“20-20” Quantitative Aptitude | Crack Dena bank 2018 Day-98

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“20-20” Aptitude Questions | Crack Dena Bank 2018 Day-98

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Directions (Q. 1 – 5) Read the information given below carefully and answer the questions that follow:

In a company there are 5200 employees in five departments – Operation, Production, Admin, IT and Marketing. Out of the total number of female employees in the company, 27% work in Operations, 22% work in Admin, and 16% work in Production and the remaining 840 female employees work in IT. There is no female in Marketing. Out of the total number of male employees in the company, 14% work in Operation, 30% work in Admin, 25% work in Production, and 11% work in IT and remaining in Marketing.

  1. If the number of male employees in Production increases by 20% , the male employees in IT increases by 25%, 60 male employees join Marketing and the number of male employees in Operation and Admin remain the same, what is the percentage increase in the number of male employees in the company?

a) 11.12 %

b) 127.22 %

c) 9.89 %

d) 6.77 %

e) 12.87 %

  1. The total number of male employees working Admin and Production together what percent of the total number of employees working in these two departments?

a) 70.50 %

b) 24.45 %

c) 50.25 %

d) 45.65 %

e) 62.80 %

  1. What is the average number of employees who work in Operation, Admin and Production department together?

a) 1112

b) 1222

c) 1989

d) 1164

e) 1275

  1. If equal number of female and male employees working in Production leave the job, the ratio of the number of male employees working in Production to the number of females working in the same department reduces to 149: 70. What is the total number of employees working in Production department who left the job?

a) 208

b) 122

c) 198

d) 102

e) 127

  1. If 92 male employees of Operation are transferred to IT and 70 female employees of IT department are transferred to Operation, then what is the ratio of the number of male employees to the number of the female employees in IT after the transfer of employees?

a) 11:12

b) 12:17

c) 19:89

d) 40:77

e) 12:75

 

Directions (Q. 6 – 10): 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. Give answer as:

  1. What is the amount of profit earned?
  1. 10% discount was offered on the labelled price?
  2. Had there been no discount, profit would have been 30%
  3. Selling price was more than the cost price by 20%

a) Only I and II

b) Only II and III

c) Only I and III

d) Any two of the three

e) All three together is not sufficient.

  1. A sum of money is placed at compound interest. In how many years will it amount to sixteen times of itself?
  1. The sum doubles itself in 4 years
  2. The sum amounts to eight times of itself in 12 years
  3. The sum amount to four times of itself in 8 years

a) Only I and II

b) Only II and III

c) Only I and III

d) Any one of the three

e) All three together is not sufficient

  1. What is the speed of the boat in still water?
  1. The speed downstream is 12 km/hr
  2. The speed upstream is 4 km/hr
  3. In a to and from journey between two points, the average speed of the boat was 6 km/hr

a) Only I and II

b) Only II and III

c) Only I and III

d) Any two of the three

e) All three together is not sufficient.

  1. At what time will the train reach city X from city Y?
  1. The length of the platform is 150 meters.
  2. The train leaves city Y at 7:15 am towards city X which is situated at a distance of 558 km.
  3. The 200 meters long train crosses a signal pole in 10 seconds.

a) Only I and II

b) Only II and III

c) Only I and III

d) Any two of the three

e) All three together is not sufficient

  1. If both the pipes are opened, how many hours will be taken to fill the tank?
  1. The capacity of the tank is 400 liters
  2. The pipe A fills the tank in 4 hours
  3.  The pipe B fills the tank in 6 hours

