5.3 math homework answers

• Solve equations and inequalities
• Simplify expressions
• Factor polynomials
• Graph equations and inequalities
• Advanced solvers
• All solvers
• Arithmetics
• Determinant
• Percentages
• Scientific Notation
• Inequalities

What can QuickMath do?

QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students.

• The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction.
• The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require.
• The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables.
• The calculus section will carry out differentiation as well as definite and indefinite integration.
• The matrices section contains commands for the arithmetic manipulation of matrices.
• The graphs section contains commands for plotting equations and inequalities.
• The numbers section has a percentages command for explaining the most common types of percentage problems and a section for dealing with scientific notation.

Math Topics

More solvers.

• Add Fractions
• Simplify Fractions

• 888-309-8227
• 732-384-0146

New User Registration

Forgot Password

enVision MATH Common Core 5, Grade: 5 Publisher: Scott Foresman Addison Wesley

Envision math common core 5, title : envision math common core 5, publisher : scott foresman addison wesley, isbn : 328672637, isbn-13 : 9780328672639, use the table below to find videos, mobile apps, worksheets and lessons that supplement envision math common core 5..

textbook resources

• Call us toll-free
• FAQs – Frequently Asked Questions
• Contact Lumos Learning – Proven Study Programs by Expert Teachers

Follow us: Lumos Learning -->

• 2024 © Lumos Learning
• Privacy Policy - Terms of Service - Disclaimers

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc... Read More

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc., the Partnership for the Assessment of Readiness for College and Careers, nor any state of the Union. Neither PARCC, Inc., nor The Partnership for the Assessment of Readiness for College and Careers, nor any member state has endorsed this product. No portion of any fees or charges paid for any products or services Lumos Learning offers will be paid or inure to the benefit of PARCC, Inc., or any state of the Union

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not aff... Read More

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not affiliated to Lumos Learning. The Regents of the University of California – Smarter Balanced Assessment Consortium, was not involved in the production of, and does not endorse these products or this site.

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC, was not... Read More

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC,was not involved in the production of, and does not endorse these products or this site.

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the... Read More

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the production of, and does not endorse these products or this site.

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the... Read More

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the production of, and does not endorse these products or this site.

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved... Read More

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved in the production of, and does not endorse these products or this site.

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the prod... Read More

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the production of, and does not endorse these products or this site.

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved... Read More

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved in the production of, and does not endorse these products or this site.

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved... Read More

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved in the production of, and does not endorse these products or this site.

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved... Read More

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved in the production of, and does not endorse these products or this site.

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved... Read More

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved in the production of, and does not endorse these products or this site.

• AI Math Solver Graphing Calculator Popular Problems Worksheets Study Guides Cheat Sheets Calculators Verify Solution
• Solutions Integral Calculator Derivative Calculator Algebra Calculator Matrix Calculator More...
• Graphing Line Graph Exponential Graph Quadratic Graph Sine Graph More...
• Calculators BMI Calculator Compound Interest Calculator Percentage Calculator Acceleration Calculator More...
• Geometry Pythagorean Theorem Calculator Circle Area Calculator Isosceles Triangle Calculator Triangles Calculator More...
• Tools Notebook Groups Cheat Sheets Worksheets Study Guides Practice Verify Solution

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Prove That Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Coterminal Angle Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

Number Line

• x^{2}-x-6=0
• -x+3\gt 2x+1
• line\:(1,\:2),\:(3,\:1)
• prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x)
• \frac{d}{dx}(\frac{3x+9}{2-x})
• (\sin^2(\theta))'
• \lim _{x\to 0}(x\ln (x))
• \int e^x\cos (x)dx
• \int_{0}^{\pi}\sin(x)dx
• \sum_{n=0}^{\infty}\frac{3}{2^n}
• Is there a step by step calculator for math?
• Symbolab is the best step by step calculator for a wide range of math problems, from basic arithmetic to advanced calculus and linear algebra. It shows you the solution, graph, detailed steps and explanations for each problem.
• Is there a step by step calculator for physics?
• Symbolab is the best step by step calculator for a wide range of physics problems, including mechanics, electricity and magnetism, and thermodynamics. It shows you the steps and explanations for each problem, so you can learn as you go.
• How to solve math problems step-by-step?
• To solve math problems step-by-step start by reading the problem carefully and understand what you are being asked to find. Next, identify the relevant information, define the variables, and plan a strategy for solving the problem.
• Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want...

We want your feedback

Please add a message.

Message received. Thanks for the feedback.

• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• Chapter 5 Arithmetic Progressions
• Exercise 5.3

Ncert Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

The NCERT Solutions for Class 10 Maths Chapter 5- Arithmetic Progression Exercise 5.3 includes detailed answers to the problems provided in Exercise 5.3. These NCERT solutions are written to guide the students in the CBSE Class 10 board examination.

In order to solve different types of questions first n terms of arithmetic progression, students are advised to study this NCERT solution thoroughly. By solving these exercise questions, using the NCERT Class 10 Maths Solutions Chapter 5 Ex 5.3 you will be better equipped to comprehend all possible types of questions that may be asked in the Class 10 board examinations.

Topics Covered in Exercise 5.3

• Sum of First n Terms of an AP
• Solved examples

How is it helpful?

• Helps in comprehending the methodology to solve the sum of first terms of an arithmetic progression
• Heps in memorising the important formulas and methodologies
• NCERT Class 10 Maths Solution assists in solving all the questions on arithmetic progression accurately

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3

carouselExampleControls112

Previous Next

Access other exercise solutions of Class 10 Maths Chapter 5- Arithmetic Progression

Students can access the exercise-wise NCERT Class 10 Solutions for Maths Chapter 5 below.

Exercise 5.1 – 4 questions (1 MCQ and 3 descriptive type questions)

Exercise 5.2 – 20 questions (1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions)

Exercise 5.3 – 20 Questions (3 fill in the blanks, 4 daily life examples, and 13 descriptive-type questions)

Exercise 5.4 – 5 Questions (5 Long answer questions)

Access Answers of Maths NCERT Class 10 Chapter 5 – Arithmetic Progression Exercise 5.3

1. Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms. (ii) − 37, − 33, − 29 ,…, to 12 terms (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms (iv) 1/15, 1/12, 1/10, …… , to 11 terms

(i) Given, 2, 7, 12 ,…, to 10 terms

For this A.P.,

First term, a = 2

And common difference, d = a 2  − a 1  = 7−2 = 5

We know that the formula for sum of nth term in AP series is,

= 5 × 49 = 245

(ii) Given, −37, −33, −29 ,…, to 12 terms

First term, a = −37

And common difference, d = a 2 − a 1

d= (−33)−(−37)

= − 33 + 37 = 4

= 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms

First term, a = 0.6

Common difference, d = a 2  − a 1  = 1.7 − 0.6 = 1.1

(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms

First term, a = 1/5

Common difference, d = a 2 –a 1 = (1/12)-(1/5) = 1/60

And number of terms n = 11

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

2. Find the sums given below :

First term, a = 7

n th term, a n = 84

Let 84 be the  n th  term of this A.P., then as per the n th term formula,

a n = a(n-1)d

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

We know that the sum of n term is;

S n  = n/2 (a + l) , l = 84 S n  = 23/2 (7+84)

S n   = (23×91/2) = 2093/2

(ii) Given, 34 + 32 + 30 + ……….. + 10

First term, a = 34

Common difference, d = a 2 −a 1  = 32−34 = −2

n th term, a n = 10

Let 10 be the  n th  term of this A.P., therefore,

a n = a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

We know that the sum of n terms is;

S n  = n/2 (a +l) , l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

(iii)  Given, (−5) + (−8) + (−11) + ………… + (−230)

First term, a = −5

nth term, a n = −230

Common difference, d = a 2 −a 1  = (−8)−(−5)

⇒d = − 8+5 = −3

Let −230 be the  n th  term of this A.P., and by the n th term formula we know,

a n =  a +( n −1) d

−230 = − 5+( n −1)(−3)

−225 = ( n −1)(−3)

( n −1) = 75

And the sum of n term,

S n  =  n /2 ( a  +  l )

3. In an AP (i) Given  a  = 5,  d  = 3,  a n  = 50, find  n  and  S n . (ii) Given  a  = 7,  a 13  = 35, find  d  and  S 13 . (iii) Given  a 12  = 37,  d  = 3, find  a  and  S 12 . (iv) Given  a 3  = 15,  S 10  = 125, find  d  and  a 10 . (v) Given  d  = 5,  S 9  = 75, find  a  and  a 9 . (vi) Given  a  = 2,  d  = 8,  S n  = 90, find  n  and  a n . (vii) Given  a  = 8,  a n  = 62,  S n  = 210, find  n  and  d . (viii) Given  a n  = 4,  d  = 2,  S n  = − 14, find  n  and  a . (ix) Given  a  = 3,  n  = 8,  S  = 192, find  d . (x) Given  l  = 28,  S  = 144 and there are total 9 terms. Find  a .

