:4. If the speed of the stream is 3km/hr and the total distance travelled upstream and downstream is 60km and 80 km respectively. Find the speed of the boat.

Quantity II: Speed of the boat when Rohit travels 40km upstream in 5 hours and the speed of the current is 2km/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

 Answers:

1). Answer: a

Quantity I:

Let the number of days taken by Amit be x

Time is taken by Arnav= 6x/5

Amit one day work= 1/x

Arnav one day work= 5/6x

1/x+ 5/6x= 1/24

6+5/6x= 1/24

6x= 24*11

x= 44 days

Quantity I= 44

Quantity II:

Total units of work= 600

Sourav one day work= 30 units

Rohit one day work= 24 units

Sumit one day work= 20 units

Work was done by them in 4 days= 74*4= 296 units

Remaining units of work= 600-296= 304 units

Units of work done by Rohit with ¾ efficiency= 24*3/4= 18 units

Units of work done by Sumit with ¾ efficiency= 20*3/4= 15 units

Time required to complete the remaining work= 304/33

Total time required= 4+304/33= 436/33 days.

Quantity I > Quantity II

2). Answer: c

Quantity I:

Let the present age of Soumen be x

Present age of Ankit = x+12

So,

(x+10)/(x+12+10) = 2/3

(x+10)/(x+22) = 2/3

3x+30= 2x+44

x= 14

Age of Ankit after 4 years= 14+12+4= 30 years

Quantity II:

Present age of Ankit= 21-5= 16 years

Age of Priyanka = 16*5/4= 20 years

Age of Madhu= 20*2= 40 years

Quantity II > Quantity I

3). Answer: a

Quantity I:

The difference in the percentage of votes received by A and B= 48-30= 18%

Total number of votes polled= 720*100/18= 4000

Quantity II:

Total number of votes received by A and B= 630*12/7= 1080

Total number of votes polled= 1080*100/90= 1200 votes.

Quantity I > Quantity II

4). Answer: a

Quantity I:

Let the actual selling price be Rs. 100

So,

Selling price when sold at one-fourth less= 100- 100*1/4= 75

Cost price= 75*100/120= Rs. 62.50

Selling price when sold at 20% more= 100*120/100= Rs. 120

Profit= 120- 62.50 = Rs. 57.50

Profit percentage= 57.50*(100/62.50) = 92 %

Quantity II:

Selling price of half of the goods= 14000*120/100= Rs. 16800

Cost price of 35% of the remaining goods= 14000*35/100= Rs. 4900

Selling price of 35% goods= 4900*5/4= Rs. 6125

Remaining cost price of the goods= 14000 – 4900= Rs. 9100

Selling price of remaining goods= 9100*110/100= Rs. 10010

Total selling price of the goods= 16800+ 6125+ 10010= Rs. 32935

Profit= 32935- 28000= 4935

Profit percentage= 4935*(100/28000) = 17.625%

Quantity I > Quantity II

5).Answer: c

Quantity I:

Let the ratio of numbers A and B be 4x and 5x

Value of A after increasing= 4x*150/100= 6x+5

Value of B after increasing= 5x*200/100= 10x+4

So,

6x+5/10x+4= 2/3

18x+ 15= 20x+8

2x= 7

x= 3.5

Original Value of A= 4*3.5= 14

Quantity II:

Let the amount of milk be added be x

Milk= 36 liters

Water= 27 liters

36/27+x= 2:3

108= 54+ 2x

2x= 54

x= 27 liters.

Quantity II > Quantity I

6). Answer: a

(12m+20w)*20= (6m+15w)*30

240m+ 400w= 180m+ 450w

60m= 50w

m / w= 50/60

m /w= 5/6

Total work= (12*5 + 20*6)*20= 3600 units

Work done by 6 men and 6 women in 25 days= (6*5+ 6*6)*25= 1650 units

Remaining units was done by 8 men and 12 women= 3600-1650= 1950

Work done by 8 men and 12 women in one day= 8*5+ 12*6= 112

Quantity I:

Value of x= 1950/112 days.

Quantity II:

Work was done by 24 men and 15 women in one day= 24*5+ 15*6= 210 units

Time required= 3600/210 days

Quantity I > Quantity II

7). Answer: a

Difference between SI and CI= P(R/100)2

80= P*100/10000

P= 8000

Quantity I:

S.I= 8000*2*12/100= 1920

Amount= 8000+1920= Rs 9920

Quantity II:

Sum Invested in C.I= 8000*90/100= 7200

Amount= P (1+r/100)2

= 7200(1+ 8/100)2

= 7200* 27*27/25*25

= 8398.08

Quantity I > Quantity II

8). Answer: c

Quantity I:

Area of the square field= 16*16= 256 sq. m

Area of the rectangular field= 2*256= 512 sq. m

Let the breadth of the rectangle be x

Length= 2x

512= x*2x

512/2= x2

256= x2

x= 16

Length = 2*16 = 32

Breadth = 16

Perimeter = 2(32+16) = 96 m

Cost of fencing = 96*8 = Rs. 768

Quantity II:

Area of the field= 7200/10= 720m2

Let the length and breadth be 5x and 4x

720= 5x*4x

20x2= 720

X2= 36

x= 6

Length= 30

Breadth= 24

Perimeter= 2(30+24) = 108 m

Cost of fencing= 108*12= Rs. 1296

Quantity II > Quantity I

9). Answer: c

Let the number of red balls be x

Probability of getting a red ball= xC1/ x+4C1

1/3= x/x+4

3x= x+4

2x= 4

x= 2

Quantity I:

Number of red balls = 2

Quantity II:

Number of green balls = 4

Quantity II > Quantity I

10). Answer: a

Quantity I:

Let the speed of the boat be x

Upstream speed= x-3

Downstream speed= x+3

Upstream time = 60/x-3

Downstream time= 80/ x+3

(60/x-3) * (x+3/80) = 5/4

3x+9/4x-12= 5/4

12x + 36 = 20x-60

8x= 96

x= 12 km/hr

Quantity II:

Upstream speed= 40/5= 8km/hr

Let the speed of the boat be x

x- 2 = 8

x= 10km/hr

Quantity I > Quantity II

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