* Let A's age be x and B's age be y
* Then average of A and B is (x + y)/2 = 20
* So, x + y = 40
* If C replaces A, average is (y + c)/2 = 19
* So, y + c = 38
* If C replaces B, average is (x + c)/2 = 21
* So, x + c = 42
* Solving the three equations, we get:
x = 22, y = 18, c = 20
The ages are:
A = 22
B = 18
C = 20
The answer is option 1.
More companies in the process of recruitment, play more emphasis in the topic of numbers in numerical aptitude. Especially for AMCAT aspirants this is very much useful.
Some standard questions asked in cognizant aptitude tests recently has been sorted with answers. it will be beneficial to other company preparation aptitude also.
Algebra is used in many field in many different ways to solve equation problems, and in business algebra is also used or in our day to day life it is also used. ... Algebra is a way of keeping track of unknown values, which can be used in equations.
More companies in the process of recruitment, play more emphasis in the topic of numbers in numerical aptitude. Especially for AMCAT aspirants this is very much useful.
Some standard questions asked in cognizant aptitude tests recently has been sorted with answers. it will be beneficial to other company preparation aptitude also.
Algebra is used in many field in many different ways to solve equation problems, and in business algebra is also used or in our day to day life it is also used. ... Algebra is a way of keeping track of unknown values, which can be used in equations.
This presentation explains Algebra in Mathematics. It includes: Introduction, Solution to Puzzle, Definition of terms, Rules in Algebra, Collecting Like Terms, Similar Terms, Expanding the Brackets, Nested Brackets, Multiplication of Algebraic Expressions of a Single Variable, Division of One Expression by another, Addition and Subtraction of Algebraic Fractions, Multiplication and Division of Algebraic Fractions, Factorisation of Algebraic Expression, Useful Products of Two Simple Factors, Examples, Trinomial Expression, Quadratic Expression as the Product of Two Simple Factors, Factorisation of Quadratic Expression ax2 + bx +c When a = 1, Factorisation of Quadratic Expression ax2 + bx +c When a ≠ 1 and Test for Simple Factors.
Vedic maths is the ancient India secret before the calculator to fast calucation with short cuts and tricks for fast easy accurate answers. GRE exam and other competative exam test students on theability to solve the complex numercials problems with efficiently and within time limits. Vedic maths helps with tricks just for same.
GREKing helping students in basic concepts.
GREking the best GRE preparation classes in Mumbai. (www.greking.com)
A Summary of Concepts Needed to be Successful in Mathematics
The following sheets list the key concepts that are taught in the specified math course. The sheets
present concepts in the order they are taught and give examples of their use.
WHY THESE SHEETS ARE USEFUL –
• To help refresh your memory on old math skills you may have forgotten.
• To prepare for math placement test.
• To help you decide which math course is best for you.
This presentation explains Algebra in Mathematics. It includes: Introduction, Solution to Puzzle, Definition of terms, Rules in Algebra, Collecting Like Terms, Similar Terms, Expanding the Brackets, Nested Brackets, Multiplication of Algebraic Expressions of a Single Variable, Division of One Expression by another, Addition and Subtraction of Algebraic Fractions, Multiplication and Division of Algebraic Fractions, Factorisation of Algebraic Expression, Useful Products of Two Simple Factors, Examples, Trinomial Expression, Quadratic Expression as the Product of Two Simple Factors, Factorisation of Quadratic Expression ax2 + bx +c When a = 1, Factorisation of Quadratic Expression ax2 + bx +c When a ≠ 1 and Test for Simple Factors.
Vedic maths is the ancient India secret before the calculator to fast calucation with short cuts and tricks for fast easy accurate answers. GRE exam and other competative exam test students on theability to solve the complex numercials problems with efficiently and within time limits. Vedic maths helps with tricks just for same.
GREKing helping students in basic concepts.
GREking the best GRE preparation classes in Mumbai. (www.greking.com)
A Summary of Concepts Needed to be Successful in Mathematics
The following sheets list the key concepts that are taught in the specified math course. The sheets
present concepts in the order they are taught and give examples of their use.
WHY THESE SHEETS ARE USEFUL –
• To help refresh your memory on old math skills you may have forgotten.
• To prepare for math placement test.
