G H Raisoni University Amravati
Anjangaon Bari Road, Amravati 444701
Standard Journal
Subject: Aptitude.
Semester / Branch:
V SEM CSE
Department of Computer Science and
Engineering
G H RAISONI UNIVERSITY, AMRAVATI
G. H. RAISONI UNIVERSITY, AMRAVATI
Department of Computer Science and Engineering
Subject- Aptitude
Semester/Branch: V SEM CSE
LIST OF PRACTICAL
Total Hrs: 20
Sr.
No
1.
2.
3.
4.
Title of Practical
To practice finding the LCM and HCF of numbers
To practice working with ratio, proportions and solving problem
that involve these concepts.
Calculate the percentage of income expenditure and increase decrease
concept.
Solve and calculate the profit or loss base on profit, loss and discount
concept.
Hrs
Allotted
Relevance to
Course Outcome
2
CO1
2
CO1
2
CO1
2
CO2
5.
Solve and calculate the time of working with two person’s
concepts and men women alternate concept
2
CO2
6.
Solve and calculate the distance between two points using
speed time and distance.
2
CO3
2
CO3
2
CO3
2
CO4
2
CO4
7.
Solve and calculate the mixture using allegation method.
9.
Solve and calculate the pipe and cisterns using simultaneous pattern
questions.
Solve and calculate the simple interest and compound
interest.
10
To practice and Solve permutation, Combination and
Probability.
8.
Practical Teacher:
Mr. Sachin Deshpande
HOD CSE
PRACTICAL No.1
AIM: To practice finding the LCM and HCF of numbers
Problem Statement: Use LCM (Least Common Multiple) and HCF (Highest Common Factor) to
solve problems related to scheduling, grouping, and synchronization. Understanding LCM helps in
finding common multiples and optimal intervals, while HCF is useful for dividing quantities into the
largest possible equal parts, ensuring efficient resource utilization.
Material: Pen and paper
Part 1: Find the LCM of following sets of numbers.
1. 12 and 18
2. 15 and 25
Solution Part 1:
1. 12 and 18
Factor of 12 = 1, 2, 3, 4, 6 and 12
Factor of 18 = 1, 2, 3, 6, 9, 18
LCM = 2*3*6 = 36
2. 15 and 25
Factor of 15 = 1, 3, 5, 15
Factor of 25= 1, 5,
LCM = 5, 15 = 75
Part 2: Find the HCF of following sets of number
1.
16 and 20
2. 54, 72 and 90.
Solution part 2:
1. 16 and 20
Factor of 16 = 1, 2, 4, 8, 16
Factor of 20 = 1, 2, 4, 5, 10, 20
HCF = 1, 2, 4= 4
2. 54, 72 and 90
Factor of 54 = 2, 3, 3, 3
Factor of 72 = 2, 2, 2, 3, 3
Factor of 90 = 2, 3, 3, 5
HCF = 2, 3, 3 = 18
Conclusion: By successfully completing this practical exercise I gain a better
Understanding of LCM and HCF. These concept of fundamental in number theory.
PRACTICAL No.2
AIM: To practice working with ratio, proportions and solving problem that involve
these concepts.
Problem statement: A problem statement involving ratios and proportions typically
revolves around comparing quantities or determining an unknown quantity based on
a given relationship. Below is a general problem statement
Material: Pen and paper
1. A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount
does B have?
Solution:
𝟒
𝟐
⇒ 𝟏𝟓 A = 𝟓 B
𝟐
𝟏𝟓
⇒A=𝟓× 𝟒 B
𝟑
⇒A= B
𝟐
⇒ AB =
𝟑
𝟐
⇒ (A: B) = (3: 2)
2
∴ B Share = Rs. 1210×5 = Rs.484
2. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a
proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of
increased seats?
Solution:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x a
nd 8x respectively.
Number of increased seats are
⇒ (140% of 5x), (150% of 7x) and (175% of 8x)
= ((140100) ×5x), ((150100) ×7x) and ((175100) ×8x)
=7x, 10.5x, and 14x
= multiply above term by 2 get exact number
∴The required ratio=7x: 10.5x: 14x
=14x: 21x: 28x
=2:3:4
Conclusion: Successfully completing this practical exercise we gain a better
understanding of ratio and proportion. These concept of fundamental in ratio and
proportion.
PRACTICAL No.3
Aim: Calculate the percentage of income expenditure and increase decrease concept
Problem Statement: percentage problems, individuals enhance their numerical literacy and become
more adept at understanding and applying quantitative information in both personal and professional
contexts.
