On this page, you can practice important Quantity Based Questions for LIC Assistant for free so you can give a better performance in your upcoming LIC Assistant exam. In this article we are also providing Quantity Based Questions PDF in the form of quizzes so practice Maths Inequality questions in offline mode too.

As you know most of the students like to solve Data Interpretation, Simplification and Number Series Etc. questions but if you practice

__Quantity Based Questions__then you will make yourself different from 60% of aspirants, you can score high in your LIC Assistant.

**Quantity Based Questions**have 50% chance to appear in your LIC Assistant Pre but they have high chance to appear in

**Exam. So in this article you will get all level of Quantity Based Questions in the form of quizzes and PDF too.**

__LIC Assistant Mains__Aspirant always keep one thing in his mind that you must practice

*Maths Inequality Questions*as much as you can either in the form of online quizzes or offline

**whatever suitable for you. This will enhance your solving skills which will help you to gain extra marks in your LIC Assistant exam.***PDF*

So, here Smartkeeda providing

**. In these**

*Maths Inequality Questions Quizzes and PDF for LIC Assistant for free**PDFs*you will see various types of Maths Inequality Questions, some are new pattern based and some are frequently asked in

*LIC Assistant*or in similar exams. So Download Maths Inequality Questions PDF for LIC Assistant.

As you know around five questions of Maths Inequality can be asked in LIC Assistant, therefore while downloading

*Quantity Based Questions PDF*for Insurance clerk level exams like LIC Assistant, you need to brush up your maths arithmetic or maths word problems skills;

Here are the some important

__Quantity Based Questions__with solution for LIC Assistant.

**Directions(Q 1-10): Each question below contains a statement followed by Quantity I and Quantity II. You have to study the information along with the question and compare the value derived from Quantity I and Quantity II, then anwer:**

**1).**In an envelope there are 5 green, 3 yellow and 4 pink tablets. 3 tablets are picked at random.

**Quantity I:**The probability that 2 tablets are yellow in colour and 1 tablet is pink in colour.

**Quantity II:**The probability that all the tablets are green in colour.

A. Quantity I > Quantity II

B. Quantity I < Quantity II

C. Quantity I = Quantity II

D. Quantity I ≤ Quantity II

E. Quantity I ≥ Quantity II

Ans:A

**Explanation:-****Quantity I:**

Favourable outcomes:

2 yellow + 1 pink tablet =

^{3}C

_{2}×

^{4}C

_{1}

= 12

Total outcomes =

^{12}C

_{3}

= 220

Probability = | 12 | = | 3 |

220 | 55 |

**Quantity II:**

Favourable outcomes:

3 green tablet =

^{5}C

_{3}

= 10

Total outcomes =

^{12}C

_{3}

= 220

Probability = | 10 | = | 1 |

220 | 22 |

Quantity I > Quantity II

Hence, option A is correct.

**2).**Find the percentage of boys in the class this year.

**Quantity I:**This year the percentage of girls in the class is 60%.

**Quantity II:**Last years out of the 300 students, 50% was girls and this year the number of girls are increased by 10% but total students remains same.

Ans: B

**Explanation:-****Quantity I:**

Because percentage of girls = 60%,

So the percentage of boys = 100 – 60% = 40%

**Quantity II:**

Last year,

Girls was 300 × 50% = 150, boys = 300 × 50% = 150

This year, because girls are increased by 10% and total students remain same.

girls = 150 × 110% = 165, boys = 300 – 165 = 135

% of boys = | 135 | × 100 = 45% |

300 |

Quantity I < Quantity II

Hence, option B is correct.

**3).**

**Quantity I :**Working alone, B can complete the work in 20 days. C is twice as efficient as B and A takes 2 days more than it takes C to complete the work. Working together, in how much time would they be able to complete 7 such works?

**Quantity II :**Working alone, A, B and C can do a work in 24, 30 and 40 days respectively. How long will it take them to complete the work if only A and B work for the first 6 days and then C joins them?

Ans: A

__Explanation:-__**Quantity : I**

Time taken by B to complete the work = 20 days

Time taken by C to complete the work = | 20 | = 10 days |

2 |

Time taken by A to complete the work alone = 10 + 2 = 12 days

Work done by A, B and C together in 1 day.