(i) Given that,  a  = 5,  d  = 3,  a n  = 50

As we know, from the formula of the nth term in an AP,

a n  =  a  +( n  −1) d ,

Therefore, putting the given values, we get,

⇒ 50 = 5+( n  -1)×3

⇒ 3( n  -1) = 45

⇒  n  -1 = 15

Now, sum of n terms,

S n  =  n /2 ( a  + a n )

S n  = 16/2 (5 + 50) = 440

(ii)  Given that,  a  = 7,  a 13  = 35

a n  =  a +( n −1) d ,

⇒ 35 = 7+(13-1) d

⇒ 12 d  = 28

⇒  d  = 28/12 = 2.33

Now,  S n  =  n /2 ( a + a n )

S 13  = 13/2 (7+35) = 273

(iii) Given that,  a 12  = 37,  d  = 3

As we know, from the formula of the n th term in an AP,

a n  =  a +( n  −1) d ,

⇒  a 12  =  a +(12−1)3

⇒ 37 =  a +33

Now, sum of nth term,

S n  =  n /2 ( a + a n )

S n  =  12 /2 (4+37)

(iv) Given that,  a 3  = 15,  S 10  = 125

a n  =  a  +( n −1) d ,

a 3  =  a +(3−1) d

15 =  a +2 d  …………………………..  (i)

Sum of the nth term,

125 = 5(2 a +9 d )

25 = 2 a +9 d  ………………………..  (ii)

On multiplying equation  (i)  by  (ii) , we will get;

30 = 2 a +4 d  ……………………………….  (iii)

By subtracting equation  (iii)  from  (ii) , we get,

From equation  (i) ,

15 =  a +2(−1)

a  = 17 = First term

a 10  =  a +(10−1) d

a 10  = 17+(9)(−1)

a 10  = 17−9 = 8

(v) Given that,  d  = 5,  S 9  = 75

As, sum of n terms in AP is,

Therefore, the sum of first nine terms are;

25 = 3( a +20)

25 = 3 a +60

3 a  = 25−60

As we know, the n th term can be written as;

a n  =  a +( n −1) d

a 9  =  a +(9−1)(5)

= -35/3+8(5)

= (35+120/3) = 85/3

(vi) Given that,  a  = 2,  d  = 8,  S n  = 90

As, sum of n terms in an AP is,

⇒ 180 =  n (4+8 n  -8) =  n (8 n -4) = 8 n 2 -4 n

⇒ 8 n 2 -4 n – 180 = 0

⇒ 2 n 2 – n -45 = 0

⇒ 2 n 2 -10 n +9 n -45 = 0

⇒ 2 n ( n  -5)+9( n  -5) = 0

⇒ ( n -5)(2 n +9) = 0

So,  n = 5 (as n only is a positive integer)

∴  a 5   = 8+5×4 = 34

(vii) Given that,  a  = 8,  a n  = 62,  S n  = 210

As, the sum of n terms in an AP is,

S n  =  n /2 ( a  +  a n )

210 =  n /2 (8 +62)

⇒ 35 n  = 210

⇒  n  = 210/35 = 6

Now, 62 = 8+5 d

⇒ 5 d  = 62-8 = 54

⇒  d  = 54/5 = 10.8

(viii) Given that, n th term, a n  = 4, common difference, d  = 2, sum of n terms, S n  = −14.

4 =  a +( n  −1)2

4 =  a +2 n −2

a +2 n  = 6

a  = 6 − 2 n  ………………………………………….  (i)

As we know, the sum of n terms is;

-14 =  n /2 ( a + 4 )

−28 =  n  ( a +4)

−28 =  n  (6 −2 n  +4) {From equation  (i) }

−28 =  n  (− 2 n  +10)

−28 = − 2 n 2 +10 n

2 n 2  −10 n  − 28 = 0

n 2  −5 n  −14 = 0

n 2  −7 n+ 2 n  −14 = 0

n  ( n −7)+2( n  −7) = 0

( n  −7)( n  +2) = 0

Either  n  − 7 = 0 or  n  + 2 = 0

n  = 7 or  n  = −2

However,  n  can neither be negative nor fractional.

Therefore,  n  = 7

From equation  (i) , we get

a  = 6−2(7)

(ix) Given that, first term, a  = 3,

Number of terms,  n  = 8

And sum of n terms, S  = 192

As we know,

(x) Given that,  l  = 28, S  = 144 and there are total of 9 terms.

Sum of n terms formula,

144 = 9/2( a +28)

(16)×(2) =  a +28

32 =  a +28

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be  n  terms of the AP. 9, 17, 25 …

First term, a  = 9

Common difference, d  =  a 2 − a 1  = 17−9 = 8

As, the sum of n terms, is;

636 =  n  (4 n  +5)

4 n 2  +5 n  −636 = 0

4 n 2  +53 n  −48 n  −636 = 0

n  (4 n  + 53)−12 (4 n  + 53) = 0

(4 n  +53)( n  −12) = 0

Either 4 n +53 = 0 or  n −12 = 0

n  = (-53/4) or  n  = 12

n  cannot be negative or fraction, therefore,  n  = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Given that,

first term, a  = 5

last term, l  = 45

Sum of the AP, S n  = 400

As we know, the sum of AP formula is;

S n  =  n /2 ( a + l )

400 =  n /2(5+45)

400 =  n /2(50)

Number of terms, n  =16

As we know, the last term of AP series can be written as;

l = a+ ( n  −1) d

45 = 5 +(16 −1) d

Common difference, d  = 40/15 = 8/3

6. The first and the last term of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?

First term, a  = 17

Last term, l  = 350

Common difference, d  = 9

Let there be  n  terms in the A.P., thus the formula for last term can be written as;

350 = 17+( n  −1)9

333 = ( n −1)9

( n −1) = 37

S 38  = 38/2 (17+350)

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in which  d  = 7 and 22 nd  term is 149. Solution:

Given, Common difference, d = 7

22 nd term, a 22  = 149

Sum of first 22 term, S 22  = ?

By the formula of nth term,

a 22  =  a +(22−1) d

149 =  a +21×7

149 =  a +147

a  = 2 = First term

Sum of n terms,

S n  =  n /2( a + a n )

S 22 = 22/2 (2+149)

8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively. Solution:

Second term, a 2  = 14

Third term, a 3  = 18

Common difference, d  =  a 3 − a 2  = 18−14 = 4

a 2  =  a + d

a  = 10 = First term

Sum of n terms;

= 51 × 220/2

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first  n  terms.

S 17  = 289

We know, sum of n terms;

7 = ( a +3 d )

a  + 3 d  = 7 ………………………………….  (i)

In the same way,

289 = 17/2 (2 a  +16 d )

17 = ( a +8 d )

a  +8 d  = 17 ……………………………….  (ii)

Subtracting equation  (i)  from equation  (ii) ,

From equation  (i) , we can write it as;

a +3(2) = 7

=  n /2(2+2 n -2)

=  n /2(2 n )

10. Show that  a 1 ,  a 2  … ,  a n  , … form an AP where  a n  is defined as below

(i)  a n  = 3+4 n (ii)  a n  = 9−5 n Also, find the sum of the first 15 terms in each case.

