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**LIC HFL Assistant Manager Quantitative Aptitude Questions – (Day-06)**

**Directions (1 – 5): Study the following table carefully and answer the questions that follow:**

**1) In 2002, if the interest was calculated semi-annually, the interest would have been Rs.2060.804 for given time period. Find the interest when compounded annually.**

a) 1960

b) 2070

c) 2040

d) 2220

e) None of these

**2) In 2003, the difference between compound interest and simple interest for the given period is Rs.40. If the sum is invested for 3 years, what will be the compound interest after 3 years?**

a) 2728

b) 2655

c) 2328

d) 2522

e) None of these

**3) In 2004, if rate of interest is increased by 3%, what will be the amount received after given period?**

a) 52000

b) 45000

c) 42000

d) 47000

e) None of these

**4) In 2005, the amount becomes Rs.57400 after 8 years, find the rate of interest.**

a) 6%

b) 10%

c) 7%

d) 9%

e) None of these

**5) In 2001, the person invested Rs.7000 at simple interest and rest at compound interest. If the total amount received after respective period is Rs.17920, then find the time period for which the amount was invested at compound interest**.

a) 1 year

b) 3 years

c) 4 years

d) 2 years

e) None of these

**6) A train engine become 10% 20% & 40% more efficient when it is supported by three other supportive engine P,Q& R respectively. Find the net saving of fuel consumed by the main engine if it is supported by all the three supportive engine independently?**

a) 20%

b) 20%

c) 80%

d) 80%

e) None of these

**7) A man buys a new mobile for Rs. 11000. After using it for an year he sell it at a loss of 15%. Then again he buys a new mobile at a price 15% higher than the previous one. He uses the new mobile for one year and sell it at a loss of 20% then find his average loss per year during this period?**

a) 1990

b) 4180

c) 2090

d) 1265

e) None of these

**8) A work can be completed by X men in 80 days. If 10 men leave before the work started then it will take 20 extra days to complete the work. Find the number of days to complete the same work if 100 people are working. Note-Consider efficiency of all the men is same.**

a) 100 Days

b) 50 Days

c) 80 Days

d) 40 Days

e) None of these

**9) The average speed of a school bus is 60km/h excluding its stoppage time and if stoppage time is included its average become 50km/hr. How many minutes does the school bus stop in an hour?**

a) 10 Minutes

b) 12 Minutes

c) 15 Minutes

d) 18 Minutes

e) 21 Minutes

**10) Average salary of 25 manager is 55000 rupees. If 5 manager excluded from it the average reduced by 5000. What is the average salary of the managers who were excluded?**

a) 3,75,000

b) 1,00,000

c) 80,000

d) 65,000

e) 75,000

**Answers :**

**Direction (1-5) :**

**1) Answer: c)**

Rate semi-annually = 4/2 = 2%

Interest = Rs2060.804 for given period, means it is not for 1 year, but the whole period.

Now, CI = 25000+2060.804 = 27060.804

- 25000 [1 + 2/100]
^{n}= 27060.804 - 25000 [1 + 2/100]
^{n}= 27060.804 - [51/50]
^{n}= 27060804/25000000 - [51/50]
^{n}= 27060804/25000000 - [51/50]
^{n}= 6765201/6250000 - 50
^{4}= 6250000 and 51^{4 }= 6765201

So, n = 4 [this is semi-annually]

Means, actual time period = 4/2 = 2 years

So, Amount compounded annually = 25000 [1 + 4/100]^{2} = Rs.27040

So, CI = 27040 – 25000 = Rs.2040

**2) Answer: d)**

SI for 1 year = Rs.800, so for 2 years = 2*800 = Rs.1600

So according to formula for 2 years, R = 2*(difference between CI and SI)/SI * 100

R = (2*40/1600) *100 = 5%

Now use,

P*R^{2}/100^{2} = (difference between CI and SI)

P * 5^{2}/ 100^{2} = 40

- P = Rs.16000

So amount after 3 years = 16000 [1 + 5/100]^{3} = Rs.18522

- CI = 18522 – 16000 = Rs.2522

**3) Answer: b)**

According to formula, Time = 1 year

(25000*r*1)/100 = 1750

- r = 7%
- New rate = 7+ 3 = 10%

Now, Interest = (15000*10*8)/100 = Rs.20000

- amount = 25000 + 20000 = Rs.45000

**4) Answer: e)**

Interest after 1 year = Rs.2800, so after 6 years = 2800*6 = Rs.16800

Let principal amount = x

So in 6 years, amount becomes (x+16800) and in 8 years it becomes 57400

So interest in 2 years = 57400 – (x+16800) = (40600-x)

So after 1 year = (40600-x)/2 à2800

- (40600-x)/2 = 2800
- x = Rs.35000

Now, (35000*r*1)/100 = 2800

- r = 8%

**5) Answer: d)**

7000 + (7000*6*5)/100 + 8000 * [1 + 5/100]^{n} = 17920

8000*[21/20]^{n} = 8820

n = 2 years

**Direction (6-10) :**

**6) Answer: c)**

Improvement in efficiency = Net fuel saving= Reduction in fuel consumed

Consumption of fuel reduces by 10%,20%, and 40% successively.

Let initial consumption of fuel be 100 Liter

10% Improvement in efficiency = 10% Net fuel saving= 10% Reduction in fuel consumed

Thus fuel used by main engine after adding supportive engine P = 100-10 = 90 Liter

20% Improvement in efficiency = 20% Net fuel saving= 20% Reduction in fuel consumed

Thus fuel used by main engine after adding supportive engine Q = 90- 18=72 Liter

40% Improvement in efficiency = 40% Net fuel saving= 40% Reduction in fuel consumed

Thus fuel used by main engine after adding supportive engine R = 72-28.8=43.2 Liter

Thus total fuel saved by main engine = 100-43.2 = 66.8 Liter

= (66.8/100)*100 = 66.80%

**7) Answer: c)**

Loss in first year = 15% 0f 11000

=1650

Cost of mobile in second year 115% of 11000

= 11000 + 1650

= 12650

Loss in second year = 20% of 12650

= 2530

Total loss = 1650 +2530

=4180

Average loss= 4180/2 = 2090

**8) Answer: d)**

If X men can be completed the work in 80 days

Then 1 men can complete the work in 80*X days ————–I

After 10 men leave the job total men = (X-10) and work completed in 80+20= 100 days

Now,

(X-10) men can complete the work in 100 days

Then 1 men can complete the work in 100*(x-10) days ————-II

Equating I and II value

80*X = 100*(X-10)

X = 50 Men

Thus 50 men can complete the work in 80 days

100 men can complete the work in (80*50)/100 = 40 days

**9) Answer: a)**

Let the distance travel by school bus is 300 km (L.C.M of 60 & 50)

Time taken when the speed is 60km/h = 300/60

=5 hours

Time taken when the speed is 50km/h = 300/50

=6 hours

In 6 hours bus travel only 5 hours and 1 hours stop on its stoppage

Thus in one hour stoppage time =1/6 hour

=(1/6*60) minutes

= 10 minutes

**10) Answer: e)**

Total salary of 25 manager = 25* 55000

=1375000

Average salary of remaining 20 manager = 55000 – 5000

= 50000

Total salary of remaining 20 manager = 50000 * 20

= 1000000

Sum of the salary or 5 manager who left = 1375000 – 1000000

=375000

Average salary or 5 manager who left 375000/5 =75000