CBSE Class 7 Mathematics Practice Worksheet Set 14

Question 106. List 4 rational numbers between \( \frac{-5}{7} \) and \( \frac{3}{8} \).
Answer: First, find the lowest common multiple of the two denominators, 7 and 8, which is 56. Convert both fractions so they share the common denominator 56:
\( \frac{-5}{7} = \frac{-5 \times 8}{7 \times 8} = \frac{-40}{56} \)
\( \frac{3}{8} = \frac{3 \times 7}{8 \times 7} = \frac{21}{56} \)
Now, pick any four rational numbers that lie between \( \frac{-40}{56} \) and \( \frac{21}{56} \). For example:
\( \frac{-10}{56}, \frac{-1}{56}, \frac{5}{56}, \frac{15}{56} \) (or in reduced form: \( -\frac{5}{28}, -\frac{1}{56}, \frac{5}{56}, \frac{15}{56} \)).
In simple words: Change both fractions so their bottom numbers match. Then choose any four numbers that sit between their top numbers.

Exam Tip: Always show the conversion step to a common denominator before picking the values to secure full working marks.

Question 107. The circumference of a circle is 31.4cm. Find the radius and area of the circle. (\( \pi = 3.14 \)).
Answer: Let the radius of the circle be \( r \).
Given that circumference \( C = 31.4\text{ cm} \).
The formula for circumference is \( C = 2\pi r \).
\( 2 \times 3.14 \times r = 31.4 \)
\( 6.28 \times r = 31.4 \)
\( \implies r = \frac{31.4}{6.28} = 5\text{ cm} \)
Next, calculate the area of the circle using the formula \( \text{Area} = \pi r^2 \):
\( \text{Area} = 3.14 \times (5)^2 = 3.14 \times 25 = 78.5\text{ cm}^2 \).
Hence, the radius is \( 5\text{ cm} \) and the area is \( 78.5\text{ cm}^2 \).
In simple words: Use the circumference formula to get the radius first. Then use that radius to find the total area.

Exam Tip: Remember to include square units (\( \text{cm}^2 \)) for area and linear units (\( \text{cm} \)) for radius.

Question 108. If \( \Delta \text{ABC} \cong \Delta \text{FED} \), under the correspondence, ABC \( \leftrightarrow \) FED, write all the corresponding congruent parts of the triangles.
Answer: Under the given correspondence \( \text{ABC} \leftrightarrow \text{FED} \), the corresponding vertices are:
\( A \leftrightarrow F \), \( B \leftrightarrow E \), and \( C \leftrightarrow D \).
The corresponding congruent sides are:
1. \( \overline{\text{AB}} \cong \overline{\text{FE}} \)
2. \( \overline{\text{BC}} \cong \overline{\text{ED}} \)
3. \( \overline{\text{AC}} \cong \overline{\text{FD}} \)
The corresponding congruent angles are:
1. \( \angle A \cong \angle F \)
2. \( \angle B \cong \angle E \)
3. \( \angle C \cong \angle D \)
In simple words: Match the letters in the exact order given. The matching sides and matching angles are equal to each other.

Exam Tip: List both the three pairs of sides and the three pairs of angles to provide a complete answer.

Question 109. Express \( 432 \times 270 \) as a product of powers of prime factors.
Answer: Find the prime factorization of each number separately:
\( 432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^4 \times 3^3 \)
\( 270 = 2 \times 3 \times 3 \times 3 \times 5 = 2^1 \times 3^3 \times 5^1 \)
Now multiply the two factorizations together:
\( 432 \times 270 = (2^4 \times 3^3) \times (2^1 \times 3^3 \times 5^1) \)
Combine the powers with the same base:
\( = 2^{4+1} \times 3^{3+3} \times 5^1 = 2^5 \times 3^6 \times 5 \).
In simple words: Break both numbers down into their prime parts. Then add the powers that have the same base.

Exam Tip: Check your additions of powers carefully (\( 4+1=5 \) and \( 3+3=6 \)) to prevent arithmetic slips.

