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Lessons

  • Age Related Problems
  • HCF and LCM
  • Linear Equations
  • Percentage Problems
  • Pipes & Cisterns
  • Profit and Loss
  • Quadratic Equations
  • Surds and Indices

HCF and LCM

Basic Concepts

Prime Factors

Prime factors are all the prime numbers that can be multiplied to obtain another number.
Example factors of 24 = 2*2*2*3 or 45 = 3*3*5

Factors

Factors are all the numbers that can be multiplied to obtain another number.

Example factors of 24 = 8*3 or 6*4 or 1*24 or 2*12

Factors of 45 = 1*45 or 3*15 or 5*9

Common Factors

For two numbers we can compare and find common factors between two numbers.

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 45 = 1, 3, 5, 9

Common factors between 24 and 45 is 3

Highest Common Factor

Highest factor obtained from common factors is Highest Common factor

Example: Find HCF of 24 and 90

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 90 = 1, 3, 5, 6, 9, 10, 15, 18, 30, 90

HCF is 6

Multiples

Multiples are the numbers that are obtained by multiplying a number with integers like 1, 2, 3, ….

Multiples of 4: 4, 8, 12, 16, ….

Multiples of 6: 6, 12, 18, 24, ….

How to Solve HCF and LCM Questions

LCM and HCF

Find Common Prime factors of the numbers. Product of the Prime factors is HCF. Next find all the common prime factors of two or more of the given numbers till no other factor exist. Product of all the common factors and other factors is LCM

Example: Find HCF and LCM of 18, 24 and 48

HCF

HCF is 2*3 = 6

LCM

LCM is 2*3*2*2*3*2 = 144

If you know factors or values higher than prime number, you can apply that also.

HCF – Long Division Method

HCF is also found by dividing larger number with smaller successively till you get 0 as remainder
Find HCF between 18 and 48.

LCM of Fractions

LCM Of fractions or decimals is given by the formula
(LCM of numerators)/(HCF of denominators)


Product of two numbers
Product of two numbers is also given by formula
P * Q = LCM (of P Q) * HCF (of P Q)
Greatest number that will divide numbers leaving same remainder
Find the difference of combination of each number and then find it HCF

Questions

Question 1: Find the HCF and LCM of 33, 44, 55
Answer
LCM is given below as 11*3*4*5 = 660

HCF is given below as 11

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Question 2: Find a number N that will divide 1305, 4665, and 6905 leaving the same remainder in each case. Find the number N.
Answer: Number N is HCF of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
=> H.C.F. of 3360, 2240 and 5600.
Number N = HCF of above numbers = 1120

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Question 3: Find the greatest number which is divisible by 15, 25, 40 and 75. Also find a smallest 4 digit number divisible by the same numbers.
Answer: Find LCM of 15, 25, 40 and 75 is

 

LCM = 5*3*5*8 = 600
Smallest 4-digit number is 1000. Smallest multiple of 600 to get 4-digit number is 1200
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Question 4: The product of two numbers is given as 4107. Find the LCM of the two numbers if their HFC is 37.
Answer 2: Using formula – Product of numbers = LCM * HCF
4107 = LCM * 37 => LCM = 4107/37 = 111
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Question 5: LCM of three numbers in the ratio of 3 : 4 : 5 is 2400. Find their HCF.
Answer: Let x be the common multiple to get the numbers 3x, 4x, 5x. Then x with the HCF.

The LCM of the three numbers is 3*4*5*x = 2400 => x = 40. The numbers are 120, 160 and 200. Thus, HCF is 40.
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Question 6: Find HCF (GCD) of 1.05, 0.35 and 0.93
Answer: To find HCF consider the numbers are 105, 35 and 91 without two decimals.

HCF is 7
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Question 7: The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 12 and 18 is _____?
Answer: We can divide my any of the numbers from 6, 9, 12, 18. The number must leave a remainder of 4 and quotient should be multiple of 7.

LCM of 6, 9, 12,18 = 3*3*2*2 = 36
Dividend = Divisor * Quotient + Remainder

Quotient = 36, Quotient = 7 and Remainder = 4
Dividend = 36*7 + 4 = 254

Question 8: Reduce 5733/23667 to its lowest terms.
Answer: Find HCF of 5733/23667

Linear Equations

Basic Concept

Linear equations have equation where the power of the variable use is 1.