• To help you decide which math course is best for you.
SINCE I CREATED IN 2010 VERSION AND SAVED IN OLD VERSION MOST OF FEATURES DISABLED. CONTENT NOT AFFECTED.
TCS written test procedure for 2014 batch.& Eligibility 60% with 2 back logs.
------------------------------------------------------------
80 minutes-30 questions,(aptitude),
10 minutes- 1 question ,(verbal)
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90 minutes-31 questions(total)
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next- round technical oral test only. Requirements for Mech,ece,eee branches (c- programme,data structures,data base concept.)
For cse ,it branches(java, operating systems, students may be asked to explain their project,)
You are leading a project team of 15 members. As you have found that the team members are not irregular in submission of weekly time sheets , you are required to stress the need to submit without fail. Using the following phrases, write an email with a minimum of 70 words and a maximum of 100 words to your team members informing the same.
can be accessed online – lead to loss of pay – every week – do not default – used to bill client – actual working hours – by friday – failure to adhere – time sheet filling application
Using the following phrases, write an email with a minimum of 70 words to a company requesting them to sponsor your college cultural festival.
pdf file – reputed institute - 10 days - 200 college – sparkling performance – extravaganza – sponsor the event – request appointment – brochure attached
This presentation uses the technology of Microsoft Multiple Mouse Mischief software. If you need assistance, visit microsoft site for multiple mouse support. Probability aptitude questions level 2
this presentation covers the topic percentage, profit and loss aptitude questions in level 1. (basic) categorywise the techniques are supported by suitable examples
This is a collection of fifty questions from important topics in Aptitude where students should pay more attention and practice. Questions taken from various net sources. Some of the answers were edited. This presentation could be run only in office 2010 or latest.
Problems on Trains is the Aptitude topic which most of the companies prefer to ask. Here students could find some examples on the different categories of problems on trains.
Time & Distance is a broader topic in aptitude. Here the moving object could be train or person or boats etc.,
Students could find useful techniques to solve time & distance aptitude problems.
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
LF Energy Webinar: Electrical Grid Modelling and Simulation Through PowSyBl -...DanBrown980551
Do you want to learn how to model and simulate an electrical network from scratch in under an hour?
Then welcome to this PowSyBl workshop, hosted by Rte, the French Transmission System Operator (TSO)!
During the webinar, you will discover the PowSyBl ecosystem as well as handle and study an electrical network through an interactive Python notebook.
PowSyBl is an open source project hosted by LF Energy, which offers a comprehensive set of features for electrical grid modelling and simulation. Among other advanced features, PowSyBl provides:
- A fully editable and extendable library for grid component modelling;
- Visualization tools to display your network;
- Grid simulation tools, such as power flows, security analyses (with or without remedial actions) and sensitivity analyses;
The framework is mostly written in Java, with a Python binding so that Python developers can access PowSyBl functionalities as well.
What you will learn during the webinar:
- For beginners: discover PowSyBl's functionalities through a quick general presentation and the notebook, without needing any expert coding skills;
- For advanced developers: master the skills to efficiently apply PowSyBl functionalities to your real-world scenarios.
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
JMeter webinar - integration with InfluxDB and GrafanaRTTS
Watch this recorded webinar about real-time monitoring of application performance. See how to integrate Apache JMeter, the open-source leader in performance testing, with InfluxDB, the open-source time-series database, and Grafana, the open-source analytics and visualization application.
In this webinar, we will review the benefits of leveraging InfluxDB and Grafana when executing load tests and demonstrate how these tools are used to visualize performance metrics.
Length: 30 minutes
Session Overview
-------------------------------------------
During this webinar, we will cover the following topics while demonstrating the integrations of JMeter, InfluxDB and Grafana:
- What out-of-the-box solutions are available for real-time monitoring JMeter tests?
- What are the benefits of integrating InfluxDB and Grafana into the load testing stack?
- Which features are provided by Grafana?
- Demonstration of InfluxDB and Grafana using a practice web application
To view the webinar recording, go to:
https://www.rttsweb.com/jmeter-integration-webinar
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Key Trends Shaping the Future of Infrastructure.pdfCheryl Hung
Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
Cheryl Hung, ochery.com
Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
5. Numbers
• The citizens of planet nigiet are 6 fingered and
have thus developed their decimal system in base
6.