Material: Pen and paper
1. The number of seats in a cinema hall is decreased by 8% and also the price of the ticket is
increased by 4 percent. What is the effect on the revenue collected?
Solution:
Let initially seats are 100 and price of each seat is 100, so total initial revenue = 10000
Now, seats are 92 and price of each seat = 104,
So total revenue = 92×104 = 9568
So percent change in revenue = (43210000)×100 = 4.32 % decrease
2. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were
invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got,
was
Solution:
Total number of votes = 7500
Given that 20% of Percentage votes were invalid => Valid votes = 80%
Total valid votes = 7500× (80/100)
1st candidate got 55% of the total valid votes.
Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = total valid votes× (45/100)
7500× (80/100) × (45/100) = 2700
Conclusion: Successfully completing this practical exercise, we gain a better
understanding of percentage base on income expenditure and increase decrease
concept. These concept of fundamental in percentage.
PRACTICAL No.4
Aim: Solve and calculate the profit or loss base on profit, loss and discount concept.
Problem statement: Calculate the profit or loss in monetary terms and as a percentage for individual
transactions and overall sales. Understand how to evaluate the financial performance of a business by
analyzing profit and loss, and apply these calculations to make informed decisions regarding pricing
and cost management.
Material: Pen and paper.
1. A loss of 20% is made by selling an article. Had it been sold for Rs 240 more, there would have been a
profit of 10%. What would be the selling price of the article if it is sold at 25% profit?
Solution:
⇒Let the cost price of the article be x.
Given that the article was sold at a loss of 20%, which means it was sold at 80% of its cost price.
Therefore, the selling price of the article can be written as 0.8x.
Had the article been sold for 240 more, there would have been a profit of 10%
This means that the selling price would have been 110% of the cost price.
Therefore, the selling price can be written as 1.1x.
1.1x = 0.8x+240
Solving this equation, we get:
0.3x = 240
x = 800
Therefore, the cost price of the article is Rs 800.
To find the selling price of the article at a 25% profit, we can use the formula:
Selling price = cost price+profitProfit is 25% of the cost price, which means profit = 0.25x
= 0.25×800
= Rs.200.So, selling price = 800+200 = Rs.1000.
Therefore, the selling price of the article at a 25% profit would be Rs 1000.
2. A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of
950gm. for a kg. His gain is____%.
Solution:
He sells 950 gm pulses and gets a gain of 50 gm (1000 gm−950 gm)
If he sells 100 gm of pulses,
he will gain =[ Error( True value )−( Error )×100]%
=( 50/(1000−50)×100)%
= 5000/ 950%
= 100/ 19%
5
= 519%
5
∴ His gain percent is 519%
Conclusion: Successfully completing this practical exercise, we gain a better understanding of how
to made profit. These concept of fundamental in profit, loss and discount.
PRACTICAL No.5
AIM: Solve and calculate the time of working with two person’s concepts and men women alternate
concept
Problem Statement: Calculate the time required to complete a task based on the number of workers
and their working rates, or determine the number of workers needed to meet a deadline. This helps in
planning and optimizing workforce allocation to meet project timelines efficiently.
MATERIAL: Pen and paper
1. Amruta can do a piece of work in 20 days. Narendra is 25% more efficient than Amruta. The
number of days taken by Narendra to do the same piece of work is:
SOLUTION:
Ratio of times taken by Amruta and Narendra
= 100 : 120
=4:5
Let’s assume Narendra takes x days to do the work
5 : 4 :: 20 : x
⇒ x= (4×20) / 5
⇒ x = 16 days
Hence, Narendra takes 16 days to complete the work
2. 1 men and 24 women can complete the work in 5 days, 6 men and 6 women in 15 days. Find in
how many days 3 men and 3 women will complete the work?
Solution:
(Number of person × time = Total work)
(6 men + 24 women) × 5 = (6 men + 6 women) × 15
M + 4W = 3Men + 3Women
⇒ 2M = W
𝑀
1
⇒𝑊 =2
---------equation (b)
Put 3 men and 3 women in equation b
𝑀
3
⇒ 𝑊 = 6 Total is 9
⇒ Total Work
54×5
9
⇒ 30 Days
CONCLUSION: Successfully completing this practical exercise, we gain a better understanding of
time and work concept.
PRACTICAL No.6
AIM: Solve and calculate the distance between two points using speed time and
distance.
