CBSE Class 8 Mathematics Practice Worksheet Set 10

SECTION - A

Select one correct answer out of the four options given.

Question 1. Point at which the two perpendicular axes meet, is known as :
(a) Centre
(b) vertex
(c) coordinate
(d) origin
Answer: (d) origin
The point of intersection of the x-axis and y-axis in a Cartesian plane is called the origin, and its coordinates are (0, 0).
In simple words: The starting point where the two graph lines cross is called the origin.

Exam Tip: Remember that axes coordinates at this point are always (0, 0).

Question 2. 20 - 30 is a class interval, the 25 is its -
(a) Lower limit
(b) Range
(c) Class mark
(d) Frequency
Answer: (c) Class mark
Class mark is calculated as \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\). For interval 20 - 30, the class mark is \(\frac{20 + 30}{2} = \frac{50}{2} = 25\).
In simple words: The exact middle value of a class interval is its class mark.

Exam Tip: Always use the formula \(\text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).

Question 3. l, m, n are lines such that \(m \perp l\), \(n \perp l\), then
(a) \(m \perp n\)
(b) \(m \parallel n\)
(c) \(m \parallel l\)
(d) \(l \parallel n\)
Answer: (b) \(m \parallel n\)
If two lines (\(m\) and \(n\)) are perpendicular to the same line (\(l\)), then both lines are parallel to each other.
In simple words: Two lines perpendicular to the same straight line never meet and are parallel to each other.

Exam Tip: Lines perpendicular to a common line are always parallel to one another.

Question 4. When K is any constant for two quantities x and y which vary inversely then :
(a) \(\frac{x}{y} = K\)
(b) \(x + y = K\)
(c) \(x - y = K\)
(d) \(xy = K\)
Answer: (d) xy = K
For inverse proportion, the product of two quantities remains constant. Thus, \(x \times y = K\).
In simple words: In inverse variation, multiplying the two quantities always gives the same constant value.

Exam Tip: Direct proportion uses \(\frac{x}{y} = K\), whereas inverse proportion uses \(xy = K\).

Question 5. Value of \(\sqrt{40 + \sqrt{81}}\) is :
(a) 11
(b) 121
(c) 7
(d) 29
Answer: (c) 7
First, evaluate the inner square root: \(\sqrt{81} = 9\).
Substitute this back: \(\sqrt{40 + 9} = \sqrt{49} = 7\).
In simple words: Simplify from the inside out by finding the square root of 81 first, add it to 40, and take the square root of the result.

Question 6. If cost price of 18 mangoes is the same as the selling price of 16 mangoes. Find the gain percent.
Answer: Let the cost price of 1 mango be Rs 1.
Cost price of 18 mangoes = Rs 18.
Selling price of 16 mangoes = Rs 18.
Cost price of 16 mangoes = Rs 16.
Profit = Selling price - Cost price = \(18 - 16 = \text{Rs } 2\).
Gain percent = \(\left(\frac{\text{Profit}}{\text{Cost Price of 16 mangoes}}\right) \times 100\)
Gain percent = \(\left(\frac{2}{16}\right) \times 100 = \frac{1}{8} \times 100 = 12.5\%\).
In simple words: By selling only 16 mangoes, we recover the cost of 18, meaning the profit from 2 mangoes is our gain percentage over 16 mangoes.

Exam Tip: Always calculate profit or loss percentage on the number of items whose selling price is equated.

Question 7. Find the smallest number by which 392 must be multiplied so that the product is a perfect cube.
Answer: Find the prime factorization of 392:
\(392 = 2 \times 196 = 2 \times 2 \times 98 = 2 \times 2 \times 2 \times 49 = 2^3 \times 7^2\)
In the prime factorization, the prime factor 2 has a triplet \(2^3\), but the prime factor 7 appears only twice (\(7^2\)).
To make it a perfect cube, we need one more 7.
Therefore, 392 must be multiplied by 7.
In simple words: Break the number into prime factors and find which factor is short of making a complete group of three.

Exam Tip: For cube roots, group identical prime factors in triplets of three.

