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πŸ“ How to Add Integers in algebraic expressions (21 MCQs)

πŸ“– From Digital SAT Algebra β€’ 1. Basics of Algebra β€’ 21 questions available

What is How to Add Integers in algebraic expressions?

Definition:
Adding integers in algebraic expressions means performing addition on signed numbers (positive and negative integers) that appear as constants or coefficients within expressions, using the rules of integer addition to combine like terms or evaluate the expression for given variable values.

Working:
When adding integers, if they have the same sign, add their absolute values and keep the sign; if they have different signs, subtract the smaller absolute value from the larger and use the sign of the larger; in expressions, combine integer constants separately from variable terms.

Example:
Simplify 5x+(βˆ’3x)+7+(βˆ’2)5x + (-3x) + 7 + (-2).
Solution: Add variable terms: 5x+(βˆ’3x)=2x5x + (-3x) = 2x; add constants: 7+(βˆ’2)=57 + (-2) = 5; result 2x+52x + 5.

Reason:
Mastering integer addition in expressions is necessary for simplifying polynomials, solving linear equations, and working with functions, as it ensures accuracy in operations that involve both positive and negative quantities.

5
Easy
9
Medium
7
Hard

πŸ“ All How to Add Integers in algebraic expressions MCQs

Q1. A student evaluates βˆ’18+25βˆ’7-18+25-7 by first combining the negative integers and then adding the positive integer. Which result should the student obtain?

A.0 βœ…
B.14
C.-50
D.32
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: The negative integers are βˆ’18-18 and βˆ’7-7, whose sum is βˆ’25-25. Adding 2525 gives 00. This checks the meaning of integer addition rather than relying on a memorized sign rule.

Q2. Which expression has the same value as βˆ’13+(βˆ’8)+21-13+(-8)+21 and best demonstrates why the answer is positive?

A.13+8βˆ’2113+8-21
B.21βˆ’(13+8)21-(13+8) βœ…
C.21+13βˆ’821+13-8
D.13βˆ’8βˆ’2113-8-21
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The two negative integers combine to βˆ’21-21, while the positive integer is 2121. Therefore, βˆ’13+(βˆ’8)+21=0-13+(-8)+21=0, and 21βˆ’(13+8)21-(13+8) also equals 00. The other choices change the signs or grouping.

Q3. A submarine is 1212 meters below sea level. It rises 77 meters, then descends 1515 meters, and finally rises 1010 meters. What is its final position relative to sea level?

A.10 meters above
B.2 meters below βœ…
C.10 meters below
D.8 meters above
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Representing the movements with integers gives βˆ’12+7βˆ’15+10-12+7-15+10. Combining step by step produces βˆ’5βˆ’15+10=βˆ’10-5-15+10=-10, so the submarine finishes 1010 meters below sea level. This models each movement using its direction.

Q4. A student claims that βˆ’24+17=βˆ’41-24+17=-41 because both numbers should be added and the negative sign makes the result negative. What is the best analysis of the error?

A.The student should subtract 1717 from 2424, giving 77. βœ…
B.The student should add absolute values and always make the result positive.
C.The student correctly added the numbers but should change 1717 to βˆ’17-17.
D.The result should be 4141 because the negative sign is ignored when adding.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The numbers have opposite signs, so their absolute values are compared and the smaller magnitude is subtracted from the larger: 24βˆ’17=724-17=7. The sign belongs to the number with greater magnitude, so the result is βˆ’7-7, not 77.

Q5. On a number line, a point starts at βˆ’6-6. It moves 99 units to the right and then 55 units to the left. At which integer does it finish?

A.βˆ’20-20
B.βˆ’2-2
C.βˆ’8-8 βœ…
D.8
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Moving right represents adding a positive integer, while moving left represents adding a negative integer. Thus the final position is βˆ’6+9+(βˆ’5)=3βˆ’5=βˆ’2-6+9+(-5)=3-5=-2. The direction of each movement is essential for interpreting the graph correctly.

Q6. Two methods are used to evaluate βˆ’16+(βˆ’9)+30-16+(-9)+30. Method A combines the negative numbers first, while Method B combines βˆ’16-16 with 3030 first. Which conclusion is correct?

A.Only Method A works because negative integers must be grouped together.
B.Only Method B works because positive integers must be used first.
C.Both methods work and produce 55. βœ…
D.Both methods work but produce different answers.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Method A gives (βˆ’16βˆ’9)+30=βˆ’25+30=5(-16-9)+30=-25+30=5. Method B gives (βˆ’16+30)βˆ’9=14βˆ’9=5(-16+30)-9=14-9=5. Both methods preserve the same total, showing that integers can be regrouped without changing their sum.

