Crack SBI Clerk Prelims 2018 Sectional Full Test (Quantitative Aptitude) Day-21

Dear Readers, SBI Clerk (Junior Associates) 2018 Preliminary Examination is scheduled to be held on 23rd, 24th, and 30th of June 2018. In that case, aspirants need to speed up the preparation as there are only few days more.

To boost up your exam preparation, Our IBPS Guide team is providing a full length Sectional Tests for English, Quantitative Aptitude and Reasoning with detailed solutions to score more marks in the prelims exam. Make use of this opportunity and recommend to your friends to achieve a successful career in Banking.

You can also practice using our SBI Clerk Test Series pack prepared by our experienced professionals as per the latest exam pattern. We wish you a great success in all forthcoming exams. Click Here for SBI Clerk Practice Mock Test

Minimum Cut Off for this Quantitative Aptitude Section Test is: 20

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Click “Start Quiz” to attend these Questions and view Solutions

Directions (Q. 1-5) what value should come in the place of question mark (?) in the following number series?

  1. 92, 119, 155, 209, 290,?
  1. 356
  2. 435
  3. 442
  4. 381
  5. 407
  1. 1872, 312, 62.4, 15.6, 5.2, ?
  1. 2.6
  2. 2.8
  3. 3
  4. 2.4
  5. 2.2
  1. 1172, 1170, 1179, 1151, 1216,?
  1. 1056
  2. 1248
  3. 1090
  4. 1224
  5. 1182
  1. 267, 133, 66, 32.5, ? , 7.375
  1. 15.5
  2. 15.75
  3. 15.25
  4. 16.25
  5. 16.5
  1. 1262, 1172, 1040, 858, 618, ?
  1. 421
  2. 358
  3. 262
  4. 312
  5. 452
  1. Two men undertook to do a piece of work for Rs. 2520. One alone could do it in 9 days and the other can do it in 6 days while with the assistance of a boy, work gets completed in 3 days. Find the share of boy?
  1. 840
  2. 480
  3. 450
  4. 420
  5. 400
  1. Three – seventh of a number is equal to five-sixth of another number. The difference of these two numbers is 68. Find the numbers?
  1. 140 and 72
  2. 120 and 52
  3. 150 and 82
  4. 150 and 218
  5. 156 and 224
  1. The simple interest on certain sum of money for 3 years at 6 % per annum is Rs. 5040. Find the principle amount?
  1. Rs. 34000
  2. Rs. 28000
  3. Rs. 36000
  4. Rs. 24000
  5. None of these
  1. Anitha purchased a computer for Rs. 35000 and sold it at a loss of 8%, with that money she purchased another computer and sold it at the profit of 9%. Find the overall profit/loss?
  1. Rs. 114
  2. Rs. 98
  3. Rs. 156
  4. Rs. 76
  5. None of these
  1. The amount obtained on Rs. 32000 at an interest 4% per annum compounded annually for certain period of time is Rs. 35995.648. Then find the time in years?
  1. 3 years
  2. 5 years
  3. 2 years
  4. 4 years
  5. None of these

Directions (Q. 11 – 15): What value should come in the place of question mark (?) in the following questions?

  1. (28/9) × (17/8) ÷ (153/6) + 1 41/54 =?
  1. 2 15/54
  2. 3 3/17
  3. 2 1/54
  4. 3 13/19
  5. None of these
  1. 28 % of 3540 + 267 % of 4500 + 24 % of 5060 =?
  1. 14220.6
  2. 13450.8
  3. 11780.4
  4. 15670.5
  5. None of these
  1. (8 1/3 ÷ 175/27) ÷ (2 5/14 – 1 1/28) =?
  1. 12/25
  2. 5/9
  3. 12/23
  4. 36/37
  5. None of these
  1. 4 1/5 of 2755 + 281 + 7 1/5 of 8000 =?
  1. 51248
  2. 69452
  3. 55426
  4. 63784
  5. None of these
  1. 7868 ÷ 14 × 37 + 175 + 332 =?
  1. 22058
  2. 28472
  3. 18564
  4. 12376
  5. None of these

Directions (Q. 16 – 20): The following questions are accompanied by three statements I, II and III. You have to determine which statement is/are sufficient to answer the questions.

