50+ Geometry Word Problems worksheets for 8th Class on Quizizz ... - Free Printable
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Step-by-step solution for: 50+ Geometry Word Problems worksheets for 8th Class on Quizizz ...
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Step-by-step solution for: 50+ Geometry Word Problems worksheets for 8th Class on Quizizz ...
Here are the step-by-step solutions for each problem on the worksheet.
1. If a number is multiplied by three-fourth of itself, the value thus obtained is 10800. What is the number?
* Step 1: Let the number be $x$.
* Step 2: "Three-fourth of itself" means $\frac{3}{4}x$.
* Step 3: The problem says the number ($x$) multiplied by this amount equals 10800. So, the equation is:
$$x \cdot \frac{3}{4}x = 10800$$
$$\frac{3}{4}x^2 = 10800$$
* Step 4: Multiply both sides by 4 to get rid of the fraction:
$$3x^2 = 43200$$
* Step 5: Divide by 3:
$$x^2 = 14400$$
* Step 6: Take the square root of 14400. Since $12 \cdot 12 = 144$, then $120 \cdot 120 = 14400$.
$$x = 120$$
2. Find the maximum number of trees which can be planted 20 meter apart on the two sides of a straight road 1760 meter long?
* Step 1: First, find out how many trees fit on just one side of the road.
* Step 2: Divide the total length by the distance between trees:
$$1760 / 20 = 88$$ spaces.
* Step 3: Remember that if you plant a tree at the very start (0 meters) and one at the very end, you need one more tree than the number of spaces. This is called the "fence post error."
Trees on one side = $88 + 1 = 89$ trees.
* Step 4: The problem asks for trees on two sides.
$$89 \cdot 2 = 178$$ trees.
3. The sum of five consecutive numbers is 190. What is the sum of the largest and smallest number?
* Step 1: Let the middle number be $n$. The five consecutive numbers are:
$(n-2), (n-1), n, (n+1), (n+2)$
* Step 2: Add them up:
$$(n-2) + (n-1) + n + (n+1) + (n+2) = 5n$$
* Step 3: We know the sum is 190, so:
$$5n = 190$$
$$n = 190 / 5 = 38$$
So the middle number is 38.
* Step 4: The numbers are 36, 37, 38, 39, 40.
Smallest = 36. Largest = 40.
* Step 5: Add the largest and smallest:
$$36 + 40 = 76$$
*(Shortcut: In any set of consecutive numbers with an odd count, the sum of the first and last is always double the middle number: $38 \cdot 2 = 76$)*.
4. If $\frac{3}{4}$ of a number is subtracted from itself, the value so obtained is 163. What is the number?
* Step 1: Let the number be $x$.
* Step 2: Subtract $\frac{3}{4}x$ from $x$:
$$x - \frac{3}{4}x = 163$$
* Step 3: Think of $x$ as $\frac{4}{4}x$.
$$\frac{4}{4}x - \frac{3}{4}x = \frac{1}{4}x$$
So, $\frac{1}{4}x = 163$.
* Step 4: To find $x$, multiply 163 by 4:
$$163 \cdot 4 = 652$$
5. If $a = 3$, then $\frac{2}{(1/7 + 1/a)} = ?$
* Step 1: Substitute $a = 3$ into the expression:
$$\frac{2}{(1/7 + 1/3)}$$
* Step 2: Solve the bottom part (the denominator) first. Find a common denominator for 7 and 3, which is 21.
$$\frac{1}{7} = \frac{3}{21}$$
$$\frac{1}{3} = \frac{7}{21}$$
$$\frac{3}{21} + \frac{7}{21} = \frac{10}{21}$$
* Step 3: Now the expression is:
$$\frac{2}{(10/21)}$$
* Step 4: Dividing by a fraction is the same as multiplying by its reciprocal (flip it):
$$2 \cdot \frac{21}{10} = \frac{42}{10}$$
* Step 5: Simplify or convert to decimal:
$$4.2$$
6. If the expression $x^3 + 2hx - 2$ is equal to 6 when $x = -2$, what is the value of h?
* Step 1: Plug in $x = -2$ and set the equation equal to 6:
$$(-2)^3 + 2h(-2) - 2 = 6$$
* Step 2: Calculate the powers and products:
$$(-2)^3 = -8$$
$$2h(-2) = -4h$$
So: $-8 - 4h - 2 = 6$
* Step 3: Combine the numbers on the left:
$$-10 - 4h = 6$$
* Step 4: Add 10 to both sides:
$$-4h = 16$$
* Step 5: Divide by -4:
$$h = -4$$
7. A solution is made of water and pure acid. If 75% of the solution is water, how many liters of pure acid are in 20 liters of this solution?
* Step 1: If 75% is water, the rest must be acid.
$$100\% - 75\% = 25\%$$ is acid.
* Step 2: Find 25% of 20 liters.
$$0.25 \cdot 20$$
*(Hint: 25% is the same as $\frac{1}{4}$)*
$$20 / 4 = 5$$ liters.
8. The diagonal of a rectangle has a measure of 13 inches and the width is 12 inches. What is the perimeter of this square?