(i)  a n  = 3+4 n

a 1  = 3+4(1) = 7

a 2  = 3+4(2) = 3+8 = 11

a 3  = 3+4(3) = 3+12 = 15

a 4  = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

a 2  −  a 1  = 11−7 = 4

a 3  −  a 2  = 15−11 = 4

a 4  −  a 3  = 19−15 = 4

Hence,  a k  + 1 − a k  is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is;

(ii)  a n  = 9−5 n

a 1  = 9−5×1 = 9−5 = 4

a 2  = 9−5×2 = 9−10 = −1

a 3  = 9−5×3 = 9−15 = −6

a 4  = 9−5×4 = 9−20 = −11

a 2  −  a 1  = −1−4 = −5

a 3  −  a 2  = −6−(−1) = −5

a 4  −  a 3  = −11−(−6) = −5

Hence, a k  + 1  −  a k  is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

= 15/2(8-70)

= 15/2(-62)

11. If the sum of the first  n  terms of an AP is 4 n  −  n 2 , what is the first term (that is  S 1 )? What is the sum of first two terms? What is the second term? Similarly find the 3 rd , the 10 th  and the  n th  terms.

S n  = 4 n − n 2

First term,  a  =  S 1  = 4(1) − (1) 2  = 4−1 = 3

Sum of first two terms =  S 2 = 4(2)−(2) 2  = 8−4 = 4

Second term,  a 2  =  S 2  −  S 1  = 4−3 = 1

Common difference, d  =  a 2 − a  = 1−3 = −2

N th term, a n  =  a +( n −1) d 

= 3+( n  −1)(−2)

= 3−2 n  +2

Therefore,  a 3  = 5−2(3) = 5-6 = −1

a 10  = 5−2(10) = 5−20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3 rd , the 10 th , and the n th  terms are −1, −15, and 5 − 2 n  respectively.

12. Find the sum of the first 40 positive integers divisible by 6.

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and the common difference is 6.

By the formula of sum of n terms, we know,

Therefore, putting n = 40, we get,

= 20(12+234)

13. Find the sum of first 15 multiples of 8.

The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore,  a  = 8

By the formula of sum of nth term, we know,

= 15(128)/2

14. Find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

First term, a = 1

Common difference, d = 2

Last term, l = 49

By the formula of last term, we know,

l  =  a +( n −1)  d

49 = 1+( n −1)2

48 = 2( n  − 1)

n  − 1 = 24

n  = 25 = Number of terms

S n  =  n /2( a  + l )

S 25  = 25/2 (1+49)

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days.

We can see that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, a  = 200 and d  = 50

The penalty that has to be paid if the contractor has delayed the work by 30 days = S 30

Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Let the cost of 1 st  prize be Rs. P .

Cost of 2 nd  prize = Rs. P  − 20

And cost of 3 rd  prize = Rs. P  − 40

We can see that the cost of these prizes are in the form of A.P., having common difference as −20 and first term as  P .

Thus, a  =  P and d  = −20

Given that,  S 7  = 700

7/2 [2 a  + (7 – 1) d ] = 700

a  + 3(−20) = 100

a  −60 = 100

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying. E.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term,  a  = 1

Common difference,  d  = 2−1 = 1

Therefore, the number of trees planted by 1 section of the classes = 78

The number of trees planted by 3 sections of the classes = 3×78 = 234

Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Perimeter of a semi-circle = π r

P 1  = π(0.5) = π/2 cm

P 2  = π(1) = π cm

P 3  = π(1.5) = 3π/2 cm

Where, P 1,   P 2 ,  P 3  are the lengths of the semi-circles.

Hence we got a series here, as,

π/2, π, 3π/2, 2π, ….

P 1   = π/2 cm

P 2  = π cm

Common difference, d  =  P 2 – P 1 = π – π/2 = π/2

First term = P 1 =  a  = π/2 cm

By the sum of n term formula, we know,

Therefor, the sum of the length of 13 consecutive circles is;

=13 /2 (7π)

=  13 /2 × 7 ×  22 /7

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

We can see that the numbers of logs in rows are in the form of an A.P. 20, 19, 18…

For the given A.P.,

First term, a  = 20 and common difference, d  =  a 2 − a 1  = 19−20 = −1

Let a total of 200 logs be placed in  n  rows.

Thus, S n  = 200

By the sum of nth term formula,

400 =  n  (40− n +1)

400 =  n  (41- n )

400 = 41 n − n 2

n 2 −41 n  + 400 = 0

n 2 −16 n −25 n +400 = 0

n ( n  −16)−25( n  −16) = 0

( n  −16)( n  −25) = 0

Either ( n  −16) = 0 or  n −25 = 0

n  = 16 or  n  = 25

By the nth term formula,

a 16  = 20+(16−1)(−1)

a 16  = 20−15

Similarly, the 25 th term could be written as;

a 25  = 20+(25−1)(−1)

a 25  = 20−24

It can be seen, the number of logs in the 16 th  row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16 th  row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2×5+2×(5+3)]

The distance of the potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.

Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, the distance to be run w.r.t distances of potatoes, could be written as;

10, 16, 22, 28, 34,……….

Hence, the first term, a  = 10 and d  = 16−10 = 6

Therefore, the competitor will run a total distance of 370 m.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

I like you byjus class

Thanks byjus class I like you byjus class

Thank you for the link

thanks byjus

It’s very good

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

• Barretts Elementary
• Bellerive Elementary
• Carman Trails Elementary
• Claymont Elementary
• Craig Elementary
• Green Trails Elementary
• Hanna Woods Elementary
• Henry Elementary
• Highcroft Ridge Elementary
• Mason Ridge Elementary
• McKelvey Elementary
• McKelvey Primary Elementary
• Oak Brook Elementary
• Pierremont Elementary
• River Bend Elementary
• Ross Elementary
• Shenandoah Valley Elementary
• Sorrento Springs Elementary
• Wren Hollow Elementary
• Central Middle
• Northeast Middle
• South Middle
• Southwest Middle
• West Middle
• Central High
• Fern Ridge High
• Early Childhood Center
• Alumni Association

Higher Expectations. Brighter Futures.

Page Navigation

• 5th Grade Home
• Clever (Parkway Apps)
• Leader In Me Parent Guide
• Social Studies
• MAP Practice
• 5th Grade Schedule
• 5th Grade Curriculum Information

Math Homework Pages and Answers

Topic 1: understand place value.

