SHSAT Math Quiz: Integer Operations
13 questions · exam conditions
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Integer OperationsQuestion 1 of 13

Find ∣−15+4∣\left|-15+4\right|.

11
-11
19
-19
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SHSAT Math Quiz

SHSAT Math Quiz: Integer Operations

Practice Integer Operations in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integer Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find ∣−15+4∣\left|-15+4\right|.

  1. 11 (correct answer)
  2. -11
  3. 19
  4. -19
Explanation: When you encounter absolute value problems, remember that absolute value measures distance from zero on the number line, so it's always non-negative. To find ∣−15+4∣\left|-15+4\right|, you must work from the inside out. First, simplify the expression inside the absolute value bars: −15+4=−11-15 + 4 = -11. Then apply the absolute value: ∣−11∣=11|-11| = 11. Since absolute value gives you the distance from zero, and −11-11 is 11 units away from zero, the answer is 11. Looking at the answer choices: A) 11 is correct for the reasons above. B) -11 represents the common error of forgetting to apply the absolute value after simplifying −15+4-15 + 4. Students often stop at −11-11 without taking the final absolute value step. C) 19 results from incorrectly adding the absolute values first: ∣−15∣+∣4∣=15+4=19|-15| + |4| = 15 + 4 = 19. This violates the order of operations since you must simplify inside grouping symbols before applying operations outside them. D) -19 combines both errors: incorrectly adding absolute values first, then forgetting that absolute value is always non-negative. Remember that absolute value bars act like parentheses for order of operations—always simplify what's inside first, then apply the absolute value. Also, your final answer can never be negative when dealing with absolute value, so you can immediately eliminate any negative answer choices.

Question 2

An elevator is at the 3rd floor. It then goes down 8 floors and finally up 2 floors.

After these moves, on which floor is the elevator?

  1. -3 (correct answer)
  2. -7
  3. -13
  4. 3
Explanation: This problem tests your ability to work with positive and negative integers on a number line, which represents floors in a building where ground level and below are negative numbers. Starting at the 3rd floor, you need to track two movements. Going "down 8 floors" means subtracting 8: 3−8=−53 - 8 = -5. This puts the elevator at the -5th floor (5 floors below ground level). Then going "up 2 floors" means adding 2: −5+2=−3-5 + 2 = -3. The elevator ends up at the -3rd floor. Looking at the wrong answers: Choice B (-7) represents a common error where students subtract the final upward movement instead of adding it: 3−8−2=−73 - 8 - 2 = -7. Choice C (-13) occurs when students add all the movements as negative values: 3−8−2=−133 - 8 - 2 = -13, misinterpreting "up 2" as "down 2." Choice D (3) happens when students incorrectly think the elevator returns to its starting position, perhaps by miscalculating 3−8+23 - 8 + 2 or assuming the movements cancel out somehow. The correct answer is A (-3). For elevator and number line problems, always establish your starting point clearly and track each movement step by step. Remember that "up" means addition and "down" means subtraction, and negative floor numbers represent basement levels below ground. Drawing a simple number line can help you visualize the movements and avoid sign errors.

Question 3

A bank account starts with a balance of $150. Over a week, the following transactions occur: withdrawal of $40, deposit of $25, withdrawal of $60, deposit of $15.

What is the account balance after these transactions?

  1. $90 (correct answer)
  2. $105
  3. $135
  4. $250
Explanation: This question tests your ability to track changes to an initial value through a series of additions and subtractions. When you see problems involving sequential transactions, deposits, or changes to a starting amount, work step-by-step through each transaction in order. Start with the initial balance of $150. Then process each transaction:
  • After withdrawing $40: $150−40=110150 - 40 = 110 $
  • After depositing $25: $$110 + 25 = 135$$
  • After withdrawing $60: $$135 - 60 = 75$$
  • After depositing $15: $75+15=9075 + 15 = 90 $
The final balance is $90, making A correct. Looking at the wrong answers: B ($105) likely comes from making an arithmetic error in one of the steps, perhaps miscalculating 135−60135 - 60 as 3030 instead of 7575. C (135)representsstoppingafterjustthefirsttwotransactions—thisisthebalancebeforethefinalwithdrawalanddeposit.D(135) represents stopping after just the first two transactions—this is the balance before the final withdrawal and deposit. D (250) results from incorrectly adding all the numbers together: 150+40+25+60+15=250150 + 40 + 25 + 60 + 15 = 250. This mistake treats all transactions as deposits rather than distinguishing between withdrawals (subtractions) and deposits (additions). For sequential transaction problems, organize your work clearly and double-check that you're adding deposits and subtracting withdrawals. Consider using a running tally or table to track each step, especially when there are many transactions. This methodical approach prevents arithmetic errors and ensures you don't miss any steps.

