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Solving LCM and HCF Puzzles: Finding the Number of Pairs

Learn how LCM and HCF concepts help you solve complex problems, find the number of pairs, determine divisors and factors, and crack math puzzles. Enhance your number theory skills and boost your chances of success in these competitive exams. Don’t miss out on this essential topic that can make a significant difference in your exam performance!

Que: How many pairs of integers (x,y) exist such that the product of x,y, and HCF(x,y)= 1080?

(a) 8

(b) 7

(c) 9

(d) 12

Total Pairs=9

Option (c) is correct.

Que: Find the least number that should be added to 2014 such that the resulting number can be divided by 3,4,5, and 6 leaving no remainder.

(a) 34

(b) 60

(c) 36

(d) 26

The LCM and HCF concepts are very useful in solving problems involving fractions, simplifying algebraic expressions, and finding solutions to equations. Some questions based on LCM and HCF concepts are as follows:

Find the LCM and HCF of two or more numbers

Simplification of fractions

Find the smallest or largest number satisfying given conditions

Calculate the time taken by two or more persons to complete a task working together.

Total number of rotations or revolutions required by two or more wheels to come back to their initial position

Find the least n-digit number which when divided by the given numbers leaves the same or different remainder.

How many pairs of numbers are there whose product is given

Arrangements in rows and columns, etc.

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The LCM and HCF concepts are very useful in solving problems involving fractions, simplifying algebraic expressions, and finding solutions t

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How to find the number of divisors

In this article, we will explore the divisor concepts, including the number of factors, prime factors, and the fascinating properties of the sum and product of factors. By understanding these concepts, you will gain valuable insights into the relationships between numbers and unlock the secrets hidden within their divisors.

The divisor or factor of any number divides it without leaving any remainder. For example, let’s consider the number 12. The divisors or factors of 12 are 1,2,3,4,6 and 12.

To find the total number of divisors of any number, we use the prime factorization method given below.

For example, the prime factorization of any number (N) is p^a×q^b×r^c, Here p,q, and r are the prime numbers (2,3,5,7,11,…etc.) and a,b, and c are their respective exponents.

The total number of divisors of N= (a+1)×(b+1)×(c+1)

Example: Find the number of divisors of 360

360=2³×3²×5¹

The number of factors= (3+1)×(2+1)×(1+1)=24

24 factors= {1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360}

Que 1: Find the number of prime factors of 360.

Solution: Three Prime factors={2,3,5}

Que 2: Find the number of odd factors of 360.

Solution: Remove 2³ from the prime factorization

Remaining Part= 3²×5¹

Number of odd factors= (2+1)×(1+1)=6

Que 3: Find the number of even factors of 360.

Solution: 360=2³×3²×5¹=2×[2²×3²×5¹]

Even Factors= 2K

K=2²×3²×5¹

Number of even factors= (2+1)×(2+1)×(1+1)=18

Que 4: Find the number of factors of 360 that are divisible by 12.

Solution: 360=2³×3²×5¹=2²×3×[2¹×3¹×5¹]

Required factor=12K

K=[2¹×3¹×5¹]

Number of required factors= (1+1)×(1+1)×(1+1)=8

Que 5: Find the number of factors of 360 which are perfect squares.

Solution: 360=2³×3²×5¹

Possible pairs={(2⁰, 2²)×(3⁰, 3²)×(5⁰)}

Number of required factors= 2×2×1=4

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In this article, we will explore the divisor concepts, including the number of factors, prime factors, and the fascinating properties of the

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Complex Percentage Money Distribution Problem: Expert Techniques for a quick solution

Que 1: If A, B, C, and D share a loot as follows: A receives a% of the total, B receives b% of the remaining (after A takes their share), C also receives c% of the remaining, and D gets the rest, where D’s share is a% less than what A gets, B and C get equal amounts, and b is equal to 2a. What percentage of what A received did C get?