1-1: Patterns with Exponents and Powers of 10

• Homework Page

1-2: Understand Whole-Number Place Value

1-3: Decimals to Thousandths

1-4: Understand Decimal Place Value

1-5: Compare Decimals

1-6: Round Decimals

Topic 2: Use Models and Strategies to Add and Subtract Decimals

2-2: Estimate Sums and Differences of Decimals

2-3: Use Models to Add and Subtract Decimals

2-4: Use Strategies to Add Decimals

2-5: Use Strategies to Subtract Decimals

2-6: Model with Math

Topic 3: Fluently Multiply Multi-Digit Whole Numbers

3-1: Multiply Greater Numbers by Powers of 10

3-2: Estimate Products

3-3: Multiply by 1-Digit Numbers

3-4: Multiply 2-Digit by 2-Digit Numbers

3-5: Multiply 3-Digit by 2-Digit Numbers

3-6: Multiply Whole Numbers with Zeros

3-7: Practice Multiplying Multi-Digit Numbers

3-8: Solve Word Problems

3-9: Critique Reasoning

Topic 4: Use Models and Strategies to Multiply Decimals

4-1:Multiply Decimals by Powers of 10

4-2: Estimate the Product of a Decimal and a Whole Number 

4-3: Use Models to Multiply a Decimal and a Whole Number

4-4: Multiply a Decimal and a Whole Number

4-5: Use Models to Multiply a Decimal and a Decimal

4-6: Multiply Decimals Using Partial Products

4-7: Use Properties to Multiply Decimals

4-8: Use Number Sense to Multiply Decimals

4-9: Model with Math

Topic 5: Use Models and Strategies to Divide Whole Numbers

Topic 5-1: Use Patterns and Mental Math to Divide

Topic 5-2: Estimate Quotients with 2-Digit Divisors

Topic 5-3: Use Models and Properties to Divide with 2-Digit Divisors

Topic 5-4: Use Partial Quotients to Divide

Topic 5-5: Use Sharing to Divide: Two Digit Divisors

Topic 5-6: Use Sharing to Divide: Greater Dividends

Topic 5-7: Choose a Strategy to Divide 

Lesson 5-8: Make Sense and Persevere

Topic 6: Use Models and Strategies to Divide Decimals

6-1: Patterns for Dividing with Decimals

6-2: Estimate Decimals Quotients

6-3: Use Models to Divide by a 1-Digit Number

6-4: Divide by a 2-digit Whole Number

6-5: Divide by a Decimal

6-6: Reasoning 

Topic 7: Use Equivalent Fractions to Add and Subtract Fractions

7-2: Find Common Denominators

•   Answers

7-3: Add Fractions with Unlike Denominators

7-4: Subtract Fractions with Unlike Denominators

7-5: Add and Subtract Fractions

7-6: Estimate Sums and Differences of Mixed Numbers

7-7: Use Models to Add Mixed Numbers

7-8: Add Mixed Numbers

7-9: Use Models to Subtract Mixed Numbers

7-10: Subtract Mixed Numbers

7-11: Add and Subtract Mixed Numbers

Topic 8: Apply Understanding of Multiplication to Multiply Fractions

8-1: Multiply a Fraction by a Whole Number

8-2: Multiply a Whole Number by a Fraction

8-3: Multiply Fractions and Whole Numbers

8-4: Use Models to Multiply Two Fractions

8-5: Multiply Two Fractions

8-6: Area of a Rectangle

8-7: Multiply Mixed Numbers

Topic 9: Apply Understanding of Division to Divide Fractions

Lesson 9-1: Fractions and Division

Lesson 9-2: Fractions and Mixed Numbers as Quotients

Lesson 9-3: Use Multiplication to Divide

Lesson 9-4: Divide Whole Numbers by Unit Fractions

Lesson 9-5: Divide Unit Fractions by Non-Zero Whole Numbers

Lesson 9-6: Divide Whole Numbers and Unit Fractions

Lesson 9-7: Solve Problems Using Division

Lesson 9-8: Repeated Reasoning

Topic 10: Represent and Interpret Data

Lesson 10-1: Analyze Line Plots

Lesson 10-2: Make Line Plots

Lesson 10-3: Solve Word Problems Using Measurement Data

Lesson 10-4: Critique Reasoning 

Topic 11: Understand Volume Concepts

Lesson 11-1: Model Volume

Lesson 11-2: Develop a Formula

Lesson 11-3: Combine Volume of Prisms

Lesson 11-4: Solve Word Problems Using Volume

Lesson 11-5: Use Appropriate Tools

5th Grade Homework Policy

We value your family time. therefore, we will be intentional with any homework we send home. students’ daily homework will be required reading of at least 30 minutes., students will have nightly math homework which supports our learning in class. there are a lot of new math concepts in 5th grade and it is important for students' growth and understanding. additionally, study guides and other assignments may be sent home periodically throughout the year., please note: if a student exhibits off-task behaviors, fails to complete an assignment, or is struggling to understand a concept, an assignment will be sent home for completion..

• Questions or Feedback? |
• Web Community Manager Privacy Policy (Updated) |

CPM Educational Program

Expert textbook solutions.

Browse your textbook to find expert solutions, hints, and answers for all exercises. The solutions are always presented as a clear and concise, step-by-step explanation with included theory and helpful figures, graphs, and diagrams. Mathleaks covers the most commonly adopted textbooks with more than 250000 expert solutions.

Mathleaks Solver

With Mathleaks, you’re not tied to your textbook for solutions. Instead, scan and solve exercises with our math solver, which instantly reads the problem by using the camera on your smartphone or tablet. Access the solver through the Mathleaks app or on our website. The Mathleaks solver works for Pre-Algebra, Algebra 1, and Algebra 2.

Mathleaks Community

Get access to the world's most popular math community with Mathleaks. You can connect with other students all over the US who are studying with the same textbook or in the same math course.

Study math more efficiently using Mathleaks for CPM Educational Program textbooks.

• Core Connections Integrated I, 2013
• Core Connections Algebra 1, 2013
• Core Connections Geometry, 2013
• Core Connections Algebra 2, 2013
• Core Connections Integrated I, 2014
• Core Connections Integrated II, 2015
• Core Connections: Course 1
• Core Connections: Course 2
• Core Connections: Course 3
• Core Connections Integrated III, 2015

• Notifications 0

• Add Friend ($5)

As a registered member you can:

• View all solutions for free
• Request more in-depth explanations for free
• Ask our tutors any math-related question for free
• Email your homework to your parent or tutor for free
• Grade 5 HMH Go Math - Answer Keys

Explanation:

308.3  ÷ 15

Estimate the quotient.

63.5 ÷ 5

57.8 ÷ 81

172.6 ÷ 39

43.6 ÷ 8

2.8 ÷ 6

467.6  ÷ 8

209.3 ÷ 48

737.5 ÷ 9

256.1  ÷ 82

Taylor uses 645.6 gallons of water in 7 days. Suppose he uses the same amount of water each day. About how much water does Taylor use each day?

On a road trip, Sandy drives 368.7 miles. Her car uses a total of 18 gallons of gas. About how many miles per gallon does Sandy’s car get?

Yes, email page to my online tutor. ( if you didn't add a tutor yet, you can add one here )

Thank you for doing your homework!

Submit Your Question

Please ensure that your password is at least 8 characters and contains each of the following:

• a special character: @$#!%*?&

Quizard AI: Homework Helper 4+

Instant math problem solver, quizard ai, inc., designed for iphone.

• #83 in Education
• 4.7 • 31.2K Ratings
• Offers In-App Purchases

iPhone Screenshots

Description.

Homework & AI Answers Solver. Are you struggling to keep up with your homework and studying? Don't worry, Quizard is here to help! Quizard is a revolutionary AI answer app designed to help students at all levels conquer their studies. Now introducing our new feature - AI answers for math questions! Just snap a photo of your math problem, and Quizard's AI will deliver the solution, along with a comprehensive explanation. This makes Quizard the perfect tool not only for general studies but also for math homework help. Whether you’re a college student, high school student, or even an adult looking to brush up on your knowledge, Quizard is the perfect tool to help you succeed. With Quizard, you can quickly and easily get help with multiple-choice questions and short answer problems. You can quickly and easily prepare for quizzes, tests, and exams, allowing you to confidently ace them. Quizard is free to use! With Quizard, you can get the help you need to understand the material and gain a better understanding of the subject. Quizard is the perfect homework helper and personal tutor, providing you with the answers you need to succeed. With Quizard, you can get help with your homework and studying, so you can get better grades and have more free time. Stop struggling with your homework and start using Quizard today! Fine print: • Payment will be charged to your Apple ID account at the confirmation of purchase. • Offers and pricing are subject to change without notice. Terms of Use: https://lovely-vault-f15.notion.site/Terms-of-Use-18a469738fd74c9bb7c919981de4bbcd Privacy policy: https://lovely-vault-f15.notion.site/Privacy-Policy-4fe9e56db4a04d88af6e1b78f2874a7d Suggestions or questions? Email us at [email protected] TikTok: @quizard.ai Instagram: @quizard.ai

Version 1.9.6

We fixed some pesky bugs!

Ratings and Reviews

31.2K Ratings

Trivia Meets Kaleidoscopic Chaos!

Quizzard might just be the hidden gem in the app store, assuming you like your learning served with a heaping spoonful of whimsy and a side of technicolor mayhem. The interface could be described as a unicorn's dream, painted in every hue of the rainbow and then some. It's so cheerfully bright, you'll forget whether you're here to learn or to play a round of digital laser tag. But it’s not all about looks; Quizzard delivers content with a zesty twist that keeps you on your toes. Trivia questions that make your brain do somersaults are the norm here. Ever wanted to decode the molecular structure of caffeine in under a minute or guess Shakespeare's favorite cheese? Quizzard makes such intellectual gymnastics delightful, and you’ll chuckle as much as you'll ponder. Plus, the ads, while frequent, are an adventure in themselves, offering you snippets of the outside world at the most unexpected moments. It’s like getting commercial breaks during your personal game show—annoying but part of the fun. So, strap in for a quiz experience that’s as entertaining as it is enlightening, where every question is a surprise party for your neurons!