Question 4

Yesterday the temperature at sunrise was −4∘C-4^{\circ}\text{C}. During the day it rose by 9∘C9^{\circ}\text{C} and then dropped by 6∘C6^{\circ}\text{C} by sunset.

What was the temperature at sunset?

  1. -1^{\circ}\text{C} (correct answer)
  2. -7^{\circ}\text{C}
  3. 1^{\circ}\text{C}
  4. 7^{\circ}\text{C}
Explanation: This problem tests your ability to work with integers and temperature changes, which involves adding and subtracting positive and negative numbers in sequence. Start with the sunrise temperature of −4°C-4°C. When the temperature "rose by 9°C9°C," you add 9 to the starting temperature: −4+9=5°C-4 + 9 = 5°C. Then the temperature "dropped by 6°C6°C," so you subtract 6 from that result: 5−6=−1°C5 - 6 = -1°C. You can also think of this as one calculation: −4+9−6=−1°C-4 + 9 - 6 = -1°C. Looking at the wrong answers: Choice B (−7°C-7°C) likely comes from adding all the numbers as if they were negative: −4+(−9)+(−6)=−19°C-4 + (-9) + (-6) = -19°C, or possibly from subtracting both changes: −4−9−6=−19°C-4 - 9 - 6 = -19°C. Choice C (1°C1°C) results from the correct calculation but with a sign error at the end—getting 5−6=15 - 6 = 1 instead of −1-1. Choice D (7°C7°C) might come from treating the starting temperature as positive: 4+9−6=7°C4 + 9 - 6 = 7°C. The correct answer is A (−1°C-1°C). Strategy tip: For temperature change problems, work step-by-step and pay careful attention to the language. "Rose by" or "increased by" means add; "dropped by" or "decreased by" means subtract. Always double-check your signs, especially when working with negative starting values.

Question 5

What is the product −6×(−4)-6\times(-4)?

  1. -24
  2. 24 (correct answer)
  3. -10
  4. 10
Explanation: When you see multiplication of two negative numbers, you need to apply the rules for multiplying signed numbers. This is a fundamental skill that appears frequently on the SHSAT. To find −6×(−4)-6 \times (-4), remember that when you multiply two numbers with the same sign, the result is positive. Since both −6-6 and −4-4 are negative, their product will be positive. Now multiply the absolute values: 6×4=246 \times 4 = 24. Since the result is positive, we get +24+24. Choice A gives −24-24, which represents the common error of thinking that multiplying two negative numbers gives a negative result. This misconception comes from confusing multiplication with addition - while adding two negative numbers does give a negative sum, multiplying them gives a positive product. Choice B is correct: 2424. Choice C gives −10-10, which you might get if you incorrectly added the numbers instead of multiplying them (−6+(−4)=−10-6 + (-4) = -10), then kept the negative sign. Choice D gives 1010, which could result from adding the absolute values instead of multiplying them (6+4=106 + 4 = 10). Remember this pattern for signed multiplication: negative × negative = positive, positive × positive = positive, but negative × positive = negative. A helpful memory trick is "same signs give positive, different signs give negative." On the SHSAT, questions involving negative number operations often test whether you correctly apply these sign rules, so practice them until they become automatic.

Question 6

Evaluate −60÷12-60\div12.

