# SBI Clerk Mains Quantitative Aptitude (Day-25)

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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

Simplification

Direction (06 – 10): What value should come in place of question mark in the following questions?

6) 35 % of 240 + 80 % of 45 + √4225 * 4 = ?

A) 320

B) 360

C) 380

D) 400

E) 420

7) 125 % of 140 – 493 ÷ √289 + ? = 18 * 12

A) 80

B) 70

C) 60

D) 40

E) 50

8) (882 ÷ 21) + 15 * 14=? + (245 ÷ 35) – 12

A) 248

B) 239

C) 257

D) 229

E) 281

9) (5 / 9) * 351 + 60 % of 150 + 45 * 8=? * 15

A) 43

B) 47

C) 41

D) 44

E) 49

10) 12.5 % of 344 + 75 % of 228 + ? % of 18 = 520

A) 1700

B) 1500

C) 1400

D) 1300

E) 1100

Directions (15) :

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

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

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

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

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) :

35 % of 240 + 80 % of 45 + √4225 * 4 = ?

84 + 36 + 260 = ?

? = 380

125 % of 140 – 493 ÷ √289 + ? = 18 * 12

175 – 29 + ? =216

? = 70

(882 ÷ 21) + 15 * 14 = ? + (245 ÷ 35) – 12

42 + 210 = ? + 7 – 12

? = 257

(5 / 9) * 351 + 60 % of 150 + 45 * 8 = ? * 15

195 + 90 + 360 = ? * 15

?= 43

12.5 % of 344 + 75 % of 228 + ? % of 18 = 520

43 + 171+ ? % of 18 = 520

?= 1700

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