* *Note: The question calls it a "square" at the end, but gives different measurements for width and diagonal, which makes it a rectangle. We will solve for the rectangle described.*
* Step 1: Use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the missing length (let's call it $L$). The diagonal ($c$) is 13, and width ($b$) is 12.
$$L^2 + 12^2 = 13^2$$
$$L^2 + 144 = 169$$
* Step 2: Subtract 144 from 169:
$$L^2 = 25$$
$$L = 5$$
* Step 3: Now we have a rectangle with Width = 12 and Length = 5.
* Step 4: Calculate the perimeter ($P = 2 \cdot (\text{length} + \text{width})$):
$$P = 2 \cdot (5 + 12)$$
$$P = 2 \cdot 17$$
$$P = 34$$ inches.
9. Jesse has \$115.00. erasers cost \$8.00 each, pencils cost \$17.00 each. How many erasers and pencils can Jesse buy?
* *Note: This problem usually implies buying the maximum amount possible or finding a combination that uses the money exactly. Since it doesn't specify "exactly," we look for a logical combination, often maximizing the more expensive item first or finding a mix that sums close to the total.*
* Let $e$ = erasers and $p$ = pencils.
$$8e + 17p \le 115$$
* Let's try to spend as much as possible.
* If he buys 0 pencils: $115 / 8 = 14$ erasers (remainder \$3). Total items: 14.
* If he buys 1 pencil (\$17): Remaining \$98. $98 / 8 = 12$ erasers (remainder \$2). Total items: 13.
* If he buys 2 pencils (\$34): Remaining \$81. $81 / 8 = 10$ erasers (remainder \$1). Total items: 12.
* If he buys 3 pencils (\$51): Remaining \$64. $64 / 8 = 8$ erasers (remainder \$0). This uses the money exactly.
* If he buys 4 pencils (\$68): Remaining \$47. $47 / 8 = 5$ erasers (remainder \$7). Total items: 9.
* If he buys 5 pencils (\$85): Remaining \$30. $30 / 8 = 3$ erasers (remainder \$6). Total items: 8.
* If he buys 6 pencils (\$102): Remaining \$13. $13 / 8 = 1$ eraser (remainder \$5). Total items: 7.
* If he buys 7 pencils (\$119): Too expensive.
The most likely intended answer for these types of problems is the combination that spends the money exactly with no change left over.
Combination: 3 Pencils and 8 Erasers.
Check: $(3 \cdot 17) + (8 \cdot 8) = 51 + 64 = 115$.
Final Answer:
1. 120
2. 178
3. 76
4. 652
5. 4.2
6. -4
7. 5 liters
8. 34 inches
9. 8 erasers and 3 pencils
1. If a number is multiplied by three-fourth of itself, the value thus obtained is 10800. What is the number?
* Step 1: Let the number be $x$.
* Step 2: "Three-fourth of itself" means $\frac{3}{4}x$.
* Step 3: The problem says the number ($x$) multiplied by this amount equals 10800. So, the equation is:
$$x \cdot \frac{3}{4}x = 10800$$
$$\frac{3}{4}x^2 = 10800$$
* Step 4: Multiply both sides by 4 to get rid of the fraction:
$$3x^2 = 43200$$
* Step 5: Divide by 3:
$$x^2 = 14400$$
* Step 6: Take the square root of 14400. Since $12 \cdot 12 = 144$, then $120 \cdot 120 = 14400$.
$$x = 120$$
2. Find the maximum number of trees which can be planted 20 meter apart on the two sides of a straight road 1760 meter long?
* Step 1: First, find out how many trees fit on just one side of the road.
* Step 2: Divide the total length by the distance between trees:
$$1760 / 20 = 88$$ spaces.
* Step 3: Remember that if you plant a tree at the very start (0 meters) and one at the very end, you need one more tree than the number of spaces. This is called the "fence post error."
Trees on one side = $88 + 1 = 89$ trees.
* Step 4: The problem asks for trees on two sides.
$$89 \cdot 2 = 178$$ trees.
3. The sum of five consecutive numbers is 190. What is the sum of the largest and smallest number?
* Step 1: Let the middle number be $n$. The five consecutive numbers are:
$(n-2), (n-1), n, (n+1), (n+2)$
* Step 2: Add them up:
$$(n-2) + (n-1) + n + (n+1) + (n+2) = 5n$$
* Step 3: We know the sum is 190, so:
$$5n = 190$$
$$n = 190 / 5 = 38$$
So the middle number is 38.
* Step 4: The numbers are 36, 37, 38, 39, 40.
Smallest = 36. Largest = 40.
* Step 5: Add the largest and smallest:
$$36 + 40 = 76$$
*(Shortcut: In any set of consecutive numbers with an odd count, the sum of the first and last is always double the middle number: $38 \cdot 2 = 76$)*.
4. If $\frac{3}{4}$ of a number is subtracted from itself, the value so obtained is 163. What is the number?
* Step 1: Let the number be $x$.