GREAT LEARNING TOOL MADE BY A GENIUS DUDE!!! Quizard reads your question and like searches the web for similar information to come up with an answer. Quizard provides a paragraph explanation along with the short answer. READ THOROUGHLY bc it is a computer and sometimes the wording makes it come up with “the wrong (short) answer” even though throughout the (longer paragraph) explanation you can figure out the true answer that Quizard happened to word incorrectly or chose a similar but wrong multiple choice option for. If you’re not trying using it for just the short answer not bothering to read and check the answer and dont read the paragraph then ya.. some of your answers are gonna be wrong. But if it doesn’t give you the right answer it’ll at least provide some explanation to put you on the right path.

DO NOT GET THIS APP!

So, I’m in 4th grade, 4TH GRADE! & IT COULDN’T HELP ME CORRECTLY! So, you know in elementary school papers, & the printer will add these little pictures to the side, well, when i took a picture of this paper, quizard told me that “ I’m sorry, there’s a object in this paper, i cannot tell you the correct answer unless you cover it” ….. IM SORRY! IS A SMALL PICTURE DISTRACTING YOU!? & HOW IS PUTTING SOMETHING OVER IT GOING TO HELP!? Mm!? Well, i cover it, & it is so stupid! It can’t even proofread! It told me that the words that were “incorrect” (by the way, i fact checked it, & the words were correct) & it tells me that the correct answer for gymnasium is “gymnasuim” …i’m sorry, did you just make up a word? WHAT KIND OF AI ARE YOU!? So, i tell it that it’s incorrect, it’s actually “gymnasium” & it tells me, “ I’m sorry for the inconvenience, but can you explain the answer’s to me?” ….. WHY DO YOU THINK I GOT THIS APP!? I mean like, it’s not like I wanted YOU to explain it to me! So, yeah, I'm not going to sit there & waist my time trying to explain a ELEMENTARY SCHOOL PAPER TO AN AI! So, I definitely DON’T recommend this app. ( it’s not even worth 1 star )

App Privacy

The developer, Quizard AI, Inc. , indicated that the app’s privacy practices may include handling of data as described below. For more information, see the developer’s privacy policy .

Data Used to Track You

The following data may be used to track you across apps and websites owned by other companies:

Data Not Linked to You

The following data may be collected but it is not linked to your identity:

• Identifiers

Privacy practices may vary, for example, based on the features you use or your age. Learn More

Information

• unlimited queue skips weekly $6.99
• quizard pro - monthly $9.99
• Quizard Pro Yearly $59.99
• quizard pro - monthly $19.99
• Quizard Pro Yearly $89.99
• Yearly Quizard Pro $39.99
• Quizard pro 6 months $49.99
• Quizard pro 6 months $69.99
• Yearly Quizard Pro $29.99
• sub_01 $2.99
• App Support
• Privacy Policy

You Might Also Like

Pic Answer - AI Solver

Answer.AI - Your AI tutor

Ai Homework Helper: Scan Solve

Nerd AI - Tutor & Math Helper

• Texas Go Math
• Big Ideas Math
• enVision Math
• EngageNY Math
• McGraw Hill My Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Go Math Answer Key for Grade K, 1, 2, 3, 4, 5, 6, 7, and 8

Go Math Answer Key:  HMH Go Math Answer Key for Grade K, 1, 2, 3, 4, 5, 6, 7, and 8 are provided helps students to have learning targets and achieve success at chapter and lesson level and makes learning visible.

Download Go Math Answer Key for Grades K-8 | HMH Go Math Solution Key for Grades Kindergarten, 1, 2, 3, 4, 5, 6, 7, 8

All the Concepts in the CCSS Go Math Answer Key for Grades Kindergarten, 1, 2, 3, 4, 5, 6, 7, 8 are given with straightforward and detailed descriptions.

Go Math Answer Key

Texas go math answer key.

• Grade 1-2 Quiz
• Grade 3-5 Quiz
• Grade 6-8 Quiz
• Grade 9-10 Quiz
• Grade 11-12 Quiz
• Big Ideas Math Answer Key
• Eureka Math Answer Key
• Envision Math Answer Key
• Engageny Math
• Everyday Mathematics Answer Key
• HMH Into Math Answer Key
• 180 Days of Math Answer Key
• Math Tables
• Math Calculator

Free Download Go Math Answer Key from Kindergarten to 8th Grade

Students can find Go Math Answer Keys right from  Primary School to High School all in one place. You just need to tap on the quick links available in order to access them and learn all the Chapters in each grade. Once you tap on the quick link you will be directed to the respective Grade Solutions Key wherein you can access the complete information. You can find Solutions for all the Go Math Textbook Questions free of cost and we don’t charge any amount.

• Go Math Kindergarten Answer Key

Go Math Grade 1 Answer Key

Go math grade 2 answer key, go math grade 3 answer key, go math grade 4 answer key, go math grade 5 answer key, go math grade 6 answer key.

• Go Math Grade 7 Answer Key

Go Math Grade 8 Answer Key

• Texas Go Math Grade 8 Answer Key
• Texas Go Math Grade 7 Answer Key
• Texas Go Math Grade 6 Answer Key
• Texas Go Math Grade 5 Answer Key
• Texas Go Math Grade 4 Answer Key
• Texas Go Math Grade 3 Answer Key
• Texas Go Math Grade 2 Answer Key
• Texas Go Math Grade 1 Answer Key
• Texas Go Math Kindergarten Answer Key

Go Math Grade K Answer Key

• Chapter 1 Represent, Count, and Write Numbers 0 to 5
• Chapter 2 Compare Numbers to 5
• Chapter 3 Represent, Count, and Write Numbers 6 to 9
• Chapter 4 Represent and Compare Numbers to 10
• Chapter 5 Addition
• Chapter 6 Subtraction
• Chapter 7 Represent, Count, and Write 11 to 19
• Chapter 8 Represent, Count, and Write 20 and Beyond
• Chapter 9 Identify and Describe Two-Dimensional Shapes
• Chapter 10 Identify and Describe Three-Dimensional Shapes
• Chapter 11 Measurement
• Chapter 12 Classify and Sort Data
• Chapter 1 Addition Concepts
• Chapter 2 Subtraction Concepts
• Chapter 3 Addition Strategies
• Chapter 4 Subtraction Strategies
• Chapter 5 Addition and Subtraction Relationships
• Chapter 6 Count and Model Numbers
• Chapter 7 Compare Numbers
• Chapter 8 Two-Digit Addition and Subtraction
• Chapter 9 Measurement
• Chapter 10 Represent Data
• Chapter 11 Three-Dimensional Geometry
• Chapter 12 Two-Dimensional Geometry
• Chapter 1 Number Concepts
• Chapter 2 Numbers to 1,000
• Chapter 3 Basic Facts and Relationships
• Chapter 4 2-Digit Addition
• Chapter 5 2-Digit Subtraction
• Chapter 6 3-Digit Addition and Subtraction
• Chapter 7 Money and Time
• Chapter 8 Length in Customary Units
• Chapter 9 Length in Metric Units
• Chapter 10 Data
• Chapter 11 Geometry and Fraction Concepts

Grade 3 HMH Go Math – Answer Keys

• Chapter 1: Addition and Subtraction within 1,000
• Chapter 2: Represent and Interpret Data
• Chapter 3: Understand Multiplication
• Chapter 4: Multiplication Facts and Strategies
• Chapter 5: Use Multiplication Facts
• Chapter 6: Understand Division
• Chapter 7: Division Facts and Strategies
• Chapter 8: Understand Fractions
• Chapter 9: Compare Fractions
• Chapter 10: Time, Length, Liquid Volume, and Mass
• Chapter 11: Perimeter and Area
• Chapter 12:Two-Dimensional Shapes