Question 110. Simplify and express each of the following in exponential form
a) \( [(5^2)^3 \times 5^4] \div 5^7 \)
b) \( \frac{3 \times 7 \times 11}{21 \times 11} \)
Answer:
a) Using the power of a power rule \( (a^m)^n = a^{mn} \):
\( (5^2)^3 = 5^{2 \times 3} = 5^6 \)
Now apply the product law \( a^m \times a^n = a^{m+n} \):
\( 5^6 \times 5^4 = 5^{6+4} = 5^{10} \)
Next, apply the quotient law \( a^m \div a^n = a^{m-n} \):
\( 5^{10} \div 5^7 = 5^{10-7} = 5^3 \).

b) Express 21 as a product of prime factors, \( 21 = 3 \times 7 \):
\( \frac{3 \times 7 \times 11}{21 \times 11} = \frac{3^1 \times 7^1 \times 11^1}{3^1 \times 7^1 \times 11^1} \)
Subtract exponents for each like base:
\( = 3^{1-1} \times 7^{1-1} \times 11^{1-1} = 3^0 \times 7^0 \times 11^0 = 1 \times 1 \times 1 = 1 = 1^1 \) (or \( 11^0 \)).
In simple words: Multiply powers of the same base by adding exponents. Divide them by subtracting exponents.

Exam Tip: State the exponent laws used at each step, such as \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^n = a^{m-n} \).

Question 111. The area of a square plot is the same of a rectangular plot. If the side of the square plot is 60m and the length of the rectangular plot is 90m, find the breadth of the rectangular plot.
Answer: First, find the area of the square plot:
\( \text{Side of square} = 60\text{ m} \)
\( \text{Area of square} = \text{side} \times \text{side} = 60 \times 60 = 3600\text{ m}^2 \)
Since both plots have equal area:
\( \text{Area of rectangle} = 3600\text{ m}^2 \)
Let the breadth of the rectangle be \( b \). Given that length \( l = 90\text{ m} \):
\( \text{Area} = l \times b \)
\( 90 \times b = 3600 \)
\( \implies b = \frac{3600}{90} = 40\text{ m} \)
Therefore, the breadth of the rectangular plot is \( 40\text{ m} \).
In simple words: Multiply 60 by 60 to find the square's area. Divide that area by 90 to get the rectangle's width.

Exam Tip: State the equality of areas clearly before equating \( l \times b \) to \( \text{side}^2 \).

Question 112. Construct a right angled triangle whose hypotenuse is 6cm long and one of the legs is 4cm long.
Answer: Follow these steps of construction:
1. Draw a line segment \( \text{BC} = 4\text{ cm} \).
2. At point \( \text{B} \), construct a right angle by drawing ray \( \text{BX} \) perpendicular to \( \text{BC} \) (\( \angle \text{CBX} = 90^\circ \)).
3. With centre \( \text{C} \) and radius \( 6\text{ cm} \) (the hypotenuse), draw an arc intersecting ray \( \text{BX} \) at point \( \text{A} \).
4. Join \( \text{AC} \).
\( \Delta \text{ABC} \) is the required right-angled triangle with leg \( \text{BC} = 4\text{ cm} \), hypotenuse \( \text{AC} = 6\text{ cm} \), and \( \angle \text{B} = 90^\circ \). In simple words: Draw a 4 cm line at the base. Make a 90-degree corner at one end, then use a compass opened to 6 cm from the other end to cut that upright line.

Exam Tip: Never erase construction arcs; examiners look for visible arc marks to award method marks.

Question 113. Simplify
a) \( 2 \times 3^4 \times 2^5 / 9 \times 4^2 \)
b) \( (25)^2 \times 7^3 / 8^3 \times 7 \)
Answer:
a) Express the numbers in the numerator and denominator using base prime factors:
Numerator: \( 2^1 \times 2^5 \times 3^4 = 2^{1+5} \times 3^4 = 2^6 \times 3^4 \)
Denominator: \( 9 \times 4^2 = 3^2 \times (2^2)^2 = 3^2 \times 2^4 = 2^4 \times 3^2 \)
Dividing powers with matching bases:
\( = \frac{2^6 \times 3^4}{2^4 \times 3^2} = 2^{6-4} \times 3^{4-2} = 2^2 \times 3^2 = 4 \times 9 = 36 \).

