Knowledges in Mathematics
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Elementary Linear Algebra with Applications
Elementary Linear Algebra with Applications
tulika goyal
Jawahar Navodaya Vidyalaya
Introduction to ode, Concept of Solutions, Applications
In mathematics, an ordinary differential equation (ODE) is a differential equation containing one or more functions of one independent variable and its derivatives. The term ordinary is used in contrast with the term partial differential equation which may be with respect to more than one independent variable
Amrendra Pratap
Solution of First Order Equations
First-Order Linear Differential Equations: A First order linear differential equation is an equation of the form y + P(x)y = Q(x). ... We can solve this equation in general but it is better to understand how to solve it than it is to just memorize the solution.
Maths Tut
Maths Tut Single Varible
Ankit Raushan
Ratio And Proportion- Basics
Ratio And Proportion- Basics What is Ratio? A ratio is a relationship between two numbers by division of the same kind. The ration of a to b is written as a: b = a / b In ratio a : b , we can say that a as the first term or antecedent and b, the second term or consequent. Example: The ratio 4 : 9 we can represent as 4 / 9 after this 4 is a antecedent and , consequent = 9 Rule of ration : In ratio multiplication or division of each and every term of a ratio by the same non- zero number does not affect the ratio. Different type of ratio problem is given in Quantitative Aptitude which is a very essential topic in banking exam. Under below given some more example for your better practice. Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks and formula are comes into action. What is Proportion? The idea of proportions is that two ratio are equal. If a : b = c : d, we write a : b : : c : d, Ex. 3 / 15 = 1 / 5 a and d called extremes, where as b and c called mean terms. Proportion of quantities the four quantities a, b, c, d said proportion then we can express it a : b = c : d Then a : b : : c : d <–> ( a x d ) = ( b x c ) product of means = product of extremes. If there is given three quantities like a, b, c of same kind then then we can say it proportion of continued. a : b = b : c the middle number b is called mean proportion. a and c are called extreme numbers. So, b2 = ac. ( middle number )2 = ( First number x Last number ). Quiz on ratio and proportion:- 1 If P : Q : R = 2 : 3 : 4 , Then P / Q : Q / R : R / P = ? A. 8: 9: 24 B. 7: 9: 24 C. 4: 6: 15 D. 8: 11: 24 E. None of these 2: If 2P = 3Q = 4R, Then P : Q : R = ? A. 2: 3: 5 B. 2: 3: 4 C. 3: 5: 6 D. 1: 2: 3 E. None of these 3: If P : Q = 2 : 3 , Q : R = 4 : 5 and R : S = 6 : 7 , then P : S = ? A. 18: 25 B. 17: 35 C. 16: 35 D. 8: 11 E. None of these 4:Rama distributes his pencil among his four friends Rakesh, Rahul, Ranjan, and Rohit in the ratio 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5 . What is the minimum number of pencils that the person should have? A. 66 B. 64 C. 72 D. 77 E. None of these 5: Two numbers are respectively 40% and 60% more than third number. Find the ration of two numbers ? A. 8: 7 B. 7: 9 C. 9: 11 D. 8: 13 E. None of these 6: Rs 1210 were divided among three person P, Q, R so that P : Q = 5 : 4 and Q : R = 9 : 10. Then R gets the amount. A. 450 B. 400 C. 500 D. 375 E. None of these 7: Share Rs.4200 among joy, sanjay and bijoy in the ration 2 : 4 : 6.Find the amount received by sanjay. A. 1200 B. 1300 C. 1400 D. 1500 E. None of these 8 :Find the mean proportional between given two number that is 64 and 49. A. 45 B. 52 C. 54 D. 56 E. None of these 9: What number has to be added to each term of 3 : 5 to make the ratio 5 : 6 . A. 7 B. 6 C. 8 D. 5 E. None of these 10:Rs. 385 were divided among P , Q , R in such a way that P had Rs 20 more than Q and R had Rs 15 more than P . How much was R’s share? A. 156 B. 145 C. 152 D. 150 E. None of these Answers with Explanation:- 1. P : Q : R = 2 : 3 : 4 . Let P = 2k, Q = 3k, R = 4k. Then, P / Q = 2k / 3k = 2 / 3 , Q / R = 3k / 4k = 3 / 4 R / P = 4k / 2k = 2 / 1. SO, P / Q : Q / R : R / P = 2 / 3 : 3 / 4 : 2 / 1 = 8 : 9 : 24. 