If the power of the variable is not 1, the equation is NOT Linear. If highest power is 2, the equation is quadratic, if highest power is 3, the equation is cubic, if highest power is 4, the equation is quartic, and if highest power is 5, the equation is quantic.

With single variable the linear equation is represented in two dimensions as y = mx + b. The equation is solved by collecting like terms and then simplifying the equation.

Why Linear Equations:

  • Linear equations are good estimations for specific assumptions and then finding unknown values of the variable.
  • Find profits if the current trend of sales continues
  • Trend estimation
  • Finding restrictions – Domain and range
  • Estimating the growth or decay rate

Definitions:

Input: x is considered as input in y = mx+b

Output: Resultant value or output is defined by y in y = mx+b. y is also written as y = f(x)

Slope: Slope is the ratio between any two distinct points on a line.

m>0 indicates the line is increasing and m < 0 m=indicates the line is decreasing.

m = 0 indicates a horizontal line and when m = ∞ indicates a vertical line

Starting Point or Y-Intercept: When input is 0 what is the output value. When x is 0, y = b is the y-intercept. When we know what our output will be, the linear equation. The goal of solving a linear equation is to find the value of the variable that will make the statement (equation) true.

How to solve Linear Equation Questions

Like Terms

Always recognize like terms and group the like terms. In the term 2x2 + 3x + 4x2 + 5,  2x2 and 4x2 are like terms

Increasing and Decreasing Slope

Must be able to see the steepness of the slope – steeper means higher slope and right handed means positive slope and left handed means negative slope

Two Lines and Equations

If both the lines have different slope, the lines will sure meet at one point.

If both lines are parallel, their slope but will have different y-intercept.

If both lines are overlapping, their slopes and y-intercepts will be same.

If both lines are perpendicular to each other then m1 * m2 = -1 i.e. the second line slope will be negative reciprocal of first line slope.

Solving simultaneous equation is done by two methods

  1. Substitution
  2. Elimination

Example – Find the point of intersection of the two lines given below using substitution method.

2x + y = 4

3x + y = 5

Step 1: In first equation isolate y => y = -2x + 4

Step 2: Substitute y in second equation => 3x –2x + 4 = 5 => x = 1

Step 3: Substitute x in first equation => y = -2 * 1 + 4 = 2

Point of intersection of two lines is (1,2)

Example – Find the point of intersection of the two lines given below using elimination method.

2x + y = 4

3x + y = 5

Step 1: In first and second equation find coefficient of x or y that can be made equal easiliy. In our equations y in both equations have same coefficient.

Step 2: Subtract or add the two equations if signs of coefficients are same – subtract and if the signs of the two coefficient is different add both the equations

2x + y = 4

3x + y = 5

– x + 0 = -1 => x = 1

Step 3: Substitute x in first equation => y = -2 * 1  + 4 = 2

Point of intersection of two lines is (1,2)

Questions:

Question 1: What is the slope of a line that goes through the points (5,3) and (7, 9)?

Answer

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Question 2: For example, a parking meter contained 80 coins made up of 1 Re and 5 Re. The total value of the coins was Rs 520. How many Re 1 coins did it contain?

Answer: Let the number of Re 1 coins be ‘a’ and number of Rs 5 coins be ‘b’.

Then                                      a + b = 120          —— (eq. 1)

Total value of coins         a*1 + b*5 = 520 —— (eq. 2)

We need to solve the two equations simultaneously

Rs 5 coins are 100 and Re 1 coins are 20

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Question 3:  Solve and find value of x

7x – 5(3 – x) = 4x + 9

Answer: Group all similar terms together and solve the answer.

7x + 5x – 4x = 15 + 9 => 8x = 24 => x = 24/8 = 3

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Question 4: Solve and find value of y

Answer: Factorize y2 – 16 = (y – 4)(y + 4)

3y = -(4+y) => 4y = -4 => y = -1

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Question 5: Solve and find value of x

Answer: Solve by adding 

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Percentage Problems

Basic Concepts:

Percent means out of 100 that is Per Cent. If Peter scores 12 out of 20, how much did he score out 100. This value out of 100 is percentage.

Why Percentage?

