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[WpProQuiz 7106]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
- 504 = 6250000 and 514 = 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*R2/1002 = (difference between CI and SI)
P * 52/ 1002 = 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 Ã 2800
- (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