Q7. Find the integer xx if x+(βˆ’14)+(βˆ’9)=22x+(-14)+(-9)=22. A student argues that x=3x=3 because 14+9=2314+9=23 and 23βˆ’22=123-22=1. Which value of xx is actually correct?

A.βˆ’1-1
B.22
C.45 βœ…
D.βˆ’45-45
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The equation is xβˆ’14βˆ’9=22x-14-9=22, so xβˆ’23=22x-23=22. Adding 2323 to both sides gives x=45x=45. The student's error comes from subtracting the target value instead of undoing both negative additions systematically.

Q8. A student has 66 positive counters and 99 negative counters. After pairing opposite counters, what integer is represented by the remaining counters?

A.-15
B.-3
C.3 βœ…
D.15
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Each positive counter cancels one negative counter. Six positive and six negative counters form six zero pairs, leaving three negative counters. Therefore, the remaining value is βˆ’3-3. This demonstrates how counters represent both magnitude and sign.

Q9. A model contains 88 positive counters and 55 negative counters. Another student adds 33 negative counters to the model. Which expression and result correctly describe the new model?

A.8+5+3=168+5+3=16
B.8+(βˆ’5)+(βˆ’3)=08+(-5)+(-3)=0 βœ…
C.8+(βˆ’5)+3=68+(-5)+3=6
D.8βˆ’5+3=68-5+3=6
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The five negative counters represent βˆ’5-5, and adding three more negative counters represents βˆ’3-3. Thus the expression is 8+(βˆ’5)+(βˆ’3)=08+(-5)+(-3)=0. All eight positive counters can be paired with eight negative counters, leaving no counters.

Q10. A learner represents βˆ’7+4-7+4 with seven negative counters and four positive counters. After removing zero pairs, the learner says the answer is 1111. What mistake was made?

A.The learner should remove only the positive counters.
B.The learner should pair opposite counters before counting the remainder. βœ…
C.The learner should turn every negative counter into a positive counter.
D.The learner should add the numbers without considering their signs.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: A positive counter and a negative counter form a zero pair, so four such pairs can be removed. The remaining three counters are negative, giving βˆ’3-3. The error was counting all counters instead of recognizing cancellation.

Q11. A game uses positive counters for points earned and negative counters for points lost. Maya has 1010 positive counters and 66 negative counters. She then loses 88 more points. How should the counters be changed, and what is the final value?

A.Add 88 positive counters; final value 1212
B.Add 88 negative counters; final value βˆ’4-4 βœ…
C.Remove 88 negative counters; final value 1212
D.Add 88 negative counters; final value βˆ’24-24
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The initial model represents 10+(βˆ’6)=410+(-6)=4. Losing 88 more points means adding eight negative counters, giving 10+(βˆ’6)+(βˆ’8)=βˆ’410+(-6)+(-8)=-4. This scenario requires interpreting the meaning of a negative quantity before performing the addition.

Q12. A counter diagram is shown conceptually from left to right as 55 positive counters, 55 negative counters, and 22 positive counters. If all possible zero pairs are removed, which value remains?

A.0
B.-2 βœ…
C.-5
D.-12
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: The five positive and five negative counters cancel completely because each opposite pair has total value zero. The two additional positive counters cannot be canceled, so the model represents 22. The important step is identifying zero pairs before counting.

Q13. Two students model βˆ’9+6-9+6. Student A creates nine negative counters and six positive counters, then removes six zero pairs. Student B creates nine negative counters and adds six positive counters without removing pairs. Which statement is most accurate?

A.Only Student A has a valid model because zero pairs must always be removed.
B.Only Student B has a valid model because counters should never be removed.
C.Both models represent βˆ’3-3, although Student A displays the simplified result more clearly. βœ…
D.Both models represent 1515 because all counters should be counted.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Both students begin with the same nine negative and six positive counters, so both models represent βˆ’3-3. Student A simplifies the physical model by removing six zero pairs, while Student B leaves equivalent zero pairs visible. Removing zero pairs changes appearance, not value.

Q14. A student claims that any collection containing equal numbers of positive and negative counters must have a value of 11, because one positive and one negative counter are always left together as a pair. Which example most strongly disproves the claim?