  • It can be used to compare each data out of 100
  • Create rate of tax etc. as percentage – 18% of sales
  • Mark fraction as percentage equivalent
  • Mark percentage profit – amount of extra money over cost price out of 100 value.
  • Conceptualize Increase or decrease in value
  • In interest calculation, it represents growth or decay with exponential function
  • In population changes, it represents growth or decay with exponential function

How to Solve Percentage Problems

Calculating percentage of a value

If you are required to find 20% of the Rs 20,000 investment, then mathematically you will apply

Practically or mentally you will do this

Converting Fraction as percentage

If you are required to convert 0.2 as percentage then you will do this

Comparing Data

If you get the following data

Physics Quiz 1 – 25 out of 30; Chemistry Quiz 1 – 10 out of 15 and Biology – 18 out of 24

To compare the performance of a student on all three subjects, it would be prudent to find how much student received marks out of 100.

   

Percentage Discount, Percentage Profit and Percentage Loss

In business following terms are used – Cost Price, Marked Price, and Sales Price

Increase in Cost Price is used to establish Marked Price. Discount on the marked price establishes Sales Price. Profit and loss are calculated on Sales price – Cost Price

Lets the figures CP = Rs 200, MP = 300, Discount = 15%. Find actual discount amount, SP and profit percentage.

   

Profit Percentage is

 

Growth and Decay

Growth and Decay are defined as follows:

Growth: M = Mo(1+r)t and Decay: M = Mo(1-r)t

Here M is final amount, Mo is the initial amount, r is the rate of increase or decrease and t is the time period.

 

Problems

Problem 1: Evaluate 30% of 80 + 60% of 270

Answer: 30% of 80 + 60% of 270 = 0.3 * 80 + 0.6 * 270 = 24 + 162 = 186

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Problem 2: What percentage is 60 out of 80?

Answer: 60 out of 80 =

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Problem 3: Anil’s marks is 20% more than Amit in English. Amit scored 30 out of 55. What is Anil’s mark?

Answer: Amit scored 30 out of 55 that is

Anil’s score is 67+20 = 87%. Anil’s score = 0.87 * 45 = 39

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Problem 4: Increase in rice production in 2019 was 7% more than 2018 when the rice production was 24,500 tonnes

Answer: Rice production = 24500 * 1.07 = 26215 tonnes

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Problem 5: Forty percent of the employees of a certain company are men, and 75 percent of the men earn more than Rs 25,000 per year. If 45 percent of the company’s employees earn more than Rs 25,000 per year, what fraction of the women employed by the company earn Rs 25,000 year or less?

Answer: Create a table. Let the number of employees be 100.

Men40Men earning more than Rs 2500075% of 40 = 30
Total100Total employees earning more than Rs 2500045% of 100 = 45
Women60Women earning more than Rs 2500045 – 30 = 15

45 women get less than Rs 25,000. 45 out of 60 means 75% of women or 3/4 as fraction.

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Problem 6: In a class there are 150 students and 40% of the students have a degree. How many students don’t have degree?

Answer: Students who don’t have degree is 60% = 0.60 * 150 = 90

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Problem 7: Amit invested Rs 12,000 in a fixed account for 3 years at the rate of 7%. What is the amount received at the end of the term?

Answer: Using growth formula M = Mo(1+r)t where Mo = Rs 12000, t = 3 years, and r = 7% or 0.07

M = 12000 (1 + 0.07)3 = 12000 * 1.073 = Rs 14700

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Problem 8: Jaya’s attendance for first two semesters out of four was 60% and 70%, respectively. What is the minimum attendance required in third semester so that her average attendance will be 80% throughout four semesters? (Assume equal number of days among the four semesters)

Answer: Create the attendance table

SemesterSemesterSemester 2Semester 3Semester 4Average in 4 Sem
Attendance60%70%80%
Total classes = 100 per Sem60 classes70 classes0.80 * 400 = 320

For Semester 3, to allow minimum attendance the student must have attended 100% of the classes in 4th Semester

SemesterSemesterSemester 2Semester 3Semester 4Average in 4 Sem
Attendance60%70%100%80%
Total classes = 100 per Sem60 classes70 classesX100 classes0.80 * 400 = 320 classes

Adding all the classes we get 60 + 70 + X + 100 = 320 => X = 320 – 230 = 90

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Problem 9: In a company, 70% of employees are above 30 years old and 60% of that is males. If the total number of male employees aged 30 years above is 4200. Find the total number of employees in the company.