  1. -5 (correct answer)
  2. 5
  3. -48
  4. 48
Explanation: When you encounter division problems involving negative numbers, you need to apply the rules for dividing integers with different signs. To evaluate −60÷12-60 \div 12, first recognize that you're dividing a negative number by a positive number. When dividing numbers with opposite signs, the result is always negative. Now perform the division: 60÷12=560 \div 12 = 5, so −60÷12=−5-60 \div 12 = -5. You can verify this by thinking about it as repeated subtraction: how many times does 12 go into -60? Since 12×(−5)=−6012 \times (-5) = -60, the answer is -5. Looking at the wrong answers: Choice B (5) represents the common error of ignoring the negative sign entirely. Students sometimes focus only on the absolute values and forget to apply the sign rules. Choice C (-48) appears to come from incorrectly subtracting instead of dividing: −60−12=−72-60 - 12 = -72, though even that calculation is off. This suggests confusion about which operation to perform. Choice D (48) combines two errors: using the wrong operation and the wrong sign, possibly from adding the absolute values: 60−12=4860 - 12 = 48. Remember this key pattern for integer division: when the signs are different (positive divided by negative, or negative divided by positive), your answer is negative. When the signs are the same (both positive or both negative), your answer is positive. This rule will help you quickly eliminate incorrect choices on similar SHSAT problems.

Question 7

Evaluate −3+6×(−2)-3+6\times(-2).

  1. -18
  2. -9
  3. 9
  4. -15 (correct answer)
Explanation: When you see an expression with both addition and multiplication, order of operations is crucial. Remember PEMDAS: you must handle multiplication before addition, even when the addition comes first in the expression. Let's work through −3+6×(−2)-3+6\times(-2) step by step. First, perform the multiplication: 6×(−2)=−126\times(-2) = -12. Now the expression becomes −3+(−12)-3 + (-12), which equals −3−12=−15-3 - 12 = -15. Looking at the wrong answers reveals common mistakes. Choice A (-18) results from incorrectly multiplying all three numbers together: −3×6×(−2)=−18-3 \times 6 \times (-2) = -18. This ignores the addition sign entirely. Choice B (-9) comes from working left to right without following order of operations: (−3+6)×(−2)=3×(−2)=−6(-3 + 6) \times (-2) = 3 \times (-2) = -6, though this doesn't quite match, suggesting a sign error as well. Choice C (9) likely stems from making multiple sign errors, perhaps calculating (−3+6)×(−2)(-3 + 6) \times (-2) but then getting the final sign wrong, or mishandling the negative signs throughout. The key strategy for order of operations problems is to identify all operations first, then work through them in the correct sequence. Always handle multiplication and division before addition and subtraction, regardless of how the expression is written. When you see mixed operations on the SHSAT, slow down and apply PEMDAS methodically—rushing through these problems often leads to the exact mistakes represented in the wrong answer choices.

Question 8

Compute −5−9-5-9.

  1. -4
  2. -14 (correct answer)
  3. 4
  4. 14
Explanation: When you're subtracting a positive number from a negative number, you're moving further left on the number line, making the result more negative. To compute −5−9-5 - 9, think of this as starting at −5-5 and then subtracting 99 more. Since you're subtracting a positive number from a negative number, you add the absolute values and keep the negative sign: ∣−5∣+∣9∣=5+9=14|-5| + |9| = 5 + 9 = 14, so the answer is −14-14. Alternatively, you can rewrite this as −5+(−9)=−14-5 + (-9) = -14. When adding two negative numbers, you add their absolute values and keep the negative sign. Choice A (−4-4) represents the common error of subtracting the smaller absolute value from the larger one: 9−5=49 - 5 = 4, then incorrectly making it negative. This would be the answer to −9−(−5)-9 - (-5) or −9+5-9 + 5, not our problem. Choice C (44) comes from the same calculation error as choice A, but forgetting the negative sign entirely. Some students mistakenly think "negative minus positive equals positive." Choice D (1414) results from correctly finding 5+9=145 + 9 = 14 but forgetting that when you start with a negative number and subtract a positive, the result must be negative. Remember this key pattern: negative minus positive always gives you a more negative result. When in doubt, think about the number line—subtracting moves you left, so starting at −5-5 and moving 9 more spaces left gets you to −14-14.

