CBSE Class 7 Mathematics Practice Assignment Set 03

Question 1. Which of the following are rational numbers? -3 , 5/7 , 0 , 5/9 , -1/2 , -11 , -3/6
Answer: A rational number is defined as a number that can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
-3 can be written as \(\frac{-3}{1}\).
5/7 is already in the form \(\frac{p}{q}\).
0 can be written as \(\frac{0}{1}\).
5/9 is already in the form \(\frac{p}{q}\).
-1/2 is already in the form \(\frac{p}{q}\).
-11 can be written as \(\frac{-11}{1}\).
-3/6 is already in the form \(\frac{p}{q}\).
Therefore, all the given numbers are rational numbers.
In simple words: All integers, fractions, and zero are rational numbers because they can all be written as fractions.
Exam Tip: Remember that any integer can be expressed as a fraction with a denominator of 1, making all integers rational numbers.

Question 2. Represent the rational numbers on the number line: -3 1/3 , 7/4 , -4 5/8
Answer: 
1. For \(-3 \frac{1}{3}\): Convert to improper fraction \(-\frac{10}{3}\). It lies between -3 and -4 on the number line. Divide the space between -3 and -4 into 3 equal parts, and take the 1st part to the left of -3.
2. For \(\frac{7}{4}\): Convert to mixed fraction \(1 \frac{3}{4}\). It lies between 1 and 2 on the number line. Divide the space between 1 and 2 into 4 equal parts, and take the 3rd part to the right of 1.
3. For \(-4 \frac{5}{8}\): Convert to improper fraction \(-\frac{37}{8}\). It lies between -4 and -5 on the number line. Divide the space between -4 and -5 into 8 equal parts, and take the 5th part to the left of -4.
In simple words: Convert mixed fractions to find out which two whole numbers each rational number sits between, then subdivide that section.

Exam Tip: Always convert mixed numbers to improper fractions or use mixed number markings to quickly identify the two integers the number falls between.

Question 3. By what number should \(3^9\) be multiplied so that the product is 3.
Answer: Let the required number be \(x\).
According to the question:
\(3^9 \times x = 3^1\)
\(x = \frac{3^1}{3^9}\)
Using the laws of exponents \(\frac{a^m}{a^n} = a^{m-n}\):
\(x = 3^{1-9}\)
\(x = 3^{-8} \text{ or } \frac{1}{3^8}\)
In simple words: Divide the target product by the given number using exponent rules to find the multiplier.

Exam Tip: Use exponent subtraction rules (\(a^m \div a^n = a^{m-n}\)) carefully, paying attention to the sign of the resulting power.

Question 4. Simplify and find the value of x: a) [(-31)^3]^0 = (-31)^x b) (3/4)^7 × (3/4)^{12} = (3/4)^x
Answer: 
a) Given: \([(-31)^3]^0 = (-31)^x\)
Using power of a power rule \((a^m)^n = a^{m \times n}\):
\((-31)^{3 \times 0} = (-31)^x\)
\((-31)^0 = (-31)^x\)
Since any non-zero number raised to the power 0 is 1, and equating exponents when bases are the same:
\(x = 0\)

b) Given: \(\left(\frac{3}{4}\right)^7 \times \left(\frac{3}{4}\right)^{12} = \left(\frac{3}{4}\right)^x\)
Using product rule of exponents \(a^m \times a^n = a^{m+n}\):
\(\left(\frac{3}{4}\right)^{7+12} = \left(\frac{3}{4}\right)^x\)
\(\left(\frac{3}{4}\right)^{19} = \left(\frac{3}{4}\right)^x\)
Equating the exponents:
\(x = 19\)
In simple words: Apply exponent laws (multiply powers for brackets, add powers for multiplication) and equate the powers.

Exam Tip: Remember that any base raised to the power 0 equals 1. For multiplication with the same base, simply add the exponents.

Question 5. Find the value of expression when x = -3: a) X^2 + 4x + 4 b) x^2 - 3x + 225
Answer: 
a) Substituting \(x = -3\) in \(x^2 + 4x + 4\):
\(= (-3)^2 + 4(-3) + 4\)
\(= 9 - 12 + 4\)
\(= 13 - 12 = 1\)

b) Substituting \(x = -3\) in \(x^2 - 3x + 225\):
\(= (-3)^2 - 3(-3) + 225\)
\(= 9 + 9 + 225\)
\(= 18 + 225 = 243\)
In simple words: Replace every \(x\) with -3 and carefully calculate following the order of operations.

Exam Tip: Always enclose negative numbers in parentheses when substituting into powers (e.g., write \((-3)^2\) instead of \(-3^2\)) to avoid sign errors.