Grade 3 HMH Go Math – Extra Practice Questions and Answers

• Chapter 1: Addition and Subtraction within 1,000 Extra Practice
• Chapter 2: Represent and Interpret Data Extra Practice
• Chapter 3: Understand Multiplication Extra Practice
• Chapter 4: Multiplication Facts and Strategies Extra Practice
• Chapter 5: Use Multiplication Facts Extra Practice
• Chapter 6: Understand Division Extra Practice
• Chapter 7: Division Facts and Strategies Extra Practice
• Chapter 8: Understand Fractions Extra Practice
• Chapter 9: Compare Fractions Extra Practice
• Chapter 10: Time, Length, Liquid Volume, and Mass Extra Practice
• Chapter 11: Perimeter and Area Extra Practice
• Chapter 12: Two-Dimensional Shapes Extra Practice

Common Core Grade 4 HMH Go Math – Answer Keys

• Chapter 1 Place Value, Addition, and Subtraction to One Million
• Chapter 2 Multiply by 1-Digit Numbers
• Chapter 3 Multiply 2-Digit Numbers
• Chapter 4 Divide by 1-Digit Numbers
• Chapter 5 Factors, Multiples, and Patterns
• Chapter 6 Fraction Equivalence and Comparison
• Chapter 7 Add and Subtract Fractions
• Chapter 8 Multiply Fractions by Whole Numbers
• Chapter 9 Relate Fractions and Decimals
• Chapter 10 Two-Dimensional Figures
• Chapter 11 Angles
• Chapter 12Relative Sizes of Measurement Units
• Chapter 13 Algebra: Perimeter and Area

Grade 4 Homework Practice FL.

Common Core – Grade 4 – Practice Book

• Chapter 1 Place Value, Addition, and Subtraction to One Million  (Pages 1- 20)
• Chapter 2 Multiply by 1-Digit Numbers  (Pages 21 – 47)
• Chapter 3 Multiply 2-Digit Numbers  (Pages 49- 65)
• Chapter 4 Divide by 1-Digit Numbers  (Pages 67 – 93)
• Chapter 5 Factors, Multiples, and Patterns  (Pages 95 – 109)
• Chapter 6 Fraction Equivalence and Comparison  (Pages 111 – 129)
• Chapter 7 Add and Subtract Fractions  (Pages 131 – 153)
• Chapter 8 Multiply Fractions by Whole Numbers  (Pages 155- 167)
• Chapter 9 Relate Fractions and Decimals  (Pages 169- 185)
• Chapter 10 Two-Dimensional Figures  (Pages 187- 204)
• Chapter 11 Angles  (Pages 205- 217)
• Chapter 12 Relative Sizes of Measurement Units  (Pages 219- 244)
• Chapter 13 Algebra: Perimeter and Area  (Pages 245- 258)

Grade 4 Homework FL. – Answer Keys

• Chapter 2 Multiply by 1-Digit Numbers Review/Test
• Chapter 3 Multiply 2-Digit Numbers Review/Test
• Chapter 4 Divide by 1-Digit Numbers Review/Test
• Chapter 5 Factors, Multiples, and Patterns Review/Test
• Chapter 6 Fraction Equivalence and Comparison Review/Test
• Chapter 7 Add and Subtract Fractions Review/Test
• Chapter 8 Multiply Fractions by Whole Numbers Review/Test
• Chapter 9 Relate Fractions and Decimals Review/Test
• Chapter 10 Two-Dimensional Figures Review/Test
• Chapter 11 Angles Review/Test
• Chapter 12 Relative Sizes of Measurement Units Review/Test
• Chapter 13 Algebra: Perimeter and Area Review/Test
• Chapter 1: Place Value, Multiplication, and Expressions
• Chapter 2: Divide Whole Numbers
• Chapter 3: Add and Subtract Decimals
• Chapter 4: Multiply Decimals
• Chapter 5: Divide Decimals
• Chapter 6: Add and Subtract Fractions with Unlike Denominators
• Chapter 7: Multiply Fractions
• Chapter 8: Divide Fractions
• Chapter 9: Algebra: Patterns and Graphing
• Chapter 10: Convert Units of Measure
• Chapter 11: Geometry and Volume
• Chapter 1: Divide Multi-Digit Numbers
• Chapter 2: Fractions and Decimals
• Chapter 3: Understand Positive and Negative Numbers
• Chapter 4: Model Ratios
• Chapter 5: Model Percents
• Chapter 6: Convert Units of Length
• Chapter 7: Exponents
• Chapter 8: Solutions of Equations
• Chapter 9: Independent and Dependent Variables
• Chapter 10: Area of Parallelograms
• Chapter 11: Surface Area and Volume
• Chapter 12: Data Displays and Measures of Center
• Chapter 13: Variability and Data Distributions

Go Math Answer Key for Grade 7

• Chapter 1: Adding and Subtracting Integers
• Chapter 2: Multiplying and Dividing Integers
• Chapter 3: Rational Numbers
• Chapter 4: Rates and Proportionality
• Chapter 5: Percent Increase and Decrease
• Chapter 6: Algebraic Expressions
• Chapter 7: Writing and Solving One-Step Inequalities
• Chapter 8: Modeling Geometric Figures
• Chapter 9: Circumference, Area, and Volume
• Chapter 10: Random Samples and Populations
• Chapter 11: Analyzing and Comparing Data
• Chapter 12: Experimental Probability
• Chapter 13: Theoretical Probability and Simulations
• Chapter 1 Real Numbers
• Chapter 2 Exponents and Scientific Notation
• Chapter 3 Proportional Relationships
• Chapter 4 Nonproportional Relationships
• Chapter 5 Writing Linear Equations
• Chapter 6 Functions
• Chapter 7 Solving Linear Equations
• Chapter 8 Solving Systems of Linear Equations
• Chapter 9 Transformations and Congruence
• Chapter 10 Transformations and Similarity
• Chapter 11 Angle Relationships in Parallel Lines and Triangles
• Chapter 12 The Pythagorean Theorem
• Chapter 13 Volume
• Chapter 14 Scatter Plots
• Chapter 15 Two-Way Tables

Give your kid the right amount of knowledge he needs as a part of your preparation by taking the help of our HMH Go Math Answer Key for Grades K-8. Resolve all your queries and assess your preparation standard using the Common Core Go Math Solution Key.

Practicing from the Go Math Answer Key for Grades K to 8 will provide a grade by grade roadmap and prepares students for College Readiness. Gradewise HMH Go Math Answer Key provided will develop problem-solving skills among students thereby helping them to Think, Explore and Grow. The Diverse Opportunities provided helps Kids to master the content with engaging activities.

Characteristics of Go Math Answer Key for Grades K to 8

Go through the below-listed features of referring to the HMH Go Math Anwer Key for Grades K to 8. They are outlined as follows

• All the Go Math Answer Key for Grades K to 8 are easy to download and we don’t charge any penny from you.
• Step by Step Solutions provided in the HMH Go Math Practice Key is aligned as per the College and Career Expectations.
• Solving from the Math 101 Practice Key helps you inculcate Higher Order Thinking Skills and you can answer any Question from your Homework, Assessment, or Review Test.
• More Rigorous Content made available meets the Common Core State Standards Initiative.
• You can gain a deeper knowledge of mathematical concepts and find solutions to all the Questions from Go Math Textbooks for Grades K, 1, 2, 3, 4, 5, 6, 7, 8

FAQs on Common Core HMH Go Math Answer Key

1. When Can I use the Go Math Answer Key for Grades K-8?

You can use the HMH Go Math Answer Key for Grades K to 8 while practicing the Go Math Textbook Questions as a part of your Homework or Assessment and make the most out of them.

2. Is there any site that provides the Common Core Go Math Solutions Key for Grades K, 1, 2, 3, 4, 5, 6, 7, 8?

Yes, you can find Go Math Answer Key for Grades K, 1, 2, 3, 4, 5, 6, 7, 8 all in one place i.e. ccssmathanswers.com a trusted and reliable portal.

3. Can I download HMH Go Math Answer Key PDF for free?

Yes, you can download the HMH Math 101 Practice Key for free on our page via quick links available and we don’t charge any amount for it.

