“20-20” Quantitative Aptitude | Crack Dena bank 2018 Day-110

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“20-20” Aptitude Questions | Crack Dena Bank 2018 Day-110

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Directions (Q. 1 – 5) The following questions are accompanied by two statements. You have to determine which statement is/are sufficient to answer the questions.

  1. Find the present age of father?

(I). The difference between the father and his son is 24 years.

(II). The present age of son is one- fifth of the present age of father.

  1. Only statement I follows
  2. Only statement II follows
  3. Either statement I or II follows
  4. Neither statement I nor II follows
  5. Both the statement I and II together follows
  1. Find the length of the train?

(I). The train crosses a 240 m long bridge in 28 sec.

(II). The speed of the train is 54 km/hr.

  1. Only statement I follows
  2. Only statement II follows
  3. Either statement I or II follows
  4. Neither statement I nor II follows
  5. Both the statement I and II together follows
  1. Area of a rectangular playground is?

(I). The perimeter of the rectangle is 150 m and its breadth is 30 m.

(II). The ratio of length and breadth of the rectangular playground is 2 : 1.

  1. Only statement I follows
  2. Only statement II follows
  3. Either statement I or II follows
  4. Neither statement I nor II follows
  5. Both the statement I and II together follows
  1. By selling an article, what is the profit % earned?

(I) 5 % discount is given on the Marked price. If no discount is given 20 % profit is gained.

(II). The cost price of the article is Rs. 400.

  1. Only statement I follows
  2. Only statement II follows
  3. Either statement I or II follows
  4. Neither statement I nor II follows
  5. Both the statement I and II together follows
  1. What is the sum of money invested by Rohit?
  1. The interest earned by Rohit after 2 years of investment is Rs. 2400.
  2. The same sum doubles itself in 10 years.
  1. Only statement I follows
  2. Only statement II follows
  3. Either statement I or II follows
  4. Neither statement I nor II follows
  5. Both the statement I and II together follows

Directions (6-10): Study the following and answer the following questions.

The floors of the college are to be renovated either with marble or with wood. All rooms, halls and office room are rectangular. The area to be renovated comprises a hall measuring 33m by 39m. The Chairman’s room measures 13m by 12m and the office room measures 14m by 12m. The principal room measures 23m by 13m and the Vice principal room measures 12m by 23m. The total area of the college is 2500 square meters. The cost of wooden flooring is Rs 210 per square meter and the cost of marble flooring is Rs 240 square meter. The vice principal room, the Principal room and the office room are to be floored with marble. The chairman’s room and the hall are to be floored with wood.

  1. If four walls and ceiling of the room (the height of the room is 12 meters) are to be painted at the cost of Rs 145 per square meter, how much will be the total cost of renovation of the Chairman’s room, including the cost of flooring?

a.Rs.773300

b.Rs.133688

c.Rs.142570

d.Rs.132830

e.Rs.142380

7. What is the difference between the total cost of wooden flooring to the total cost of marble flooring?

  1. Rs 124620
  2. Rs 124710
  3. Rs 184510
  4. Rs 145810
  5. Rs 122450

8. What is the total cost of renovation of the hall and vice principal room?

  1. Rs 336510
  2. Rs 329810
  3. Rs 436510
  4. Rs 237690
  5. Rs 215790

9. If all the remaining area of the college is to be carpeted at the rate of Rs 280 per square meter, by how much will the cost of renovation of college will increase?

  1. Rs 89530
  2. Rs 21420
  3. Rs 87210
  4. Rs 87920
  5. Rs 87450

10. What is the approximate percentage area of the college that is not to be renovated?

a.31%

b.15%

c.13%

d.62%

e.42%

 