2. Let 2P = 3Q = 4R = k , Then , P = k / 2, Q = k / 3 , R = k / 4. SO , P : Q : R = k / 2 : k / 3 : k / 4 = 6 : 4 : 3. 3. (C) 4. Rakesh : Rahul : Ranjan : Rohit = 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5 Step 1: At First we need to do is LCM of 2,3,4 and 5 is 60. Step 2: Then pencil are distributed in ratio among friends, Rakesh = ( 1 / 2 x 60 ) = 30. Rahul = ( 1 / 3 x 60 ) = 20. Ranjan = ( 1 / 4 x 60 ) = 15. Rohit = ( 1 / 5 x 60 ) = 12. Step 3: Total number of pencils are ( 30 x + 20 x + 15 x + 12 x) = 77 x. For minimum number of pencils x = 1 . The person should have atleast 77 pencils. 5.Step 1: Let the third number is A Then first number is 140% of A = 140 x A / 100 = 7A / 5 and second number is 160% of B = 160 x B / 100 = 8B /5. Step 2: now ratio of first and second number is 7A / 5 : 8B / 5 = 35A : 40B = 7 : 8. 6. P : Q = 5 : 4, Q : R = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9. So, P : Q : R = 5 : 4 : 40 /9 = 45 : 36 : 40 Sum of ratio terms is = ( 45 + 36 + 40 ) =121. R share of amount is Rs (1210 x 40 / 121) = Rs. 400. 7. Amount received by sanjay. 4 / 12 X 4200 = 1400= ( related ratio / sum of ratio ) x Total amount So, the Amount received by sanjay is 1400. 8. The mean proportion of two numbers is Root of 64 and 49 is √8 x √ 7 = 8 x 7 = 56. So, the mean proportional is 56. 9.Let the number to be added x , Then 3 + x / 5 + x = 5 / 6 6 ( 3 + x ) = 5 ( 5 + x ) x = ( 25 – 18 ) = 7 So , the number to be added is 7 . 10. Let Q gets Rs x. Then We can say P gets Rs (x + 20 ) and R gets Rs ( x + 35) . x + 20 + x + x + 35 = 385 3x = 330 x = 110 . R’s share = Rs ( 110 + 35 ) = Rs 145 .
Pratyaksh Mangal
Fourier Series
Fourier series basics
shiva singh
Formulae Quant
Formulae Quant
Vaibhav Kumar
Mathematics Tutor and Assignment help provider
I have more than 15 years of teaching experience.I provide a hand on approach to help students to understand the basics
Mansi Bhatia
GRAPHS OF COMMON FUNCTIONS
Sharing notes of GRAPHS OF COMMON FUNCTIONS in this Attachment. Watch the related video on this link - https://www.youtube.com/watch?v=39qGDqH6Gj0
Learn Everyone
Maths Tut Solution 2
Maths Tut Solution 2
Hypothesis Testing
Hypothesis testing is an important application of statistics. A thesis is something that has been proven to be true. A hypothesis is something that has not yet been proven to be true. It is a proposition that is empirically testable. In other words, a hypothesis (prediction) is your best guess about what you think will happen in the investigation based on some research or some experience you have had. The first step in a hypothesis test is to formalize it by specifying the null hypothesis. A null hypothesis (H0) is an assertion about the value of a population parameter. It is an assertion that we hold as true unless we have enough statistical evidence to conclude otherwise. The alternate hypothesis (H1) is the negation of the null hypothesis. Example – A vendor claims that his company fills any accepted order, on the average, in at most six working days. You suspect that the average is greater than six working days and want to test the claim. Here the claim is the null hypothesis and the suspicion is the alternate hypothesis. Thus, with µ denoting the average time to fill an order, Null hypothesis H0 : µ ≤ 6 days Alternate hypothesis H1 : µ > 6 days
Subham Sekhar