CBSE Class 9 Mathematics Number System Worksheet Set 03

Question. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisiors be reversed.
(A) 6, 5, 1
(B) 6, 4, 2
(C) 2, 4, 6
(D) 1, 3, 3
Answer: B

Question. If the LCM of first 100 natural numbers is P then the LCM of first 105 natural numbers would be :
(A) P
(B) P + 105
(C) P(P + 101)(P + 103)
(D) 103 × 101 × P
Answer: D

Question. If (232 + 1) is divisible by a certain number then which of the following is also divisible by that number.
(A) (216 – 1)
(B) 216 + 1
(C) 296 + 1
(D) None of these
Answer: C

Question. There are four prime numbers written in ascending order. The product of the first three is 385 and that of the last three is 1001. The lest number is :
(A) 11
(B) 13
(C) 17
(D) 19
Answer: C

Question. How many three-digit numbers would you find, which when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2, 3, 4 and 5 respectively ?
(A) 4
(B) 3
(C) 2
(D) 1
Answer: C

Question. An operation defined as the product of non zero digits of x. e.g. An operation x defined as the product of non zero digits of x. e.g. 
25 = 2 × 5 = 10, then the sum of all the possible n where n is a two digit number formed by the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 is :
(A) 2350
(B) 2007
(C) 2025
(D) 4014
Answer: C

Question. What will be the last digit of (73)⁷⁵⁶⁴⁷⁶ .
(A) 9
(B) 7
(C) 3
(D) 1
Answer: C

Question. A number when divided by 259 leaves a remainder 139. What will be the remainder when the same number is divided by 37 ?
(A) 21
(B) 23
(C) 27
(D) 28
Answer: D

Question. Three pieces of cakes of weights 4(1/2) lbs, 6(3/4) lbs and 7(1/5) lbs respectively are to be divided into parts of equal weights. Further, each must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained ?
(A) 54
(B) 72
(C) 20
(D) 41
Answer: D

Question. How many zeros at the end of first 100 multiples of 10.
(A) 100
(B) 124
(C) 920
(D) 1225
Answer: B

1. Express the following numbers in the form of p : q

(a) -25.6875

(b) 0.621621621..

(c) 15.712121212..

2. Write one rational & one fractional number between 0.515115111…. and 0.535335333….

3. Write one example each of two irrational numbers whose:

(a) Sum is a rational number

(b) Product is an irrational number

(c) Difference is an irrational number

(d) Quotient is a rational number

4. Simplify : [ { (625) ^-1/2 } ^-1/4 ] ²

5. (a) Find the value of x if 2x-7 x 5x-4 =1250

(b) Find the value of x and y if:

3^x-1 = 9 & 4^y+2 = 64

6. Find the value of a & b if 5 + 2√3/7 + 4√3 = a + b√3

7. If x = 3 + √8, then find the value of x² + 1x²

Question 1. Visualize 3.756 on the number line, using successive magnification
Answer: To represent 3.756 on the real number line, we apply successive magnification in the following stages:
1. The given number 3.756 lies between the integers 3 and 4. We divide the unit interval between 3 and 4 into 10 equal parts.
2. Next, 3.756 falls in the section between 3.7 and 3.8. We magnify the segment from 3.7 to 3.8 and partition it into 10 equal subdivisions.
3. Observing the third decimal place, 3.756 lies between 3.75 and 3.76. We magnify the portion from 3.75 to 3.76 and divide it into 10 equal parts.
4. Finally, 3.756 corresponds precisely to the 6th subdivision mark to the right of 3.75.
In simple words: Zoom in between 3 and 4 to see the tenths, zoom into 3.7 and 3.8 to see the hundredths, and zoom into 3.75 and 3.76 to pinpoint 3.756 at the sixth tick mark.

Exam Tip: Always state the boundaries clearly at every zooming step (between 3 and 4, then 3.7 and 3.8, then 3.75 and 3.76) to secure full marks.

Question 2. Represent \( \sqrt{3.5} \) on the number line
Answer: We construct \( \sqrt{3.5} \) geometrically using the following steps:
1. Draw a horizontal straight line. Mark a point \( A \) and another point \( B \) such that \( AB = 3.5 \) units.
2. From point \( B \), measure a distance of 1 unit in the same direction to point \( C \), making \( AC = 3.5 + 1 = 4.5 \) units.
3. Construct the perpendicular bisector of segment \( AC \) to determine its midpoint \( O \).
4. With \( O \) as center and radius \( OA = OC = 2.25 \) units, draw a semicircle above \( AC \).
5. Erect a perpendicular to the line \( AC \) at point \( B \), meeting the semicircle at point \( D \). By the geometric mean property, the length of segment \( BD = \sqrt{3.5} \) units.
6. Taking \( B \) as the origin (representing 0 on the number line) and using a compass with radius \( BD \), draw an arc cutting the number line at point \( E \). Point \( E \) denotes \( \sqrt{3.5} \).
In simple words: Draw a line of length 3.5, add 1 unit to make it 4.5, find the center to draw a semicircle, and make a vertical line up from the 3.5 mark to the curve. That vertical length is the square root of 3.5.

