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Quant practice set 5 for SBI PO PRELIM



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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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 πr<sup>2</sup> = 77 [area of semicircle]
So r = 7 cm
So circumference of semicircle = πr = 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 = 2πrh = 528cm<sup>2</sup>
Hence I < II