16). What is Manita’s share in the profit earned at the end of 2 years in a joint business run by Manita, Sunita and Yamini?

  1. Manita invested Rs 75000 to start the business.
  2. Sunita and Yamini joined Manita after six months by investing the amount in the ratio of 3:5.
  3. The total amount invested by Sunita and Yamini is Rs 2.4 lakh.
  1. Only I and II together
  2. Only I
  3. Only I and III
  4. Either I and II or I and III together
  5. None of these

17). In a room there are two windows and one door. What will be the cost of papering the walls of that room, whose length, width and height are 6m, 5m and 8m respectively?

  1. The cost of papering the walls is Rs 55 per square meter.
  2. Windows and the door have dimensions 1(1/2)m *1m and 2m*1(1/2)m respectively.
  3. The area of the windows is 1.50 m2, which is half of the area of the door.
  1. Only I and II together
  2. Only I
  3. Only I and III
  4. Either I and II or I and III together
  5. All are required.

18). How many marks did Bipin get in English?

  1. Bipin secured an average of 75 marks in four subjects, including English.
  2. He secured a total of 170 in English and Mathematics together.
  3.  He secured a total of 145 in Mathematics and Science together.
  1. Only I and II together
  2. Only II and III are required
  3. Only I and III
  4. All are required.
  5. None of the above

19). What is the rate of interest per annum?

  1. The amount doubles itself in 8 years on simple interest.
  2. The difference between the CI and SI earned on this amount in two years is Rs. 600.
  3. Simple interest per annum is Rs 1500.
  1. Only I or II and III together
  2. Only II and III are required
  3. Only I and III
  4. All are required.
  5. None of the above

20). What is the total surface area of the cone?

  1. The area of the base of the cone is 154 Sq. cm.
  2. The curved surface area of the cone is 550 Sq. cm
  3.  The volume of the cone is 1232 Cu. cm
  1. Only I and either II or III
  2. Only II and III are required
  3. Only I and III
  4. All are required.
  5. None of the above

Directions (Q. 21 – 25): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

  1. Quantity I: The area of rectangular garden is 1715 Sq m. The length of the rectangular garden is 40 % more than the breadth. Find the perimeter of the garden?

Quantity II: The area of a square park is 1936 Sq m.  Find the perimeter of the park?

  1. Quantity I > Quantity II
  2. Quantity I ≥ Quantity II
  3. Quantity II > Quantity I
  4. Quantity II ≥ Quantity I
  5. Quantity I = Quantity II or Relation cannot be established
  1. Quantity I: The present age of Mano is 42 years. After 4 years, the age of Julee, if Julee is 3 years younger than Mano?

Quantity II: The difference between the age of Ravi and Banu is 20 years. 16 years hence, the ratio of their ages is 2: 1. After 8 years, the age of Ravi is?

  1. Quantity I > Quantity II
  2. Quantity I ≥ Quantity II
  3. Quantity II > Quantity I
  4. Quantity II ≥ Quantity I
  5. Quantity I = Quantity II or Relation cannot be established
  1. Quantity I: Find the simple interest on Rs. 26000 for 2 years at 13 % per annum?

Quantity II: Find the compound interest on Rs. 26000 for 2 years at 12 % per annum?

  1. Quantity I > Quantity II
  2. Quantity I ≥ Quantity II
  3. Quantity II > Quantity I
  4. Quantity II ≥ Quantity I
  5. Quantity I = Quantity II or Relation cannot be established
  1. Quantity I: The shopkeeper sold an article at 10 % discount on marked price and he gains 20 %. If the marked price of the article is Rs. 450, then the cost price is?