  1. A, B and C started a business by investing Rs. 50000, Rs. 35000 and Rs. 55000 respectively. After 6 months, A invested 5000 more and C withdraw 10000. If the total profit at the end of the year is Rs. 26785, then find the share of B?
  1. Rs. 7236
  2. Rs. 9172
  3. Rs. 5690
  4. Rs. 6818
  5. None of these
  1. 9 men can complete a work in 10 days, 6 women can complete it in 24 days and 10 children can complete the same work in 18 days. In how many days can 10 men, 6 women and 15 children complete the same work?
  1. 4 4/17 days
  2. 6 3/13 days
  3. 5 6/11 days
  4. 3 5/7 days
  5. None of these
  1. 4 years ago, the ratio of age of A and B is 3: 5. After 12 years, the sum of the ages of A and B is 72. C is 8 years older than D. The average of present ages of A, B, C and D is 25 years. Find the present age of C and D?
  1. 32 years and 24 years
  2. 26 years and 18 years
  3. 30 years and 22 years
  4. 28 years and 20 years
  5. None of these
  1. In a bag, there are 4 white balls, 6 pink balls, 8 violet balls. Two balls are picked at random. What is the probability that both the balls are same colour?
  1. 53/171
  2. 49/153
  3. 34/163
  4. 17/92
  5. None of these
  1. The perimeter of a rectangle is 80 m and the difference between the length and the breadth of the rectangle is 8 m then find the perimeter of square whose area is 16 m more than the area of the rectangle?
  1. 80 m
  2. 125 m
  3. 134 m
  4. 96 m
  5. None of these

Directions (Q. 16 – 20) In the following questions, two equations I and II are given. You have to solve both the equations and give answer as,

  1. If x > y
  2. If x ≥ y
  3. If x < y
  4. If x ≤ y
  5. If x = y or the relation cannot be established
  1. I. 3/√x + 7/√x = √x

II.y2 = 125/2/ √y

  1. I. x3 = 2744

II.y2 – 11y – 60 = 0

  1. I. 3x = 5y – 4

II. 5x + 7y – 70 = 0

  1. I. x2+ 12x + 32 = 0

II. 3y2 – 10y + 7 = 0

20.  I. x2 = 14161

II. y = √14161

Answers:

1. Answer e

From (I),

Father – son = 24

From (II),

Present age of son = (1/5)*present age of father

Present age of son/present age of father = (1/5)

The ratio of present age of son and father = 1 : 5(x, 5x)

Father – son = 24

5x – x = 24

4x = 24

X = 6

The present age of father = 5x = 30 years

Both (I) and (II) follows.

2. Answer e

From (I) and (II),

Let the length of the train be x,

Speed = Distance / Time

Distance = Train length + Bridge length = x + 240

Speed of the train = 54 km/hr

54*(5/18) = (x + 240)/28

54*(5/18)*28 = x + 240

420 = x + 240

X= 420 – 240 = 180 m

Length of the train = 180 m

3. Answer a

From I,

Perimeter of rectangle = 300 m

2(l + b) = 150

l + b = 75

l + 30 = 75

l = 45 m

Area of rectangle = lb = 30*45 = 1350 Sq m

From II,

We can’t find the area of rectangle.

4. Answer a

From I,

Let the cost price be Rs. 100

If no discount is given 20 % profit is gained.

MP = 100*(120/100) = Rs. 120

SP = 120*(95/100) = Rs. 114

P % = (Profit/CP)*100

= > (14/100)*100 = 14 %

From II,

The cost price of the article only given, So, we can’t find the profit %.

5. Answer e

From both the statements,

From I,

Let the principal be x,

Amount= 2x

Simple interest= 2x-x= x

Rate= 100*x/10*x= 10%

From II,

Sum of money invested by Rohit

= 2400*100/2*10

= Rs. 12000

Directions (6-10):

Solution:

Area of hall= 33*39= 1287m2

Area of Chairman’s room=13*12=156m2

Area of Principal room= 23*13= 299m2

Area of office room= 14*12= 168 m2

Area of Vice principal room= 12*23= 276 m2

Thus total area to be floored= 2186 m2

6. Answer: e

Area of wall=2(12*13+12*12)= 2(300)= 600 m2

We know that Area of the Chairman’s room= 156 m2

Now cost of paining of four walls=145*(600+156)= Rs 109620

And cost of flooring= 210*156=Rs 32760

Hence total cost= 109620+32760=Rs 142380

7. Answer: b

Total flooring area with wood= 1287+156= 1443 square meter

Cost of flooring area with wood= 1443*210= Rs 303030

Now total flooring area with marble= 743*240=Rs 178320

Hence required ratio= 303030 – 178320= Rs 124710

8. Answer: a

Total cost of renovation of the hall= 1287*210=Rs 270270

The cost of renovating the Vice principal’s room= 276*240= Rs 66240

Now required total cost= 270270+ 66240= Rs 336510

9. Answer: d

Given that, total area of the college= 2500 square meter

Now, remaining area= 2500-2186= 314 square meter

Thus cost of renovation of the remaining area= 314* 280=Rs 87920

10. Answer: c

Total area of the college that is not to be renovated= 2500-2186=314 square meter

Thus required percentage= 314/2500*100= 13%

 

11. Answer d

The share of A, B and C

= > [(50000*6) + (55000*6)]: [35000*12]: [55000*6 + 45000*6]

= > 630000: 420000: 600000

= > 21: 14: 20

Total profit = Rs. 26785

55’s = 26875

1’s = (26875/55) = 487

The share of B = 14’s = Rs. 6818

12. Answer a

1 man’s 1 day’s work= 1/90

1 woman’s 1 day’s work= 1/144

1 child’s 1day’s work= 1/180

So,

Required days = (10/90 +6/144 +15/180)

= 1/9 + 1/24 + 1/12

= 17/72

Required days = 72/17 = 4 4/17 days

13. Answer c

4 years ago, the ratio of age of A and B = 3 : 5 (3x, 5x)

After 12 years, the sum of the ages of A and B = 64

= > 3x+ 16 + 5x + 16 = 72

= > 8x + 32 = 72

= > 8x = 40

= > x = 5

The present age of A and B = 3x + 4, 5x + 4 = 19, 29

C = D + 8

C – D = 8 —> (1)

The average of present ages of A, B, C and D = 25 years

The total present ages of A, B, C and D = 25*4 = 100

= > 19 + 29 + C + D = 100

= > C + D = 100 – 48

= > C + D = 52 –> (2)

By solving of equation (1) and (2), we get

C = 30, D = 22

The present age of C and D = 30 years and 22 years

14. Answer b

Total probability n(S) = 18C2

Required probability n(E) = 4C2 or 6C2 or 8C2

P(E) = n(E)/n(S) = (4C2 or 6C2 or 8C2)/ 18C2

= > (6+15+28)/153

= > 49/153

15). Answer a

The perimeter of a rectangle = 80 m

2(l + b) = 80

l + b = 40 –> (1)

l – b = 8 –> (2)

By solving the equation (1) and (2), we get,

L = 24 m, b = 16 m

The area of rectangle + 16 = The Area of square

The area of rectangle = lb = 24*16 = 384 Sq m

The area of square = 384 + 16 = 400 Sq m

(a2) = 400

Side (a) = 20 m

The perimeter of square = 4*20 = 80 m

 

Direction (16-20)

16. Answer c

I. 3/√x + 7/√x = √x

(3 + 7)/ √x = √x

X = 10

II. y2 = 125/2/ √y

y2 × √y = 125/2

y2 + (1/2) = 125/2

y5/2 = 125/2

y = 12

x < y

17. Answer e

I. x3 = 2744

X = 14

II. y2 – 11y  – 60 = 0

(y – 15)(y + 4) = 0

Y = 15, -4

Can’t be determined

18. Answer a

3x – 5y = 4–> (1)

5x + 7y = 70–>(2)

By solving the equation (1) and (2),

X = 7, y = 5

x > y

19. Answer c

I. x2+ 12x + 32 = 0

x2+ 8x + 4x + 32 = 0

x(x+ 8)+4 (x +8) = 0

(x+ 4)(x+8) = 0

X = -4, -8

II. 3y2 – 10y + 7 = 0

3y2 – 3y – 7y + 7 = 0

3y(y – 1) – 7(y – 1) = 0

(3y – 7)(y – 1) = 0

Y = 7/3, 1 = 2.33, 1

x < y

20. Answer d

I. x2 = 14161

X = +119, -119

II. y = √14161

Y = 119

x y

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