Question 9

What is the value of −7+12-7+12?

  1. 5 (correct answer)
  2. -19
  3. -5
  4. 19
Explanation: When adding a negative and positive number, you're essentially finding the difference between their absolute values, then applying the sign of the number with the larger absolute value. To solve −7+12-7 + 12, think of this as starting at -7 on a number line and moving 12 units to the right (since adding a positive means moving right). You'll land at 5. Alternatively, you can rewrite this as 12−7=512 - 7 = 5 since adding a negative is the same as subtracting its absolute value. The answer is A) 5, which correctly represents this calculation. Looking at the wrong answers: B) -19 comes from adding the absolute values and keeping the negative sign (−7+(−12)=−19-7 + (-12) = -19), which would be correct if both numbers were negative, but 12 is positive here. C) -5 results from subtracting incorrectly—finding 7−12=−57 - 12 = -5—which happens when you mistakenly apply the negative sign from the smaller absolute value. D) 19 comes from simply adding the absolute values without considering signs at all (7+12=197 + 12 = 19). Strategy tip: When adding numbers with different signs, always subtract the smaller absolute value from the larger one, then use the sign of the number with the larger absolute value. This method works every time and helps you avoid the common trap of just adding absolute values or applying the wrong sign.

Question 10

The product of three consecutive even integers is 1680. What is the sum of these three integers?

  1. 30
  2. 36 (correct answer)
  3. 42
  4. 48
Explanation: Let the three consecutive even integers be n−2n-2, nn, and n+2n+2, where nn is even. Then (n−2)⋅n⋅(n+2)=1680(n-2) \cdot n \cdot (n+2) = 1680. This gives us n(n2−4)=1680n(n^2 - 4) = 1680, so n3−4n=1680n^3 - 4n = 1680. We need to find nn such that n3−4n−1680=0n^3 - 4n - 1680 = 0. Testing even values: if n=12n = 12, then 123−4(12)=1728−48=168012^3 - 4(12) = 1728 - 48 = 1680. So n=12n = 12, and the three integers are 10,12,1410, 12, 14. Their sum is 10+12+14=3610 + 12 + 14 = 36. Choice A (30) might result from using 8,10,128, 10, 12 (product = 960). Choice C (42) might result from using 12,14,1612, 14, 16 (product = 2688). Choice D (48) might result from using 14,16,1814, 16, 18 (product = 4032).

Question 11

The sum 13+23+33+⋯+n31^3 + 2^3 + 3^3 + \cdots + n^3 equals 1296. What is the value of 1+2+3+⋯+n1 + 2 + 3 + \cdots + n?

  1. 36 (correct answer)
  2. 64
  3. 72
  4. 81
Explanation: We use the formula for the sum of cubes: 13+23+⋯+n3=(n(n+1)2)21^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2. We're told this sum equals 1296, so (n(n+1)2)2=1296\left(\frac{n(n+1)}{2}\right)^2 = 1296. Taking the square root: n(n+1)2=1296=36\frac{n(n+1)}{2} = \sqrt{1296} = 36. Therefore, 1+2+3+⋯+n=n(n+1)2=361 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} = 36. To verify, we can solve for nn: n(n+1)=72n(n+1) = 72, so n2+n−72=0n^2 + n - 72 = 0. Using the quadratic formula: n=−1+1+2882=−1+2892=−1+172=8n = \frac{-1 + \sqrt{1 + 288}}{2} = \frac{-1 + \sqrt{289}}{2} = \frac{-1 + 17}{2} = 8. So n=8n = 8, and indeed 8×92=36\frac{8 \times 9}{2} = 36. Choice B (64) might result from incorrectly computing 1296\sqrt{1296} as 6464 instead of 3636. Choice C (72) is the value of n(n+1)n(n+1). Choice D (81) might result from other computational errors.

Question 12

If aa, bb, and cc are integers such that a<b<ca < b < c and a+b+c=0a + b + c = 0, what is the minimum possible value of a2+b2+c2a^2 + b^2 + c^2?