b) Note that standard textbook exercises write the first factor as \( (2^5)^2 \). Simplifying both interpretations:
If read as \( (2^5)^2 \):
\( \frac{(2^5)^2 \times 7^3}{8^3 \times 7} = \frac{2^{10} \times 7^3}{(2^3)^3 \times 7^1} = \frac{2^{10} \times 7^3}{2^9 \times 7^1} = 2^{10-9} \times 7^{3-1} = 2^1 \times 7^2 = 2 \times 49 = 98 \).
If read literally as \( (25)^2 = (5^2)^2 = 5^4 \):
\( \frac{5^4 \times 7^3}{(2^3)^3 \times 7^1} = \frac{5^4 \times 7^{3-1}}{2^9} = \frac{625 \times 49}{512} = \frac{30625}{512} \).
In simple words: Rewrite composite numbers into prime bases like 2 and 3. Then subtract powers on the bottom from powers on top.

Exam Tip: Always convert numbers like 4, 8, and 9 into powers of prime numbers (2 and 3) before simplifying exponents.

Question 114. Find x and y
Answer: Solving for each given diagram:

(i) Triangle ABC: In \( \Delta \text{ABC} \), \( \text{AB} = \text{AC} = 5\text{ cm} \).
Angles opposite to equal sides are equal, so \( x = y \).
Using the angle sum property:
\( 70^\circ + x + y = 180^\circ \)
\( 70^\circ + 2x = 180^\circ \)
\( \implies 2x = 180^\circ - 70^\circ = 110^\circ \)
\( \implies x = 55^\circ \)
Since \( y = x \), \( y = 55^\circ \).

(ii) Right Triangle DEF: In \( \Delta \text{DEF} \), \( \angle \text{E} = 90^\circ \) and sides \( \text{DE} = \text{EF} = 3\text{ cm} \).
Opposite angles are equal, so \( x = y \).
\( x + y + 90^\circ = 180^\circ \)
\( 2x = 90^\circ \)
\( \implies x = 45^\circ \), and \( y = 45^\circ \).

(iii) Triangle ABC with intersecting lines at B: The line through the base and line AB intersect at B:
Angles \( 133^\circ \) and \( x \) lie on a straight line and form a linear pair:
\( x + 133^\circ = 180^\circ \)
\( \implies x = 180^\circ - 133^\circ = 47^\circ \)
Angle \( y \) and angle \( 133^\circ \) are vertically opposite angles formed by the intersecting lines:
\( y = 133^\circ \).

(iv) Triangle PQR: At vertex Q, the angle \( 150^\circ \) and interior angle \( x \) form a linear pair:
\( x + 150^\circ = 180^\circ \)
\( \implies x = 180^\circ - 150^\circ = 30^\circ \)
At vertex R, the exterior angle \( y \) equals the sum of its two interior opposite angles:
\( y = \angle \text{P} + x = 50^\circ + 30^\circ = 80^\circ \).
In simple words: Equal sides give equal base angles. Straight lines add up to 180 degrees, and an outside angle equals the sum of two opposite inside angles.

Exam Tip: Mention the specific geometric property applied, such as "angles opposite equal sides" or "exterior angle theorem", for each step.

Question 115. Find the angles x,y,z
Answer:
(i) Left Diagram: In right triangle ABC, \( \angle \text{A} = 90^\circ \) and \( \angle \text{B} = 30^\circ \):
\( x = 180^\circ - (90^\circ + 30^\circ) = 180^\circ - 120^\circ = 60^\circ \)
Lines AD and BE intersect at point C, so angles \( x \) and \( y \) are vertically opposite angles:
\( y = x = 60^\circ \)
In right triangle CDE, \( \angle \text{D} = 90^\circ \) and the third angle is \( z \):
\( z = 180^\circ - (90^\circ + y) = 180^\circ - (90^\circ + 60^\circ) = 30^\circ \).
Hence, \( x = 60^\circ \), \( y = 60^\circ \), and \( z = 30^\circ \).

