Lesson Miscellaneous evaluating problems
Algebra
->
Customizable Word Problem Solvers
->
Evaluation
-> Lesson Miscellaneous evaluating problems
Log On
Ad:
Over 600 Algebra Word Problems at edhelper.com
Word Problems: Evaluation, Substitution
Word
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Source code of 'Miscellaneous evaluating problems'
This Lesson (Miscellaneous evaluating problems)
was created by by
ikleyn(52780)
:
View Source
,
Show
About ikleyn
:
<H2>Miscellaneous evaluating problems</H2> <H3>Problem 1</H3>If a-b = 4 and ab+c^2+4 = 0, find a+b. <B>Solution</B> <pre> From a-b = 4 you have a = b+4. Sunstitute it into the second equation, replacing "a". You will have (b+4)*b + c^2 + 4 = 0, or b^2 + 4b + c^2 = 0, (b^2 + 4b + 4) + c^2 = 0 (b+2)^2 + c^2 = 0. The sum of two squares of real numbers can be equal to zero if and only iff both addends are zeros. Thus (b+2)^2 = 0, which implies b= -2. Then from very first equation, a = 4 + b = 4 + (-2) = 2. Then a + b = 2 + (-2) = 0, and the solution is completed. </pre> <H3>Problem 2</H3>If (2006-a)(2004-a)=2005, find (2006-a)^2 + (2004-a)^2. <B>Solution</B> <pre> From (2006-a)(2004-a)=2005 you have 2006*2004 - 2006a - 2004a + a^2 = 2005, 2006*2004 - 2*2005*a + a^2 = 2005, -2*2005*a + a^2 = 2005 - 2006*2004. (1) Next, (2006-a)^2 + (2004-a)^2 = 2006^2 -2*2006*a + a^2 + 2004^2 -2*2004*a + a^2 = = 2006^2 + 2004^2 -4*2005a + 2a^2 = 2006^2 + 2004^2 + (2*(-2*2005a + a^2)) = Now replace the expression (-2*2005a + a^2) by 2005 - 2006*2007, based on (1), and then you get = 2006^2 + 2004^2 + 2*(2005 - 2006*2004) = (2006-2004)^2 + 2*2005 = 2^2 + 4010 = 4014. <U>Answer</U>. 4014. </pre> <H3>Problem 3</H3>If a-b = -6 and b+c = 9, find the value of {{{3a^2-b^2-2c^2}}}. <B>Solution</B> <pre> If a-b = -6 and b+c = 9, then a = -6 + b, c = 9 - b, 3a^2 - b^2 - 2c^2 = 3*(-6+b)^2 - b^2 - 2*(9-b)^2 = = 3*(36 - 12b + b^2) - b^2 - 2*(81 - 18b + b^2) = 108 - 36b + 3b^2 - b^2 - 162 + 36b - 2b^2 = = combine like terms = -54. <U>ANSWER</U>. If a-b = -6 and b+c = 9, then the value of {{{3a^2-b^2-2c^2}}} is -54. </pre> <H3>Problem 4</H3>Evaluate the following sums a) 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 b) 300 - 299 + 298 - 297 + ... + 100 - 99 <B>Solution</B> a) 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 <pre> Group the numbers in pairs, using parentheses: 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 = = (1-3) + (5-7) + (9-11) + . . + (201-203) = = -2 + (-2) + (-2) + . . . + (-2) = 51 times (-2) = 51*(-2) = -102. </pre> b) 300 - 299 + 298 - 297 + ... + 100 - 99 <pre> The same idea works: Group the numbers in pairs, using parentheses 300 - 299 + 298 - 297 + . . . + 100 - 99 = = (300-299) + (298-297) + . . . + (100-99) = = 1 + 1 + . . . + 1 = 101 times 1 = 101*1 = 101. </pre> <H3>Problem 5</H3>Express the following product as reduced fraction {{{(1-1/2)(1-1/3)(1-1/4)*ellipsis*(1-1/2008)}}}. <B>Solution</B> <pre> {{{(1-1/2)(1-1/3)(1-1/4)......(1-1/2008)}}} = {{{(1/2)*(2/3)*(3/4)*ellipsis*(2007/2008)}}} = Notice that the denominator of the first fraction is the numerator of the second fraction. Next, the denominator of the second fraction is the numerator of the third fraction. This pattern continues: the denominator of the third fraction is the numerator of the fourth fraction. And so on . . . Cancel all common factors in numerators and denominators to get final answer = {{{1/2008}}}. </pre> <H3>Problem 6</H3>Find the value of the product {{{(1-1/2^2)(1-1/3^2)(1-1/4^2)*ellipsis*(1-1/999^2)}}}. <B>Solution</B> <pre> First, it is clear that P = {{{(1-1/2^2)(1-1/3^2)(1-1/4^2)*ellipsis*(1-1/999^2)}}} = {{{(1-1/2)*(1-1/3)*(1-1/4)*ellipsis*(1-1/999)}}}*({{{(1+1/2)*(1+1/3)*(1+1/4)*ellipsis*(1+1/999)}}} Now, consider Q = {{{(1-1/2)*(1-1/3)*(1-1/4)*ellipsis*(1-1/999)}}}. Notice that the