Solving Problems On Set Theory

 

Grade 10 Set Theory Practice Quiz

Test your understanding of set theory concepts. Click "Show Solution" to check your answers.

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1. If Set $A = \{2, 3, 5, 7\}$ and Set $B = \{1, 3, 5, 7, 9\}$, find $A \cap B$.

• A) $\{3, 5, 7\}$
• B) $\{1, 2, 3, 5, 7, 9\}$
• C) $\{2\}$
• D) $\{1, 9\}$

Correct Answer: A

Solution: The intersection ($\cap$) represents elements that belong to both sets. The numbers 3, 5, and 7 are found in both Set A and Set B.

2. What is the cardinality of the empty set $\emptyset$?

• A) 1
• B) 0
• C) Infinite
• D) Undefined

Correct Answer: B

Solution: Cardinality ($n$) is the total count of components within a set. Since an empty set contains no elements, its cardinality is exactly 0.

3. If the Universal Set $U = \{x \mid 1 \le x \le 6, x \in \mathbb{W}\}$ and Set $M = \{2, 4, 6\}$, what is the complement set $M'$?

• A) $\{2, 4, 6\}$
• B) $\{0, 1, 3, 5\}$
• C) $\{1, 3, 5\}$
• D) $\{\}$

Correct Answer: C

Solution: The complement ($M'$) contains all elements in the universal set $U = \{1, 2, 3, 4, 5, 6\}$ that are missing from set $M$. Removing 2, 4, and 6 leaves behind $\{1, 3, 5\}$.

4. Which of the following pairs represents disjoint sets?

• A) $A = \{even\ numbers\}$, $B = \{prime\ numbers\}$
• B) $A = \{multiples\ of\ 3\}$, $B = \{multiples\ of\ 5\}$
• C) $A = \{positive\ integers\}$, $B = \{negative\ integers\}$
• D) $A = \{vowels\}$, $B = \{letters\ in\ "math"\}$

Correct Answer: C

Solution: Disjoint sets share zero elements in common. Positive and negative integers are separate entirely, while option A shares {2}, B shares {15}, and D shares {a}.

5. In a grade 10 class of 30 students, 18 take Science, 14 take Art, and 6 take both. How many students take neither subject?

• A) 2
• B) 4
• C) 8
• D) 12

Correct Answer: B

Solution: Using the principle of inclusion-exclusion: Total unique students taking subjects = $n(S \cup A) = 18 + 14 - 6 = 26$. Students taking neither = Total class $-$ Union = $30 - 26 = 4$.

Solving Problems On Provability

 

Understanding Probability

Definition: Probability is the mathematical measure of the likelihood that a specific event will occur. It quantifies uncertainty on a scale from 0 (impossible event) to 1 (certain event), often expressed as a fraction, decimal, or percentage.

Necessary Formulas

1. Theoretical Probability: Used when all outcomes in a sample space are equally likely.

P(A) = (Number of favorable outcomes) / (Total number of possible outcomes)

2. Complement of an Event: The probability of an event not happening.

P(A') = 1 - P(A)

3. Experimental Probability: Based on actual trials or historical data.

P(E) = (Number of times event occurs) / (Total number of trials)

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Grade 10 Probability Practice Quiz

Test your knowledge with these 10 multiple-choice questions aligned with the Canadian curriculum.

Question 1: A standard Canadian deck of 52 playing cards is well-shuffled. What is the theoretical probability of randomly drawing a Heart?

• A) 1/2
• B) 1/4
• C) 1/13
• D) 1/52

Correct Answer: B) 1/4
Solution: There are 4 suits in a deck (Hearts, Diamonds, Clubs, Spades), each with 13 cards.
P(Heart) = 13 / 52 = 1/4 (or 25%).

Question 2: A fair six-sided die is rolled. What is the probability of rolling a number greater than 4?