a) Only I and II

b) Only II and III

c) Only I and III

d) Any two of the three

e) All three together is not sufficient

  1. If the numerator of a fraction is increased by 130% and the denominator of the fraction is increased by 150%, the resultant fraction becomes 1/2. Then what is the original fraction?
  1. 25/46
  2. 13/19
  3. 15/26
  4. 27/32
  5. None of these
  1. Ajay gave 35% of his salary to his wife. From the remaining amount 40% he kept in his bank account. The sum of the amount he kept in bank and he gave to his wife was Rs 87840. What was his monthly salary?
  1. 98000
  2. 85000
  3. 116000
  4. 144000
  5. None of these
  1. A can finish a work in 24 days. B can finish the same work in 9 days and C in 12 days. B and C start the work but they are forced to leave after 3 days. How many days A will take to finish the remaining work?
  1. 14 days
  2. 10 days
  3. 18 days
  4. 12 days
  5. None of these
  1. A train crosses a platform of length 150 m in 12 sec and the same train crosses a pole in 8 sec. Find the length of the train?
  1. 175 m
  2. 225 m
  3. 300 m
  4. 275 m
  5. None of these
  1. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is?
  1. 1750
  2. 2500
  3. 3150
  4. 3750
  5. None of these

Directions (Q. 16 – 20): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer

  1. if x > y
  2. if x ≥ y
  3. if x < y
  4. if x ≤ y
  5. if x = y or no relation can be established between x and y

16).

I. 10x2 -17x +7 =0

II. 15y2 -19y +6 =0

17).

I. 12x2 +19x +5 =0

II. 5y2 + 16y +3 =0

 18). 

I. 6x2 + 31x + 35 =0

II. 2y2 + 3y + 1=0

19).

I. X2 – 264 = 361

II. y3 – 878 = 453

20).

I. 4x2– 30x +56 =0

II. y2 – 7y+12 =0

Answers:

Directions (1-5):

Number of females employees in IT department= 840

Percentage of female employees in IT department

= 100- (27+22+16) = 100-65= 35%

So,

35%= 840

100%= 840/35*100= 2400

So,

Male employees in the company= 5200- 2400= 2800

Operations department:

Female = 27/100*2400= 648

Male = 14/100* 2800= 392

Admin Department:

Female= 22/100* 2400= 528

Male= 30/100* 2800= 840

Production Department

Female= 16/100* 2400= 384

Male= 25/100* 2800= 700

IT Department:

Female = 840

Male= 11/100* 2800= 308

Marketing

Female = 0

Male = 2800- (392+ 840+ 700+ 308) = 560

Department Male Female
Operation 392 648
Production 700 384
Admin 840 528
IT 308 840
Marketing 560 _

 

  1. Answer c

Total increase in employees= 20% of 700 + 25% of 308 + 60= 277

Percentage Increase= 277/2800* 100= 9.89%

  1. Answer e

Total number of male employees in Admin and Production= 840+700= 1540

Total number of employees in these two departments

= > 840+528+700+384= 2452

Required percentage= (1540/2452)* 100= 62.8 %

  1. Answer d

Average = (648+528+384+392+840+700)/3 = 3492/3=1164

  1. Answer a

Let the number of male employees who left the job be x

Then,

700-x/384-x= 149/70

149x- 70x= 149 * 384 – 700*70

79x= 57216- 49000

X= 8216/79= 104

So,

Total number of employees who left= 104*2= 208

  1. Answer d

Number of male employees in IT after the transfer of 92 males = 308+92= 400

Number of females in IT= 840 -70= 770

Required ratio = 400:770= 40:77

 

Directions (Q. 6 – 10):

  1. Answer e

Let the marked price be x

From I

Selling Price= 90% of X= 9x/10

From II

If Selling price be x then gains= 30%

Cost Price= 100/130*x= 10x/13

From III

Gain= 20%

So, the answer cannot be find out.

  1. Answer d

From I

P (1+R/100)4 = 2P

= (1+R/100)4 = 2

From II

P (1+R/100)12 = 8P

= (1+R/100)12 = 8

From III

P (1+R/100)8 = 4P

= (1+R/100)8 = 4

Let the given sum become 16 times in n years

P (1+R/100)n = 16P

= (1+R/100)n = 16

So, Any of the three statement is sufficient

  1. Answer d

From I and II

Speed of the boat in still water= ½*12= 6km/hr

From II and III

Using average speed,

2*4*y/4+y= 6

8y= 24+6y

y= 12

Required speed= ½(12+4) = 8 km/hr

Similarly I and III also give the answer.