Problem Statement: Solve problems involving speed, time, and distance to
determine travel times, required speeds, and distances covered. This helps in
planning routes, scheduling deliveries, and optimizing travel efficiency in realworld scenarios
MATERIAL: Pen and paper
1. A man covers the first half of the journey at the speed of 40kmph and the remaining at the speed of
60kmph. Find his average speed?
SOLUTION:
Let's assume that the total distance of the journey is’d’ km.
𝑑
The first half of the journey would then be 2 km.
Time taken to cover the first half of the journey, t1 = (d/2) km ÷ 40 kmph
= (d/80) hours.
Time taken to cover the second half of the journey, t2 = (d/2) km ÷ 60 kmph = (d/120
) hours.
Therefore, the total time taken to complete the journey, t = t1+t2 = (d/80)+(d/120)
⇒ (3d/240)+(2d/240) = (5d/240) = (d/48) hours.
Average speed = Total distance ÷ Total time = d km ÷ (d/48) hours = 48 kmph.
Therefore, the man's average speed for the entire journey is 48 kmph.
2. A man completes a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr
and the second half at the rate of 24 km/hr. Find the total journey in km.
Solution:
Let us assume that the total journey is "d" km.
The man travelled the first half of the journey at the rate of 21 km/hr.
Therefore, he would have covered half the distance in: time = distance/speed
time = (d/2)/21 time = d/42
Similarly, he travelled the second half of the journey at the rate of 24 km/hr.
Therefore, he would have covered the other half of the distance in:time = distance/speed
time = (d/2)/24
time = d/48
The total time taken for the journey is given as 10 hours.
Thus, we can write the equation :( d/42) + (d/48)
= 10
Simplifying this equation,
we get:Multiplying both sides by the least common multiple of the denominators, which is 10080:
24d+21d = 10080
45d = 10080
d = 224
Therefore, the total journey is 224 km.
CONCLUSION: Successfully completing this practical exercise, we gain a better
understanding distance, speed and time.
PRACTICAL No.7
AIM: Solve and calculate the mixture using allegation method.
Problem statement: Mixture and Allegation is a mathematical method used to solve
problems related to the mixing of different quantities with varying concentrations,
prices, or other attributes. This method is particularly useful in problems where two or
more components with different characteristics are mixed together to achieve a desired
result.
MATERIAL: Pen and paper
1. A milkman has 80 liters mixture of milk and water in the ratio of 4:1. If he added
X liters of water in the mixture, then the ratio of the milk to water becomes 8:5.
Find the value of X.
SOLUTION:
Total
Milk
4
Water = 64
New amount
64
*8
New amount ratio 8
= 40 liters – 16 liters
X = 24
80
Water
1
16= milk
40
*8
5
2. A grocer has two types of rice and their price are 38 Rs/kg and 26 Rs/kg. He want to make a new
type of rice by mixing old types. In what ratio old types should be mixed so that the price of new type
would be 30 Rs/kg.
Solution:
Let x be the ratio of the more expensive rice (Rs 38/kg) to the total mixture.
Then, the ratio of the less expensive rice (Rs 26/kg) to the total mixture
To obtain a mixture that costs Rs 30/kg, we can set up the following equation based on the weighted average of
the prices:
(38x+26y)/(x+y) = 30
We can solve for the ratio of x to y by using the given equation and simplifying:
38x+26y = 30(x+y)
38x+26y = 30x+30y
8x = 4y
x/y = 4/8 = ½
Therefore, the ratio of x to y that satisfies the given equation is 1:2.
Another methods is Allegation method
38kg
26kg
30kg
4
8
= 1:2
CONCLUSION: Successfully completing this practical exercise, we gain a better
understanding of mixture of using allegation method.
PRACTICAL No.8
AIM: Solve and calculate the pipe and cisterns using simultaneous pattern questions.
Problem Statement: Pipe and Cistern problems involve calculating the time required to fill or
empty a tank (cistern) when different pipes are involved. Pipes can either fill the tank (inlet pipes) or
empty it (outlet pipes). These problems typically require an understanding of rates and how they
combine.
MATERIAL: Pen and paper
1. Pipe A and B can fill a tank in 15 minutes and 10 minutes respectively. Pipe C can empty the
tank in 5 minutes. Pipe A and B are opened simultaneously and after 4 minute pipe C is also
opened. In what time the tank will be emptied?