• A certain street in nigiet contains 1000 (in base 6)
buildings numbered 1 to 1000.
• How many 3s are used in numbering these
buildings? a) 256 b) 54 c) 192 d) 108
• (N-1)*(BASE)(N-2)
N=NUMBER OF LAST DIGIT=4(1000 contains 4 digits)
• base=6
• (4-1)*62 = 3*36 = 108.
6. Numbers
• The citizens of planet nigiet are 8 fingered and
have thus developed their decimal system in base
8. A certain street in nigiet contains 1000 (in base
8) buildings numbered 1 to 1000. How many 3s
are used in numbering these buildings? a) 54 b)
64 c) 265 d) 192
• (N-1)*(BASE)(N-2)
• N=NUMBER OF LAST DIGIT=4(1000 contains 4
digits)
• base=8
• (4-1)*8(4-2) =192
7. NUMBER
• If a and b are odd numbers, then which of the
following is even ?
• A. a + b
• B. a + b + 1
• C. ab
• D. ab + 2
• E. None of these
•
• always add two odd number it gives even number
8. FRACTIONS
• One-third of the marks obtained by Naveen in maths
is the same as the marks he obtained in English. If the
total marks obtained by him in both the subjects is
160, the marks he obtained in English is:
• A) 40 B) 30 C) 50 D) None of the options
• Let the marks obtained in English be x.
• Marks obtained in Maths = 3x.
• Total marks obtained 4x = 160.
• Hence marks obtained in English x = 40.
• Answer : a) 40
10. Decimal Number
• In which of the system, decimal number 184 is
equal to 1234?
• 184= 1*x^3 + 2*x^2 + 3*x + 4
• X3 + 2x2 + 3x – 180 = 0.
• Has no integral solution.
• Hence answer choose none of these.
12. DECIMALS & FRACTIONS
• Which pair of rational numbers lie between
1/5 and 2/5 -
a. 262/1000, 275/1000 b. 362/1000, 562/1000
c. 451/1000, 552/1000 d. 121/1000,131/1000
• BETWEEN 0.2 AND 0.4
• ANS A.
16. ALGEBRA
• Three consecutive whole numbers are such
that the square of the middle number is
greater than the product of the other two by
1. Find the middle number.
• a. 6 b. 18 c. 12 d. All of these
• Let the numbers be x–1, x, x + 1.
• Since x2 = (x–1) x (x+1) + 1 it is true for any
values of x.
17. A.M & G.M
• The arithmetic mean of 2 numbers is 34 and
their geometric mean is 16. One of the
numbers will be
• a. 4 b. 16 c. 18 d. 12
• a + b = 68 and ab = 256.
• Hence choice a is correct.
18. DIVISIBILITY
• What least number must be subtracted from
9400 to get a number exactly divisible by 65?
• a. 40 b. 20 c. 80 d. none of these
• 65 = 5 X 13
• Last digit should be zero or 5.
• 9400–40=9360 divisible by 13 & 5
• 9400–20=9380, 9400–80=9320. none of these
divisible by 13.
• Hence answer is A.
20. Divisiblity
• 12 divides, ab313ab (in decimal notation, where a,b
are digits>0, the smallest value of a+b is a)7 b)6 c)2 d) 4
• If a number is divisible by 12 then it should be divisible
by 4&3
• for divisible by 4 last [2 digit]no's should be divisible by
4
• Also sum of digits should be divisible by 3.
• Hence 2(a+b)+1 should be 3k and ab = 4m.
• That is a + b = (3k-1)/2 . from the given choices only
first choice is possible k = 5 and smallest should be 7.
21. Divisibility
• If a number 774958A96B is to be divisible by 8 and 9,
the values of A and B, respectively, will be:
• Using the divisibility rules,
• For 8, the last three digits have to divisible by 8,
therefore the number 774958A96B is divisible by 8 if
96B is divisible by 8.
• 96B is divisible by 8 if it is 960 or 968 thus B is either 0
or 8.
• For 9, the sum of the number has to be divisible by 9,
therefore (55 + A + B) is divisible by 9 if (A + B) is 8.
• Now either of A or B could be 8, but the other has to
be zero.