Answer: Let there be 100 employees in the company

Total EmployeesX
Above 30 years70% 0f X = 0.70X
Males60% 0f 0.70X= 0.42X

0.42X  = 4200 => X = 4200/0.42 = 10,000

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Problem 10: A man loses 12(1/2)% of his money and after spending 70% of the remainder, is left with Rs. 210. Find the amount of money he had.

Answer: Let the man have X amount.

Initial amountX
After losing 12.5%, remaining amount(1-0.125)X = 0.875X
Money spent70% of 0.0875X
Money left is 30% 0f 0.875X =0.30*0.875X = 0.2625X

0.2625X = 210 => X = Rs 800

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Pipes & Cisterns

Basic Concept

While considering the questions on Pipes and Cisterns – Always think of Unit Time

If a Tank can be filled by Pipe A in 4 hours, then how much of tank will be filled up in 1 hour. Using Ratios and Proportion we get 1/4th

If a Tank can be filled by Pipe B in 5 hours, then how much of tank will be filled up in 1 hour. Using Ratios and Proportion we get 1/5th

When Both Pipes A and B are opened together the amount of water filled in tank in 1 hour is 1/4 + 1/5 = 9/20

Remember to compare values ¼ = 25%, 1/5 = 20% and 9/20 = 45%

How much time will the two pipes take together to fill 90% of the tank? – 2 hours

How to calculate Problems:

Method 1: Find how much water tank is filled in 1 hour and then add the amount of percentage of tank filled.

To find time to fill complete tank invers the values. For the above problem, we know that pipes A and B fills 9/20 tank in 1 hour. Thus, 1 tank will be filled in 20/9 hours = 2 hours and 13.4 minutes.

Method 2: Use the formula

Questions

Question 1: Peter fills tank using pipe A in 1.3 hours. He changed the pipe A and now the tank is filled in 1 hour time. What will be the change in time if pipe B fills the tank in 2 hours?

Answer:

Pipe A takes 1.3 hours and Pipe B takes 2 hours to fill the same tank. Total time taken to fill the tank when both pipes are used =

When the pipe A is changed the time taken to fill the tank =

The saving in time to fill tank = 65 – 40 = 25 min.

NOTE: Convert hours to minutes – Multiply decimal values by 60

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Question 2: Ann has a water tank above her house with Pipe A that can fill the tank in 3 hours and the pipe that supplies water to her house and takes 4 hours to empty the tank. How many hours will it take to fill the tank if Ann left the drain pipe 50% open?

Answer:

Pipe A takes 3 hours and Pipe B takes 4 hours to empty the same tank. Since the drain pipe (the second pipe) was left 50% open, only 50% of water will be drained from tank. This pipe B will take 8 hours to drain the tank Total time taken to fill the tank when both pipes are used =

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Question 3: Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Answer:

Case I: 4 min

14 min – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 min – Pipe A –> Fills 1/15 tank => 4min –> 4/15 tank
  • 1 min – Pipe B –> Fills 1/20 tank => 4 min –> 4/20 = 1/5 tank

In 4 min. both the pipes will fill up 4/15 + 1/5 = 7/15 tank. Remaining to be filled up is

1 – 7/15 = 8/15

Case II: filling up rest of the tank by tank B only.

  • 1 min – Pipe B –> Fills 1/20 tank

=> Time taken by pipe B to fill up = Remaining tank ÷ rate of filling up by pipe B = 8/15 ÷ 1/20

= 8/15 * 20 = 32/3 = 10 min 40 sec

Total time required to fill up the tank = 4+10:40 = 14m 40s

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Question 4: Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled?

Answer:

5 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/5 tank
  • 1 hours – Pipe B –> Fills 1/10 tank
  • 1 hours – Pipe C –> Fills 1/30 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/5 + 1/10 + 1/30 = (6 + 3 + 1) /30 = 1/3 tank

So 1/3 tank will get filled in 1 hour

⸫ 1 tank will get filled in 3 hours (flip the 1/3 from previous step to get answer)

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Question 5: Pipe A can fill a tank in 3 hours, pipe B in 6 hours and pipe C in 18 hours. If all the pipes are open, in how many hours will the tank be filled?