• A) 1/3
• B) 1/2
• C) 2/3
• D) 1/6

Correct Answer: A) 1/3
Solution: The possible outcomes are {1, 2, 3, 4, 5, 6} (Total = 6). The numbers greater than 4 are 5 and 6 (Favorable = 2).
P(>4) = 2 / 6 = 1/3.

Question 3: The probability that it will snow in Calgary tomorrow is 0.35. What is the probability that it will NOT snow tomorrow?

• A) 0.35
• B) 0.55
• C) 0.65
• D) 1.35

Correct Answer: C) 0.65
Solution: Using the complement formula: P(Not Snow) = 1 - P(Snow).
P(Not Snow) = 1 - 0.35 = 0.65.

Question 4: A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If one marble is drawn at random, what is the probability that it is green?

• A) 1/5
• B) 1/2
• C) 1/10
• D) 2/5

Correct Answer: A) 1/5
Solution: Total marbles = 5 + 3 + 2 = 10. Green marbles = 2.
P(Green) = 2 / 10 = 1/5 (or 20%).

Question 5: Two fair coins are tossed at the same time. What is the probability of getting exactly one Head and one Tail?

• A) 1/4
• B) 1/2
• C) 3/4
• D) 1/3

Correct Answer: B) 1/2
Solution: The sample space for two coins is {HH, HT, TH, TT} (Total = 4). The favorable outcomes are HT and TH (Favorable = 2).
P(One H, One T) = 2 / 4 = 1/2.

Question 6: In a standard Toronto high school class, 12 students play soccer, 8 play basketball, and 4 play both sports. If there are 26 students in total, what is the probability that a randomly chosen student plays soccer OR basketball?

• A) 20/26
• B) 16/26
• C) 24/26
• D) 12/26

Correct Answer: B) 16/26
Solution: Use the Principle of Inclusion-Exclusion: P(A or B) = P(A) + P(B) - P(A and B).
Favorable outcomes = 12 + 8 - 4 = 16.
P(Soccer or Basketball) = 16 / 26 (which simplifies to 8/13).

Question 7: A spinner is divided into 8 equal sectors numbered 1 through 8. What is the probability of spinning a prime number?

• A) 1/2
• B) 3/8
• C) 5/8
• D) 1/4

Correct Answer: A) 1/2
Solution: The prime numbers between 1 and 8 are 2, 3, 5, and 7 (Total of 4 numbers).
P(Prime) = 4 / 8 = 1/2.

Question 8: A box contains 4 red pens and 6 blue pens. A pen is drawn at random, its color is recorded, and it is not replaced. A second pen is then drawn. What is the probability that both pens are red?

• A) 16/100
• B) 2/15
• C) 12/90
• D) 4/25

Correct Answer: B) 2/15
Solution: For the first draw, P(Red 1) = 4/10. Since it's not replaced, there are now 3 red pens left out of 9 total pens.
P(Red 2 given Red 1) = 3/9.
P(Both Red) = (4/10) * (3/9) = 12/90 = 2/15.

Question 9: During a hockey practice, a player takes 50 shots on goal and scores 15 times. What is the experimental probability that the player will score on their next shot?

• A) 3/10
• B) 7/10
• C) 15/35
• D) 1/50

Correct Answer: A) 3/10
Solution: Experimental probability is based on past trials.
P(Score) = 15 / 50 = 3/10 (or 30%).

Question 10: Which of the following values cannot represent the mathematical probability of an event?

• A) 0%
• B) 4/3
• C) 0.99
• D) 2/7

Correct Answer: B) 4/3
Solution: Probability values must always fall strictly within the range of 0 to 1 (inclusive). The fraction 4/3 is equal to 1.33, which is greater than 1 and therefore impossible.

Solving Problems On Factorial Permutation And Combination

 

Factorial, Permutation & Combination Quiz

Grade 10 Level Practice Problems

Question 1: Evaluate the expression: 6!

