# SBI Clerk Mains Quantitative Aptitude (Day-29)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

Start Quiz

Ensure Your Ability Before the Exam – Take SBI Clerk 2020 Mains Free Mock Test

Quantity I and II

Directions (1 – 5): Each question contains two quantities as Quantity I and Quantity II. Read the contents clearly and answer the questions accordingly.

1)  Quantity I: Rahul invests Rs.X at 15% simple interest for 4 years and Sam Invests the same amount at 25% per annum of simple interest for 5years. If the difference between the interest received by Sam and Rahul Rs.1300, then find the value of X?

Quantity II: Difference between the compound interest and Simple interest is Rs.45 for 2 years at 15% per annum. What is the Sum?

A) Quantity I > Quantity II

B) Quantity II ≥ Quantity I

C) Quantity I ≥ Quantity II

D) Quantity I = Quantity II or Relation cannot be established

E) Quantity II > Quantity I

2)  Quantity I: If the ratio of the length of the rectangle to the side of the square is 5 : 8 and the ratio of the breadth of the rectangle to the side of the square is 1:2. If the difference between the perimeter of the rectangle and the square is 56cm, then what is the area of the rectangle?

Quantity II: If the radius of the sphere is 4.9 cm, what is the surface area of the sphere?

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

3) Quantity I: If the shopkeeper sold the article for Rs.1200 and he earned X% of the profit. If he increase the selling price of the article by Rs.300 and he can makes the profit of (X + 30)% . Find the value of X?

Quantity II: If the marked price of the article is Rs.2800 and the Shopkeeper offers X% of discount. If he sold at the article at Rs.2400, then find the value of X?

A) Quantity II > Quantity I

B) Quantity I ≥ Quantity II

C) Quantity I > Quantity II

D) Quantity I = Quantity II or Relation cannot be established

E) Quantity II ≥ Quantity I

4) Quantity I: A milkman has 72litres of mixture of water and milk in the ratio of 4:5.Then he adding some amount of water and the volume of the mixture is increased by 88 liters. The quantity of the milk is how much percentage more or less than of water in the final mixture?

Quantity II: 15%

A) Quantity I = Quantity II or Relation cannot be established

B) Quantity II > Quantity I

C) Quantity I ≥ Quantity II

D) Quantity II ≥ Quantity I

E) Quantity I > Quantity II

5) Quantity I: Ratio of the number of red, blue and yellow balls in the box is 4: 3: 5. If 18 yellow balls taken out and 10 blue balls is added, then the ratio of red, blue and yellow balls become 6:7:3. What is the total number of balls in the box at initially?

Quantity II: The income of A is Rs.80,000 and he spends Rs.32000 and save the remaining amount. What percent of salary he saves?

A) Quantity II > Quantity I

B) Quantity I ≥ Quantity II

C) Quantity I = Quantity II or Relation cannot be established

D) Quantity I > Quantity II

E) Quantity II ≥ Quantity I

Application Sums

6) A can do the piece of work in 20days. The efficiency of B is 50% more than that efficiency of C. C can do a work in X number of days. If A started and after 5 days he left from the work and the remaining work is completed by B and C together. Find the value of X?

A) 20

B) 15

C) 18

D) Cannot be determine

E) None of these

7) If the difference between the selling price and the cost price of the article is Rs.1200 and the difference between the marked price and the selling price of the article is Rs. 2000, If the shopkeeper allows the discount of X% and also he earned X % of Profit then find the value of X?

A) 20%

B) 25%

C) 15%

D) Can’t be determined

E) None of these

8) Car X is Starting from Chennai to Trichy at 8.am and Car Y is started from Chennai to Trichy at 11 a.m. Both the cars are reached at Trichy at the same time. At what time Car Y reached at Trichy from Chennai, if Car X is started at Chennai to Trichy and Car Y is started from Trichy to Chennai at the same time, both the cars will meet at after 2hours?

A) 1.p.m.

B) 2.p.m

C) 3.p.m

D) 4.p.m

E) Cannot be determined

9) Total time taken by the Boat travel from A to B in upstream and downstream in 10hours. The ratio of the speed of the Boat in still water to the speed of the stream is 5: 1. If the difference between the upstream and downstream speed is 10kmph, then find the distance between A and B?

A) 90 km

B) 60 km

C) 180 km

D) 120 km

E) None of these

10) Rahul invest Rs. X at simple interest for 4 years and after 4 years he received the total amount of Rs.2x. Suresh invest Rs. X at simple interest for 2 years and Sam invest Rs. X at compound interest for 2 years. All the three persons are investing the same rate of interest. If the difference between the interest received by both Sam and Suresh together and Rahul is Rs.832, then find the value of X?