A.2 positive and 2 negative counters
B.4 positive and 1 negative counter
C.5 positive and 5 negative counters βœ…
D.6 positive and 2 negative counters
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Five positive counters and five negative counters can be arranged into five zero pairs, so their total value is 00, not 11. The example directly disproves the claim and reinforces that opposite counters cancel completely rather than leaving a leftover pair.

Q15. Which expression has a positive result even though it contains both positive and negative integers?

A.βˆ’14+9-14+9
B.18+(βˆ’7)18+(-7) βœ…
C.βˆ’12+(βˆ’5)-12+(-5)
D.βˆ’20+13-20+13
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: When integers have different signs, compare their absolute values and subtract the smaller from the larger. Since 1818 is greater than 77, 18+(βˆ’7)=1118+(-7)=11, which is positive. The other choices produce negative results.

Q16. A student evaluates βˆ’17+(βˆ’8)-17+(-8) as 99 because the student subtracts the smaller absolute value from the larger and chooses the sign of the larger number. What is the correct result?

A.-25 βœ…
B.-9
C.9
D.25
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Both integers are negative, so their absolute values must be added rather than subtracted. Thus 17+8=2517+8=25, and the result keeps the negative sign. Therefore, βˆ’17+(βˆ’8)=βˆ’25-17+(-8)=-25. The student's method incorrectly treats same-sign integers as different-sign integers.

Q17. A bank account changes by -\<span class="katex-error" title="ParseError: KaTeX parse error: Can&#x27;t use function &#x27;' in math mode at position 3: 35\Μ²)Μ², then +\" style="color:#cc0000">35, then +\</span>50+\</span>50, and then βˆ’$20-\$20. What is the overall change in the account balance?

A.Increase of $5\$5
B.Decrease of $5\$5 βœ…
C.Increase of $105\$105
D.Decrease of $105\$105
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The changes can be represented as βˆ’35+50+(βˆ’20)-35+50+(-20). Combining the negative changes gives βˆ’55-55, and adding 5050 gives βˆ’5-5. Therefore, the account experiences an overall decrease of $5\$5.

Q18. A student says βˆ’24+15=βˆ’39-24+15=-39 because the negative sign should remain and the absolute values should always be added. Which reasoning correctly identifies the error?

A.The absolute values should be multiplied.
B.The integers have different signs, so their absolute values should be subtracted and the sign of the larger absolute value retained. βœ…
C.Both integers should be made positive before adding.
D.The result should always have the sign of the second integer.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The integers have opposite signs, so their absolute values are compared: 24βˆ’15=924-15=9. Because 2424 has the greater absolute value and is negative, the result is βˆ’9-9. Adding absolute values would incorrectly give βˆ’39-39.

Q19. A point starts at βˆ’7-7 on a number line, moves 1212 units right, and then moves 99 units left. Which statement correctly describes its final position?

A.It finishes at 1414 because all distances are added.
B.It finishes at βˆ’4-4 because the net movement is 33 units right from βˆ’7-7. βœ…
C.It finishes at 44 because 12βˆ’9=312-9=3.
D.It finishes at βˆ’28-28 because both movements affect the negative starting value.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Moving right represents +12+12, while moving left represents βˆ’9-9. The expression is βˆ’7+12+(βˆ’9)-7+12+(-9). The two movements have a net effect of 33 units right, so the final position is βˆ’7+3=βˆ’4-7+3=-4.

Q20. Three students simplify βˆ’18+27+(βˆ’14)-18+27+(-14). Student A gets 2323, Student B gets βˆ’5-5, and Student C gets βˆ’59-59. Which student is correct, and why?

A.Student A, because 18+27βˆ’14=2318+27-14=23.
B.Student B, because 27βˆ’18βˆ’14=βˆ’527-18-14=-5. βœ…
C.Student C, because all three absolute values must be added.
D.Students A and B, because different grouping methods give different valid answers.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: The expression can be evaluated by combining the positive and negative amounts: 27βˆ’(18+14)=27βˆ’32=βˆ’527-(18+14)=27-32=-5. Student A ignores the negative sign of 1818, while Student C adds all absolute values. Only Student B correctly accounts for the signs.

Q21. A contest problem requires three integers to be added: a=βˆ’16a=-16, b=23b=23, and cc is negative. The final sum must be 00. Which value of cc satisfies the condition?

A.-7
B.7 βœ…
C.-39
D.39
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: First combine the known integers: βˆ’16+23=7-16+23=7. To make the total 00, the remaining integer must contribute βˆ’7-7. Thus c=βˆ’7c=-7, giving βˆ’16+23+(βˆ’7)=0-16+23+(-7)=0. This requires reasoning backward from the required sum.

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