Answer:

3 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/3 tank
  • 1 hours – Pipe B –> Fills 1/6 tank
  • 1 hours – Pipe C –> Fills 1/18 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/3 + 1/6 + 1/18 = (6 + 3 + 1) /18 = 10/18 tank

So 10/18 tank will get filled in 1 hour

⸫ 1 tank will get filled in 18/10 = 1.8 hours (flip the 10/18 from previous step to get answer)

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Question 6: Pipe A can fill a tank in 4 hours, pipe B in 5 hours and pipe C in 20 hours. If all the pipes are open, in how many hours will the tank be filled?

Answer:

5 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/4 tank
  • 1 hours – Pipe B –> Fills 1/5 tank
  • 1 hours – Pipe C –> Fills 1/20 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/4 + 1/5 + 1/20 = (5 + 4 + 1) /20 = 1/2 tank

So 1/2 tank will get filled in 1 hour

⸫ 1 tank will get filled in 2 hours (flip the 1/2 from previous step to get answer)

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Question 7: Two pipes P and Q can fill a cistern in 8 and 12 minutes respectively. Both are opened together, but at the end of 4 minutes, the first is turned off. How much longer will the cistern take to fill?

Answer:

Case I: 4 min

8 min – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 min – Pipe A –> Fills 1/8 tank => 4 min –> 4/8 = 1/2 tank
  • 1 min – Pipe B –> Fills 1/12 tank => 4 min –> 4/12 = 1/3 tank

In 4 min. both the pipes will fill up 1/2+1/3 = 5/6 tank. Remaining to be filled up is

1 – 5/6 = 1/6

Case II: filling up rest of the tank by tank B only.

  • 1 min – Pipe B –> Fills 1/12 tank

=> Time taken by pipe B to fill up = Remaining tank ÷ rate of filling up by pipe B = 1/6 ÷ 1/12

= 1/6 * 12 = 2m

Total time required to fill up the tank = 4+2 = 6m

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Question 8: Two pipes P and Q can fill a cistern in 5 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes, pipe A is turned off. How much total time will the cistern take to fill?

*a. 6 min

  1. 7 min
  2. 8 min
  3. 3 min

@AF Explanation:

Case I: 3 min

5 min – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 min – Pipe A –> Fills 1/5 tank => 3 min –> 3/5 tank
  • 1 min – Pipe B –> Fills 1/15 tank => 3 min –> 3/15 = 1/5 tank

In 4 min. both the pipes will fill up 3/5+1/5 = 4/5 tank. Remaining to be filled up is

1 – 4/5 = 1/5

Case II: filling up rest of the tank by tank B only.

  • 1 min – Pipe B –> Fills 1/15 tank

=> Time taken by pipe B to fill up = Remaining tank ÷ rate of filling up by pipe B = 1/5 ÷ 1/15

= 1/5 * 15 = 3m

Total time required to fill up the tank = 3+3 = 6m

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Question 9: Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?

Answer:

Case I: 4 min

12 min – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 min – Pipe A –> Fills 1/12 tank => 3 min –> 3/12 = 1/4 tank
  • 1 min – Pipe B –> Fills 1/15 tank => 3 min –> 3/15 = 1/5 tank

In 3 min. both the pipes will fill up 1/4+1/5 = 9/20 tank. Remaining to be filled up is

1 – 9/20 = 11/20

Case II: filling up rest of the tank by tank B only.

  • 1 min – Pipe B –> Fills 1/12 tank

=> Time taken by pipe B to fill up = Remaining tank ÷ rate of filling up by pipe B = 11/20 ÷ 1/12

= 11/20 * 12 = 6.6m = 6m 36s

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Question 10: Pipe A can fill a tank in 39 hrs and pipe B can fill it in 26 hrs. If both the pipes are opened in the empty tank. In how many hours will it be full?

Answer:

39 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/39 tank
  • 1 hours – Pipe B –> Fills 1/26 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/39 + 1/26 = (3+2) /78 = 5/78 tank

So 5/78 tank will get filled in 1 hour

⸫ 1 tank will get filled in 78/5 = 15.6 hours (flip the 5/78 from previous step to get answer)

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Question 11: Pipe A can fill a tank in 20 hrs; pipe B can fill it in 15 hrs and pipe C can fill it in 45 hrs. If all the pipes are opened in the empty tank. In how many hours will it be full?