(ii) Right Diagram: On the straight line LM, the angles at the interior point form a linear pair:
\( x + 70^\circ = 180^\circ \)
\( \implies x = 180^\circ - 70^\circ = 110^\circ \)
In the right-hand smaller triangle:
\( y + x + 60^\circ = 180^\circ \)
\( y + 110^\circ + 60^\circ = 180^\circ \)
\( \implies y + 170^\circ = 180^\circ \)
\( \implies y = 10^\circ \)
In the left-hand smaller triangle:
\( z + 50^\circ + 70^\circ = 180^\circ \)
\( \implies z + 120^\circ = 180^\circ \)
\( \implies z = 60^\circ \).
Hence, \( x = 110^\circ \), \( y = 10^\circ \), and \( z = 60^\circ \).
In simple words: The three angles in any triangle always total 180 degrees, and angles across from each other on crossing lines match.

Exam Tip: Solve the linear pair first to find \( x \); this immediately gives you the key angle needed for the remaining triangles.

Question 116. Construct triangle PQR if PQ = 5cm, measure of ∠PQR = 105° and measure of ∠QRP = 40°.
Answer: First, find the measure of \( \angle \text{QPR} \) (or \( \angle \text{P} \)) using the angle sum property of triangles:
\( \angle \text{P} + \angle \text{Q} + \angle \text{R} = 180^\circ \)
\( \angle \text{P} + 105^\circ + 40^\circ = 180^\circ \)
\( \angle \text{P} + 145^\circ = 180^\circ \)
\( \implies \angle \text{P} = 180^\circ - 145^\circ = 35^\circ \)

Steps of Construction:
1. Draw a line segment \( \text{PQ} = 5\text{ cm} \).
2. At point \( \text{P} \), use a protractor to draw a ray making an angle of \( 35^\circ \) with \( \text{PQ} \).
3. At point \( \text{Q} \), draw another ray making an angle of \( 105^\circ \) with \( \text{QP} \).
4. Let the two rays intersect at point \( \text{R} \).
\( \Delta \text{PQR} \) is the required triangle with \( \angle \text{QRP} = 40^\circ \).
In simple words: Calculate the missing angle at P by taking 145 away from 180. Then draw the 5 cm base and build both corner angles so the rays meet at R.

Exam Tip: You cannot construct this triangle without first calculating \( \angle \text{P} \); show the angle sum step clearly.

Question 117. Express as a rational number p/q :
a) \( (3/7)^2 \)
b) \( (-5/9)^3 \)
Answer:
a) Apply the power to both the numerator and denominator:
\( \left(\frac{3}{7}\right)^2 = \frac{3^2}{7^2} = \frac{9}{49} \).

b) Apply the power to both terms:
\( \left(\frac{-5}{9}\right)^3 = \frac{(-5)^3}{9^3} = \frac{-125}{729} \).
In simple words: Raise both the top and bottom numbers to the power shown. An odd power keeps the negative sign.

Exam Tip: Keep in mind that a negative number raised to an odd exponent stays negative, while an even exponent yields a positive value.

Question 118. Simplify
a) \( (-2 / 3)^4 \times (-3 / 4)^3 \)
b) \( (-2)^5 \div (-1 / 3)^3 \)
Answer:
a) Expand the powers:
\( \left(-\frac{2}{3}\right)^4 = \frac{(-2)^4}{3^4} = \frac{16}{81} \)
\( \left(-\frac{3}{4}\right)^3 = \frac{(-3)^3}{4^3} = \frac{-27}{64} \)
Multiply the two fractions:
\( \frac{16}{81} \times \frac{-27}{64} = \frac{16 \times (-27)}{81 \times 64} \)
Cancel common factors (\( 16 \) with \( 64 \) gives 4; \( -27 \) with \( 81 \) gives \( -\frac{1}{3} \)):
\( = \frac{1 \times (-1)}{3 \times 4} = -\frac{1}{12} \).

b) Evaluate each term first:
\( (-2)^5 = -32 \)
\( \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} \)
Division by a fraction is multiplication by its reciprocal:
\( (-32) \div \left(-\frac{1}{27}\right) = (-32) \times (-27) = 864 \).
In simple words: Work out each power first. Flip the dividing fraction to multiply, and cancel common factors where possible.

