CBSE Class 7 Mathematics Practice Worksheet Set 02

Question 1. Find the value of (−20 + 45) ÷ [−50 × (−10 − 40)].
Answer: Follow the order of operations (BODMAS):
First, evaluate the expression in the first bracket:
\( -20 + 45 = 25 \).
Next, simplify inside the square brackets:
\( -10 - 40 = -50 \).
Now, multiply inside the square brackets:
\( -50 \times (-50) = 2500 \).
Finally, perform the division:
\( 25 \div 2500 = \frac{25}{2500} = \frac{1}{100} = 0.01 \).
Thus, the value is \( \frac{1}{100} \) (or 0.01).
In simple words: The first part gives 25. The second part gives 2500. Dividing 25 by 2500 gives 1/100 or 0.01.

Exam Tip: Work from the innermost brackets outward, and remember that multiplying two negative numbers yields a positive product.

Question 2. The area of rectangular field is 321 5/8 m². If its length is 20 3/4 m, find its breadth.
Answer: Convert the mixed fractions into improper fractions:
Area = \( 321\frac{5}{8}\text{ m}^2 = \frac{321 \times 8 + 5}{8} = \frac{2573}{8}\text{ m}^2 \).
Length = \( 20\frac{3}{4}\text{ m} = \frac{20 \times 4 + 3}{4} = \frac{83}{4}\text{ m} \).
Formula for breadth of a rectangle:
\( \text{Breadth} = \text{Area} \div \text{Length} \)
\( \text{Breadth} = \frac{2573}{8} \div \frac{83}{4} = \frac{2573}{8} \times \frac{4}{83} \)
Simplify by cancelling common factors:
\( = \frac{2573}{2 \times 83} = \frac{31}{2} = 15.5\text{ m} = 15\frac{1}{2}\text{ m} \).
Hence, the breadth of the rectangular field is \( 15\frac{1}{2}\text{ m} \) (or 15.5 m).
In simple words: Change both mixed numbers into improper fractions. Divide the area by the length to get a breadth of 15.5 metres.

Exam Tip: To divide fractions, multiply the first fraction by the reciprocal of the second fraction.

Question 3. Rita bought a dozen notebooks for Rs.136.20. If she has to buy 5 more notebooks, then how much more money will she have to pay to the shopkeeper?
Answer: We know that 1 dozen = 12 notebooks.
Cost of 12 notebooks = Rs. 136.20.
Find the cost of 1 notebook using the unitary method:
\( \text{Cost of 1 notebook} = \frac{136.20}{12} = \text{Rs. } 11.35 \).
Now calculate the cost of 5 additional notebooks:
\( \text{Cost of 5 notebooks} = 5 \times 11.35 = \text{Rs. } 56.75 \).
Therefore, Rita will have to pay Rs. 56.75 more to the shopkeeper.
In simple words: One notebook costs Rs. 11.35. Multiplying this by 5 means she must pay Rs. 56.75 more.

Exam Tip: Divide by 12 first to get the price of a single notebook, then multiply by 5.

Question 4. Find the mean of all the multiples of 20 before 150.
Answer: The multiples of 20 less than 150 are:
20, 40, 60, 80, 100, 120, 140.
Total number of values (\( n \)) = 7.
Find the sum of these multiples:
\( \text{Sum} = 20 + 40 + 60 + 80 + 100 + 120 + 140 = 560 \).
Calculate the mean:
\( \text{Mean} = \frac{\text{Sum of values}}{n} = \frac{560}{7} = 80 \).
Thus, the mean of the multiples of 20 before 150 is 80.
In simple words: Add up the seven multiples of 20 to get 560. Divide 560 by 7 to find the average, which is 80.

Exam Tip: For evenly spaced numbers, the mean is always the exact middle number in the list.

Question 5. If 85 1/2 − 64 x/8 = 21 1/8 , then find the value of x.
Answer: Rearrange the equation to isolate the term containing \( x \):
\( 85\frac{1}{2} - 21\frac{1}{8} = 64\frac{x}{8} \)
Express \( \frac{1}{2} \) with a common denominator of 8:
\( 85\frac{4}{8} - 21\frac{1}{8} = 64\frac{x}{8} \)
Subtract the whole numbers and the fractions separately:
\( (85 - 21) + \left(\frac{4}{8} - \frac{1}{8}\right) = 64\frac{x}{8} \)
\( 64 + \frac{3}{8} = 64\frac{x}{8} \)
\( 64\frac{3}{8} = 64\frac{x}{8} \).
Comparing both sides gives:
\( x = 3 \).
In simple words: Subtract 21 1/8 from 85 1/2 to get 64 3/8. Matching this with 64 x/8 shows that x = 3.

Exam Tip: Subtracting whole numbers and fractional parts separately is faster than converting large mixed numbers to improper fractions.