Answer:

20 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/20 tank
  • 1 hours – Pipe B –> Fills 1/15 tank
  • 1 hours – Pipe C –> Fills 1/45 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/20 + 1/15 + 1/45 = (27+36+12) /540 = 75/540 tank

So, 75/540 = 5/36 tank will get filled in 1 hour

⸫ 1 tank will get filled in 36/5 = 7.2 hours (flip the 36/5 from previous step to get answer)

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Question 12: Pipe A can fill a tank in 6 hrs; pipe B can fill it in 24 hrs and pipe C can fill it in 72 hrs. If all the pipes are opened in the empty tank. In how many hours will it be full?

Answer:

6 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/6 tank
  • 1 hours – Pipe B –> Fills 1/24 tank
  • 1 hours – Pipe C –> Fills 1/72 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/6 + 1/24 + 1/72 = (12+3+1) /72 = 16/72 = 2/9 tank

So, 2/9 tank will get filled in 1 hour

⸫ 1 tank will get filled in 9/2 = 4.5 hours (flip the 2/9 from previous step to get answer)

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Question 13: Pipe A can fill a tank in 6 hrs; pipe B can fill it in 24 hrs and pipe C drains water in 72 hrs. If all the pipes are opened in the empty tank. In how many hours will it be full?

Answer:

7 hours – Pipe A –> Fills 1 tank

Finding amount of tank filled in 1 hour

  • 1 hours – Pipe A –> Fills 1/7 tank
  • 1 hours – Pipe B –> Fills 1/14 tank
  • 1 hours – Pipe C –> Fills -1/28 tank

Together when all three pipes are working in 1 hour will fill up tank

= 1/7 + 1/14 – 1/28 = (4+2-1) /72 = 5/72  tank

So, 5/72 tank will get filled in 1 hour

⸫ 1 tank will get filled in 72/5 = 14.4 hours (flip the 5/72 from previous step to get answer)

Profit and Loss

Basic Concepts

Items or raw materials are bought, stored and used in Finished goods. Finished goods are sold in market. Cost price and Selling Prices are calculated on running basis and updated frequently. For preparing Profit and Loss account and balance sheet these are used to find the assets at the end of the year. When items are sold, there are three terms commonly used – Cost Price (CP), Marked Price (MP), and Selling Price (SP).

Profit: Compare the height of each block under best possible conditions. In this situation Cost price has been marked up and a discount on marked price is given

Loss: Compare the height of each block under best possible conditions. In this situation Cost price has been marked up and a deep discount on marked price is given

Profit and loss involve comparing SP and CP.

How to Solve Problems?

Use the given formulas to calculate various terms

  • Marked UP = MP – CP
  • Percentage Marked UP = (MP-CP)/CP×100

  • Discount = MP – SP
  • Percentage Discount = Percentage of MP
  • SP = MP – Percentage Discount => Percentage Discount = (MP-SP)/MP×100
  • Profit = SP – CP (SP is greater than CP)

  • Profit Percentage = (SP-CP)/CP×100 => SP = CP(1+Percentage Profit/100)
  • Loss = CP-SP (CP is greater than SP)
  • Loss Percentage = (CP-SP)/CP×100 => SP = CP(1-Percentage Loss/100)

  • Short Cuts: If a person sells two items at same price. He gains of x% in one of the items and loss of x% in the second item, then the person will always suffer loss percentage of (x/10)^2
  • Short Cut: If a trader sells items at cost price but uses false wights, then he will always gain by Profit Percentage= (Amount of less weight)/(Original Weight-Amount of less weight)×100

  • Successive Discounts: Multiply the discounts to obtain a single equivalent discount.