⇒ | ( | 1 | + | 1 | + | 1 | ) | = | 3 + 6 + 5 | = | 14 |

20 | 10 | 12 | 60 | 60 |

this means if would take | 30 | days |

7 |

for A, B and C to complete the work together

∴ Time taken A, B and C to complete

7 works = 7 × | 30 | days = 30 days |

7 |

**Quantity : II**Time taken by A to complete the work = 24 days

Time taken by B to complete the work = 30 days

Time taken by C to complete the work = 40 days

Work done by A and B in 1 day

⇒ | ( | 1 | + | 1 | ) | = | 5 + 4 | = | 9 |

24 | 30 | 120 | 120 |

Work done by A and B in 6 days

= 6 × | 9 | = | 9 |

120 | 20 |

Remaining work = 1 – | 9 | = | 11 |

20 | 20 |

Work done by A, B and C together in 1 day

= | ( | 1 | + | 1 | + | 1 | ) |

24 | 30 | 40 |

= | 5 + 4 + 3 | = | 1 |

120 | 10 |

Hence, A, B and C will complete the work in 10 days

Now, time taken by A, B and C to complete | 11 | of the work |

20 |

= | 11 | × 10 = 5.5 days |

20 |

Hence, Quantity I > Quantity II

Therefore, option (A) is correct.

**4).**

**Quantity I :**A milkman sells one - third quantity of milk what he has at a profit of 6% and the remaining milk at a profit of 15%. What is the total profit per cent for the milkman?

**Quantity II :**A boat covers a distance of 13.5 km in upstream in 1.5 hours while in downstream it covers the same distance in 54 minutes. What is the speed of boat in still water?

Ans: E

**Explanation:-****Quantity I :**Let total profit per cent = x%

By the rule of allegation-

⇒ | 15 – x | = | 1 |

x – 6 | 2 |

⇒ x = 12%

**Quantity II :**Let speed of boat is ‘a’ km/h and speed of stream is ‘b’ km/h.

According to question-

⇒ | 13.5 | = | 54 |
.. (1) |

a + b | 60 |

⇒ | 13.5 |
= 1.5 .......(2) |

a – b |

From (1) and (2)-

⇒ a = 12 and b = 3

Hence, option (E) is correct.

**5).**

**Quantity I:**The distance travelled by a bus in 4 hours is 320 km. If the speed of bus is increased by 20% then what will be the time taken by bus to cover the triple of distance?

**Quantity II:**A bus which is travelling from point A to point B which are 150 km apart, covers half of distance with 25 km/h and rest of distance with 30 km/hr and take rest of 30 minutes after travelling half of distance then what will be the total time taken by bus to reach destination?

A. Quantity : I > Quantity : II

B. Quantity : I ≥ Quantity : II

C. Quantity : I < Quantity : II

D. Quantity : II ≥ Quantity : I

E. Quantity I = Quantity II or relation can not be established

Ans: A

Quantity : I

__Explanation:-__Quantity : I

Distance travelled by the bus = 320 km

Time taken = 4 hours

∴ Speed of the bus = | 320 | = 80 km/hr |

4 |

Speed after 20% increment = 80 × | 120 | = 96 km/hr |

100 |

Now distance need to travel = 320 × 3 = 960 km

∴Time take = | 960 | = 10 hours |

96 |

**Quantity : II**

Distance between point A and B = 150 km

half of distance = 75 km

Time taken to cover 75 km at a speed of 25 km/hr

= | 75 | = 3 hours |

25 |

Time taken to cover 75 km at a speed of 30 km/hr

= | 75 | = 2.5 hours |

30 |

And we also know that bus took 0.5 hour rest in between.

Therefore, total time = 3 hr + 2.5 hr + 0.5 hr = 6 hours

Here, Quantity I > Quantity II

Therefore, option (A) is correct.

**6).**

**Quantity I:**Speed of boat in still water, if a man can travel 54 km downstream in 9 hours and 40 km upstream in 10 hours.

**Quantity II:**Speed of boat in still water, if a man can travel 45 km downstream in 9 hours and the speed of stream is 1 kmph.

A. Quantity I ≤ Quantity II

B. Quantity I ≥ Quantity II

C. Quantity I > Quantity II

D. Quantity I < Quantity II

E. Quantity I = Quantity II

Ans: C

__Explanation:-__**Quantity I:**

Let speed of boat = x kmph, speed of stream = y kmph

upstream speed = x – y, downstram speed = x + y

x + y = | 54 |

9 |

x + y = 6 ....1

x – y = | 40 |

10 |

x – y = 4 ....2

equation 1 + equation 2

x = 5

speed of boat = 5 kmph

**Quantity II:**

Let speed of boat = x kmph, speed of stream = 1 kmph

downstream speed = x + 1

45 | = x + 1 |

9 |

5 – 1 = x

x = 4

Speed of boat = 4 kmph

Quantity I > Quantity II

Hence, option C is correct.