A) 36

B) 120

C) 720

D) 5040

Correct Answer: C
Solution: A factorial means multiplying a sequence of descending natural numbers down to 1.
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Question 2: Simplify the fraction: 8! / 6!

A) 2

B) 1.33

C) 14

D) 56

Correct Answer: D
Solution: Expand the larger factorial until it matches the smaller one, then cancel out common factors:
(8 × 7 × 6!) / 6! = 8 × 7 = 56

Question 3: In how many different ways can a student council elect a President, Vice-President, and Secretary from a group of 7 candidates?

A) 210

B) 35

C) 840

D) 7

Correct Answer: A
Solution: Because the roles (positions) are distinct, order matters, making this a Permutation problem.
We use _7P_3 = 7! / (7-3)! = 7 × 6 × 5 = 210 ways.

Question 4: A student needs to choose 3 books to read from a summer reading list of 9 books. How many different selections can be made?

A) 504

B) 84

C) 27

D) 362,880

Correct Answer: B
Solution: The order in which the books are selected does not matter, so this is a Combination problem.
We use _9C_3 = 9! / (3!(9-3)!) = (9 × 8 × 7) / (3 × 2 × 1) = 504 / 6 = 84 ways.

Question 5: Which of the following situations represents a permutation?

A) Choosing 4 toppings for a pizza from a list of 10.

B) Selecting 5 cards out of a standard deck of 52 cards.

C) Creating a 4-digit bank PIN using numbers 0-9 without repetition.

D) Picking 3 students to help clean up the science lab.

Correct Answer: C
Solution: In a bank PIN, the arrangement order of the numbers changes the code (e.g., 1234 is different from 4321). All other options represent combinations because the order of selection changes nothing.

Question 6: Evaluate the permutation notation expression: _5P_2

A) 10

B) 20

C) 60

D) 120

Correct Answer: B
Solution: Using the permutation formula _nP_r = n! / (n-r)!:
_5P_2 = 5! / (5-2)! = 5! / 3! = 5 × 4 = 20.

Question 7: Evaluate the combination notation expression: _6C_4

A) 15

B) 30

C) 360

D) 24

Correct Answer: A
Solution: Using the combination formula _nC_r = n! / (r!(n-r)!):
_6C_4 = 6! / (4! × 2!) = (6 × 5) / (2 × 1) = 30 / 2 = 15.
Tip: Note that _6C_4 is equal to _6C_2!

Question 8: How many unique ways can the letters of the word "CANADA" be rearranged?

A) 720

B) 360

C) 120

D) 24

Correct Answer: C
Solution: This involves permutations with repeating items. "CANADA" has 6 total letters, where 'A' repeats 3 times.
Formula: n! / a! = 6! / 3! = (6 × 5 × 4 × 3!) / 3! = 6 × 5 × 4 = 120 unique ways.

Question 9: A hockey coach must choose 3 players out of 8 available players to take part in a post-game shootout. If the order of the shooters matters, how many ways can they be selected?

A) 56

B) 336

C) 40,320

D) 24

Correct Answer: B
Solution: Because the order of shooters determines who goes first, second, and third, it is a permutation.
_8P_3 = 8! / (8-3)! = 8 × 7 × 6 = 336 arrangements.

Question 10: An ice cream shop offers 8 different flavors. An individual wants to buy a take-home tub containing exactly 2 different flavors mixed together. How many combinations are possible?

A) 16

B) 56

C) 28

D) 64

Correct Answer: C
Solution: Since the flavors are mixed in a single tub, order does not matter (Chocolate + Vanilla is the same as Vanilla + Chocolate).
_8C_2 = 8! / (2! × 6!) = (8 × 7) / 2 = 56 / 2 = 28 combinations.

Question 11: What is the value of 0!?

A) 0

B) 1

C) Undefined

D) -1

Correct Answer: B
Solution: By mathematical definition, 0! = 1. This convention ensures that formulas for permutations and combinations consistently yield correct values when selecting all items (e.g., _nP_n = n! / 0! = n!).