Problems:

Problem 1: If the SP of a table is Rs 2200, then what is the Percentage Profit if the Cost Price is Rs 1800?
Answer: Profit Percentage = (SP-CP)/CP×100= (2200-1800)/1800×100= 400/1800×100=22%


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Problem 2: If the SP of a table is Rs 2200, then what is the Percentage Loss if the Cost Price is Rs 2500?
Answer: Loss Percentage = (CP-SP)/CP×100= (2500-2200)/2500×100= 300/2500×100=12%

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Problem 3: If by selling an item for Rs 6500, a trader gained 22%, what is the actual profit?
Answer: Profit Percentage = (SP-CP)/CP×100=>25%= (CP-6500)/CP×100
CP = 6500/0.75 = Rs 4875

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Problem 4: The profit percentage of an item sold at Rs 200 is same Loss percentage when sold at 140. What the cost price of the item?
Answer: Let %Profit = X and cost price = CP, then SP = CP(1+Percentage Profit/100) => 200 = CP*(1+X) and 140 = CP(1-X)
Adding both the equations 200 + 140 = 2*CP => CP = Rs 170
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Problem 5: Tina bought a clock for Rs 2200 and spent Rs 250 on repairs. She then sold it for Rs 3185. What was his Gain percentage?
Answer: Total cost of clock = 2200+250 = 2450. SP = 3000.
Profit Percentage = (3185-2450)/2450×100= 735/2450×100=30%

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Problem 6: A trader sells 2 boxes for Rs 1980 each, one at 10% gain and other at 10% loss. What is his total profit or loss?
Answer: Find cost price of the original boxes as the cost price are different.
Let the price of first box be X, gain is 10% => SP = X(1+0.1) => 1980 = 1.1X => X = 1800
Let the cost price of first box = Y, there is a loss of 10% => SP = X(1-0.1) => 1980 = 0.90X => X = 2200
Total cost price = 1800+2200 = 4000
SP of two items = 2*1980 = 3960
Loss = 4000-3960 = 40
%Loss = 40/4000 * 100 = 1%
Alternately using shortcut %loss = (x/10)^2= (10/10)^2=1%

Answer: 1%
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Problem 7: If selling price is doubled, the profit is tripled. What is the profit percentage?
Answer: Before change Profit = SP-CP and after change 3*Profit = 2*SP – CP
Substituting Profit in second equation we get 3*(SP-CP) = 2SP-CP => 3SP-3CP = 2SP-CP
=> SP = 2CP
Profit Percentage = (2SP-CP)/CP×100= (2*2CP-CP)/CP×100=300%


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Problem 8: The successive discounts on a product with Marked Price of Rs 10000 in a store is given as 5%, 10% and 15%. If the CP is Rs 5500 what is the profit percentage?
Answer: SP = MP-Discount = MP(1-%discount)
SP = MP(1-0.05)(1-0.10)(1-0.15) = 0.72675MP = 0.72675*10000 = Rs 7267.5
Profit Percentage = (SP-CP)/CP×100= (7267.5-5500)/5500×100=32.13%


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Quadratic Equations

Basic Concepts

Quadratic equations have format y = x2 and general format is Ax2 + Bx + C = 0. Most common format is a(x-h)2 + k = 0. Problem in quadratic involves formatting of the equation.

How to Format Quadratic Equation

Method I:

Consider = x2 + (m+n)x + mn = 0

Find numbers m and n such a way that m + n is the middle term, then the quadratic equation can be written in factor form.

From the equation Ax2 + Bx + C = 0 where product A*C will result in factors and sum or difference of one of the combinations is = B

Example: 2x2 + 9x + 4 = 0. Here product is 2*4 = 8. The factors of 8 are

Addition or subtraction of factors is

Replacing 9 = 8+1 in the equation 2x2 + 9x + 4 = 0 we get 2x2 + 8x + x + 4 = 0

Grouping the terms and taking common from the equation 2x(x + 4) + 1(x + 4) = 0

(2x + 1)(x + 4) = 0

The product 2x+1 and x+4 is 0. Means one or both of the terms are equivalent to 0.

2x + 1 = 0 => 2x = 1 => x = ½

X + 4 = 0 => x = -4

Method II: For equation x2 + Bx + C = 0. Example x2 + 9x + 8 = 0

X + 8 = 0 => x = -8 and x + 1 = 0 => x = -1

Perfect Squares

Prefect squares are conditions when we the two factors are identical. Consider the following:

Middle term is twice the product of square root of the first and the last terms

Sum and Difference

Sum of squares

 

Questions

Question 1: Solve the quadratic equation and get the value of x.

x2 – x – 12

Answer:

x + 3 = 0 => x = – 3 and

x – 4 = 0 => x = 4

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Question 2: Two integers differ by 12 and the sum of their squares is 74. Find the integers.