Exam Tip: Remember to specify point B as the zero origin when transferring length BD onto the number line, not point A or O.

Question 3. Express \( 1.\overline{32} + 0.\overline{35} \) as a fraction in simplest form.
Answer: Let us convert each repeating decimal into its rational form \( \frac{p}{q} \):
For \( 1.\overline{32} \):
Let \( x = 1.323232\ldots \)
\( \implies 100x = 132.323232\ldots \)
\( \implies 100x - x = 132.3232\ldots - 1.3232\ldots \)
\( \implies 99x = 131 \)
\( \implies x = \frac{131}{99} \)
For \( 0.\overline{35} \):
Let \( y = 0.353535\ldots \)
\( \implies 100y = 35.353535\ldots \)
\( \implies 100y - y = 35.3535\ldots - 0.3535\ldots \)
\( \implies 99y = 35 \)
\( \implies y = \frac{35}{99} \)
Now, sum the two fractions:
\( x + y = \frac{131}{99} + \frac{35}{99} = \frac{131 + 35}{99} = \frac{166}{99} \)
In simple words: Turn both repeating decimals into fractions with 99 on the bottom, then simply add the top numbers together to get 166/99.

Exam Tip: Because both decimals have two recurring digits, multiplying by 100 aligns the decimal parts perfectly so they cancel upon subtraction.

Question 4. Express \( 0.12\overline{54} \) in the form \( p/q \)
Answer: Let \( x = 0.12545454\ldots \)
Multiply both sides by 100 so only the repeating block remains after the decimal point:
\( \implies 100x = 12.545454\ldots \) (Equation 1)
Next, multiply Equation 1 by 100 to shift one full repeating period:
\( \implies 10000x = 1254.545454\ldots \) (Equation 2)
Subtract Equation 1 from Equation 2:
\( \implies 10000x - 100x = 1254.5454\ldots - 12.5454\ldots \)
\( \implies 9900x = 1242 \)
\( \implies x = \frac{1242}{9900} \)
Reducing the fraction to its lowest terms by dividing both numerator and denominator by 18:
\( \implies x = \frac{69}{550} \)
In simple words: Shift the non-repeating digits out of the way, shift the repeating block across the decimal point, subtract the two equations, and simplify the fraction down to 69/550.

Exam Tip: Divide by the greatest common divisor (18 here) step-by-step using common factors like 9 and 2 to avoid errors while reducing the fraction.

Question 5. If \( x = 3+2\sqrt{2} \), find the value of \( x^2 + 1/x^2 \)
Answer: Given \( x = 3 + 2\sqrt{2} \).
First, find the reciprocal \( \frac{1}{x} \) by rationalising the denominator:
\( \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \)
Next, find the sum \( x + \frac{1}{x} \):
\( x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \)
Using the algebraic identity \( x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \):
\( x^2 + \frac{1}{x^2} = 6^2 - 2 = 36 - 2 = 34 \)
In simple words: Rationalise 1/x to get 3 - 2√2. Add x and 1/x to get 6, then square 6 and subtract 2 to arrive at 34.

Exam Tip: Never square x and 1/x directly; finding (x + 1/x) and using the identity (x + 1/x)² - 2 is much faster and less prone to algebraic mistakes.

Question 6. If \( x = 2 + \sqrt{5} \), Prove that \( x^2 + \frac{1}{x^2} = 18 \)
Answer: Given \( x = \sqrt{5} + 2 \).
Determine the reciprocal \( \frac{1}{x} \) by rationalisation:
\( \frac{1}{x} = \frac{1}{\sqrt{5} + 2} = \frac{\sqrt{5} - 2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \)
Now, calculate the sum \( x + \frac{1}{x} \):
\( x + \frac{1}{x} = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \)
Squaring both sides gives:
\( \left(x + \frac{1}{x}\right)^2 = (2\sqrt{5})^2 \)
\( \implies x^2 + \frac{1}{x^2} + 2 = 4 \times 5 = 20 \)
\( \implies x^2 + \frac{1}{x^2} = 20 - 2 = 18 \)
Hence, it is proved that \( x^2 + \frac{1}{x^2} = 18 \).
In simple words: Write x as √5 + 2 so its reciprocal is √5 - 2. Their sum is 2√5, and squaring that sum minus 2 gives 18.

Exam Tip: Writing 2 + √5 as √5 + 2 avoids negative signs in the denominator when calculating (√5)² - 2² = 1.