denominator of the first fraction is the numerator of the second fraction. Next, the denominator of the second fraction is the numerator of the third fraction. This pattern continues: the denominator of the third fraction is the numerator of the fourth fraction. And so on . . . Cancel all common factors in numerators and denominators to get the value Q = {{{1/999}}}. Next, consider R = {{{(1+1/2)*(1+1/3)*(1+1/4)*ellipsis*(1+1/999)}}}. Notice that the numerator of the first fraction is the denominator of the second fraction. Next, the numerator of the second fraction is the denominator of the third fraction. This pattern continues: the numerator of the third fraction is the denominator of the fourth fraction. And so on . . . Cancel all common factors in numerators and denominators to get the value R = {{{(1/2)*1000}}} = 500. So, the final answer is P = Q*R = {{{(1/999)*500}}} = {{{500/999}}}. <U>ANSWER</U>. </pre> <H3>Problem 7</H3>Find the sum of the sequences 8, 88, 888, . . . up to n terms. <B>Solution</B> <pre> So, we want to find the sum R = 8 + 88 + 888 + . . . + 888...8, (1) where the last number is written using "n" digits of 8. Consider another, closely related sum T = {{{(9/8)*R}}} = 9 + 99 + 999 + . . . + 999...9, (2) where the last number is written using "n" digits of 9. Add n to T. It is the same as to add "1" to each term of the sum (2). Therefore, T + n = 10 + 100 + 1000 + . . . + 10^n. (3) The sum (3) is the sum of n first terms of the geometric progression with the first term 10 and the common ratio of 10. So, the sum (3) is equal to T + n = {{{10*((10^n-1)/(10-1))}}} = {{{(10/9)*(10^n-1)}}}. Thus you have T = {{{(10/9)*(10^n-1) -n}}}, or {{{(9/8)*R}}} = {{{(10/9)*(10^n-1) -n}}}, which implies R = {{{(80/81)*(10^n-1) - (8/9)*n}}}. It is the final formula and the <U>ANSWER</U> : R = 8 + 88 + 888 + . . . + 888...8 = {{{(80/81)*(10^n-1) - (8/9)*n}}}. <U>Check</U>. Let us take n = 2. Then from one side, R = 8 + 88 = 96. From the other side, R = {{{(80/81)*(100-1) - (8/9)*2}}} = {{{(80/81)*99 - (8/9)*2}}} = {{{(80*11)/9 - 16/9}}} = {{{(80*11-16)/9}}} = {{{864/9}}} = 88. ! Correct ! Let us take n = 3. Then from one side, R = 8 + 88 + 888 = 984. From the other side, R = {{{(80/81)*(1000-1) - (8/9)*3}}} = {{{(80/81)*999 - (8/9)*3}}} = {{{(80*111)/9 - 24/9}}} = {{{(80*111-24)/9}}} = {{{864/9}}} = 984. ! Correct ! </pre> <H3>Problem 8</H3>Find the sum of all possible 4-digit numbers that can be formed using the digits 2, 4, 5, 6, 7, and 8, with no repeated digits. <B>Solution</B> <pre> In forming one of those 4-digit numbers, there are 6 choices for the first digit, then 5 for the second, 4 for the third, and 3 for the fourth. So, there are 6*5*4*3 = 360 four-digit numbers that can be formed in this way. Imagine the list of those 360 numbers, ready to be added. Consider, for example the digit 4 in the "hundreds" position. There are 5*4*3 = 60 numbers among all 360 numbers that contain the digit "4" in this position. (The factor 5 corresponds to any of 5 remaining digits in the "thousands" position; factor 4 corresponds to any of 4 remaining digits in the "tens" position; factor 3 corresponds to any of 3 remaining digits in the "ones" position). It is true for any of 6 digits in any of four positions: Each of the 6 digits is used 60 times in each of four positions. It means that in each column of the list of the 360 numbers, each of the given digits is used 60 times. The sum of the given digits is 2+4+5+6+7+8 = 32. So the sum of the digits in each column is 60*32 = 1920. And then the sum of the 360 numbers is 1920*(1000+100+10+1) = 2133120. <U>Answer</U>. The sum is 2133120. </pre> <H3>Problem 9</H3>If x + y + z = 20 and x^2 + y^2 + z^2 = 100, find the value of xy + xz + yz. <B>Solution</B> <pre> Use an identity {{{(x+y+z)^2}}} = {{{x^2 + y^2 + z^2 + 2xy + 2xz + 2yz}}}. Substitute the given data. You will get {{{20^2}}} = 100 + 2*(xy + xz + yz), which implies xy + xz + yz = {{{(400-100)/2}}} = {{{300/2}}} = 150. <U>ANSWER</U> <U>Answer</U>. Then the value of xy + xz + yz is equal to 150. </pre> <H3>Problem 10</H3>If F(x) = ax^2 +bx+c and F(x+5) = x^2+9x-7, find the sum of a+b+c. <B>Solution</B> <pre> You have two sources of information: 1) F(x) = ax^2 + bx + c (a general definition of the polynomial F(x)). From it, a + b + c = F(1). 