Question 8. Expand \((-3x + 4y - 5)^2\)
Answer: Use the algebraic identity: \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).
Here, let \(a = -3x\), \(b = 4y\), and \(c = -5\).
\((-3x + 4y - 5)^2 = (-3x)^2 + (4y)^2 + (-5)^2 + 2(-3x)(4y) + 2(4y)(-5) + 2(-5)(-3x)\)
\(= 9x^2 + 16y^2 + 25 - 24xy - 40y + 30x\)
In simple words: Square each term individually, then add twice the product of every possible pair of terms with their proper signs.

Exam Tip: Be very careful with negative signs when multiplying pairs for the 2ab, 2bc, and 2ca terms.

Question 9. Find the greatest number of 4 digits which is a perfect square.
Answer: The greatest 4-digit number is 9999.
Using long division to find the square root of 9999:
\(9900\) has root around 99.
Let us check \(99^2 = 9801\).
Let us check \(100^2 = 10000\) (which is a 5-digit number).
Therefore, the greatest 4-digit perfect square is 9801.
In simple words: The largest 4-digit number is 9999, and checking numbers near 100 shows that 99 squared gives 9801.

Exam Tip: Always test squares around the boundary where digits transition from 4 to 5 (i.e., 100).

SECTION - C

Question 10. Simplify \((a^2 - b^2)(a^2 + b^2) - (a^2 - b^2)^2\)
Answer: Expand each part using standard algebraic identities:
1. \((a^2 - b^2)(a^2 + b^2) = (a^2)^2 - (b^2)^2 = a^4 - b^4\)
2. \((a^2 - b^2)^2 = (a^2)^2 - 2(a^2)(b^2) + (b^2)^2 = a^4 - 2a^2b^2 + b^4\)
Now substitute back into the expression:
\(= (a^4 - b^4) - (a^4 - 2a^2b^2 + b^4)\)
\(= a^4 - b^4 - a^4 + 2a^2b^2 - b^4\)
\(= 2a^2b^2 - 2b^4\)
In simple words: Expand both parts using algebraic formulas and then combine like terms by changing signs inside the subtracted bracket.

Exam Tip: Distribute the negative sign carefully to all terms inside the parenthesis when opening a subtracted bracket.

Question 11. A shopkeeper purchased 200 bulbs for Rs 10 each. However 5 bulbs were fused and had to be thrown away. The remaining was sold at Rs 12 each. Find the gain or loss percent.
Answer: Total number of bulbs purchased = 200
Cost price of 1 bulb = Rs 10
Total Cost Price (CP) = \(200 \times 10 = \text{Rs } 2000\)
Number of fused bulbs = 5
Remaining bulbs sold = \(200 - 5 = 195\)
Selling price of 1 bulb = Rs 12
Total Selling Price (SP) = \(195 \times 12 = \text{Rs } 2340\)
Since SP > CP, there is a profit.
Profit = SP - CP = \(2340 - 2000 = \text{Rs } 340\)
Gain percent = \(\left(\frac{\text{Profit}}{\text{CP}}\right) \times 100\)
Gain percent = \(\left(\frac{340}{2000}\right) \times 100 = \frac{34}{2} = 17\%\)
In simple words: Subtract broken bulbs from total stock, calculate total money spent and total money earned, then find the percentage profit.

Exam Tip: Always calculate profit or loss percentage based on the total cost price, not the selling price.

Question 12. Find the value of \(\sqrt{2}\), correct upto two decimal places.
Answer: Using long division method for square root of 2.0000:
\(1 \times 1 = 1\) (remainder 1)
Bring down 00 to make 100. Double 1 to get 2_.
\(24 \times 4 = 96\) (remainder 4)
Bring down another 00 to make 400. Double 14 to get 28_.
\(281 \times 1 = 281\) (remainder 119)
Thus, \(\sqrt{2} \approx 1.414\)
Correct up to two decimal places, \(\sqrt{2} = 1.41\).
In simple words: Use the long division method for finding square roots to calculate digits after the decimal point.

Exam Tip: To find square root up to two decimal places, calculate up to three decimal places and round off.