  1. 2 (correct answer)
  2. 5
  3. 6
  4. 14
Explanation: Since a+b+c=0a + b + c = 0, we have c=−(a+b)c = -(a + b). We want to minimize a2+b2+c2=a2+b2+(a+b)2=a2+b2+a2+2ab+b2=2a2+2b2+2ab=2(a2+ab+b2)a^2 + b^2 + c^2 = a^2 + b^2 + (a + b)^2 = a^2 + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 + 2ab = 2(a^2 + ab + b^2). To minimize this, we need to minimize a2+ab+b2a^2 + ab + b^2 subject to a<b<−(a+b)a < b < -(a + b) (since c=−(a+b)c = -(a + b) and b<cb < c). The condition b<−(a+b)b < -(a + b) gives 2b<−a2b < -a, so a<−2ba < -2b. Since aa and bb are integers with a<ba < b, let's try small values. If b=−1b = -1, then a<−1a < -1 and a<2a < 2, so a≤−2a \leq -2. Try a=−2,b=−1a = -2, b = -1: then c=−(−2+(−1))=3c = -(-2 + (-1)) = 3. Check: −2<−1<3-2 < -1 < 3 ✓. We get a2+b2+c2=4+1+9=14a^2 + b^2 + c^2 = 4 + 1 + 9 = 14. If b=0b = 0, then a<0a < 0 and a<0a < 0, so a≤−1a \leq -1. Try a=−1,b=0a = -1, b = 0: then c=1c = 1. Check: −1<0<1-1 < 0 < 1 ✓. We get a2+b2+c2=1+0+1=2a^2 + b^2 + c^2 = 1 + 0 + 1 = 2. If b=1b = 1, then a<1a < 1 and a<−2a < -2, so a≤−3a \leq -3. Try a=−3,b=1a = -3, b = 1: then c=2c = 2. Check: −3<1<2-3 < 1 < 2 ✓. We get a2+b2+c2=9+1+4=14a^2 + b^2 + c^2 = 9 + 1 + 4 = 14. The minimum value is 2, achieved with (a,b,c)=(−1,0,1)(a, b, c) = (-1, 0, 1).

Question 13

If xx and yy are integers such that x2−y2=77x^2 - y^2 = 77, how many different ordered pairs (x,y)(x, y) are possible?

  1. 6
  2. 8 (correct answer)
  3. 10
  4. 12
Explanation: We have x2−y2=(x+y)(x−y)=77x^2 - y^2 = (x+y)(x-y) = 77. Since 77=7×11=1×77=(−1)×(−77)=(−7)×(−11)77 = 7 \times 11 = 1 \times 77 = (-1) \times (-77) = (-7) \times (-11), we need to find all ways to factor 77 as a product of two integers. The factor pairs are: (1,77),(7,11),(11,7),(77,1),(−1,−77),(−7,−11),(−11,−7),(−77,−1)(1, 77), (7, 11), (11, 7), (77, 1), (-1, -77), (-7, -11), (-11, -7), (-77, -1). For each pair (a,b)(a, b) where a⋅b=77a \cdot b = 77, we set x+y=ax + y = a and x−y=bx - y = b, giving x=a+b2x = \frac{a+b}{2} and y=a−b2y = \frac{a-b}{2}. For (x,y)(x, y) to be integers, both a+ba + b and a−ba - b must be even, which happens when aa and bb have the same parity. Checking: (1,77)(1, 77): both odd, x=39,y=−38x = 39, y = -38. (7,11)(7, 11): both odd, x=9,y=−2x = 9, y = -2. (11,7)(11, 7): both odd, x=9,y=2x = 9, y = 2. (77,1)(77, 1): both odd, x=39,y=38x = 39, y = 38. (−1,−77)(-1, -77): both odd, x=−39,y=38x = -39, y = 38. (−7,−11)(-7, -11): both odd, x=−9,y=2x = -9, y = 2. (−11,−7)(-11, -7): both odd, x=−9,y=−2x = -9, y = -2. (−77,−1)(-77, -1): both odd, x=−39,y=−38x = -39, y = -38. All 8 pairs work, giving us 8 different ordered pairs (x,y)(x, y).