Question 12: A classroom has 10 boys and 12 girls. A committee consisting of 2 boys and 2 girls needs to be formed. How many different committees can be made?

A) 2,970

B) 44

C) 11,880

D) 240

Correct Answer: A
Solution: Calculate the options for boys and girls independently using combinations, then apply the Fundamental Counting Principle to multiply them together:
Boys: _10C_2 = (10 × 9) / 2 = 45
Girls: _12C_2 = (12 × 11) / 2 = 66
Total Committees = 45 × 66 = 2,970.

Solving Problems On Mode Median And Mean

 

Grade 10 Math: Data Management Quiz
Measures of Central Tendency & Spread

1. A student records their math quiz scores out of 10: 6, 8, 7, 9, 8, 10, 6, 8. What is the mode of these scores?

• A) 6
• B) 8
• C) 7.75
• D) 4

Click for Solution

Formula: Mode = The value that occurs most frequently

Brief Solution: Count the frequency of each number. 6 appears twice, 7 appears once, 8 appears three times, 9 appears once, and 10 appears once.

Correct Answer: B (8)

2. Find the median of the following set of daily high temperatures (in Celsius): -3, 2, -1, 5, 0, 4, -2.

• A) -1 °C
• B) 0.71 °C
• C) 0 °C
• D) 5 °C

Click for Solution

Formula: Median = The middle value when data is arranged in order

Brief Solution: 1. Arrange the data from least to greatest: -3, -2, -1, 0, 2, 4, 5.
2. Locate the middle position. Since there are 7 data points (an odd number), the 4th position is the exact middle.

Correct Answer: C (0 °C)

3. Consider the following table showing the number of goals scored by a hockey team over 6 games:

Game	Goals Scored
1	2
2	5
3	1
4	4
5	3
6	3

• A) 3.0
• B) 3.5
• C) 4.0
• D) 18.0

Click for Solution

Formula: Mean = (Sum of all values) ÷ (Total number of data points)

Brief Solution:
1. Sum = 2 + 5 + 1 + 4 + 3 + 3 = 18
2. Total games (n) = 6
3. Mean = 18 ÷ 6 = 3.0

Correct Answer: A (3.0)

4. The range of a set of data is 24. If the lowest value (minimum) in the dataset is 11, what is the highest value (maximum)?

• A) 13
• B) 24
• C) 35
• D) 46

Click for Solution

Formula: Range = Maximum value - Minimum value

Brief Solution: Rearrange the formula to solve for the Maximum:
Maximum = Range + Minimum
Maximum = 24 + 11 = 35

Correct Answer: C (35)

5. A local coffee shop tracks the number of espresso shots added to specialty drinks over an hour:

Shots Added (x)	Frequency (f)
1	5
2	12
3	4
4	2

• A) 2
• B) 12
• C) 2.13
• D) 3

Click for Solution

Formula: Mode = The data value with the highest frequency

Brief Solution: Look down the "Frequency" column to find the highest number, which is 12. The corresponding data value in the "Shots Added" column is 2.

Correct Answer: A (2)

6. An even number of data points are already arranged in order: 12, 15, 17, 21, 25, 30. How do you calculate the median?

• A) Take the difference between 30 and 12.
• B) Select 17 because it is on the left side of the middle.
• C) Add all values and divide by 6.
• D) Find the mean of 17 and 21.

Click for Solution

Formula: Median (for even n) = (Sum of the two middle values) ÷ 2

Brief Solution: There are 6 items. The two middle values are the 3rd and 4th terms, which are 17 and 21. You must average them: (17 + 21) ÷ 2 = 19.

Correct Answer: D (Find the mean of 17 and 21.)

7. A small retail business has five employees who earn hourly wages of $16, $16, $18, $20, and $45. Which measure of central tendency is distorted the most by the $45 outlier?