2) You also are given the second piece of information: F(x+5) = x^2 + 9x - 7. To get F(1), you need to take x = -4 in the last identity, and then you will get a + b + c = F(1) = F(-4 + 5) = (-4)^2 + 9*(-4) - 7 = 16 - 36 - 7 = -27. <U>ANSWER</U>. a + b + c = -27. </pre> <H3>Problem 11</H3>For positive numbers a, b, and c, if 2ab = 1, 3bc = 2, and 4ca = 3, find the value of a + b + c. <B>Solution</B> <pre> From the given info, we have these equations ab = {{{1/2}}} (1) bc = {{{2/3}}} (2) ac = {{{3/4}}} (3) Multiply all these three equations (their left sides and their right sides separately). You will get a^2*b^2*c^2 = {{{(1/2)*(2/3)*(3/4)}}} = {{{1/4}}}, or (a*b*c)^2 = {{{1/4}}}. Take the square root of both sides. Since "a", "b", and "c" are positive, it gives you a*b*c = {{{1/2}}}. (4) At this point, you may find "a", "b" and "c" separately. To find "a", divide (4) by (2); To find "b", divide (4) by (3); To find "c", divide (4) by (1). You will get a = {{{((1/2))/((2/3))}}} = {{{3/4}}}; b = {{{((1/2))/((3/4))}}} = {{{2/3}}}; c = {{{((1/2))/((1/2))}}} = 1. Now a + b + c = {{{3/4}}} + {{{2/3}}} + 1 = {{{9/12}}} + {{{8/12}}} + {{{1}}} = {{{17/12}}} + {{{1}}} = {{{29/12}}}. <U>ANSWER</U> </pre> <H3>Problem 12</H3>All students in Ms. Fay's Spanish class are either going on the Spain trip, the Mexico trip, or both. 1/4 of the students going to Spain are also going to Mexico, and 2/7 of the students going on the Mexico trip are also going to Spain. Which of the following could be the total number of students in Ms. Fay's Spanish class? (A) 26; (B) 27; (C) 28; (D) 29; (E) 30. <B>Solution</B> <pre> Let x be the number of students going to the Spain trip; y going to the Mexico trip; z going to both trips. We are given {{{(1/4)x}}} = z, {{{(2/7)y}}} = z. or, equivalently, x = 4z, 2y = 7z. The total is x + y - z = 4z + {{{(7/2)z}}} - z = 3z + {{{(7/2)z}}}. So we should check the appropriate values of z, if they can produce the values of the option's list. For short, let's make a Table T A B L E ----------------------------- z 3z + (7/2)z 3 19.5 4 26 5 32.5 6 39 7 45.5 8 52 Looking in this Table, we see only ONE opportunity for possible solution: 26 students. <U>ANSWER</U>. Under given conditions, the only possible solution is 26 students in the class. </pre> <H3>Problem 13</H3>If {{{2^x}}} = {{{4^y}}} = {{{8^z}}} and {{{1/2x}}} + {{{1/4y}}} + {{{1/8z}}} = {{{22/7}}}, show that x = {{{7/16}}}, y = {{{7/32}}} and z = {{{7/48}}}. <B>Solution</B> <pre> From the given part {{{2^x}}} = {{{4^y}}} = {{{8^z}}}, we have {{{2^x}}} = {{{2^(2y)}}} = {{{2^(3z)}}}. It gives us x = 2y = 3z, or y = {{{x/2}}}, z = {{{x/3}}}. Now substitute it into equation {{{1/(2x)}}} + {{{1/(4y)}}} + {{{1/(8z)}}} = {{{22/7}}}. You will get {{{1/(2x)}}} + {{{1/(4*(x/2))}}} + {{{1/(8*(x/3))}}} = {{{22/7}}}, or {{{1/(2x)}}} + {{{2/(4x)}}} + {{{3/(8x)}}} = {{{22/7}}}, {{{4/(8x)}}} + {{{4/(8x)}}} + {{{3/(8x)}}} = {{{22/7}}}, {{{(4+4+3)/(8x)}}} = {{{22/7}}}, {{{11/(8x)}}} = {{{22/7}}}, x = {{{(11*7)/(22*8)}}} = {{{7/16}}}. Then y = {{{x/2}}} = {{{7/32}}}, z = {{{x/3}}} = {{{7/48}}}, QED. </pre> My other lessons on Evaluating expressions in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-expressions-involving-x%2Binv%28x%29-x2%2Binv%28x2%29-and-x%5E3%2Binv%28x%5E3%29.lesson>HOW TO evaluate