Exam Tip: When multiplying two negative numbers, remember that the final product becomes positive (\( -32 \times -27 = +864 \)).

Question 119. Express in power notation:
\( -625/14641 \)
Answer: Find the prime factors of numerator and denominator:
\( 625 = 5 \times 5 \times 5 \times 5 = 5^4 \)
\( 14641 = 11 \times 11 \times 11 \times 11 = 11^4 \)
Since 4 is an even power, the negative sign must be placed outside the power:
\( -\frac{625}{14641} = -\frac{5^4}{11^4} = -\left(\frac{5}{11}\right)^4 \).
In simple words: 625 is 5 multiplied 4 times, and 14641 is 11 multiplied 4 times. Write it as minus 5 over 11 raised to power 4.

Exam Tip: Do not write \( \left(-\frac{5}{11}\right)^4 \) without the external minus sign, because an even power makes a negative base positive.

Question 120. Find the reciprocal of :
\( (3/7)^3 \times (7/3)^5 \)
Answer: First, simplify the given expression:
\( \left(\frac{3}{7}\right)^3 \times \left(\frac{7}{3}\right)^5 = \frac{3^3}{7^3} \times \frac{7^5}{3^5} \)
Using the quotient law of exponents:
\( = \frac{7^{5-3}}{3^{5-3}} = \frac{7^2}{3^2} = \left(\frac{7}{3}\right)^2 = \frac{49}{9} \)
The reciprocal of a number \( \frac{a}{b} \) is \( \frac{b}{a} \):
\( \text{Reciprocal of } \frac{49}{9} = \frac{9}{49} = \left(\frac{3}{7}\right)^2 \).
In simple words: Simplify the problem down to 49 over 9 first. Then flip the fraction upside down to get 9 over 49.

Exam Tip: A common mistake is stopping at \( \frac{49}{9} \); remember to take the reciprocal as the final step.

Question 121. Simplify:
a) \( (3^2 - 2^2) \div (1/5)^2 \)
b) \( \{(1/3) - (1/4)\} \times 2^3 \)
Answer:
a) Evaluate terms inside the brackets first:
\( 3^2 - 2^2 = 9 - 4 = 5 \)
Now evaluate the divisor:
\( \left(\frac{1}{5}\right)^2 = \frac{1}{25} \)
Divide:
\( 5 \div \frac{1}{25} = 5 \times 25 = 125 \).

b) Find a common denominator for the terms in brackets (LCM of 3 and 4 is 12):
\( \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12} \)
Evaluate \( 2^3 = 8 \):
\( \frac{1}{12} \times 8 = \frac{8}{12} = \frac{2}{3} \)
(Note: If the first fraction is read as \( \frac{1}{2} \), then \( (\frac{1}{2} - \frac{1}{4}) \times 8 = \frac{1}{4} \times 8 = 2 \)).
In simple words: Solve what is inside brackets first. Then multiply or divide according to standard rules.

Exam Tip: Follow BODMAS/PEMDAS order of operations strictly by simplifying inside parentheses before multiplying or dividing.

Question 122. Find the reciprocal of : (i) \( (- 3 )^5 \) (ii)\( ( 3/4)^4 \) (iii) \( ( -1/5)^8 \div ( 1/5 )^2 \)
Answer:
(i) \( (-3)^5 = -243 \).
The reciprocal is:
\( \frac{1}{(-3)^5} = \left(-\frac{1}{3}\right)^5 = -\frac{1}{243} \).

(ii) \( \left(\frac{3}{4}\right)^4 = \frac{3^4}{4^4} = \frac{81}{256} \).
The reciprocal is obtained by swapping the numerator and denominator:
\( \left(\frac{4}{3}\right)^4 = \frac{256}{81} \).

(iii) Since 8 is an even power, \( \left(-\frac{1}{5}\right)^8 = \left(\frac{1}{5}\right)^8 \).
Simplifying the division:
\( \left(\frac{1}{5}\right)^8 \div \left(\frac{1}{5}\right)^2 = \left(\frac{1}{5}\right)^{8-2} = \left(\frac{1}{5}\right)^6 \)
The reciprocal of \( \left(\frac{1}{5}\right)^6 \) is:
\( 5^6 = 15625 \).
In simple words: The reciprocal of any fraction is just that fraction turned upside down.