* Step 2: Subtract $\frac{3}{4}x$ from $x$:
$$x - \frac{3}{4}x = 163$$
* Step 3: Think of $x$ as $\frac{4}{4}x$.
$$\frac{4}{4}x - \frac{3}{4}x = \frac{1}{4}x$$
So, $\frac{1}{4}x = 163$.
* Step 4: To find $x$, multiply 163 by 4:
$$163 \cdot 4 = 652$$
5. If $a = 3$, then $\frac{2}{(1/7 + 1/a)} = ?$
* Step 1: Substitute $a = 3$ into the expression:
$$\frac{2}{(1/7 + 1/3)}$$
* Step 2: Solve the bottom part (the denominator) first. Find a common denominator for 7 and 3, which is 21.
$$\frac{1}{7} = \frac{3}{21}$$
$$\frac{1}{3} = \frac{7}{21}$$
$$\frac{3}{21} + \frac{7}{21} = \frac{10}{21}$$
* Step 3: Now the expression is:
$$\frac{2}{(10/21)}$$
* Step 4: Dividing by a fraction is the same as multiplying by its reciprocal (flip it):
$$2 \cdot \frac{21}{10} = \frac{42}{10}$$
* Step 5: Simplify or convert to decimal:
$$4.2$$
6. If the expression $x^3 + 2hx - 2$ is equal to 6 when $x = -2$, what is the value of h?
* Step 1: Plug in $x = -2$ and set the equation equal to 6:
$$(-2)^3 + 2h(-2) - 2 = 6$$
* Step 2: Calculate the powers and products:
$$(-2)^3 = -8$$
$$2h(-2) = -4h$$
So: $-8 - 4h - 2 = 6$
* Step 3: Combine the numbers on the left:
$$-10 - 4h = 6$$
* Step 4: Add 10 to both sides:
$$-4h = 16$$
* Step 5: Divide by -4:
$$h = -4$$
7. A solution is made of water and pure acid. If 75% of the solution is water, how many liters of pure acid are in 20 liters of this solution?
* Step 1: If 75% is water, the rest must be acid.
$$100\% - 75\% = 25\%$$ is acid.
* Step 2: Find 25% of 20 liters.
$$0.25 \cdot 20$$
*(Hint: 25% is the same as $\frac{1}{4}$)*
$$20 / 4 = 5$$ liters.
8. The diagonal of a rectangle has a measure of 13 inches and the width is 12 inches. What is the perimeter of this square?
* *Note: The question calls it a "square" at the end, but gives different measurements for width and diagonal, which makes it a rectangle. We will solve for the rectangle described.*
* Step 1: Use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the missing length (let's call it $L$). The diagonal ($c$) is 13, and width ($b$) is 12.
$$L^2 + 12^2 = 13^2$$
$$L^2 + 144 = 169$$
* Step 2: Subtract 144 from 169:
$$L^2 = 25$$
$$L = 5$$
* Step 3: Now we have a rectangle with Width = 12 and Length = 5.
* Step 4: Calculate the perimeter ($P = 2 \cdot (\text{length} + \text{width})$):
$$P = 2 \cdot (5 + 12)$$
$$P = 2 \cdot 17$$
$$P = 34$$ inches.
9. Jesse has \$115.00. erasers cost \$8.00 each, pencils cost \$17.00 each. How many erasers and pencils can Jesse buy?
* *Note: This problem usually implies buying the maximum amount possible or finding a combination that uses the money exactly. Since it doesn't specify "exactly," we look for a logical combination, often maximizing the more expensive item first or finding a mix that sums close to the total.*
* Let $e$ = erasers and $p$ = pencils.
$$8e + 17p \le 115$$
* Let's try to spend as much as possible.
* If he buys 0 pencils: $115 / 8 = 14$ erasers (remainder \$3). Total items: 14.
* If he buys 1 pencil (\$17): Remaining \$98. $98 / 8 = 12$ erasers (remainder \$2). Total items: 13.
* If he buys 2 pencils (\$34): Remaining \$81. $81 / 8 = 10$ erasers (remainder \$1). Total items: 12.
* If he buys 3 pencils (\$51): Remaining \$64. $64 / 8 = 8$ erasers (remainder \$0). This uses the money exactly.
* If he buys 4 pencils (\$68): Remaining \$47. $47 / 8 = 5$ erasers (remainder \$7). Total items: 9.
* If he buys 5 pencils (\$85): Remaining \$30. $30 / 8 = 3$ erasers (remainder \$6). Total items: 8.
* If he buys 6 pencils (\$102): Remaining \$13. $13 / 8 = 1$ eraser (remainder \$5). Total items: 7.
* If he buys 7 pencils (\$119): Too expensive.
The most likely intended answer for these types of problems is the combination that spends the money exactly with no change left over.
Combination: 3 Pencils and 8 Erasers.
Check: $(3 \cdot 17) + (8 \cdot 8) = 51 + 64 = 115$.
Final Answer:
1. 120
2. 178
3. 76
4. 652
5. 4.2
6. -4
7. 5 liters
8. 34 inches
9. 8 erasers and 3 pencils
Parent Tip: Review the logic above to help your child master the concept of 8th grade math word problems worksheet.