• A) Mode
• B) Median
• C) Mean
• D) Range

Click for Solution

Concept: The Mean is sensitive to outliers because it incorporates every exact value.

Brief Solution:
- Without $45, the mean is $17.50. With $45, it jumps to $23.00 (higher than what 4 out of 5 people make).
- The median remains $18 and the mode remains $16 regardless of the outlier value.

Correct Answer: C (Mean)

8. A teacher organizes the class project percentages into a frequency table:

Mark (%)	Number of Students (f)
65	3
75	7
85	8
95	2

• A) 4
• B) 20
• C) 30
• D) 80

Click for Solution

Formula: n = Σf (Sum of all frequencies)

Brief Solution: Add up the number of students in each category to find the total sample size: 3 + 7 + 8 + 2 = 20.

Correct Answer: B (20)

9. A set of four numbers has a mean of 10. If three of the numbers are 8, 11, and 12, what is the fourth number?

• A) 7
• B) 9
• C) 10
• D) 11

Click for Solution

Formula: Required Total Sum = Mean × n

Brief Solution:
1. Calculate required total sum: 10 × 4 = 40
2. Sum the known numbers: 8 + 11 + 12 = 31
3. Subtract to find the missing number: 40 - 31 = 9.

Correct Answer: B (9)

10. You are given the dataset: 5, 5, 7, 8, 10. If a new number, 15, is added to this dataset, which of the following statements is true?

• A) The mode changes.
• B) The mean decreases.
• C) The range increases.
• D) The median changes to 8.5.

Click for Solution

Formula: Range = Maximum - Minimum

Brief Solution:
- Original Range: 10 - 5 = 5
- New Range (with 15 included): 15 - 5 = 10.
Since the new value is higher than the previous maximum, the gap between the largest and smallest numbers expands.

Correct Answer: C (The range increases.)

Solving LCM Problems

 

Nigerian Navy Secondary Schools Entrance Examination

Mathematics & Quantitative Reasoning Past Questions Guide

What is LCM?

LCM (Lowest Common Multiple) is the smallest positive integer that is perfectly divisible by a given set of numbers without leaving any remainder.

Example: The LCM of 4 and 6 is 12 because 12 is the smallest number that both 4 and 6 can divide into evenly.

Practice Questions & Solutions

1. Find the Lowest Common Multiple (L.C.M) of 12, 18, and 24.

• A) 36
• B) 48
• C) 72
• D) 144

Solution:
Prime factors: 12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3.
LCM = 2³ × 3² = 8 × 9 = 72.
Correct Answer: C) 72

2. Simplify: (0.2 × 0.01) / 0.0002

• A) 1
• B) 10
• C) 100
• D) 2

Solution:
0.2 × 0.01 = 0.002.
0.002 / 0.0002 = 20 / 2 = 10.
Correct Answer: B) 10

3. What is the difference between 100 and 99.991?

• A) 0.009
• B) 0.01
• C) 0.09
• D) 0.001

Solution:
100.000 - 99.991 = 0.009.
Correct Answer: A) 0.009

4. A line that divides a circle into two equal halves is called a _______.

• A) Chord
• B) Radius
• C) Segment
• D) Diameter

Solution:
The diameter passes through the center of a circle and divides it into two equal semicircles.
Correct Answer: D) Diameter

5. Change 1 hour 5 minutes 10 seconds to seconds.

• A) 3600 seconds
• B) 3910 seconds
• C) 3900 seconds
• D) 4110 seconds

Solution:
1 hr = 3600s; 5 mins = 300s; Total = 3600 + 300 + 10 = 3910 seconds.
Correct Answer: B) 3910 seconds

6. What is the place value of 7 in 45.372?

• A) Tens
• B) Tenths
• C) Hundredths
• D) Thousandths

Solution:
The first digit after the decimal point is tenths (3), the second digit is hundredths (7).
Correct Answer: C) Hundredths

7. A car uses 1 litre of petrol for every 14km. If 1 litre costs ₦250.00, how far can the car go with ₦1,500.00 worth of petrol?