(a) 160%

(b) 80%

(c) 175%

(d) 50%

Solution: Let the total amount= 100 rs

A⇒a

B⇒(100-a).b/100= (100-a).2a/100

C⇒(100-a).2a/100

D⇒a.(100-a)/100

A+B+C+D=100

5a.(120-a)=10000

a=20

A=20%=20

B=40% of 80=32

C=32

D= 80% of A=16

C received= 160% of the A

Option (a) is correct

Que 2: What is the range of values x can take if an alloy of copper and aluminum has 40% copper, an alloy of copper and zinc has a copper-to-zinc ratio of 2:7, and when these two alloys are mixed in such a way that the overall alloy has more aluminum than zinc, with copper constituting x% of this alloy?

a) 30% ≤x ≤40%

b) 32.5% ≤x ≤42%

c) 33.33% ≤x ≤42%

d) 32.25% ≤x ≤40%

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Challenging Simple and Compound Interest Mix Problems: Tricky Solution

Que: A person borrowed a certain amount of money from a private organization at a simple interest rate of 5% per annum. He lent out 50% of this amount to another individual at a compound interest rate of 10% per annum. As a result, he earned a profit of Rs. 3,205 in 4 years. What was the initial amount borrowed by the person?

(a) 80,000

(b) 1,00,000

(c) 1,20,000

(d) 1,50,000

Que: Given a rate of 5% per annum, if the difference between the compound interest and simple interest earned on a sum of money over a period of 3 years is 36.60 rs, what is the value of the sum of money?

(a) 8000 rs

(b) 8400 rs

(c) 4400 rs

(d) 4800 rs

In this intriguing problem, we dive into the world of simple and compound interest. Join us as we unravel the solution to a challenging scenario involving borrowing, lending, and calculating profits.

In this YouTube video, we present a step-by-step breakdown of the problem and its tricky solution. Discover how a person borrowed a certain amount of money at a simple interest rate of 5% per annum and then lent out 50% of this amount to another individual at a compound interest rate of 10% per annum. After 4 years, the person earned a profit of Rs. 3,205.

Through detailed explanations and strategic problem-solving techniques, we unravel the mystery of the initial amount borrowed by the person. You’ll gain valuable insights into simple interest, compound interest, rates, and profits.

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Shaded Area Ratios in Parallelograms: Practice Problems and Solutions| Geometry tricks

Que: In the parallelogram ABCD, where IJ is parallel to AD, and ED and HF are perpendicular to DC, what is the ratio of the areas of quadrilateral AGJI to triangle HFC?

Solution: In △ADE, GJ∥AE

So, ar(△GJD): ar(△AED)=GD²: AD²=9:25

ar(AEJG)= 25–9= 16 unit …………..(i)

△IEJ ~△GJD (Similar Triangles)

ar(△IEJ): ar(△GJD)=IJ²:GD²=4:9

ar(△IEJ)=4 unit ………(ii)

ar(△HFC)=ar(△GJD)=9 unit

ar(AGJI): ar(△HFC)=16–4: 9=12:9= 4:3

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Home Work Question

In the rectangle ABCD, the diagonal AC is extended to point E such that CE measures 5 cm. If the lengths of sides AB and BC are 8 cm and 6 cm respectively, what is the length of DE?

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Profit and Loss Assignment:2

Que 1: What is the cost price of an article if the profit on selling it for Rs. 425 is equal to the loss on selling it for Rs. 355?

(a) 410

(b) 380

(c) 400

(d) 390

Solution: CP-355=425-CP

CP=390

Option (d) is correct

Que 2: If the cost price of 10 articles is equal to the selling price of 15 articles, what is the percentage of profit or loss incurred on the transaction?

(a) 25.5%

(b) 35%

(c) 10%

(d) 33.33%

Solution: CP×10=SP×15

CP: SP=3:2

Loss=33.33%

Option (d) is correct

Que 3: If a trader incurred a loss of 10% on selling a cycle and could have gained 6% had the selling price been increased by Rs. 200, what was the cost price of the cycle?

(a) 1200 rs

(b) 1205 rs

(c) 1250 rs

(d) 1275 rs

Solution: 90%+200=106%

16%=200

1%=25/2

100%=1250

Option (c) is correct

Que 4: If A sold an article to B at a profit of 20% and B sold it to C at a loss of 15%, what profit or loss would A make if he had sold the article directly to C at the same selling price as B?

(a) 5% profit

(b) 2% profit

(c) 2% loss

(d) 5% profit

Solution: A: B=5: 6

B: C=20: 17

A: B: C= 100: 120: 102

A makes 2% profit

Option (b) is correct

Que 6: If a shopkeeper sells rice at a profit of 10% while using a weight that is 30% less than the actual measure, what is the percentage of profit that the shopkeeper earns?