Quantity II: The shopkeeper marks the price of the book Rs. 250 and his profit % is 25 %. Find the cost price of the book, if he allows a discount 15 %?

  1. Quantity I > Quantity II
  2. Quantity I ≥ Quantity II
  3. Quantity II > Quantity I
  4. Quantity II ≥ Quantity I
  5. Quantity I = Quantity II or Relation cannot be established
  1. Quantity I: If (x1/4 / 16)2 = 144 / x3/2, then find the value of x?

Quantity II: If y1/3 × y2/3  × 3104 = 16 × y2, then find the value of y?

  1. Quantity I > Quantity II
  2. Quantity I ≥ Quantity II
  3. Quantity II > Quantity I
  4. Quantity II ≥ Quantity I
  5. Quantity I = Quantity II or Relation cannot be established

Direction (26-30): In the following questions, two equations I and II are given. You have to solve both the equations.

Give Answer

A) If p > q
B) If p < q
C) If p ≥ q
D) If p ≤ q
E) If p = q or relation cannot be established

26) I. p2 – 18p + 72= 0

II. 5q2– 18q + 9 = 0 

27) I. p2 = 4

II. 3q2– 4q – 4 = 0

28) I. 6p2 – 5p – 6 = 0

II. 2q2– 13q + 20 = 0

29) I. 2p2 – 5p = 0

II. 3q2– 7q – 6 = 0

30) I. 2p2 + 5p + 2= 0

II. 2q2+ 19q + 45 = 0

Directions (Q. 31-35) Study the following information carefully and answer the given questions:

Following table shows the total number of candidates attended the interview in different companies and the ratio of male and female also given.

  Total candidates Male : Female
P 1100 6 : 5
Q 1400 3 : 4
R 1600 5 : 3
S 2200 4 : 7
T 900 1 : 1

 

  1. Total number of female candidates attended the interview in company T is approximately what percentage of total number of male candidates attended the interview in Company R?
  1. 60 %
  2. 45 %
  3. 70 %
  4. 35 %
  5. 65 %
  1. Find the ratio of male candidates in company P to that of female candidates in company S?
  1. 4 : 9
  2. 5 : 11
  3. 3 : 7
  4. 13 : 17
  5. None of these
  1. Find the difference between the total number of candidates attended the interview in company Q and R together to that of total number of candidates attended the interview in company P and R together?
  1. 300
  2. 450
  3. 350
  4. 500
  5. None of these
  1. If the percentage of selected candidates in company S and T is 72 % and 65 % respectively, then find the sum of selected candidates in Company S and T?
  1. 2896
  2. 3123
  3. 3567
  4. 2169
  5. None of these
  1. Find the average number of female candidates attended the interview in company R and T together?
  1. 615
  2. 575
  3. 525
  4. 655
  5. None of these

Answers:

1). Answer e

The difference of difference is, 9, 18, 27, 36…

The answer is, 407

2). Answer a

The pattern is, ÷ 6, ÷ 5, ÷ 4, ÷ 3, ÷ 2,…

The answer is, 2.6

3). Answer c

The pattern is, -(13 +1), +(23 +1), -(33 +1), +(43+1), -(53 +1),….

The answer is, 1090

4). Answer b

The pattern is, ÷2 – 0.5

The answer is, 15.75

5). Answer d

The answer is, 312

6). Answer d

 (1/9) + (1/6) + 1 boy = (1/3)

1 boy = (1/3) – [(1/9) + (1/6)]

1 boy = (1/3) – (5/18) = (6 – 5)/18 = 1/18

The share of two men and a boy = (1/9): (1/6): (1/18)