A) Rs.13312

B) Rs.12890

C) Rs.13229

D) Rs.12762

E) None of these

Answers :

1) Answer: D

From quantity I,

SI = P * N * R/100

Rahul interest amount = x * 4 * 15/100 = 3x/5

Sam interest amount = x * 5 * 25/100 = 5x/4

5x/4 – 3x/5 = 1300

25x – 12x = 20 * 1300

x = 2000

From quantity II,

Difference = P * R2/1002

45 = P * 15 * 15/(100 * 100)

P = 2000

Quantity I = quantity II

2) Answer: A

From quantity I,

Ratio of length of rectangle to side of the square = 5: 8

Ratio of the breadth of the rectangle to side of the square = 1: 2

Ratio of the length and breadth of the rectangle and side of the square = 5: 4: 8

Perimeter of the rectangle = 2 * (l + b) = 2 * (9x) = 18x

Perimeter of the square = 4 * a = 4 * 8x = 32x

32x – 18x = 56

x = 4 cm

Area of the rectangle = l * b

= (5 * 4) * (4 * 4)

= 320 cm2

From quantity II,

Radius of the sphere = 4.9 cm

Surface area of the square = 4 * 22/7 * r * r

= 4 * 22/7 * 4.9 * 4.9 = 301.84 cm2

Quantity I > quantity II

3) Answer: C

From quantity I,

CP * (X + 100)/100 = 1200

CP * (X + 130)/100 = 1500

120000/(X + 100) = 150000/(X + 130)

4X + 520 = 5X + 500

X = 20%

From quantity II,

2800 * (100 – X)/100 = 2400

100 – X = 600/7

X = 100/7

X = 14.28%

Quantity I > quantity II

4) Answer: E

From Quantity I:

Milk = 5/9 * 72 = 40

Water = 4/9 * 72 = 32

Added water quantity = 88 – 72 = 16

Total quantity of water = 16 + 32 = 48

Required percentage = [(48 – 40)/48] * 100 = 16.67%

From Quantity II:

Quantity II = 15%

Quantity I > quantity II

5) Answer: D

From quantity I,

Ratio of the red, blue and yellow balls = 4: 3: 5

(5x – 18)/(3x + 10) = 3/7

9x + 30 = 35x – 126

26x = 156

X = 6

Total number of balls = 12x = 12 * 6 = 72

From quantity II,

Income = 80000

Expenditure = 32000

Savings = 80000 – 32000 = 48000

Required percentage = 48000/80000 * 100 = 60%

Quantity I > quantity II

Directions (6-10) :

6) Answer: D

Work done by A in one day,

A = 1/20

A work for 5 days = 5/20 = 1/4

Remaining work = 3/4

B/C = 150/100 = 3/2

We cannot find the answer.

7) Answer: D

SP – CP = 1200

MP – SP = 2000

CP + 1200 = MP – 2000

MP – CP = 3200

We can’t find the CP, so we can’t find the value of X.

8) Answer: B

Car Y reached in Trichy = t

Car X reached in Trichy = t + 3

Speed ratio of X and Y = t : (t + 3)

Distance = t * (t + 3)

t * (t + 3)/(t + t + 3) = 2

t2 + 3t = 4t + 6

t2 – t – 6 = 0

t2 – 3t + 2t – 6 = 0

t(t – 3) + 2(t – 3) = 0

(t + 2)(t -3) = 0

t = -2, 3(negative value neglect)

t = 3

Car Y reached Trichy at after three hours, in time = 11 am + 3

= 2 p.m

9) Answer: D

Downstream speed = 5x + x = 6x

Upstream speed = 5x – x = 4x

Difference = 6x – 4x = 10

2x = 10

x = 5 kmph

Downstream speed = 6 * 5 = 30 kmph

Upstream speed = 4 * 5 = 20 kmph

Distance between A and B = d

d/30 + d/20 = 10

2d + 3d = 600

5d = 600

d = 120 km

10) Answer: A

SI = P * N * R/100

CI = P * (1 + R/100)2 – P

Rahul received the interest = 2x – x = x

x = x * R * 4/100

R = 25%

Suresh received the interest = x * 25 * 2/100 = x/2

Sam received the interest = x * (1 + 25/100)2 – x

= 9x/16

(9x/16 + x/2) – x = 832

(9x + 8x) – 16x = 832 * 16

x = 13312

 Check Here to View SBI Clerk Mains 2020 Quantitative Aptitude Questions Day 28 Day 27 Day 26 Click Here for SBI Clerk 2020 – Detailed Exam Notification
0 0 votes
Rating 0 Comments
Inline Feedbacks
View all comments