Answer: Let the integers be x and y. Thus x – y = 12 or x = y + 12

And x2 + y2 = 74. Replacing x = y + 12, we get

(y + 12)2 + y2 = 74 => y2 + 24y + 144 + y2 = 74

=> 2y2 + 24y + 70 = 0 => y2 + 12y + 35 = 0 => 35 = 7*5, splitting the middle term

(y + 7)(y + 5) = 0 => y = -7 and -5

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Question 3: The sum of two natural numbers and its square is 210. Find the numbers.

Answer: Let the two natural number be m then

m2 + m = 210 => m2 + m – 210 = 0 Factors of 210 are 15 and 14

(m+15)(m-14) = 0 => m = -15 (not possible as natural number) and m = 14

m = 14 is correct answer

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Question 4: The length of a rectangle is 3 cm longer than its width. The rectangle has an area of 28 cm2. Find its width.

Answer: Area of a triangle is given by L*W = 28.

L = W + 4. Substituting length in area formula we get,

(W+4)W = 26 => W2 + 4W – 28 = 0

28 has factors of 7 and 3 => (W + 3)(W – 7) = 0

W = 7 cm is the only valid answer

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Question 5: Is it possible to bend a 22 cm length of wire into the shape of a rectangle which has an area of 30 cm2?

Answer: When the wire of 20 cm is bent into a rectangle, then rectangles perimeter = 2(L+W) = 22

=> L = 11-W

Also, area of rectangle = L*W = 30

Substituting value of L in area we get

(10-W)W = 30 => -W2 +10W – 30 = 0 => W2 -11W + 30 = 0

(W-5)(W-6) = 0 => W = 5 or 6 in that case L = 6 or 5

Taking L = 6 and W = 5, it is possible to construct the rectangle

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Question 6: Two trains travel a 160 km track each day. The express travels 14 kmph faster and takes 30 minutes less time than the normal train. Find the speed of the express.

Answer:

Speed = Distance/Time => S = 160/T => T = 160/S

Speed = Distance/Time => S + 14 = 160/(T-0.5) => T = 160/(S+10) + 0.5

Substituting T from Normal Speed into Express train speed

160/S = 160/(S+14) + 0.5 => Multiplying both side by S(S+10)

160(S+14) = 160S + 0.5(S+14)S => 160S + 1600 = 160S + 0.5S2 + 7S

=> S2 + 14S – 3200 = 0 =>

Express train speed is S + 14 = 50+14 = 64 kmph

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Surds and Indices

Basic Concept

Surds are numbers in the format that has no equivalent of Rational number in decimal form or whole number . Some of the numbers can be written in equivalent of whole number or decimal form – example √4 = 2 and ∛8=2. But ∛7 has no equivalent value.

Rules – Surds

  • When conjugates are multiplied the root is eliminated –> (a+c√b) ×(a-c√b)=a2 – c2b

Rules – Indices

How to Solve Surds and Indices Problems

  • Find factors of the numbers inside the root.
    Example –  
  • Try to raise power to easily simplify the power of the base
    Example 
    Compare the above solution with the factors below

Try to equalize the base
Example: If  then what is the value of x

Problems

Problem solving involves surds in denominator where it is preferred to have surds in numerator only. To remove surds without from denominator multiply both numerator and denominator with same surd or its conjugate

Problem 1: Solve √32 − √18
Answer: √32 − √18 = 4√2 − 3√2 = √2
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Problem 2: Rationalize the following expression:
  
Answer:


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Problem 3: Solve the following expression:

Answer:
taking LCM of denominator and adding

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Problem 4: If √(0.64X) = ∛512, what is the value of x?
Answer: √(0.64X) = ∛512 => √0.64*√X = ∛83 => √X * 0.8 = 8
=> √X = 8/0.8 => √X = 10 => X = 100
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Problem 5: Simplify (625)0.17 × (625)0.08:
Answer: (625)0.17 × (625)0.08. Add the powers. (625)0.25 = (625)1/4 = (54) 1/4 = 5
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Problem 6: Simplify
Answer:

Rationalizing denominator by using its conjugate √3+1

= 5-√3+√3+1=5+1=6
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