SBI Clerk and SBI PO Prelims exam are going to be held in the upcoming months. We have already provided you with the PDFs of many topics of Quantitative Aptitude like Simplification/Approximation, Number Series,datainterpretation.important Arithmetic questions to prepare for SBI Clerk and PO Pre exam. Practicing these questions will help you to know about the level of the questions. To increase your speed and accuracy, enhance your calculations.

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**TELEGRAM**

__START TEST__
1) I. 3x2– 2x – 8 = 0,

II. 6y2– 17y + 10 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

2) I. 32+ 11x + 6 = 0,

II. 5y2+ 16y + 3 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

3)I. 4x2– 11x + 6 = 0,

II. 6y2– 29y + 28 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

4)I.x2– 25x + 52 = 0,

II. 3y2– 8y – 16 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

5) I. 8x2+ 10x + 3 = 0,

II. 3y2+ 70y + 40 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

Directions (6 – 10): Study the following table carefully and answer the

questions that follow:

The premises of a bank are to be renovated. The renovation is in terms of flooring. Certain areas are to be floored either with marble or wood. All rooms/halls and pantry are rectangular. The area to be renovated comprises of a hall for customer transaction measuring 23 m by 29 m, branch manager’s room measuring 13 m by 17 m, a pantry measuring 14 m by 13 m, a record keeping cum server room measuring 21m by 13 m and locker area measuring

29 m by 21 m.

The total area of the bank is 2000 square meters. The cost of wooden flooring

is Rs 170/- per square meter and the cost of marble flooring is Rs 190/- per square meter. The locker area, record keeping cum server room and pantry are to be floored with marble. The branch manager’s room and the hall for customer transaction are to be floored with wood. No other area is to be renovated in terms of flooring.

Q6) What is the ratio of the total cost of wooden flooring to the total cost of

marble flooring?

A) 1733 : 2544

B) 1224 : 2586

C) 1987 : 2260

D) 1887 : 2521

E) 1535 : 2677

Q7) If the four walls and ceiling of the branch manger’s room (the height of

the room is 12 m) are to be painted at the cost of Rs. 190 per sq m how much

will be the total cost of renovation of the branch manger’s room, including the

cost of flooring?

A) Rs 212870

B) Rs 215580

C) Rs 212960

D) Rs 216360

E) Rs 214320

Q8) If the remaining area of bank is to be carpeted at the rate of Rs. 110 per

m2, how much will be the increment in the total cost of renovation of bank

premises?

A) Rs 5510

B) Rs 5460

C) Rs 5670

D) Rs 5280

E) Rs 5340

Q9) What is the percentage of the bank that is not to be renovated?

A) 2.7%

B) 2.8%

C) 3.2%

D) 3.9%

E) 2.4%

Q10) What is the total cost of renovation of the hall for customer transaction

and the locker area?

A) Rs 227200

B) Rs 225700

C) Rs 228200

D) Rs 229100

E) Rs 223400

11)Directions: Each question below contains a statement followed by Quantity I

and Quantity II. Find both to find the relationship among them. Mark your

answer accordingly.

Q11) The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.

Quantity I: Cost Price of Item X.

Quantity II: Selling Price of item Y after two successive discount of 10% and 15% is provided on the Marked Price of Rs 10,000.

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

Q12) Quantity I: Simple Interest on a sum of Rs 6200 at the rate of 6% per

annum for 5 years

Quantity II: Compound Interest on a sum of Rs 6400 at rate of 10%

compounded semi-annually for one year.

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity

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

Q13) The price of item X is marked at Rs 10,000. After giving a discount of 4%, a gain of 20% is achieved.

Quantity I: Cost Price of an article if it is sold making a loss of 10% given that if the cost price was 20% less, a profit of Rs 12 could be made.

Quantity II: Labeled Price of an article if a discount of 30% is given on it and then sold for Rs 105 at a loss of 20%

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

Q14) Given that x < 0 and y > 0

Quantity I: 14x3y2

Quantity II: 28x2y3

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

Q15) Quantity I: Area of rectangle whose sides are in the ratio 3 : 1 and perimeter twice the circumference of a semi-circle with area 77 cm<sup>2</sup>

Quantity II: Curved surface area of a cylinder of radius 12 cm and height equal to the side of square of area 49cm<sup>2</sup>

A) Quantity I > Quantity II

B) Quantity I ≥ Quantity II

C) Quantity II > Quantity I

D) Quantity II ≥ Quantity I

__SOLUTIONS__:
1.View Answer

Option D

Solution:

3x2– 2x – 8 = 0

3x2– 6x + 4x – 8 = 0

So x = -4/3, 2

6y2– 17y + 10 = 0

6y2– 12y – 5y + 10 = 0

So y = 5/6, 2

Put all values on number line and analyze the relationship

-4/3… 2… 5/6

2.View Answer

Option E

Solution:

3x2+ 11x + 6 = 0

3x2+ 9x + 2x + 6 = 0

So x = -3, -2/3

5y2+ 16y + 3 = 0

5y2+ 15y + y + 3 = 0

So y = -1/5, -3

Put all values on number line and analyze the relationship

-3 …. -2/3 ….-1/5

Since the common value (-3) is not in between other 2 values, there is no

relationship between x and y.