• A) 70km
• B) 84km
• C) 98km
• D) 112km

Solution:
Litres bought = 1500 / 250 = 6 litres.
Distance = 6 × 14 = 84km.
Correct Answer: B) 84km

8. Correct 0.002473 to 3 significant figures.

• A) 0.002
• B) 0.00247
• C) 0.00248
• D) 0.0025

Solution:
Count significant figures starting from the first non-zero digit (2, 4, 7). The next digit is 3, so round down.
Correct Answer: B) 0.00247

9. Quantitative Reasoning: If 4 Δ 2 = 10 and 5 Δ 3 = 13, find 7 Δ 4.

• A) 11
• B) 18
• C) 19
• D) 22

Solution:
The logic pattern is: (First number × 2) + Second number.
For 7 Δ 4: (7 × 2) + 4 = 14 + 4 = 18.
Correct Answer: B) 18

10. An angle on a straight line is equal to ______ degrees.

• A) 90°
• B) 180°
• C) 270°
• D) 360°

Solution:
Angles on a straight line always sum up to exactly 180 degrees.
Correct Answer: B) 180°

Word Problems On Agez

 

Mastering Age Word Problems: Concepts, Hints, and Examples

Age problems are a classic staple of algebra. They might look like riddles at first, but once you break them down into variables and equations, they become simple puzzles to solve.

Introduction to the Concept of Ages

At its core, an age problem is just a linear equation in disguise. The fundamental concept relies on a single, unchangeable truth: time moves at the same rate for everyone.

If you are 5 years older than your sister today, you will still be 5 years older than her 20 years from now. When solving these problems, we generally deal with three distinct time frames:

• The Past: indicated by phrases like "5 years ago" or "in 2018".
• The Present: indicated by phrases like "currently", "is", or "now".
• The Future: indicated by phrases like "in 10 years", "hence", or "will be".

Key Hints for Solving Age Calculations

To translate a word problem into a workable mathematical equation, keep these handy rules in mind:

• Define the Present First: Always let the present age of a person be your primary variable (e.g., J for John). It makes navigating backwards or forwards in time much easier.
• "Years Ago" means Subtraction: If a person's current age is x, their age n years ago was x - n.
• "Years Hence" means Addition: If a person's current age is x, their age n years from now will be x + n.
• Apply Time Changes to Everyone: If a problem moves 5 years into the future, you must add 5 to every single person's age.
• Watch the Verbs: "Is/Was/Will be" translate to equals (=), and "Times" translates to multiplication (×).

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5 Simple Practice Examples

Part 1: Age of One Person (Past vs. Future)

Example 1: Finding Present Age from the Future

Problem: In 12 years, John will be three times as old as he is right now. How old is John today?

Setup: Let John's current age be J. In 12 years, his age will be J + 12.

Equation: J + 12 = 3J

Calculation:
12 = 2J
J = 6

Answer: John is 6 years old today.

Example 2: Finding Present Age from the Past

Problem: Five years ago, Sarah was half the age she will be next year. How old is Sarah now?

Setup: Let Sarah's current age be S. Five years ago she was S - 5. Next year she will be S + 1.

Equation: S - 5 = 0.5(S + 1)

Calculation:
2(S - 5) = S + 1
2S - 10 = S + 1
S = 11

Answer: Sarah is 11 years old now.

Part 2: Relationship Between Three People

Example 3: Sum of Ages with Ratios

Problem: The total age of Amy, Brad, and Charlie is 45 years. Brad is twice as old as Amy, and Charlie is 5 years older than Brad. How old is Amy?

Setup: Let Amy's age be A. Therefore, Brad is 2A, and Charlie is 2A + 5.

Equation: A + 2A + (2A + 5) = 45

Calculation:
5A + 5 = 45
5A = 40
A = 8

Answer: Amy is 8 years old.