22. Divisibility
• 311311311311311311 is divisible by:
a)3 and 11. b)11 but not 3. c)3 but not 11.
d)none of the above.
• a.3 and 11
• the sum of digits is 30 which is divisible by 3
• the differences of odd place digits - even place
is divisible by 11
23. DIVISIBILITY
• What least number be added to 5200 to get a
number exactly divisible by 180?
(1) 160 (2) 60 (3) 20 (4) 180
• 180 = 5 X 4 X 9.
• 5200 + 160 = 5360 NOT DIVISIBLE BY 9.
• 5200 + 60 = 5260 NOT DIVISIBLE BY 9
• 5200 +20 = 5220 DIVISIBLE BY 9
• 5200 + 180 NOT DIVISIBLE BY 9.
24. DIVISIBILITY
• A number when divided by 32 leaves the
remainder 29. This number when divided by 8
will leave the remainder (1) 3 (2) 5 (3) 7 (4) 29
• Let the number be 32k + 29.
• 32k + 29 = 32k + 24 + 5
• 8(4k+3) + 5.
• The remainder is 5.
25. Powers
• The unit digit in the expression
• ( 36234 x 33512 x 39180) – (5429 x 25123 x 31512) will be
• (a) 2 (b) 6 (c) 8 (d) 4
• Power of 6 end in 6
• Fourth power of 3 end in 1
• Even power of 9 end in 1
• Odd power of 4 end in 4
• Power of 5 end in 5
• Power of 1 end in 1
• Ans: 6– 4 = 2.
26. POWERS
• If x increases linearly, how will a-x behave
(a>1) ?
• a. Increase linearly
• b. Decrease linearly
• c. Increase exponentially
• d. Decrease exponentially
• Since x increases linearly, –x decreases linearly
and a–x decreases exponentially. Ans d.
27. POWERS
• What is x if
xx
3
3
1
9
81
1
3
27
1 100
Rewrite each term as a power of three, then combine exponents:
12932
4
100
3
333
3
1
3
3
1 xxx
Since the bases are the same, the exponents must be equal:
1293 xx
which has the solution x = –94
28. LOGARITHM
• What is the value of the following expression:
2 log105 + log104 ?
a. 2 b. 2.5 c. 3 d. None of these
• Log10 25*4 = log10100 = 2.
29. LCM
• Find the greatest number of five digits, which is
exactly divisible by 7, 10, 15, 21 and 28.
• The number should be exactly divisible by 15 (3,
5), 21 (3, 7), 28 (4, 7).
• Hence, it is enough to check the divisibility for 3,
4, 5 and 7.
• Lcm of 3,4,5 and 7 is 420.
• 105/420 = 238 is quotient
• 238* 420 = 99960 is the only number which
satisfies the given condition.
30. HCF
• The ratio of two numbers is 3 : 4 and their H.C.F.
is 4.
• Their L.C.M. is: A. 12 B. 16 C. 24 D. 48
• Let the numbers be 3x and 4x.
• Then, their H.C.F. = x.
• So, x = 4(given)
• the numbers are 12 and 16.
• LCM of 12,16 is 48.
• So, ans is OPT (D)
32. HCF & LCM
• Four different electronic devices make a beep after every 30
minutes, 1 hour, 3/2 hour and 1 hour 45minutes respectively.
• All the devices beeped together at 12 noon.
• They will again beep together at:
• i) 12 Midnight ii) 3 a.m iii) 6 a.m. iv) 9 a.m
Interval after which the devices will beep together
= (L.C.M. of 30, 60, 90, 105) min.
= 1260 min.
= 21 hrs.
So, the devices will again beep together 21 hrs. after 12 noon i.e., at 9 a.m.
34. Factorization
• The prime factorization of integer N is A*A*B*C where A, B
and C are all distinct prime integers. How many factors
does N have? a)12 b)24 c)4 d)6
• n is A*A*B*C =A^2*B*C
no. of factors=(2+1)(1+1)(1+1)
=12
• A HAS POWER OF 2,B HAS POWER OF 1,C HAS POWER 1
SO PRIME FACTORIZATION CAN BE CALCULATED AS:
IF A^P+B^Q+C^R,THEN
PRIME FACTORIZATION IS (P+1)*(Q+1)*(R+1)
SO IN THIS CASE P=2,Q=1,R=1
SO (2+1)(1+1)(1+1)=12 FACTORS
35. RACES & GAMES
• An athlete decides to run the same distance
in 1/4th less time that she usually took. By
how percent will she have to increase her
average speed
36.