3.View Answer

Option E

Solution:

4x2– 11x + 6 = 0

4x2– 8x – 3x + 6 = 0

So x = 3/4, 2

6y2– 29y + 28 = 0

6y2– 8y – 21y + 28 = 0

So y = 4/3, 7/2

Put all values on number line and analyze the relationship

3/4 …. 4/3….. 2…. 7/2

4)View Answer

Option C

Solution:

3x2– 25x + 52 = 0

3x2– 12x – 13x + 52 = 0

So x = 4, 13/3

3y2– 8y – 16 = 0

3y2– 12y + 4y – 16 = 0

So y = 4, -4/3

Put all values on number line and analyze the relationship

-4/3 …. 4…. 13/3

5)View Answer

Option A

Solution:

8x2+ 10x + 3 = 0

8x2+ 4x + 6x + 3 = 0

So x = -3/4, -1/2

3y2+ 70y + 40 = 0

3y2+ 30y + 40y + 40 = 0

So y = -10, -4/3

Put all values on number line and analyze the relationship

-10 … -4/3 …. -3/4 …. -1/2

6)View Answer

Option C

Solution:

Area of hall = 23*29 = 667 m2

Area of Manager room = 221 m2

, of pantry= 182 m2

Area of record room = 273 m2

Area of locker room = 609 m2

Total area to be floored with marble = locker area + record keeping + pantry = 182

+ 273 + 609 = 1064

So cost = 190*1064

Total area to be floored with marble = 221 + 667 = 888

So cost = 170*888

So required ratio = 170*888 : 190*1064 = 1887 : 2527

7)View Answer

Option D

Solution:

Manager room flooring cost = 170 * 221 = Rs 37570

So cost of painting = [ 2 (12*17 + 12*13) + 17*13] * 190

= Rs 178790

So required total = 178790 + 37570 = Rs 216360

8)Option B

Solution:

Total area including all rooms = 2000 m2

Total area to be floored = 1952 = sq m

So remaining area = 2000 – 1952 = 48

So increased cost = 48*100 = Rs 5280

9)View Answer

Option E

Solution:

Area not to be renovated = 48 m2

So required % = 48/2000 * 100 = 2.4%

10)Option D

Solution:

Cost of hall + locker area = 170 * 667 + 190*609= Rs 229100

11)VIew Answer

Option A

Solution:

I : Mp =Rs 10,000; SP after 4% discount = Rs 9600; CP = Rs 8000

II : SP= 10,000*90/100 * 85/100 = Rs 7650

Hence I > I

12)View Answer

Option A

Solution:

I: SI = 6200*6*5/100 = Rs 1860

II: CI = 6400 [1 – 5/100]<sup>2</sup> – 6400 = Rs 1376

Hence I > II

13)View Answer

Option E

Solution:

I: Let CP = Rs 100, then at 10% loss, SP = Rs 90

Now if CP is 20% less, means CP = Rs 80, then profit is = 90 – 80 = Rs 10

but given profit is Rs 12

So if profit is Rs 10, then CP = Rs100

If profit is Rs 12, then CP is 100/10 * 12 = Rs 120

II: Use formula MP = (100-loss%)/(100-dicount%) * CPS

So MP = (100-20)/(100-30) * 105 = Rs 120

Hence I = II

14)View Answer

Option C

Solution:

I: Since x < 0, so I will always be negative

II: It will be always positive

Hence I < II

15)View Answer

Option C

Solution:

I: ½ Ï€r<sup>2</sup> = 77 [area of semicircle]

So r = 7 cm

So circumference of semicircle = Ï€r = 22 cm

So perimeter of rectangle = 2*22 = 44 cm

So 3x+x = 44 cm, x = 11,

So area of rectangle = 3x<sup>2<sup> = 363cm<sup>2</sup>

II: height of cylinder = √49 = 7 cm

So CSA = 2Ï€rh = 528cm<sup>2</sup>

Hence I < II