Example 4: Moving into the Past with Three People

Problem: Tom is 10 years older than Jerry, and Jerry is 4 years older than Spike. Three years ago, the sum of their ages was 30. How old is Spike now?

Setup: Let Spike's age be S. Jerry is S + 4. Tom is S + 14.

Equation (Three years ago): (S - 3) + (S + 4 - 3) + (S + 14 - 3) = 30

Calculation:
(S - 3) + (S + 1) + (S + 11) = 30
3S + 9 = 30
3S = 21
S = 7

Answer: Spike is 7 years old.

Example 5: Future Multipliers

Problem: Liam is twice as old as Noah, and Noah is twice as old as Mason. In 2 years, the sum of Liam and Mason's ages will equal 3 times Noah's current age. How old is Noah?

Setup: Let Mason = M. Noah = 2M. Liam = 4M.

Equation: (4M + 2) + (M + 2) = 3(2M)

Calculation:
5M + 4 = 6M
M = 4
Noah's age = 2(4) = 8

Answer: Noah is 8 years old.

Problems On Profit And Loss

 

Profit and Loss: The Basics

• Profit: Occurs when a business sells a product or service for more than it cost to buy. It represents financial gain.
• Loss: Occurs when a business sells a product or service for less than its cost price. It represents financial deficit.

The Key Formulas

• Profit = Selling Price (SP) - Cost Price (CP)
• Loss = Cost Price (CP) - Selling Price (SP)
• Profit Percentage (%) = (Profit / CP) × 100
• Loss Percentage (%) = (Loss / CP) × 100

8 Simple Practice Questions & Solutions

Q1. Simple Profit

A shopkeeper buys a book for $200 and sells it for $250. Find the profit.

Solution:
CP = $200, SP = $250
Profit = SP - CP = 250 - 200 = 50
Answer: The profit is $50.

Q2. Simple Loss

A cycle bought for $1,200 is sold for $1,000. Find the loss.

Solution:
CP = $1,200, SP = $1,000
Loss = CP - SP = 1200 - 1000 = 200
Answer: The loss is $200.

Q3. Profit Percentage

An item bought for $80 is sold for $100. Find the profit percentage.

Solution:
CP = $80, SP = $100
Profit = 100 - 80 = 20
Profit % = (20 / 80) × 100 = 25%
Answer: The profit percentage is 25%.

Q4. Loss Percentage

A toy bought for $50 is sold for $40. Find the loss percentage.

Solution:
CP = $50, SP = $40
Loss = 50 - 40 = 10
Loss % = (10 / 50) × 100 = 20%
Answer: The loss percentage is 20%.

Q5. Finding Selling Price (Profit)

A man buys a watch for $500 and wants to make a 10% profit. What should the selling price be?

Solution:
CP = $500, Profit % = 10%
Profit Amount = 10% of 500 = $50
SP = CP + Profit = 500 + 50 = 550
Answer: The selling price should be $550.

Q6. Finding Selling Price (Loss)

A table bought for $300 is sold at a loss of 15%. Find the selling price.

Solution:
CP = $300, Loss % = 15%
Loss Amount = 15% of 300 = $45
SP = CP - Loss = 300 - 45 = 255
Answer: The selling price is $255.

Q7. Finding Cost Price (Profit)

An article is sold for $120 at a profit of 20%. Find its cost price.

Solution:
SP = $120, Profit % = 20%
SP = CP × 1.20 → 120 = CP × 1.20
CP = 120 / 1.20 = 100
Answer: The cost price is $100.

Q8. Finding Cost Price (Loss)

By selling a bag for $90, a shopkeeper incurs a loss of 10%. Find the cost price.

Solution:
SP = $90, Loss % = 10%
SP = CP × 0.90 → 90 = CP × 0.90
CP = 90 / 0.90 = 100
Answer: The cost price is $100.