37. RACES & GAMES
• In August, a cricket team that played 120 matches won 20% of the
games it played. After a continuous winning streak, this team
raised its average to 52%. How many matches did the team win to
attain this average?
• A) 40 B) 52 C) 68 D) 80
• (24+x) / (120+x) = 0.52, where x is the number of games played and
24 is 20% of 120
• 24 + x = 0.52(120 + x)
24+x = 62.4 + 0.52x
0.48x = 62.4-24
0.48x=38.4
x=38.4/0.48
x=80
42. PERCENTAGE
• If 7% of A is 42, then find 27% of A.
• A) 162 B) 62 C) 600 D) 126
• Ans: 162
• Another method
• 7 is prime number and the last digit of given
number is 2. Hence 27% of A also should have
last digit is 2 and so C and D are not possible.
• B is also not possible because it is not divisible by
9(ie 27).
45. PERCENTAGE
• 4% is equal to one out of every:
• A) 12 B) 20 C) 40 D) 25
• Answer
• 25
46. PERCENTAGE
• The difference between a discount of 35% and
two successive discounts of 20% and 20% on a
certain bill was Rs. 22. Find the amount of the
bill.
(1) Rs. 1100 (2) Rs. 200 (3) Rs. 2200 (4) Data
inadequate
• Successive 20% discount means 36% discount.
• The difference between 35% and 36% is 1%.
• 1% = 22 means the answer is 2200.
47. Percentage
• An athlete decides to run the same distance in 1/4th less time that she
usually took.
• By how much percentage will she have to increase her average speed?
• let original speed be s1 and time be t1
• then s1=d/t1 ---eqn 1
• new speed be s2 and time given is 3t1/4
• therefore s2=d/(3t1/4) -----eqn 2
• dividing eqn 2 by eqn 1
• s2=4s1/3
• increased speed = 4s1/3-s1
• =1s1/3
• percent increase=[(1s1/3)/s1]*100
• =33.33%
49. NEGATIVE POWERS
• What is the value of 0-10?
• i) 0 ii) 1 iii) -10 iv) None of these
• 0.
50. • 2525 is divided by 24, the remainder is
• 1) 23 2) 22 3) 1 4) 2
• 2525 = (24 + 1)25 = 24K + 1.
• CHOICE 3.
51. AVERAGE
• The average age of a group of nine men today is
the same as it was six years ago, with one of the
men having been replaced by another much
younger man. The difference in ages of the man
being replaced and the man replacing him is how
many years?
(a) 53 (b) 54 (c) 55 (d) 56
• Six years ago the average is x and now also x. the
number is same. Hence
• No of persons * time elapsed = 9 * 6 = 54.
52. AGE
• The average age of A and B is 20 years. If C
were to replace A, the average would be 19
and if C were to replace B, the average would
be 21. What are the ages of A, B and C?
(1) 22, 18, 20 (2) 18, 22, 20
(3) 22, 20, 18 (4) 18, 20, 22
53. RATIO
• If a:b=3:4 and b:c=2:7, then what is the value
of a:b:c?
• A) 3:4:14 B) 3:4:7 C) 6:4:7 D) 6:8:12
• A) 3:4:14
54. RATIO
• The prices of a refrigerator and a cell phone
are in the ratio 3:2. If the refrigerator costs
Rs. 6000 more than the cell phone, what is
• the price of the cell phone?
• A) Rs. 1200 B) Rs. 12000 C) Rs. 8000 D) Rs.
9000
• Let the price of refrigerator be 3x and
cellphone be 2x.
• X =6000 price of cellphone = 12000.
56. PROFIT / LOSS
• A man buys a watch for Rs.135. His overhead
expenses were Rs.50. If he sells the watch for
Rs.254, what is his approximate profit/loss in
percent?
• A) 39% profit B) 37% profit C) 37% loss D) 39%
loss
• B) 37% profit
• Wild guess 185 divisible by 37. and S.P > C.P.
