Dear Readers, Find the aptitude test questions to crack latest bank exams. We regularly provide 20 aptitude test questions daily for students. Aspirants practice these questions on a regular basis to improve your score in aptitude section. Aspirants preparing for the exams can make use of this “20-20” Quantitative Aptitude Questions. Here we have started New Series of Practice Materials specially for Crack NIACL Assistant Prelims 2018. Aspirants those who are preparing for the exams can use this **“20-20” Quantitative Aptitude Questions.**

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**“20-20” Aptitude Questions | Crack NIACL Assistant prelims 2018 Day-190**

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**Directions (1 – 5): In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,**

a) If x > y

b) If x ≥ y

c) If x < y

d) If x ≤ y

e) If x = y or the relation cannot be established

** ****1) I.** 8x^{2} + 28x + 12 = 0

II. 9y^{2} + 27y + 14 = 0

**2) I.** 2x^{2} + 11x + 14 = 0

II. 3y^{2} + 14y + 15 = 0

**3) I.** 4x^{2 }+ 16x + 7 = 0

II. 5y^{2} + 32y + 35 = 0

** ****4) I. **7x – 5y = 55

II. 3x + 7y = – 13

** ****5) I.** x^{2} – 3x – 28 = 0

II. y^{2} – 17y + 72 = 0

** ****Directions (6 – 10): what value should come in place of question mark (?) in the following questions?**

**6) 6 × (?) ÷ (24 × 9) – 125 = 23**

a) 4756

b) 3274

c) 5328

d) 6120

e) None of these

** ****7) ****? – (21 ^{2} ÷ 525) × 1725 = (57900 ÷ 15)**

a) 4277

b) 3135

c) 4851

d) 5309

e) None of these

**8) ****(5/9) of 3402 – 16 × 78 ÷ 24 =?**

a) 2364

b) 1838

c) 3160

d) 2742

e) None of these

**9) (256 × 10 ^{-3})^{2} × 1024 ÷ (4^{13} × 10^{-4}) = 10^{?}**

a) -2

b) 3

c) -4

d) 5

e) None of these

** ****10) ****∛****110592 = 11 ^{3} + 24 % of 1800 – ? **

a) 2180

b) 2565

c) 1570

d) 1715

e) None of these

**Directions (11 – 15): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.**

**11) Quantity I: **200 m long train crosses a tunnel of length 175 m at the speed of 75 km/hr. Then find the time taken to cross the tunnel?

**Quantity II: **A train of length 150 m crosses a pole in 15 sec. Find the time taken by another train of length 250 m running in the opposite direction at the speed of 44 km/hr?

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

** ****12) Quantity I: **The S.I on a certain sum of money for 4 years at 6 % per annum is Rs. 6000. Then find the principle?

**Quantity II: **The C.I on a certain sum of money for 3 years at 10 % per annum is Rs. 6620. Then find the principle?

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

** ****13) (x + y)/(x – y) = 4 and x – (y/3) = 4**

**Quantity I: **The value of x is?

**Quantity II: **The value of y is?

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

** ****14) Quantity I: **3 years ago, the ratio of ages of P and Q is 2: 3. P’s age after 6 years is equal to present age of Q. Find the present age of Q?

**Quantity II: **5 years ago, the ratio of ages of A and B is 5: 3. Sum of the ages of A and B, after 6 years is 62 years. Find the present age of C, if C is 5 years younger than A?

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

** ****15) **A box contains 8 white balls, 6 violet balls and 4 yellow balls.

**Quantity I: **If 4 balls are drawn randomly, then the probability of getting at least one violet balls?

**Quantity II: **If 3 balls are drawn randomly, then the probability of getting 2 white balls?

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

** ****Directions (16 – 20): Study the following information carefully and answer the given questions. **

The following bar graph shows the price of gold per gram, diamond per gram and platinum per gram in different weeks in a certain month.