**7).**

**Quantity I:**Age of mother, if the age of Smiti is 1/7th of her mother's age and after 5 years Smiti's age will be 12 years.

**Quantity II:**Age of mother, if the ratio of the ages of Sukriti and her mother is 3 : 7 and after 3 years the ratio of their ages will be 6 : 13.

Quantity I:

__Explanation:-__Quantity I:

Smiti's age after 5 years = 12 years

Present age of Smiti = 7 years

Mother's age | = Smiti's age |

7 |

Mother's age = 7 × 7 = 49 years

**Quantity II:**Let Sukriti's age = 3x, Mother's age = 7x

According to the statement,

3x + 3 : 7x + 3 = 6 : 13

13 (3x + 3) = 6 (7x + 3)

39x + 39 = 42x + 18

39 – 18 = 42x – 39x

21 = 3x

x = 7

Mother's age = 7 × 7 = 49 years

Quantity I = Quantity II

Hence, option E is correct.

**8).**What is the area of the rectangle?

**Quantity I:**The area of the rectangle is equal to the area of the square whose side is 24 cm.

**Quantity II:**The perimeter of the rectangle is 88cm, if the ratio of the length and the breadth of the rectangle is 6 : 5.

Ans:

Quantity I:

__Explanation:-__Quantity I:

Area of the square = 24 × 24

= 576 cm

^{2}

Area of the square = Area of the rectangle

Area of the rectangle = 576 cm

^{2}

**Quantity II:**

Length of the rectangle = 6x, Breadth = 5x

Perimeter = 88

2 (l + b) = 88

6x + 5x = 44

11x = 44

x = 4

Length = 24cm, breadth = 20cm

Area of the rectangle = 24 × 20

= 480 cm

^{2}

Quantity I > Quantity II

Hence, option E is correct.

**9).**

**Quantity I:**Selling price, if cost price is Rs. 50,000 and profit is 20%

**Quantity II:**Selling price, if cost price is Rs. 50,000 and shopkeeper gained 50/3 % after giving discount of 25%

**Quantity I :**

SP = | 120 | × 50000 = Rs.60000 |

100 |

**Quantity II :**

Discount = 25% = | 1 | (MP = 4, SP = 4 – 1 = 3) |

4 |

Gain = | 50 | % = | 1 | (CP = 6, SP = 6 + 1 = 7) |

3 | 6 |

Make SP same

CP…………SP…………..MP

18………….21………….28

18 = 50000

So 21 = Rs. 58333.33

Hence, Quantity I > Quantity II

Therefore, option (A) is correct.

**10).**

**Quantity :I**5 years ago, the ratio of Seema and Meera's age was 8 : 5. Two years hence, the ratio of their ages will becomes 5 : 4 Find the present age of Meera.

**Quantity :II**The age of son is half the age of mother. His sister’s age is 20 years. Age of the mother is twice the age of his sister. Find the age of the son.

A. Quantity : I > Quantity : II

B. Quantity : I ≥ Quantity : II

C. Quantity : I < Quantity : II

D. Quantity : II ≥ Quantity : I

E. Quantity I = Quantity II or relation can not be established

Ans: C

Quantity : I

__Explanation:-__Quantity : I

Let the present ages of seema and meean are s and m respectively.

Now,

s – 5 | = | 8 |

m – 5 | 5 |

⇒ 8m – 40 = 5s – 25

⇒ | 8m –15 | = s ......(i) |

5 |

And,

s + 2 | = | 5 |

m + 2 | 4 |

⇒ 5m + 10 = 4s + 8

⇒ s = | 5m + 2 | .......(ii) |

4 |

equating (i) and (ii), we get

8m – 15 | = | 5m + 2 |

5 | 4 |

⇒ m = 10 Hence, the present age of meera is 10 years.

**Quantity : II**

Let, the present ages of Son, mother and sister are s, m and sis.

Now, 2s = m ............(i)

And, sis = 20 years ............(ii)

m = 2 sis ...........(iii)

Using equation (ii) and (iii), we can say that

m = 40 years ............(iv)

Using (iv) and (i),

we can say that s = 20 years.

Here, Quantity: I < Quantity II

Hence, option (C) is correct.

Quiz 1 | Quiz 2 | Quiz 3 |

Quiz 4 | Quiz 5 | Quiz 6 |

Quiz 7 | Quiz 8 | Quiz 9 |

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