(a) 57.67%

(b) 57.14%

(c) 57.40%

(d) 57.43%

Solution: SP= Price per unit× No. of Units

Weight⇒ 10: 7

SP⇒ 10: 11

Price= 10/10: 11/7= 7: 11

Percentage Profit= 400/7= 57(1/7)=57.14%

Option (b) is correct

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Simple and Compound Interest Assignment:1

Que 1: What is the annual interest rate at which an investment of 3000 rs will grow to 3993 rs after 3 years, assuming the interest is compounded annually?

(a) 9%

(b) 10%

(c) 11%

(d) 13%

Solution: 3000: 3993

1000: 1331

Cube root both sides⇒ ∛1000: ∛1331= 10:11

Annual interest rate= 10%

Option (b) is correct.

Que 2: What will be the compound interest earned on an investment of 10,000 rs at a rate of 4% per annum, compounded half-yearly, over a period of 2 years?

(a) 636.80 rs

(b) 824.32 rs

(c) 912.86 rs

(d) 828.82 rs

Solution: 4% per annum= 2% half yearly

10000⇒ 10000+200=10200

10200⇒ 10200+204=10404

10404⇒ 10404+208.**=10612.**

10612⇒ 10612+212.**= 10824.**

Option (b) is correct.

Que 3: If a certain amount of money grows to 10648 rs in 3 years and 9680 rs in 2 years with compound interest, what is the annual interest rate?

(a) 5%

(b) 10%

(c) 15%

(d) 20%

Solution: 9680: 10648

1210: 1331⇒ Difference or CI= 121

Option (b) is correct.

Que 4: What is the annual interest rate at which an investment of 2304 rs will grow to 2500 rs in 2 years with compound interest?

(a) 4(1/2)%

(b) 4(1/3)%

(c) 4(1/5)%

(d) 4(1/6)%

Solution: 2304:2500

Square root both sides⇒ √2304: √2500= 48: 50

Interest rate= 200/48= 25/6= 4(1/6)%

Option (d) is correct.

Que 5: Assuming a compound interest rate of 4% per annum for the first year, 5% per annum for the second year, and 6% per annum for the third year, what will be the total compound interest earned on an investment of 10,000 rs for 3 years?

(a) 1600

(b) 1625.80

(c) 1575.20

(d) 2000

Solution: 10000⇒ 10000+400=10400

10400⇒ 10400+520= 10920

10920⇒ 10920+ 654.**=11574.**

CI= 1574.**

Option (c) is correct.

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The Concepts of Surds and Indices

Surds and indices are two important mathematics concepts often studied in high school and college-level mathematics courses. They are used extensively in algebraic and geometric computations and play a crucial role in many areas of science and engineering.

Surds are irrational numbers that cannot be expressed as a ratio of two integers. They are typically expressed in the form of √n, where n is a non-square positive integer. For example, the square root of 2 (√2) is a surd, as is the cube root of 3 (∛3) or the fifth root of 7. Surds can be added, subtracted, multiplied, and divided just like other numbers, but their irrational nature can sometimes produce unexpected results when manipulated.

Indices, also known as exponents or powers, are shorthand notations used to express the repeated multiplication of a number by itself. For example, 3² (read as “3 to the power of 2”) is shorthand for 3 × 3 = 9. Similarly, 4³ (read as “4 to the power of 3”) is shorthand for 4 × 4 × 4 = 64. Indices can also be negative or fractional, which allows them to represent roots and reciprocal powers. For example, 5⁻² (read as “5 to the power of negative 2”) is shorthand for 1/5² or 1/25, and ∛2 (read as “the cube root of 2”) is shorthand for 2^(1/3).

Surds and indices are often used together in mathematical expressions, particularly when working with roots and fractional exponents. For example, the expression √2 × 2^(-1/2) can be simplified by recognizing that √2 is the same as 2^(1/2) and that 2^(-1/2) is the reciprocal of 2^(1/2), so the expression simplifies to 1.

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Surds and indices chapter plays a crucial role in simplification problems of arithmetic. Here we would learn about its basic concepts and tr

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