Exam Tip: Simplify complex expressions using exponent quotient rules before inverting the fraction for the reciprocal.

Question 123. Distinguish between the rational numbers \( 3^2/4 \) and \( (3/4)^2 \), which is smaller of the two ?Insert 12 rational numbers between \( 3^2/4 \) and \( (3/4)^2 \).
Answer: Evaluating each rational number:
\( \frac{3^2}{4} = \frac{9}{4} \) (here, the power 2 applies only to the numerator 3).
\( \left(\frac{3}{4}\right)^2 = \frac{3^2}{4^2} = \frac{9}{16} \) (here, the power 2 applies to both the numerator and the denominator).
To compare them, convert \( \frac{9}{4} \) to an equivalent fraction with denominator 16:
\( \frac{9}{4} = \frac{9 \times 4}{4 \times 4} = \frac{36}{16} \)
Comparing \( \frac{9}{16} \) and \( \frac{36}{16} \): since \( 9 < 36 \), \( \left(\frac{3}{4}\right)^2 \) is smaller than \( \frac{3^2}{4} \).

Now, insert 12 rational numbers between \( \frac{9}{16} \) and \( \frac{36}{16} \):
Any 12 fractions with denominator 16 and numerators between 9 and 36 will work, such as:
\( \frac{10}{16}, \frac{11}{16}, \frac{12}{16}, \frac{13}{16}, \frac{14}{16}, \frac{15}{16}, \frac{16}{16}, \frac{17}{16}, \frac{18}{16}, \frac{19}{16}, \frac{20}{16}, \frac{21}{16} \).
In simple words: Without brackets, the power only squares the top number. With brackets, it squares both top and bottom.

Exam Tip: Clearly state the effect of brackets in your explanation: brackets apply exponents to both numerator and denominator.

Question 124. Express -1/243 in power notation.
Answer: Find the prime factorization of 243:
\( 243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5 \)
Since 5 is an odd exponent, a negative base raised to power 5 gives a negative result:
\( (-1)^5 = -1 \)
Therefore, \( -\frac{1}{243} = \frac{(-1)^5}{3^5} = \left(-\frac{1}{3}\right)^5 \).
In simple words: 243 is 3 multiplied five times. Since 5 is odd, keep minus 1 over 3 inside brackets to the power 5.

Exam Tip: Include the negative sign inside the parenthesis because an odd exponent correctly preserves the negative value.

Question 125. Write true or false for the following statements
i. (-3/5 ) 100 = -3100 /5100
ii. The reciprocal of (5/9 ) 30 is (9/5)30 .
iii. [(10)10]10 =(100)10
iv. (30 + 30) 30 = 30 30 + 30 30
v. (-1 )99 = ( - 99 )
Answer:
i. False. Since 100 is an even exponent, a negative base raised to an even power becomes positive: \( \left(-\frac{3}{5}\right)^{100} = \frac{3^{100}}{5^{100}} \neq -\frac{3^{100}}{5^{100}} \).
ii. True. The reciprocal of \( \left(\frac{5}{9}\right)^{30} \) is \( \frac{1}{(5/9)^{30}} = \left(\frac{9}{5}\right)^{30} \).
iii. False. \( [(10)^{10}]^{10} = 10^{10 \times 10} = 10^{100} \), whereas \( (100)^{10} = (10^2)^{10} = 10^{20} \).
iv. False. Exponentiation does not distribute over addition: \( (30 + 30)^{30} = 60^{30} \neq 30^{30} + 30^{30} \).
v. False. \( -1 \) raised to any odd power is always \( -1 \), so \( (-1)^{99} = -1 \neq -99 \).
In simple words: Even powers turn negatives into positives, powers multiply together, and minus one to an odd power stays minus one.

Exam Tip: Always verify powers of \( -1 \) quickly: even powers give \( +1 \), and odd powers give \( -1 \).