Solution:
Pipe can fill the tank in 15 minutesQuantity of tank filled by first pipe in 1 minute = 1/15 units
Second pipe can fill the tank in 10 minutesQuantity of tank filled by second pipe in 1 minute =1/10 un
its
Third pipe can empty the tank in 5 minutesQuantity of tank emptied by third pipe in 1 minute =1/5uni
ts
According to the question, Quantity of tank filled by both pipe in 4 min = 4(1/15+1/10) = 2/3
2/3 of tanks work done by both the pipe working together
Quantity of tank filled by all the three pipes in 1 minute = (1/15+1/10-15) = 1/30 units
Here, negative sign shows the tank is emptying1/30 tank is emptied in 1 min
Total mins = 30 mins
2/3 of the tank will be emptied in 2×(30/3) = 20 min
The tank will be emptied in 20 minutes.
2. It takes two pipes A and B, running together, to fill a tank in 6 minutes. It takes A 5 minutes less than B
to fill the tank, then what will be the time taken by B alone to fill the tank?
SOLUTION:
Let the time taken by pipe A to fill the tank be x minutes
Time is taken by pipe B to fill the tank = x+5 minutes
So, 1/x + 1/(x+5) = 1/6
⇒ x = 10
Thus, time taken by B alone to fill the tank is 10+5, i.e., 15 minutes
CONCLUSION: Successfully completing this practical exercise, we gain a better understanding of
pipe and cisterns using simultaneous concept.
PRACTICAL No.9
AIM: Solve and calculate the simple interest and compound interest.
Problem statement: Simple and Compound Interest problems involve calculating the interest earned
or paid on a principal amount over time. The key difference between simple and compound interest is
how the interest is calculated and added to the principal.
MATERIAL: Pen and paper
1. A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of
simple interest. The rate of interest per annum is
Solution:
=S.I. for 3 years=Rs.(12005−9800)
=Rs. 2205
S.I. for 5 years=Rs.((2205/3)×5)
=Rs.3675
∴Principal=Rs.(9800−3675)
=Rs.6125
Hence, rate=((100×3675)/(6125×5))%=12%
2. A sum fetched a total simple interest of Rs. 3570 at the rate of 8.5% per annum in 6 years. What is
the sum (in Rs.)?
SOLUTION: consider principle is 100%,
= Rate 8.5% * time
= 8.5% * 6 = 51%
= 51% = 3570
= 100% =?
𝟑𝟓𝟕𝟎∗𝟏𝟎𝟎
=
= Rs. 7000
𝟓𝟏
3. Find the compound interest on Rs. 25,600 invested at 37.5%per annum for 3 years?
𝟑
SOLUTION: 37.5% = 𝟖
𝟑
= 𝟖 means 8 per 3 = 8+3 =11
𝟏𝟏
𝟏𝟏
𝟏𝟏
CA= 25,600 * 𝟖 * 𝟖 * 𝟖
𝟏𝟑,𝟑𝟏𝟎𝟎
=
= 66,550
𝟐
= Compound interest = 66,550 – 25,600 = 40,950.
4. The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:
Solution:
Amount of Rs.100 for 1 year
when compounded half−yearly=Rs.[(1 + 3/1002 )]=Rs.106.09
∴Effective rate=(106.09−100)% = 6.09%
CONCLUSION: Successfully completing this practical exercise, we gain a better understanding of
compound interest
PRACTICAL No.10
AIM: To practice and Solve permutation, Combination and Probability.
MATERIAL: Pen and paper.
1. How many ways the letter of the word “ROOM” be arranged?
SOLUTION: 1. In ROOM total 4 element = 4!
But O is repeated two times = 2!
4! 4∗3∗2∗1
=
2!
2∗1
24
= 2
= 12
So the word of ROOM be arranged 12 ways
2. If 3 particular boys are never to be selected, then in how many ways can 5 boys be chosen from a
total of 12 Boys?
SOLUTION: Total 12-3 = 9
Select
=5
n
= C
𝑛!
= 𝑟! (𝑛−𝑟)! = 1) n = number of objects, 2) r = Number of Selection
r
9
C
=
=
9∗8∗7∗6
= 126
4∗3∗2
5
3) A Coin is tossed 8 times, the probability of getting exactly 5 heads is
SOLUTION: = 𝟐𝟖
8
=
C
5
=
𝟖∗𝟕∗𝟔
𝟑∗𝟐
=
𝟕
𝟑𝟐
CONCLUSION: Successfully completing this practical exercise, we gain a better understanding of
Permutation, Combination and Probability.