** **

**16) Total price of gold per gram of 1 ^{st} and 3^{rd} week is what percentage more/less than the total price of platinum per gram of 2^{nd} and 4^{th} week?**

a) 50 %

b) 35 %

c) 25 %

d) 30 %

e) None of these

** ****17) Find the average number of diamond per gram in the month?**

a) Rs. 7000

b) Rs. 6000

c) Rs. 6500

d) 7500

e) None of these

** ****18) Find the difference between the total number of gold per gram in the month to that of the total number of platinum per gram in the month?**

a) 2500

b) Rs. 4000

c) Rs. 3000

d) Rs. 3500

e) None of these

** ****19) The price of diamond per gram in 4 ^{th} week is approximately what percentage of the price of gold per gram in 3^{rd} week?**

a) 185 %

b) 172 %

c) 243 %

d) 224 %

e) None of these

** ****20) Find the ratio between the total number price per gram of all the given type of jewels in 2 ^{nd} week to that of 3^{rd} week?**

a) 5: 7

b) 7: 8

c) 3: 5

d) 11: 13

e) None of these

**Answers:**

**Direction (1-5)**

**1) Answer: e)**

I. 8x^{2} + 28x + 12 = 0

8x^{2} + 24x + 4x + 12 = 0

8x (x + 3) + 4(x + 3) = 0

(8x + 4) (x + 3) = 0

X = -4/8, -3 = -0.5, -3

II. 9y^{2} + 27y + 14 = 0

9y^{2} + 6y + 21y + 14 = 0

3y (3y + 2) + 7 (3y + 2) = 0

(3y + 7) (3y + 2) = 0

Y = -7/3, -2/3 = – 2.33, – 0.667

Can’t be determined

**2) Answer: e)**

I. 2x^{2} + 11x + 14 = 0

2x^{2} + 4x + 7x + 14 = 0

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

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

X = -7/2, -2 = -3.5, -2

II. 3y^{2} + 14y + 15 = 0

3y^{2} + 9y + 5y + 15 = 0

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

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

Y = -5/3, -3 = -1.667, -3

Can’t be determined

**3) Answer: e)**

I. 4x^{2 }+ 16x + 7 = 0

4x^{2 }+ 2x + 14x + 7 = 0

2x (2x + 1) + 7(2x + 1) = 0

(2x + 7)(2x + 1) = 0

X = -7/2, -1/2 = -3.5, -0.5

II. 5y^{2} + 32y + 35 = 0

5y^{2} + 25y + 7y + 35 = 0

5y(y + 5) + 7(y + 5) = 0

(5y + 7) (y + 5) = 0

Y = -7/5, -5 = – 1.4, -5

Can’t be determined

**4) Answer: a)**

7x – 5y = 55 –>(1)

3x + 7y = – 13–>(2)

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

X = 5, y = -4

X > y

**5) Answer: c)**

I. x^{2} – 3x – 28 = 0

(x + 4) (x – 7) = 0

x = -4, 7

II. y^{2} – 17y + 72 = 0

(y – 8) (y – 9) = 0

y = 8, 9

x < y

**Direction (6-10)**

**6) Answer: c)**

6*(x) ÷ 216 – 125 = 23

(x) / 36 = 23 + 125

(x) = 148*36

X = 5328

**7) Answer: d)**

X – (21*21*1725)/525 = (57900/15)

X – 1449 = 3860

X = 3860 + 1449

X = 5309

**8) Answer: b)**

(5/9)*3402 – (16*78)/24 = x

X = 1890 – 52

X = 1838

**9) Answer: a)**

(4^{4} × 10^{-3})^{2} × 4^{5} ÷ (4^{13} × 10^{-4}) = 10^{x}

4^{8} × 10^{-6} × 4^{5} ÷ (4^{13} × 10^{-4}) = 10^{x}

(4^{13} × 10^{-6}) / (4^{13} × 10^{-4}) = 10^{x}

10^{-6 + 4} = 10^{x}

10^{-2} = 10^{x}

X = -2

**10) Answer: d)**

48 = 1331 + (24/100)*1800 – x

48 = 1331 + 432 – x

x = 1331 + 432 – 48

x = 1715

**Direction (11-15)**

**11) Answer: e)**

Quantity I:

Time = D/S = (200 + 175)/(75*(5/18))

= > (375*18)/(75*5) = 18 sec

Quantity II:

Speed of first train = 150/15 = 10 m/sec = 10*(18/5) = 36 km/hr

Time = D/S = (150 + 250)/[(36 + 44)*(5/18)]

= > (400*18)/(80*5) = 18 sec

Quantity I = Quantity II

**12) Answer: a)**

Quantity I:

S.I = (P*n*r)/100

6000 = (P*4*6)/100

Principle = (6000*100)/24 = Rs. 25000

Quantity II:

C.I = P*[(1 + (r/100))^{n} – 1]

6620 = P*(1 + (10/100))^{3} – 1]

6620 = P*[(110/100)^{3} – 1]

6620 = P*[(1331/1000) – 1]

6620 = P*(331/1000)

P = 6620*(1000/331) = Rs. 20000

Quantity I > Quantity II

**13) Answer: a)**

(x + y)/(x – y) = 4

X + y = 4x – 4y

3x – 5y = 0 –à (1)

x – (y/3) = 4

3x – y = 12 –à (2)

Solving (1) and (2), we get,

X = 5, y = 3

Quantity I > Quantity II

**14) Answer: c)**

Quantity II:

3 years ago, the ratio of ages of P and Q = 2: 3 (2x, 3x)

Present age of P and Q = 2x + 3, 3x + 3

P’s age after 6 years = Present age of Q

2x + 3 + 6 = 3x + 3

X = 6

Present age of Q = 3x + 3 = 21 years

Quantity II:

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

Present age of A and B = 5x + 5, 3x + 5

Sum of the ages of A and B, after 6 years = 62 years

According to the question,

5x + 11 + 3x + 11 = 62

8x = 62 – 22

8x = 40 = > x = 5

Present age of A = 5x + 5 = 30 years

Present age of C = Present age of A – 5 = 30 – 5 = 25 years

Quantity II > Quantity I

**15) Answer: a)**

Quantity I:

Total probability n(S) = 18C_{4}

Required probability n(E) = 1 – P(none is violet)

Probability of getting none is violet balls,

P(E) = n(E)/n(S) = 12C_{4}/18C_{4}

= > 11/68

Required probability = 1 – (11/68) = 57/68

Quantity II:

Total probability n(S) = 18C_{3}

Required probability n(E) = 8C_{2} and 10C_{1}

P(E) = n(E)/n(S) = (8C_{2} and 10C_{1})_{} / 18C_{3}

= > 35/102

Quantity I > Quantity II

**Direction (16-20)**

**16) Answer: b)**

Total price of gold per gram of 1^{st} and 3^{rd} week

= > 3250 + 3500 = 6750

Total price of platinum per gram of 2^{nd} and 4^{th} week

= > 2750 + 2250 = 5000

Required % = [(6750 – 5000)/5000]*100 = 35 %

**17) Answer: d)**

The average number of diamond per gram in the month

= > (8000 + 6500 + 7000 + 8500)/4

= > 30000/4 = Rs. 7500

**18) Answer: a)**

The total number of gold per gram in the month

= > 3250 + 3000 + 3500 + 3750 = Rs. 13500

The total number of platinum per gram in the month

= > 2500 + 2750 + 3500 + 2250 = Rs. 11000

Required difference = 13500 – 11000 = Rs. 2500

**19) Answer: c)**

The price of diamond per gram in 4^{th} week = 8500

The price of gold per gram in 3^{rd} week = 3500

Required % = (8500/3500)*100 = 242.85 % = 243 %

**20) Answer: b)**

The total number price per gram of all the given type of jewels in 2^{nd} week

= > 3000 + 6500 + 2750 = Rs. 12250

The total number price per gram of all the given type of jewels in 3^{rd} week

= > 3500 + 7000 + 3500 = Rs. 14000

Required ratio = 12250: 14000 = 7: 8

**Daily Practice Test Schedule | Good Luck**

Topic |
Daily Publishing Time |

Daily News Papers & Editorials |
8.00 AM |

Current Affairs Quiz |
9.00 AM |

Quantitative Aptitude “20-20” |
11.00 AM |

Vocabulary (Based on The Hindu) |
12.00 PM |

General Awareness “20-20” |
1.00 PM |

English Language “20-20” |
2.00 PM |

Reasoning Puzzles & Seating |
4.00 PM |

Daily Current Affairs Updates |
5.00 PM |

Data Interpretation / Application Sums (Topic Wise) |
6.00 PM |

Reasoning Ability “20-20” |
7.00 PM |

English Language (New Pattern Questions) |
8.00 PM |