  1. Answer b

From III,

Speed = 200/10= 20m/s

In km/hr=20*18/5= 72km/hr

From II,

Time taken= 558/72= 31/4 hours = 7 hours 45 minutes

So, the train will reach city X at 3 pm

So, statement II and III is required.

  1. Answer b

From II and III,

Total units of work= 12

A’s one hour work= 3 units

B’s one hour work= 2 units

Working together they will take= 12/5 hours

 

  1. Answer a

The original fraction be (X/Y)

[X*(230/100)]/[Y*(250/100)] = (1/2)

23X/25Y = ½

(X/Y) = (1/2)*(25/23) = 25/46

  1. Answer d

Let the monthly salary of Ajay be x

Amount given to her wife= x*35/100= 7x/20

Remaining amount= x-7x/20= 13x/20

Amount kept in bank

= (13x/20)*40/100= 13x/50

So,

13x/50 + 7x/20= 87840

26x+35x/100= 87840

61x= 87840*100

x= 8784000/61

x= Rs. 144000

  1. Answer b

A’s one day work = (1/24)

B’s one day work = (1/9)

C’s one day work = (1/12)

(B + C)’s one day work = (1/9) + (1/12) = 21/(9*12) = 7/36

(B + C)’s 3 day work = (7/36)*3 = 7/12

Remaining work = 5/12

A can finish the remaining work in,

= > (5/12)*24 = 10 days

  1. Answer c

According to the question,

(T.L + 150)/12 = T.L/8

2T.L + 300 = 3T.L

300 = T.L

The length of the train = 300 m

  1. Answer a

Compound Interest:

4000*(10/100) = 400

4400*(10/100) = 440

C.I = 840

S.I = (1/2)*C.I = 840/2 = 420

According to the question,

= > S.I = PNR/100

= > 420 = P*3*8/100

= > P = (420*100)/24

= > P = Rs. 1750

 

Directions (Q. 16 – 20):

  1. Answer a

I. 10x2 -17x +7 =0

10x2 -10x-7x +7 =0

10x(x-1) -7(x-1) =0

(10x-7) (x-1) =0

X= 7/10, 1

II. 15y2 -19y +6 =0

15y2 -10y-9y +6 =0

5y(3y-2)-3 (3y-2) =0

(5y-3) (3y-2) =0

Y= 3/5, 2/3

x>y

  1. Answer e

I. 12x2 +19x +5 =0

12x2 +4x +15x +5 =0

4x(3x+1)+5 (3x+1) =0

(4x+5)(3x+1) =0

X=-5/4, -1/3

II. 5y2 + 16y +3 =0

5y2 + y+15y +3 =0

Y(5y+1)+3 (5y+1) =0

(y+3) (5y+1) =0

Y=-3, -1/5

Can’t be determined

  1. Answer c

I. 6x2 + 31x + 35 =0

6x2 + 21x+10x + 35 =0

3x(2x+7)+5 (2x+7)=0

(3x+5)(2x+7) =0

X= -5/3, -7/2

II. 2y2 + 3y + 1=0

2y2 + 2y +y+ 1=0

2y(y+1) +1(y+1) =0

(2y+1) (y+1) =0

Y= -1/2, -1

X<y

  1. Answer e

I. X2 – 264 = 361

X2 = 361+264

X2 = 625

X= 25, -25

II. y3 – 878 = 453

Y3 = 453 + 878

Y3 = 1331

Y =11

Can’t be determined

  1. Answer e

I. 4x2– 30x +56 =0

4x2– 16x-14x +56 =0

4x(x-4) -14(x-4) =0

(4x-14) (x-4) =0

X= 14/4, 4= 7/2, 4

II. y2 – 7y+12 =0

y2 – 4y-3y+12 =0

y(y-4)-3 (y-4) =0

(y-3) (y-4) =0

Y =3, 4

Can’t be determined

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