59. PROFIT / LOSS
• After giving a discount of Rs.45 the shopkeeper
still gets a profit of 20%, if the cost price is
Rs.180. Find the mark up percent.
• A) 0.4 B) 0.55 C) 0.45 D) 0.48
• LET THE S.P = X
• C.P = 180. S.P = 180 + 36 = 216.
• M.P = S.P + 45 = 216+45 = 261.
• M.P% = 0.48
PROFIT
COST PRICE
DISCOUNT
MARKED PRICESELLING PRICE
61. PROFIT / LOSS
• In a shop 80% of the articles are sold at a
profit of 10% and the remaining at a loss of
40%.what is the overall profit/loss?
a.10% profit b.10% loss c.15% profit d. no
profit, no loss
• Option d no profit,no loss
• 80*1.1+20*0.6/100=1
62. PROFIT / LOSS
• If a pen is being sold at 4% profit instead of 4%
loss the actual profit is Rs 16. What is the
actual cost price of the pen ?
• Let x be the CP.
• (104/100)x - (96/100)x = 16
• Solving we get x = Rs.200.
64. PROFIT / LOSS
• John buys a cycle for 31 dollars and given a cheque of
amount 35 dollars. Shop Keeper exchanged the cheque
with his neighbor and gave change to John. After 2 days, it
is known that cheque is bounced. Shop keeper paid the
amount to his neighbor. The cost price of cycle is 19 dollars.
What is the profit/loss for shop keeper? a) loss 23 b) gain
23 c) gain 54 d) Loss 54
• CP of cycle = 19$
• SP of cycle = 31$
• Profit = 31$-19$ = 12$
• Again, shopkeeper gave 35$ to neighbour.
• Loss = 35$
• So, net loss = 35$-12$ = 23$
66. • A man can do a job in 25 days. He worked at it
for 15 days and then gave up. After that, B
completed the work in 10 days. Together, both
can finish the job in ________days.
• a) 12 ½days b) 25 days c) 6 days d) 12 days
67. TIME & WORK
• If 50 people finish a job in 10 days, how long
will it take 20 people to finish the same job?
• A) 25 B) 50 C) 4 D) None of the options
• Answer : A) 25.
69. TIME – WORK
• Ten men can paint a room in 3 hours. How
many hours would 5 men take to paint the
room, if they work at the same rate?
• A) 6 hours B) 1.5 hours C) 10 hours D) None of
the options
• A) 6 hours.
70. PIPES & CISTERNS
• Pipe A takes 16 min to fill a tank. Pipes B and C, whose cross-
sectional circumferences are in the ratio 2:3, fill another tank
twice as big as the first. If A has a cross-sectional circumference
that is one-third of C, how long will it take for B and C to fill the
second tank? (Assume the rate at which water flows through a
unit cross-sectional area is same for all the three pipes.)
• a. 66/13 b. 40/13 c. 16/13 d. 32/13
• If A has a cross – sectional circumference is x and area will be
proportional to x2 cross section of C is 3x and area will be
proportional to 9x2 .Hence cross sectional circumference of B will
be (2/3)*3x = 2x so area is proportional to 4x2.
• Pipe A fills in 16 min. Pipe B and C sum of area is proportional to
13x2. will fill the same tank in 16/13 min.
• The second tank is twice big in size require 2 * 16 / 13 = 32/13 min
71. AGES
• 20) If the ages of a husband and a wife are in
the ratio 5:4 and the sum of their ages is 45,
what is the difference between their ages?
• A) 4 years B) 5 years C) 3 years D) 6 years
• Let the ages be 5x and 4x.
• Sum 9x = 45 x = 5 years.
• Difference between their ages is 5 years.
72. CLOCK
• Samarth asked his friend: .What will be the
time after 2 hours 30 mins, if the time was
3:45 p.m. half an hour ago?" What should
• be his friend’s reply?
• A) 6:30 p.m. B) 6 p.m. C) 6:45 p.m. D) None of
the options
• Time now = 3:45 + 0:30 = 4:15 min
• After 2:30 time will be 6:45 min.
73. • Four bells ring at intervals of 10 min, 12 min,
15 min, and 20 min, respectively. If they ring
together at 8 a.m., find out the interval of
time after which they will ring together again.