• Big Ideas Math Answers

Elementary School Big Ideas Math Answers

• Big Ideas Math Answers Grade K
• Big Ideas Math Answers Grade 1
• Big Ideas Math Answers Grade 2
• Big Ideas Math Answers Grade 3
• Big Ideas Math Answers Grade 4
• Big Ideas Math Answers Grade 5

Middle School Big Ideas Math Solutions

• Big Ideas Math Answers Grade 6
• Big Ideas Math Answers Grade 6 Advanced
• Big Ideas Math Answers Grade 7
• Big Ideas Math Answers Grade 7 Advanced
• Big Ideas Math Answers Grade 7 Accelerated
• Big Ideas Math Answers Grade 8

High School Big Ideas Math Answers

• Big Ideas Math Algebra 1 Answers
• Big Ideas Math Algebra 2 Answers
• Big Ideas Math Geometry Answers
• Envision Math Common Core Grade K Answer Key
• Envision Math Common Core Grade 1 Answer Key
• Envision Math Common Core Grade 2 Answer Key
• Envision Math Common Core Grade 3 Answer Key
• Envision Math Common Core Grade 4 Answer Key
• Envision Math Common Core Grade 5 Answer Key
• Envision Math Common Core Grade 6 Answer Key
• Envision Math Common Core Grade 7 Answer Key
• Envision Math Common Core Grade 8 Answer Key

Engage NY Eureka Math Answer Key

Engage NY Math Answer Key Pre K – 12

• engage ny math answer key
• Eureka Math Pre K Answer Key
• Eureka Math Kindergarten Answer Key
• Eureka Math Grade 1 Answer Key
• Eureka Math Grade 2 Answer Key
• Eureka Math Grade 3 Answer Key
• Eureka Math Grade 4 Answer Key
• Eureka Math Grade 5 Answer Key
• Eureka Math Grade 6 Answer Key
• Eureka Math Grade 7 Answer Key
• Eureka Math Grade 8 Answer Key
• Eureka Math Algebra 1 Answer Key
• Eureka Math Geometry Answer Key
• Eureka Math Pre Calculus Answer Key
• Eureka Math Algebra 2 Answer Key
• Everyday Mathematics Kindergarten Answer Key
• Everyday Mathematics Grade 1 Answer Key
• Everyday Mathematics Grade 2 Answer Key
• Everyday Mathematics Grade 3 Answer Key
• Everyday Mathematics Grade 4 Answer Key
• Everyday Mathematics Grade 5 Answer Key
• Everyday Mathematics Grade 6 Answer Key
• Math Expressions Grade 5 Homework and Remembering Answer Key
• Math Expressions Grade 4 Homework and Remembering Answer Key
• Math Expressions Grade 3 Homework and Remembering Answer Key
• Math Expressions Grade 2 Homework and Remembering Answer Key
• Math Expressions Grade 1 Homework and Remembering Answer Key
• Math Expressions Grade K Homework and Remembering Answer Key

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Eureka Math Grade 5 Module 3 Lesson 13 Answer Key

Engage ny eureka math 5th grade module 3 lesson 13 answer key, eureka math grade 5 module 3 lesson 13 problem set answer key.

Question 1. Are the following expressions greater than or less than 1? Circle the correct answer. a. \(\frac{1}{2}\) + \(\frac{2}{7}\)    greater than 1    less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)   greater than 1     less than 1

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   greater than 1     less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\) lcm of 8 and 5 is 40. \(\frac{25}{40}\) + \(\frac{24}{40}\)   = \(\frac{49}{40}\) = 1\(\frac{9}{40}\)

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   = \(\frac{5}{4}\) – \(\frac{1}{3}\) lcm of 4 and 3 is 12. \(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\)

d. 3\(\frac{5}{8}\) – 2\(\frac{5}{9}\)  = \(\frac{29}{8}\) – \(\frac{23}{9}\) lcm of 8 and 9 is 72 . \(\frac{261}{72}\) – \(\frac{184}{72}\)  = \(\frac{77}{72}\) = 1\(\frac{5}{72}\) .

Question 2. Are the following expressions greater than or less than \(\frac{1}{2}\) ? Circle the correct answer. a. \(\frac{1}{4}\) + \(\frac{2}{3}\)      greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\)        greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\)       greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\) lcm of 7 and 8 is 56. \(\frac{24}{56}\) – \(\frac{7}{56}\) =  \(\frac{17}{56}\) less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\) = \(\frac{8}{7}\) – \(\frac{7}{8}\) lcm of 7 and 8 is 56. \(\frac{64}{56}\) – \(\frac{49}{56}\) = \(\frac{15}{56}\) less than \(\frac{1}{2}\)

d. \(\frac{3}{7}\) + \(\frac{2}{6}\) lcm of 7 and 6 is 42. \(\frac{18}{42}\) + \(\frac{14}{42}\)  = \(\frac{32}{42}\) = \(\frac{16}{21}\) greater than \(\frac{1}{2}\) .

Question 3. Use > , < , or = to make the following statements true. a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) _______ 8\(\frac{2}{3}\) b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) _______ 1\(\frac{5}{8}\) + \(\frac{2}{5}\) c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) _______ 6 + \(\frac{13}{14}\) d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) _______ 4\(\frac{4}{7}\) + \(\frac{2}{5}\) Answer: a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = 8\(\frac{2}{3}\) b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) < 1\(\frac{5}{8}\) + \(\frac{2}{5}\) c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = 6 + \(\frac{13}{14}\) d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) > 4\(\frac{4}{7}\) + \(\frac{2}{5}\) Explanation : a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = \(\frac{17}{3}\) + \(\frac{12}{4}\) lcm of 3 and 4 is 12. \(\frac{68}{12}\) + \(\frac{36}{12}\) = \(\frac{104}{12}\) = \(\frac{26}{3}\) =8\(\frac{2}{3} \)

b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) = \(\frac{37}{8}\) – \(\frac{17}{5}\) lcm of 8 and 5 is 40 . \(\frac{185}{40}\) – \(\frac{136}{40}\) = \(\frac{49}{40}\) =1\(\frac{9}{40}\) 1\(\frac{5}{8}\) + \(\frac{2}{5}\) = \(\frac{13}{8}\) + \(\frac{2}{5}\) lcm of 8 and 5 is 40 . \(\frac{65}{40}\) + \(\frac{16}{5}\) = \(\frac{81}{40}\) = 2\(\frac{1}{40}\)

c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = \(\frac{11}{2}\) + \(\frac{10}{7}\) lcm of 2 and 7 is 14. \(\frac{77}{14}\) + \(\frac{20}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\) 6 + \(\frac{13}{14}\) =\(\frac{84}{14}\) + \(\frac{13}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\)

d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) =\(\frac{109}{7}\) – \(\frac{57}{5}\) lcm of 7 and 5 is 35 . \(\frac{545}{35}\) – \(\frac{399}{35}\) = \(\frac{944}{35}\) = 26\(\frac{34}{35}\) . 4\(\frac{4}{7}\) + \(\frac{2}{5}\) = \(\frac{32}{7}\) + \(\frac{2}{5}\) lcm of 7 and 5 is 35 . \(\frac{160}{35}\) + \(\frac{14}{35}\) = \(\frac{174}{35}\)= 4\(\frac{34}{35}\)

Question 4. Is it true that 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\)? Prove your answer. Answer: No it is wrong 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) < 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\) Explanation : 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = \(\frac{23}{5}\) – \(\frac{11}{3}\) lcm of 5 and 3 is 15 . \(\frac{69}{15}\) – \(\frac{55}{15}\) = \(\frac{14}{15}\)

1 + \(\frac{3}{5}\) + \(\frac{2}{3}\) lcm of 5 and 3 is 15 . \(\frac{15}{15}\) + \(\frac{9}{15}\) + \(\frac{10}{15}\) = latex]\frac{34}{15}[/latex] = 2latex]\frac{4}{15}[/latex] .