Question 13. If \(5x - 2y = 7\) and \(xy = 2\), find the value of \((5x + 2y)^2\)
Answer: We know the algebraic identity connecting sum and difference:
\((5x + 2y)^2 = (5x - 2y)^2 + 4(5x)(2y)\)
Substitute the given values \(5x - 2y = 7\) and \(xy = 2\):
\((5x + 2y)^2 = (7)^2 + 4 \times 10 \times xy\)
\((5x + 2y)^2 = 49 + 40(2)\)
\((5x + 2y)^2 = 49 + 80 = 129\)
In simple words: Use the formula that links the square of a sum to the square of a difference and the product of the terms.

Exam Tip: Memorize the identity \((a + b)^2 = (a - b)^2 + 4ab\) to quickly switch between sum and difference.

SECTION - D

Question 14. Sangeeta allows \(8\%\) discount on the marked price of a suit and still makes a profit of \(15\%\). If her gain over the sale of a suit is Rs 156. Find the marked price of the suit.
Answer: Let the Cost Price (CP) be Rs \(x\).
Gain = \(15\%\) of CP = Rs 156.
\(\frac{15}{100} \times x = 156\)
\(x = \frac{156 \times 100}{15} = \frac{15600}{15} = \text{Rs } 1040\)
So, Cost Price = Rs 1040.
Selling Price (SP) = \(\text{CP} + \text{Gain} = 1040 + 156 = \text{Rs } 1196\).
Let Marked Price (MP) be \(y\).
Discount = \(8\%\) of MP.
\(\text{SP} = \text{MP} - \text{Discount}\)
\(1196 = y - \frac{8}{100}y\)
\(1196 = \frac{92}{100}y\)
\(y = \frac{1196 \times 100}{92} = 13 \times 100 = \text{Rs } 1300\).
Marked price of the suit = Rs 1300.
In simple words: First find the cost price using the profit amount, then find the selling price, and finally use the discount percentage to find the marked price.

Exam Tip: Cost price and marked price are always connected through the selling price.

Question 15. A train 120 m long is running at a speed of 50 km/hr. What time will it take to cross a 130 m long bridge?
Answer: Length of the train = 120 m
Length of the bridge = 130 m
Total distance to be covered = Length of train + Length of bridge
Distance = \(120 + 130 = 250\) m
Speed of the train = \(50 \text{ km/hr}\)
Convert speed to m/s: \(50 \times \frac{5}{18} = \frac{250}{18} \text{ m/s}\)
Time taken = \(\frac{\text{Distance}}{\text{Speed}}\)
Time = \(\frac{250}{\frac{250}{18}} = 250 \times \frac{18}{250} = 18\) seconds.
In simple words: Total distance is the sum of the train and bridge lengths. Convert km/hr to m/s and divide distance by speed to get the time in seconds.

Exam Tip: When a train crosses a bridge, always add the length of the train to the length of the bridge to find the total distance.

Question 16. Bharti bought two fans for Rs 1200 each. She sold one at a loss of \(5\%\) and the other at a profit of \(10\%\). Find the total profit or loss.
Answer: Cost price of each fan = Rs 1200.
Total Cost Price of two fans = \(1200 + 1200 = \text{Rs } 2400\).
For the first fan (sold at \(5\%\) loss):
Loss = \(5\%\) of 1200 = \(\frac{5}{100} \times 1200 = \text{Rs } 60\)
Selling Price of first fan = \(1200 - 60 = \text{Rs } 1140\).
For the second fan (sold at \(10\%\) profit):
Profit = \(10\%\) of 1200 = \(\frac{10}{100} \times 1200 = \text{Rs } 120\)
Selling Price of second fan = \(1200 + 120 = \text{Rs } 1320\).
Total Selling Price of both fans = \(1140 + 1320 = \text{Rs } 2460\).
Since Total SP > Total CP, there is a total profit.
Total Profit = \(2460 - 2400 = \text{Rs } 60\).
In simple words: Calculate individual selling prices for both fans, find the total cost and total selling price, and compare them to find the overall profit or loss.

Exam Tip: Find the total selling price of all items combined to find the net profit or loss easily.