Question 6. In the given figures (i) & (ii) find angles x,y and z (Give reasons)
Answer:
(i) In Figure (i):
Two straight lines intersect at point O with a perpendicular ray between them.
1. Vertically opposite angles are equal:
The angle vertically opposite to \( 2x \) is \( y \):
\( y = 2x \) ... (Equation 1)
2. Angles along the straight line on the bottom side sum to \( 180^\circ \):
\( 90^\circ + x + y = 180^\circ \)
\( x + y = 180^\circ - 90^\circ = 90^\circ \).
Substitute \( y = 2x \) into this equation:
\( x + 2x = 90^\circ \)
\( 3x = 90^\circ \)

\( \implies x = \frac{90^\circ}{3} = 30^\circ \).
3. Now find \( y \):
\( y = 2x = 2 \times 30^\circ = 60^\circ \).
4. Find angle \( z \):
Angle \( z \) and angle \( 2x \) form a linear pair on a straight line:
\( z + 2x = 180^\circ \)
\( z + 60^\circ = 180^\circ \)

\( \implies z = 180^\circ - 60^\circ = 120^\circ \).
Therefore, in figure (i): \( x = 30^\circ \), \( y = 60^\circ \), and \( z = 120^\circ \).
(ii) In Figure (ii):
Three horizontal lines are parallel to each other.
1. For angle \( y \):
The angle \( 120^\circ \) and angle \( y \) are consecutive interior angles on the same side of the transversal:
\( y + 120^\circ = 180^\circ \)

\( \implies y = 180^\circ - 120^\circ = 60^\circ \).
2. For angle \( x \):
The angle \( 130^\circ \) and angle \( x \) are consecutive interior angles on the same side of the second transversal:
\( x + 130^\circ = 180^\circ \)

\( \implies x = 180^\circ - 130^\circ = 50^\circ \).
3. For angle \( z \):
The middle parallel line and the bottom parallel line are intersected by the transversal, so alternate interior angles are equal:
\( z = 35^\circ \).
Therefore, in figure (ii): \( x = 50^\circ \), \( y = 60^\circ \), and \( z = 35^\circ \).In simple words: In figure (i), x = 30°, y = 60°, and z = 120°. In figure (ii), co-interior angles give x = 50° and y = 60°, while alternate angles give z = 35°.

Exam Tip: State the exact geometric theorem (such as "co-interior angles sum to 180°" or "alternate interior angles are equal") alongside each calculation.

Question 7. Represent \( \frac{8}{3} \) and \( \frac{-8}{3} \) on the number line.
Answer: Convert both improper fractions into mixed numbers:
\( \frac{8}{3} = 2\frac{2}{3} \) (lies between 2 and 3).
\( -\frac{8}{3} = -2\frac{2}{3} \) (lies between \( -2 \) and \( -3 \)).
Divide each unit gap on the number line into 3 equal parts.
- To mark \( \frac{8}{3} \), move 2 units right from 0, then take 2 parts out of 3 towards 3.
- To mark \( -\frac{8}{3} \), move 2 units left from 0, then take 2 parts out of 3 towards \( -3 \).In simple words: Divide each whole section into 3 equal steps. Count 8 steps left for -8/3 and 8 steps right for 8/3.

Exam Tip: Since the denominator is 3, divide the space between consecutive integers into 3 equal parts using 2 tick marks.

Question 8. If \( \left(\frac{27}{8}\right)^{-8} \times \left(\frac{27}{8}\right)^n \div \left(\frac{2}{3}\right)^{-18} = 1 \), then find the value of n.
Answer: Express all terms with a common base of \( \frac{3}{2} \):
Notice that \( \frac{27}{8} = \left(\frac{3}{2}\right)^3 \).
Therefore:
\( \left(\frac{27}{8}\right)^{-8} = \left[\left(\frac{3}{2}\right)^3\right]^{-8} = \left(\frac{3}{2}\right)^{-24} \)
\( \left(\frac{27}{8}\right)^n = \left[\left(\frac{3}{2}\right)^3\right]^n = \left(\frac{3}{2}\right)^{3n} \)
Now rewrite \( \left(\frac{2}{3}\right)^{-18} \):
\( \left(\frac{2}{3}\right)^{-18} = \left[\left(\frac{3}{2}\right)^{-1}\right]^{-18} = \left(\frac{3}{2}\right)^{18} \).
Substitute these into the equation:
\( \left(\frac{3}{2}\right)^{-24} \times \left(\frac{3}{2}\right)^{3n} \div \left(\frac{3}{2}\right)^{18} = 1 \)
Apply the laws of exponents \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^k = a^{m-k} \):
\( \left(\frac{3}{2}\right)^{-24 + 3n - 18} = \left(\frac{3}{2}\right)^0 \)
\( \left(\frac{3}{2}\right)^{3n - 42} = \left(\frac{3}{2}\right)^0 \).
Equating the exponents gives:
\( 3n - 42 = 0 \)
\( 3n = 42 \)

\( \implies n = \frac{42}{3} = 14 \).
Thus, the value of \( n \) is 14.
In simple words: Change all parts to the base 3/2. Combine the powers to get 3n - 42 = 0, which solves to n = 14.

Exam Tip: Remember that any non-zero number raised to the power 0 equals 1, so write 1 as \( \left(\frac{3}{2}\right)^0 \).