expressions involving {{{(x + 1/x)}}}, {{{(x^2+1/x^2)}}} and {{{(x^3+1/x^3)}}}</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-lesson-on-evaluating-expressions.lesson>Advanced lesson on evaluating expressions</A> - <A HREF=http://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-square-equation.lesson>HOW TO evaluate functions of roots of a square equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-cubic-and-quartic-equation.lesson>HOW TO evaluate functions of roots of a cubic and quartic equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Problems-on-Vieta%27s-formulas.lesson>Problems on Vieta's formulas</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-problems-on-Vieta%27s-formulas.lesson>Advanced problems on Vieta's theorem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Miscellaneous-problems-on-Vieta%27s-theorem.lesson>Miscellaneous problems on Vieta's theorem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-expressions-that-contain-infinitely-many-square-roots.lesson>Evaluating expressions that contain infinitely many square roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Solving-equations-that-contain-infinitely-many-radicals.lesson>Solving equations that contain infinitely many radicals</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Problems-on-evaluating-in-Geometry.lesson>Problems on evaluating in Geometry</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-trigonometric-expressions-.lesson>Evaluating trigonometric expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluate-the-sum-of-the-coefficients-of-a-polynomial.lesson>Evaluate the sum of the coefficients of a polynomial</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-evaluating-problems.lesson>Advanced evaluating problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Lowering-a-degree-method.lesson>Lowering a degree method</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Find-the-number-of-factorable-quadratic-polynomials-of-special-form.lesson>Find the number of factorable quadratic polynomials of special form</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-a-function-defined-by-functional-equation.lesson>Evaluating a function defined by functional equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-circle-level-problems-on-evaluating-expressions.lesson>Math circle level problems on evaluating expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-circle-level-problem-on-finding-polynomial-with-assigned-roots.lesson>Math circle level problems on finding polynomials with prescribed roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-Olympiad-level-problem-on-evaluating-of-a-9-degree-polynomial.lesson>Math Olympiad level problem on evaluating a 9-degree polynomial</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Upper-league-problem-on-evaluating-the-sum.lesson>Upper league problem on evaluating the sum</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Finding-coefficients-of-decomposition-of-a-rational-function.lesson>Finding coefficients of decomposition of a rational function</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Upper-level-problem-on-evaluating-an-expression-of-polynomial-roots.lesson>Upper level problem on evaluating an expression of polynomial roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/A-truly-miraculous-evaluating-problem.lesson>A truly miraculous evaluating problem with a truly miraculous solution</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Entertainment-problems-on-evaluating-expressions.lesson>Entertainment problems on evaluating expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/OVERVIEW-of-lessons-on-Evaluation-Substitution.lesson>OVERVIEW of lessons on Evaluating expressions</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