= > 2: 3: 1

6’s = 2520

1’s = 420

The share of boy = Rs. 420

7). Answer a

 (3/7) of a number = (5/6) of another number

(3/7)*x = (5/6)*y

(x/y) = (7/3)*(5/6) = 35/18

X : y = 35 : 18

17’s = 68

1’s = 4

The numbers are = 35*4 and 18*4 = 140 and 72

8). Answer b

S.I = PNR/100

5040 = P*3*6/100

P = (5040*100)/18

P = Rs. 28000

9). Answer b

C. P1 = 35000, Loss = 8 %

S.P1 = 35000 * (92/100) = 32200

With that money she purchased another computer and sold it at the profit of 9%,

C.P2 = 32200

S.P2 = 32200*(109/100) = 35098

The overall profit = 35098 – 35000 = Rs. 98

10). Answer a

Amount = P (1 + r/100)n

35995.648 = 32000(1 + (4/100))n

35995.648/32000 = (104/100)n

35995648/32000000 = (26/25)n

8998912/8000000 = (25/26)n

(208/200)3 = (25/26)n

(26/25)3 = (25/26)n

n = 3 years

Direction (11-15)

11). Answer c

(28/9)*(17/8)*(6/153) + (95/54) = x

X = (7/27) + (95/54)

X = (14 + 95)/54 = 109/54 = 2 1/54

12). Answer a

(28/100)*3540 + (267/100)*4500 + (24/100)*5060 = x

X = 991.20 + 12015 + 1214.4

X = 14220.6

13). Answer d

[(25/3) * (27/175)] ÷ (33/14 – 29/28) = x

X = (9/7) ÷ (37/28)

X = (9/7)*(28/37) = 36/37

14). Answer b

(21/5)*2755 + 281 + (36/5)*8000 = x

X = 11571 + 281 + 57600

X = 69452

15). Answer a

(7868/14)*37 + 175 + 1089 = x

X = 20794 + 175 + 1089

X = 22058

Directions (Q. 16 – 20):

16). Answer e

In order to find the share of profit for Manita total profit is required which is not provided in the question. So the answer is can’t be determined.

17). Answer d

From I and II

Total cost of papering the walls

= 55*{2*8(6+5)- (2* 3/2*1 + 2*3/2)

= 55(16*11 -6)

= 55*170

= Rs 9350

From I and III

Total cost of papering

= 55{2*8(6+5)-(2*1.5+2*3)}

= 55(16*11-9) = Rs 9185

18). Answer e

From I

We can find the total marks in four subjects

From II and III

The third subject is not given so we can’t find the answer.

19). Answer a

From I

Rate = x*100/x*8= 12.5%

From II and III:

SI per annum = 1500

Difference between SI and CI for 2 yrs = 600

Rate of interest = 600/1500*100 = 40%

20). Answer a

Total surface area of the cone = (πrl + πr2) cm2

From I

πr2= 154 we can get r.

From II

πrl= 550

From I and II can get the answer.

From III

1/3πr2h= 1232

From I and III we can get h

So surface area can be determined

Directions (Q. 21 – 25):

21). Answer c

Quantity I:

The area of rectangular garden= 1715 Sq m

Length = (140/100)*breadth

l/b = 7/5 = > l : b = 7 : 5

7x*5x = 1715

35x2 = 1715

X2 = (1715/35) = 49

X = 7

Length = 49 m, Breadth = 35 m

Perimeter of the garden = 2*(l + b) = 2*(49 + 35) =2*84 = 168 m

Quantity II:

The area of a square park = 1936 Sq m

Area (a2) = 1936

Side (a) = 44

Perimeter of the park = 4a = 4*44 = 176 m

Quantity II > Quantity I

22). Answer a

Quantity I:

The present age of Mano = 42 years

Julee = Mano – 3 = 42 – 3 = 39 years

After 4 years, the age of Julee = 39 + 4 = 43 years

Quantity II:

16 years ago, the ratio of their ages = 2 : 1(2x, x)

According to the question,

X = 20

16 years hence, the age of Ravi and Banu = 40 and 20

Present age of Ravi and Banu = 24 and 4

After 8 years the age of Ravi = 32 years

Quantity I > Quantity II

23). Answer a

Quantity I:

S.I = P*n*r/100

S.I = (26000*2*13)/100 = Rs. 6760

Quantity II:

C.I:

26000*(12/100) = 3120

29120*(12/100) = 3494.4

C.I = 3120 + 3494.4 = Rs. 6614.4

Quantity I > Quantity II

24). Answer a

Quantity I:

Selling price of the article = 450*(90/100) = Rs. 405

Cost price of the article

= > CP*(120/100) = 405

= > CP = 405*(100/120)

= > CP = 337.50

Quantity II:

Selling price of the book = 250*(85/100) = Rs. 212.50

Cost price of the book = 212.50*(100/125) = Rs. 170

Quantity I > Quantity II

25). Answer c

Quantity I:

(x1/4 / 16)2 = (144 / x3/2 )  = (x1/2 / 256) = (144 / x3/2 )

(x1/2)  × (x3/2) = 256 × 144

x2 = (256 × 144)

x = √(256 × 144)

x = ± (16 × 12) = ±192

Quantity II:

y1/3 × y2/3 × 3104 = 16y2

y × 3104 = 16y2

3104 = 16y

Y = 3104 / 16 = 194

x < y

Quantity II > Quantity I

Direction (26-30):

26). Answer: a

p2 – 18p + 72= 0
(p-12)(p-6) = 0
Gives p = 6, 12
5q2 – 18q + 9 = 0
5q2 – 15q – 3q + 9 = 0

5q (q-3) – 3(q-3) =0

(5q-3) (q-3) =0
Gives q = 3/5, 3
Put on number line
3/5   3    6  12

P > q

27). Answer: e

p2 = 4
Gives p = 2 , -2
3q2 – 4q – 4 = 0
3q2 – 6q + 2q – 4 = 0
Gives q = -2/3, 2
Put on number line
-2      -2/3        2
So no relation

28). Answer: b

6p2 – 5p – 6 = 0
6p2 – 9p + 4p – 6 = 0
Gives p = -2/3, 3/2
2q2 – 13q + 20 = 0
2q2 – 8q – 5q +20 = 0
Gives q = 4, 5/2
Put on number line
-2/3   3/2    5/2    4

P < q

29). Answer: e

2p2 – 5p = 0
p(2p-5) = 0
Gives p = 0, 5/2
3q2 – 7q – 6 = 0
3q2 – 9q + 2q – 6 = 0
Gives q = -2/3, 3
Put on number line
-2/3    0   5/2         3

There is no relation

30). Answer: a

2p2 + 5p + 2= 0
2p2 + 4p + p + 2= 0
Gives p = -1/2, -2
2q2 + 19q + 45 = 0
2q2 + 10q + 9q + 45 = 0
Gives q= -10/2, -9/2
Put on number line
-10/2     -9/2    -2     -1/2

P > q

Directions (Q. 31-35)

31). Answer b

Total number of female candidates attended the interview in company T

= > 900*(1/2) = 450

Total number of male candidates attended the interview in Company R

= > 1600*(5/8) = 1000

Required % = (450/1000)*100 = 45 %

32). Answer c

Male candidates in company P

= > 1100*(6/11) = 600

Female candidates in company S

= > 2200*(7/11) = 1400

Required ratio = 600: 1400 = 3: 7

33). Answer a

The total number of candidates attended the interview in company Q and S together

= > 1400 + 1600 = 3000

The total number of candidates attended the interview in company P and R together

= > 1100 + 1600 = 2700

Required difference = 3000 – 2700 = 300

34). Answer d

Total selected candidates in Company S

= > 2200*(72/100) = 1584

Total selected candidates in Company T

= > 900*(65/100) = 585

Required sum = 1584 + 585 = 2169

35). Answer c

The total number of female candidates attended the interview in company R and T together

= > 1600*(3/8) + 900*(1/2)

= > 600 + 450 = 1050

Required average = 1050/2 = 525

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