CBSE Class 8 Mathematics Practice Worksheet Set 03

Question 1. Gunjan sells two watches for Rs. 2400 each, gaining 20% on one and losing 20% on the other. Find gain or loss percent on the whole transaction.
Answer: Selling price of each watch = Rs. 2400
Total selling price of both watches = \(2400 + 2400 = 4800\) Rs.
For the first watch (gain of 20%):
Let the cost price of the first watch be \(C_1\).
\(C_1 \times \left(1 + \frac{20}{100}\right) = 2400\)
\(C_1 \times 1.2 = 2400\)
\(C_1 = \frac{2400}{1.2} = 2000\) Rs.
For the second watch (loss of 20%):
Let the cost price of the second watch be \(C_2\).
\(C_2 \times \left(1 - \frac{20}{100}\right) = 2400\)
\(C_2 \times 0.8 = 2400\)
\(C_2 = \frac{2400}{0.8} = 3000\) Rs.
Total cost price of both watches = \(2000 + 3000 = 5000\) Rs.
Since Total Cost Price (Rs. 5000) is greater than Total Selling Price (Rs. 4800), there is a loss.
Loss = \(5000 - 4800 = 200\) Rs.
Loss % = \(\left(\frac{\text{Loss}}{\text{Total Cost Price}}\right) \times 100 = \left(\frac{200}{5000}\right) \times 100 = 4\%\)
In simple words: When an article is sold at equal percentage of profit and loss, there is always a net loss in the transaction.

Exam Tip: For equal percentage gain and loss on same selling price, use the direct formula: Loss % = \(\left(\frac{\text{Common Profit/Loss %}}{10}\right)^2 = \left(\frac{20}{10}\right)^2 = 4\%\).

Question 2. A shopkeeper marks his goods at such a price that after allowing a discount of 12.5% for cash payment, he still makes a profit of 10%. Find marked price of an article which costs him Rs. 245
Answer: Cost Price (CP) = Rs. 245
Profit = 10%
Selling Price (SP) = \(\text{CP} \times \left(1 + \frac{\text{Profit } \%}{100}\right)\)
\(\text{SP} = 245 \times \left(1 + \frac{10}{100}\right) = 245 \times 1.1 = 269.50\) Rs.
Let the Marked Price be MP.
Discount = 12.5%
Selling Price = \(\text{MP} \times \left(1 - \frac{\text{Discount } \%}{100}\right)\)
\(269.50 = \text{MP} \times \left(1 - \frac{12.5}{100}\right)\)
\(269.50 = \text{MP} \times 0.875\)
\(\text{MP} = \frac{269.50}{0.875} = 308\) Rs.
In simple words: First find the selling price using cost price and profit, then use the discount percentage to find the marked price.

Exam Tip: Remember the fundamental relation: Selling Price = Cost Price + Profit = Marked Price - Discount.

Question 3. A car can finish a certain journey in 10 hours at a speed of 24km/hr. By how much should it's speed be increased so that it may take only 8 hours to cover the same distance ?
Answer: Time taken initially = 10 hours
Speed initially = 24 km/hr
Total distance = \(\text{Speed} \times \text{Time} = 24 \times 10 = 240\) km.
New time taken = 8 hours
Required new speed = \(\frac{\text{Total Distance}}{\text{New Time}} = \frac{240}{8} = 30\) km/hr.
Increase in speed = New speed - Initial speed
Increase in speed = \(30 - 24 = 6\) km/hr.
In simple words: Calculate the total distance first, then divide it by the new time to find the required speed, and subtract the original speed from it.

Exam Tip: This is an inverse proportion problem: \(\text{Speed}_1 \times \text{Time}_1 = \text{Speed}_2 \times \text{Time}_2\).

Question 4. A can do a piece of work in 20 days and B can finish it in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Answer: A's 1 day work = \(\frac{1}{20}\)
B's 1 day work = \(\frac{1}{15}\)
Combined 1 day work of A and B = \(\frac{1}{20} + \frac{1}{15} = \frac{3 + 4}{60} = \frac{7}{60}\)
Work done by both in 2 days = \(2 \times \frac{7}{60} = \frac{14}{60} = \frac{7}{30}\)
Remaining work = \(1 - \frac{7}{30} = \frac{23}{30}\)
Time taken by B to finish the remaining work = \(\frac{\text{Remaining Work}}{\text{B's 1 day work}} = \frac{\frac{23}{30}}{\frac{1}{15}} = \frac{23}{30} \times 15 = \frac{23}{2} = 11\frac{1}{2}\) days.
In simple words: Find how much work A and B did together in 2 days, subtract it from the total work (1), and divide the leftover work by B's daily work rate.

Exam Tip: Always convert individual days into 1-day work rates before adding or subtracting work fractions.

Question 5. Find the smallest no which must be added to 2300 so that it becomes a perfect square.
Answer: Let us find the square root of 2300 using long division method:
\(48^2 = 2304\)
\(47^2 = 2209\)
Since \(2209 < 2300 < 2304\), the next perfect square after 2300 is \(48^2 = 2304\).
Smallest number to be added = \(2304 - 2300 = 4\).
In simple words: Find the square of the next integer immediately greater than the square root of 2300, and find the difference.

Exam Tip: For "what must be added" problems, always find the square of the next higher integer (\(\text{quotient} + 1\)).