• a) 9 a.m. b) 10 a.m. c) 11 a.m. d) 1 p.m.
74. PERMUTATION
• The letters of the word WOMAN are
written in all possible orders and these
words are written out as in a dictionary
,then the rank of the word 'WOMAN' is
• a. 117b. 120 c. 118 d. 119
• 5!=120
120/5=24
• Since W is the last in rank WOMAN is
in the last 24.
• O is in the last rank of the remaining 4
letters. 24/4 = 6
• Therefore WOMAN is in the last 6 in
rank.
• M is between A and N so WOMAN is in
rank 117 or 118.
• A is before N so WOMAN's rank is 117.
There are 120 possible
arrangements of the letters in
WOMAN. Since WOMAN is near the
end, I'll list out the ones after
WOMAN:
WONMA
WONAM
WOMNA
There are only three, so these
account for the 118th, 119th, 120th
words, and it places WOMAN
117th.
75. PROBABILITY
• What is the probability of getting the sum 5
in two throws of the dice?
• a. 1/12
• b. 1/5
• c. 1/9
• d. None of these
76. PROBABILITY
• The probability of A solving a problem is 2/3.
The probability of B solving a problem is ¾. If
the same problem is given to both of them,
what is the probability of neither of them
solving it?
(a) 3/4 – 2/3 (b) (1- ¾ ) x 2/3 (c) 1/3 + 1/4
(d) 1/3 x 1/4
•
77. CUBES
• A 4 inch cube is taken and it is painted in
green colour and then it is cut into 1 inch
cubes. How many cubes are 2 faces painted?
(a) 12 (b) 24 (c) 36 (d) 48
78. TRAINS
• Two trains start from stations Chennai and
Villupuram spaced 150 km apart at the same time
and speed.
• As the trains start, a bird flies from one train
towards the other and on reaching the second
train; it flies back to the 1st train.
• This is repeated till the trains collide. If the speed
of the train is 75kmph and that of the bird is
100kmph.
• How much did the bird travel till collision?
• (a) 100km (b) 120km (c) 220km (d) 175km
79. • distance between trains is 150 km.
one train travels 75 km / hr.
other train travels 75 km / hr.
they will meet when the total distance
traveled by both of them is equal to
150 km.
rate * time = distance.
We know that the time that each train
travels will be the same.
We let h = the time each train travels.
train 1 travels 75 km / hr * h hours.
train 2 travels 75km / hr * h hours.
the total distance they travel is 150
km.
75 * h + 75 * h = 150
h = 1
They will meet in 1 hour.
the bird was flying at 100 km/h.
in the same 1 hour the bird flew a
total of 100 km.
bird flies 50 km in one direction, then
travels back 50 km when the trains
meet.
81. • The difference between the simple interest
and compound interest obtained on a
principal amount at 5% per annum after two
years is Rs.35. What is the principal amount?
• a) Rs.15,000 b) Rs.10,000 c) Rs.14,000 d)
Rs.13,000
82. ATTENTION TO DETAIL
• Follow the directions given below to answer the questions that follow.
Your answer for each question below would be:
• A. If all the THREE items given in the question are exactly ALIKE
• B. If only the FIRST and SECOND items are exactly ALIKE.
• C. If only the FIRST and THIRD items are exactly ALIKE.
• D. If none of the above
•
• 49 A0067DF667016 A0067DF667016 A0067DF667106
• (a) A (b) B (c) C (d) D
•
• 50 Janmasthami Janmasthhami Jamnasthami
• (a) A (b) B (c) C (d) D
•
83. • If * stands for +, + stands for -, and - stands for /, then 27/5*8+11-7=?
• (a) 141.4 (b)38.4 (c) 16.4 (d)36.1
•
• If * stands for -, / stands for +, + stands for / and – stands for *, then which
of the following is TRUE?
• A)8/12*4+20-8=18.4 B)12*8/4+20-8= -5.6 C)8*4/12+8-8= -12.2
D)8*8/4+20-8= -4
• (a) A (b) B (c) C (d) D
•
• Identify the correct match 345611242627
• A) 345612414627 B) 345611426227 C)345611624227 D) 345616242227
• (a) A (b) B (c) C (d) D