Question 6. Divide 243 into three parts such that half of the first part , one - third of the second part and one - fourth of the third part shall be equal.
Answer: Let the three parts be \(x\), \(y\), and \(z\).
According to the given condition:
\(x + y + z = 243\) ---(1)
And \(\frac{1}{2}x = \frac{1}{3}y = \frac{1}{4}z = k\) (where \(k\) is a constant)
Therefore:
\(x = 2k\)
\(y = 3k\)
\(z = 4k\)
Substitute these values in equation (1):
\(2k + 3k + 4k = 243\)
\(9k = 243\)
\(k = \frac{243}{9} = 27\)
Now, find the three parts:
First part \(x = 2 \times 27 = 54\)
Second part \(y = 3 \times 27 = 81\)
Third part \(z = 4 \times 27 = 108\)
In simple words: Assume a common ratio constant \(k\) for equal fractions, express each part in terms of \(k\), find \(k\), and solve for the parts.

Exam Tip: When parts are proportional to fractions, equating them to a single variable \(k\) makes solving simultaneous equations much easier.

Question 7. A is 4 years older than B , and half of A's age exceeds one - sixth of B's age by 8 years. Find their ages.
Answer: Let B's age be \(x\) years.
Then A's age = \(x + 4\) years.
According to the second condition:
\(\frac{1}{2}(x + 4) - \frac{1}{6}x = 8\)
Multiply the entire equation by 6 to clear the denominators:
\(3(x + 4) - x = 48\)
\(3x + 12 - x = 48\)
\(2x + 12 = 48\)
\(2x = 48 - 12\)
\(2x = 36\)
\(x = 18\)
Therefore:
B's age \(x = 18\) years.
A's age \(x + 4 = 18 + 4 = 22\) years.
In simple words: Set up algebraic expressions for both ages, translate the sentence into an equation, and solve for the unknown variable.

Exam Tip: Clearing denominators early by multiplying by the LCM saves time and reduces calculation errors with fractions.

Question 8. Aneeta bought a dress for her sister at discount of 25%, the previous marked price is Rs1245. Find the amount she has to pay after the discount.
Answer: Marked Price (MP) = Rs. 1245
Discount percentage = 25%
Discount amount = 25% of Rs. 1245
\(\text{Discount} = \frac{25}{100} \times 1245 = \frac{1}{4} \times 1245 = \text{Rs. } 311.25\)
Amount to be paid = Marked Price - Discount
\(\text{Amount paid} = 1245 - 311.25 = \text{Rs. } 933.75\)
Alternatively, Amount paid = \(75\% \text{ of } 1245 = \frac{75}{100} \times 1245 = \text{Rs. } 933.75\)
In simple words: Subtract 25% of the marked price from the original marked price to find the final discounted price.

Exam Tip: You can calculate the final price directly by finding (100 - Discount)% of the marked price.

Question 9. Find the simple interest and amount of: a) Rs 200 for 2 years @ 4% b) Rs 4000 for 3 \(\frac{1}{2}\) years @6%
Answer: 
a) Principal (\(P\)) = Rs. 200, Time (\(T\)) = 2 years, Rate (\(R\)) = 4% per annum
\(\text{Simple Interest (SI)} = \frac{P \times R \times T}{100}\)
\(\text{SI} = \frac{200 \times 4 \times 2}{100} = \text{Rs. } 16\)
\(\text{Amount} = P + \text{SI} = 200 + 16 = \text{Rs. } 216\)

b) Principal (\(P\)) = Rs. 4000, Time (\(T\)) = \(3\frac{1}{2}\) years = \(\frac{7}{2}\) years, Rate (\(R\)) = 6% per annum
\(\text{SI} = \frac{4000 \times 6 \times \frac{7}{2}}{100}\)
\(\text{SI} = \frac{4000 \times 6 \times 7}{200} = 20 \times 6 \times 7 = \text{Rs. } 840\)
\(\text{Amount} = P + \text{SI} = 4000 + 840 = \text{Rs. } 4840\)
In simple words: Use the formula SI = (P × R × T) / 100 to find interest, and add it to the principal to get the total amount.

Exam Tip: Always convert mixed fraction time periods into improper fractions before substituting them into the Simple Interest formula.

Question 10. Draw an equilateral acute triangle , draw a perpendicular from each vertex. Measure the angles of each altitude.
Answer: 1. Construction Steps:
- Draw an equilateral triangle \(ABC\) where each side is of equal length and each interior angle is 60°.
- From vertex \(A\), draw a perpendicular line segment meeting side \(BC\) at point \(D\).
- From vertex \(B\), draw a perpendicular line segment meeting side \(AC\) at point \(E\).
- From vertex \(C\), draw a perpendicular line segment meeting side \(AB\) at point \(F\).
2. Measurement of angles of each altitude:
- In an equilateral triangle, the altitudes drawn from the vertices to the opposite sides are also angle bisectors and medians.
- Each altitude meets the base at a right angle (90°).
- Specifically, looking at the right-angled triangles formed inside (e.g., \(\triangle ABD\)): angle \(\angle ADB = 90^\circ\), angle \(\angle BAD = 30^\circ\), and angle \(\angle B = 60^\circ\).
In simple words: In an equilateral triangle, dropping a perpendicular from any vertex creates 90-degree angles where it meets the opposite side.

Exam Tip: In an equilateral triangle, all three altitudes are equal in length and perpendiculars always form 90° angles with the sides.