Question 5. Jackson needs to be 1\(\frac{3}{4}\) inches taller in order to ride the roller coaster. Since he can’t wait, he puts on a pair of boots that add 1\(\frac{1}{6}\) inches to his height and slips an insole inside to add another \(\frac{1}{8}\) inch to his height. Will this make Jackson appear tall enough to ride the roller coaster? Answer: Fraction of Height required to ride a roller coaster for Jackson = 1\(\frac{3}{4}\) inches. Fraction of his height = 1\(\frac{1}{6}\) inches = \(\frac{7}{6}\) Fraction of his boots length = \(\frac{1}{8}\) inches Total fraction of his height with boots = \(\frac{7}{6}\) + \(\frac{1}{8}\) = \(\frac{28}{24}\) + \(\frac{3}{24}\) = \(\frac{31}{24}\) = 1\(\frac{7}{24}\) . 1\(\frac{3}{4}\) = multiply by 6 both numerator and denominator = 1\(\frac{18}{24}\) therefore, 1\(\frac{18}{24}\) > is greater than 1\(\frac{7}{24}\)  So, he is not taller enough to ride roller coaster . So, he cant ride the roller coaster .

Question 6. A baker needs 5 lb of butter for a recipe. She found 2 portions that each weigh 1\(\frac{1}{6}\) lb and a portion that weighs 2\(\frac{2}{7}\) lb. Does she have enough butter for her recipe? Answer: Fraction of butter required for a recipe = 5 lb Fraction of 2 portions that weigh = 2 × \(\frac{7}{6}\) lb = \(\frac{7}{3}\) Fraction of portions that weighs = 2\(\frac{2}{7}\) lb. = \(\frac{16}{7}\) lb. Fraction of butter of portions = \(\frac{7}{3}\) + \(\frac{16}{7}\) = \(\frac{49}{21}\) + \(\frac{48}{21}\) = \(\frac{97}{21}\) = 4\(\frac{13}{21}\) Therefore, she doesnot have enough butter for the recipe =  4\(\frac{13}{21}\)

Eureka Math Grade 5 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1. Circle the correct answer. a. \(\frac{1}{2}\) +\(\frac{5}{12}\)        greater than 1          less than 1

b. 2\(\frac{7}{8}\) – 1\(\frac{7}{9}\)      greater than 1         less than 1

c. 1\(\frac{1}{12}\) – \(\frac{7}{10}\)     greater than \(\frac{1}{2}\)      less than \(\frac{1}{2}\)

Question 2. Use > , < , or = to make the following statement true. 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\) Answer: 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\) Explanation : 4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) = \(\frac{24}{5}\) + \(\frac{11}{3}\) lcm of 5 and 3 is 15 . \(\frac{72}{15}\) + \(\frac{55}{15}\) = \(\frac{127}{15}\) = 8 \(\frac{7}{15}\)

Eureka Math Grade 5 Module 3 Lesson 13 Homework Answer Key

Question 1. Are the following expressions greater than or less than 1? Circle the correct answer. a. \(\frac{1}{2}\) + \(\frac{4}{9}\)        greater than 1          less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)        greater than 1          less than 1

c. 1\(\frac{1}{5}\) – \(\frac{1}{3}\)       greater than 1           less than 1

Question 2. Are the following expressions greater than or less than \(\frac{1}{2}\)? Circle the correct answer. a. \(\frac{1}{5}\) + \(\frac{1}{4}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

b. \(\frac{6}{7}\) – \(\frac{1}{6}\)         greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{5}{6}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

Question 3. Use > , < , or = to make the following statements true. a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) _______ 8\(\frac{3}{4}\) b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) _______ 1\(\frac{4}{7}\) + \(\frac{3}{5}\) c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) _______ 5 + \(\frac{13}{18}\) d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) _______ 3\(\frac{3}{8}\) + \(\frac{3}{5}\) Answer: a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) < 8\(\frac{3}{4}\) b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) < 1\(\frac{4}{7}\) + \(\frac{3}{5}\) c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) < 5 + \(\frac{13}{18}\) d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) > 3\(\frac{3}{8}\) + \(\frac{3}{5}\) Explanation : a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) = \(\frac{29}{5}\) + \(\frac{8}{3}\) lcm of 5 and 3 is 15 . \(\frac{57}{15}\) + \(\frac{40}{15}\) = \(\frac{97}{15}\) = 6\(\frac{7}{15}\) .

b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) = \(\frac{25}{7}\) – \(\frac{13}{5}\) . lcm of 7 and 5 is 35 . \(\frac{125}{35}\) – \(\frac{91}{35}\) = \(\frac{34}{35}\) 1\(\frac{4}{7}\) + \(\frac{3}{5}\) = \(\frac{11}{7}\) + \(\frac{3}{5}\) lcm of 5 and 7 is 35 . \(\frac{55}{35}\) + \(\frac{21}{35}\) = \(\frac{76}{35}\) = 2 \(\frac{6}{35}\)

c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) = \(\frac{9}{2}\) + \(\frac{13}{9}\) lcm of 2 and 9 is 18 . 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) 5 + \(\frac{13}{18}\) = \(\frac{90}{18}\) + \(\frac{13}{18}\) = \(\frac{103}{18}\) = 5\(\frac{13}{18}\)

d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) = \(\frac{83}{8}\) – \(\frac{38}{5}\) lcm of 8 and 5 is 40 . \(\frac{415}{40}\) – \(\frac{304}{40}\) = \(\frac{311}{40}\) = 7\(\frac{31}{40}\) 3\(\frac{3}{8}\) + \(\frac{3}{5}\) = \(\frac{27}{8}\) + \(\frac{3}{5}\) lcm of 8 and 5 is 40 . \(\frac{135}{40}\) + \(\frac{24}{40}\) = \(\frac{159}{40}\)= 3\(\frac{39}{40}\)

Question 4. Is it true that 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\)? Prove your answer. Answer: It is not true . 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) < 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\) Explanation : 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = \(\frac{17}{3}\) – \(\frac{15}{4}\) lcm of 3 and 4 is 12. \(\frac{68}{12}\) – \(\frac{45}{12}\) = \(\frac{23}{12}\) = 1\(\frac{11}{12}\)

1 + \(\frac{2}{3}\) + \(\frac{3}{4}\) lcm of 3 and 4 is 12 . \(\frac{12}{12}\) + \(\frac{8}{12}\) + \(\frac{9}{12}\) = \(\frac{29}{12}\) = 2\(\frac{5}{12}\) .

Question 5. A tree limb hangs 5\(\frac{1}{4}\) feet from a telephone wire. The city trims back the branch before it grows within 2 \(\frac{1}{2}\) feet of the wire. Will the city allow the tree to grow 2\(\frac{3}{4}\) more feet? Answer: Fraction of height at which telephone wire is hung = 5\(\frac{1}{4}\) =\(\frac{21}{4}\)  feet Fraction of height city allow the tree to grow = 2\(\frac{3}{4}\) = \(\frac{11}{4}\) feet . Fraction of height city trims back the branch before it grows = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) feet Fraction of height of telephone wire can be hang = \(\frac{21}{4}\)  – \(\frac{11}{4}\)  = \(\frac{10}{4}\) = \(\frac{5}{2}\) both are equal that means the tree will be trim back .

Question 6. Mr. Kreider wants to paint two doors and several shutters. It takes 2\(\frac{1}{8}\) gallons of paint to coat each door and 1\(\frac{3}{5}\) gallons of paint to coat all of his shutters. If Mr. Kreider buys three 2-gallon cans of paint, does he have enough to complete the job? Answer: Fraction of cost of paint to coat each door = 2\(\frac{1}{8}\) gallons = \(\frac{17}{8}\) Fraction of cost of paint to coat all his shutters = 1\(\frac{3}{5}\) gallons = \(\frac{8}{5}\) Fraction of cost to paint 2 doors and shutters = 2 × \(\frac{17}{8}\) + \(\frac{8}{5}\) = \(\frac{17}{4}\) + \(\frac{8}{5}\) = \(\frac{85}{20}\) + \(\frac{32}{20}\) = \(\frac{117}{20}\) = 5latex]\frac{17}{20}[/latex] Total paint = three 2-gallon cans of paint = 3 × 2 = 6 gallons. Therefore Kreider doesn’t have sufficient amount of paint .

Leave a Comment Cancel Reply

You must be logged in to post a comment.