Question 9. Simplify: (9x² − x + 15)(x² − 3) − (3x² + 5x − 9)(3x − 9). Then find the value of the expression when x=1
Answer: Expand the first product:
\( (9x^2 - x + 15)(x^2 - 3) = 9x^2(x^2 - 3) - x(x^2 - 3) + 15(x^2 - 3) \)
\( = 9x^4 - 27x^2 - x^3 + 3x + 15x^2 - 45 \)
\( = 9x^4 - x^3 - 12x^2 + 3x - 45 \).
Expand the second product:
\( (3x^2 + 5x - 9)(3x - 9) = 3x(3x^2 + 5x - 9) - 9(3x^2 + 5x - 9) \)
\( = 9x^3 + 15x^2 - 27x - 27x^2 - 45x + 81 \)
\( = 9x^3 - 12x^2 - 72x + 81 \).
Now subtract the second expansion from the first:
\( (9x^4 - x^3 - 12x^2 + 3x - 45) - (9x^3 - 12x^2 - 72x + 81) \)
\( = 9x^4 - x^3 - 12x^2 + 3x - 45 - 9x^3 + 12x^2 + 72x - 81 \)
Combine like terms:
\( = 9x^4 - 10x^3 + 75x - 126 \).
Now find the value at \( x = 1 \):
\( 9(1)^4 - 10(1)^3 + 75(1) - 126 \)
\( = 9 - 10 + 75 - 126 \)
\( = -1 + 75 - 126 = 74 - 126 = -52 \).
Hence, the simplified expression is \( 9x^4 - 10x^3 + 75x - 126 \), and its value at \( x = 1 \) is \( -52 \).
In simple words: Multiply out both parts and subtract like terms to get 9x⁴ - 10x³ + 75x - 126. Putting in x = 1 gives -52.

Exam Tip: Check your final evaluation by substituting \( x = 1 \) directly into the original expression: \( (23)(-2) - (-1)(-6) = -46 - 6 = -52 \).

Question 10. On a school picnic, a group of girls agree to pay equally for the use of a boat and each pays Rs.25. If there had been 3 more girls in the group, each would have paid Rs. 5 less. How many girls were there in the group?
Answer: Let the original number of girls be \( n \).
Each girl pays Rs. 25, so the total hire cost of the boat = \( 25n \).
If 3 more girls join, the total number becomes \( n + 3 \).
Each girl would then pay Rs. 5 less: \( 25 - 5 = \text{Rs. } 20 \).
Total hire cost with 3 extra girls = \( 20(n + 3) \).
Since the total boat cost remains unchanged:
\( 20(n + 3) = 25n \)
\( 20n + 60 = 25n \)
\( 25n - 20n = 60 \)
\( 5n = 60 \)

\( \implies n = \frac{60}{5} = 12 \).
Therefore, there were 12 girls originally in the group.
In simple words: 12 girls paying Rs. 25 gives Rs. 300. If 15 girls paid Rs. 20, it still equals Rs. 300. There were 12 girls.

Exam Tip: Set up the equation by equating the total cost of the boat in both scenarios: \( 25n = 20(n + 3) \).

Question 11. If A: B=5:6 and B: C=4:7, then find A: B: C.
Answer: Given ratios:
\( A : B = 5 : 6 \)
\( B : C = 4 : 7 \)
To combine the ratios, make the term for B equal in both ratios by finding the LCM of 6 and 4, which is 12.
Multiply the first ratio by 2:
\( A : B = (5 \times 2) : (6 \times 2) = 10 : 12 \).
Multiply the second ratio by 3:
\( B : C = (4 \times 3) : (7 \times 3) = 12 : 21 \).
Therefore, the combined ratio is:
\( A : B : C = 10 : 12 : 21 \).
In simple words: Make B equal to 12 in both ratios. That gives A = 10, B = 12, and C = 21, so the ratio is 10:12:21.

Exam Tip: Always equalize the common quantity (B) using the LCM of its two given values.

Question 12. Find the mean weight of the following data:

Answer: Let the weights be \( x_i \) and the number of students be \( f_i \).
Calculate the product \( f_i \times x_i \) for each weight category:
- \( 50 \times 8 = 400 \)
- \( 52 \times 10 = 520 \)
- \( 54 \times 15 = 810 \)
- \( 56 \times 20 = 1120 \)
- \( 58 \times 8 = 464 \)
- \( 60 \times 6 = 360 \)
- \( 62 \times 2 = 124 \)
- \( 64 \times 1 = 64 \)
Find the total number of students (\( \sum f_i \)):
\( \sum f_i = 8 + 10 + 15 + 20 + 8 + 6 + 2 + 1 = 70 \).
Find the sum of all products (\( \sum f_i x_i \)):
\( \sum f_i x_i = 400 + 520 + 810 + 1120 + 464 + 360 + 124 + 64 = 3862 \).
Calculate the mean weight:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3862}{70} = \frac{1931}{35} \approx 55.17 \).
Hence, the mean weight is approximately 55.17.
In simple words: Multiply each weight by the number of students, add them all up to get 3862, and divide by 70 to get 55.17.

Exam Tip: Set up a clear frequency distribution table with columns for \( x \), \( f \), and \( f \times x \) to avoid addition mistakes.