Question 6. Divide \(3y^5 + 6y^3 + 7y^2 + 6y^4 + 8y + 9\) by \(3y^3 + 1\). Find Quotient and Remainder.
Answer: First, write the dividend in standard descending order of powers:
Dividend = \(3y^5 + 6y^4 + 6y^3 + 7y^2 + 8y + 9\)
Divisor = \(3y^3 + 1\)
Perform polynomial long division:
1. Divide leading term \(3y^5\) by \(3y^3\) to get \(y^2\).
Multiply \(y^2\) by \((3y^3 + 1)\) = \(3y^5 + y^2\).
Subtract from dividend: \((3y^5 + 6y^4 + 6y^3 + 7y^2 + 8y + 9) - (3y^5 + y^2) = 6y^4 + 6y^3 + 6y^2 + 8y + 9\).
2. Divide leading term \(6y^4\) by \(3y^3\) to get \(2y\).
Multiply \(2y\) by \((3y^3 + 1)\) = \(6y^4 + 2y\).
Subtract: \((6y^4 + 6y^3 + 6y^2 + 8y + 9) - (6y^4 + 2y) = 6y^3 + 6y^2 + 6y + 9\).
3. Divide leading term \(6y^3\) by \(3y^3\) to get \(2\).
Multiply \(2\) by \((3y^3 + 1)\) = \(6y^3 + 2\).
Subtract: \((6y^3 + 6y^2 + 6y + 9) - (6y^3 + 2) = 6y^2 + 6y + 7\).
Quotient = \(y^2 + 2y + 2\)
Remainder = \(6y^2 + 6y + 7\)
In simple words: Arrange the terms in decreasing order of exponents and perform standard algebraic long division step by step.

Exam Tip: Always re-arrange the polynomial terms in standard descending power form before starting division to avoid sign errors.

Question 7. If \(x + y = 4\) and \(xy = 2\), find the value of \(x^2 + y^2\).
Answer: We know the algebraic identity:
\((x + y)^2 = x^2 + y^2 + 2xy\)
Substitute the given values \(x + y = 4\) and \(xy = 2\) into the identity:
\((4)^2 = x^2 + y^2 + 2(2)\)
\(16 = x^2 + y^2 + 4\)
\(x^2 + y^2 = 16 - 4\)
\(x^2 + y^2 = 12\)
In simple words: Use the expansion formula for the square of a binomial sum to easily find the sum of squares.

Exam Tip: Memorize basic algebraic identities like \((x+y)^2 = x^2 + y^2 + 2xy\) thoroughly as they frequently appear in exams.

Question 8. 100 soldiers had enough food for 20 days. But some soldiers were transferred to another place and food lasted for 25 days. How many soldiers were transferred .
Answer: Total food available in terms of soldier-days = \(100 \times 20 = 2000\) soldier-days.
Let the remaining number of soldiers after transfer be \(N\).
Since the same food lasted for 25 days:
\(N \times 25 = 2000\)
\(N = \frac{2000}{25} = 80\) soldiers.
Number of soldiers transferred = Initial soldiers - Remaining soldiers
Number of soldiers transferred = \(100 - 80 = 20\).
In simple words: Total food units remain constant. Equate initial (soldiers times days) to the new (soldiers times new days) to find the remaining soldiers.

Exam Tip: This is an inverse variation problem: more soldiers mean food lasts for fewer days.

Question 9. Pipe A can fill a tank in 4 hours and Pipe B can empty it in 6 hours. If both the pipes are opened in the empty tank, then in how many hours will it be full ?
Answer: Part of tank filled by Pipe A in 1 hour = \(\frac{1}{4}\)
Part of tank emptied by Pipe B in 1 hour = \(\frac{1}{6}\)
Net part of tank filled in 1 hour when both pipes are open = \(\frac{1}{4} - \frac{1}{6}\)
\(\frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12}\)
Total time taken to fill the empty tank = \(\frac{1}{\frac{1}{12}} = 12\) hours.
In simple words: Subtract the emptying rate of pipe B from the filling rate of pipe A to find the net filling speed per hour.

Exam Tip: Inlet pipes have a positive work rate while outlet/drain pipes have a negative work rate.

Question 10. Using laws of exponents - Simplify : \(\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}\)
Answer: Given expression: \(\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}\)
Express all base numbers in terms of prime factors 2 and 3:
\(12 = 2^2 \times 3 \implies 12^4 = (2^2 \times 3)^4 = 2^8 \times 3^4\)
\(9 = 3^2 \implies 9^3 = (3^2)^3 = 3^6\)
\(4 = 2^2\)
\(6 = 2 \times 3 \implies 6^3 = (2 \times 3)^3 = 2^3 \times 3^3\)
\(8 = 2^3 \implies 8^2 = (2^3)^2 = 2^6\)
\(27 = 3^3\)
Substitute prime factorizations into the numerator and denominator:
Numerator = \((2^8 \times 3^4) \times (3^6) \times (2^2) = 2^{8+2} \times 3^{4+6} = 2^{10} \times 3^{10}\)
Denominator = \((2^3 \times 3^3) \times (2^6) \times (3^3) = 2^{3+6} \times 3^{3+3} = 2^9 \times 3^6\)
Now divide the expression using laws of exponents (\(\frac{a^m}{a^n} = a^{m-n}\)):
\(= \frac{2^{10} \times 3^{10}}{2^9 \times 3^6}\)
\(= 2^{10-9} \times 3^{10-6}\)
\(= 2^1 \times 3^4\)
\(= 2 \times 81 = 162\)
In simple words: Break down all composite numbers into their prime factors (2 and 3), combine exponents with the same base, and subtract denominator powers from numerator powers.

Exam Tip: Always convert bases into prime numbers (like 2, 3, 5) before applying exponent laws.