Question 7. Rationalise the denominator \( \frac{1}{\sqrt{6} + \sqrt{5} - \sqrt{11}} \)
Answer: Group the first two radical terms in the denominator as \( (\sqrt{6} + \sqrt{5}) - \sqrt{11} \).
Multiply both the numerator and denominator by the conjugate \( (\sqrt{6} + \sqrt{5}) + \sqrt{11} \):
\( \frac{1}{(\sqrt{6} + \sqrt{5}) - \sqrt{11}} \times \frac{(\sqrt{6} + \sqrt{5}) + \sqrt{11}}{(\sqrt{6} + \sqrt{5}) + \sqrt{11}} = \frac{\sqrt{6} + \sqrt{5} + \sqrt{11}}{(\sqrt{6} + \sqrt{5})^2 - (\sqrt{11})^2} \)
Expand the denominator:
\( (\sqrt{6} + \sqrt{5})^2 - 11 = (6 + 5 + 2\sqrt{30}) - 11 = 11 + 2\sqrt{30} - 11 = 2\sqrt{30} \)
The expression is now:
\( \frac{\sqrt{6} + \sqrt{5} + \sqrt{11}}{2\sqrt{30}} \)
Rationalise the remaining radical in the denominator by multiplying top and bottom by \( \sqrt{30} \):
\( \frac{(\sqrt{6} + \sqrt{5} + \sqrt{11})\sqrt{30}}{2\sqrt{30} \times \sqrt{30}} = \frac{\sqrt{180} + \sqrt{150} + \sqrt{330}}{2 \times 30} = \frac{6\sqrt{5} + 5\sqrt{6} + \sqrt{330}}{60} \)
In simple words: Group two roots together and multiply by the conjugate to turn the bottom into 2√30, then multiply by √30 once more to remove all roots from the denominator.

Exam Tip: Look for combinations where the sum of two radicands equals the third (6 + 5 = 11) because the integer terms cancel out cleanly in the first rationalisation step.

Question 8. If a and b are rational numbers, find a and b
a) \( \frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a + b\sqrt{6} \)
b) \( \frac{\sqrt{5} - 2}{\sqrt{5} + 2} - \frac{\sqrt{5} + 2}{\sqrt{5} - 2} = a + b\sqrt{5} \)
Answer:
(a) Rationalise the left-hand side denominator by multiplying numerator and denominator by \( (3\sqrt{2} + 2\sqrt{3}) \):
\( \frac{(\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})} = \frac{3(\sqrt{2})^2 + 2\sqrt{6} + 3\sqrt{6} + 2(\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2} \)
\( = \frac{3(2) + 5\sqrt{6} + 2(3)}{18 - 12} = \frac{6 + 5\sqrt{6} + 6}{6} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6} \)
Equating with \( a + b\sqrt{6} \):
\( a = 2 \), \( b = \frac{5}{6} \)
(b) Take the common denominator on the left-hand side:
\( \frac{(\sqrt{5} - 2)^2 - (\sqrt{5} + 2)^2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \frac{(5 + 4 - 4\sqrt{5}) - (5 + 4 + 4\sqrt{5})}{(\sqrt{5})^2 - 2^2} \)
\( = \frac{(9 - 4\sqrt{5}) - (9 + 4\sqrt{5})}{5 - 4} = \frac{-8\sqrt{5}}{1} = 0 - 8\sqrt{5} \)
Equating with \( a + b\sqrt{5} \):
\( a = 0 \), \( b = -8 \)
In simple words: Multiply each side out so there are no square roots below the fraction line, then compare the rational part to a and the root coefficient to b.

Exam Tip: Be sure to write the rational part as 0 when no constant term remains, so the value of a = 0 is clearly credited.

Question 9. Simplify: a) \( \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} \)
Answer: Rationalise each term separately by multiplying numerator and denominator by its conjugate:
First term:
\( \frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \)
Second term:
\( \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} \)
Third term:
\( \frac{1}{\sqrt{4} + \sqrt{3}} = \frac{\sqrt{4} - \sqrt{3}}{(\sqrt{4} + \sqrt{3})(\sqrt{4} - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \)
Now, sum all three terms:
\( (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) = -1 + 2 = 1 \)
In simple words: Rationalise each term so each denominator becomes 1. The intermediate square root terms cancel out in pairs, leaving only -1 + 2 = 1.

Exam Tip: Write terms in descending order under the root (e.g., √2 + 1 rather than 1 + √2) so the denominator becomes a positive 1 without negative sign errors.

Question 10. If \( a = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \) and \( b = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \) , find the value of \( a^2 + b^2 \)
Answer: First, simplify \( a \) and \( b \) by rationalising their denominators:
\( a = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{3 + 2 + 2\sqrt{6}}{3 - 2} = 5 + 2\sqrt{6} \)
\( b = \frac{(\sqrt{3} - \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{3 + 2 - 2\sqrt{6}}{3 - 2} = 5 - 2\sqrt{6} \)
Calculate their sum and product:
\( a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10 \)
\( ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1 \)
Now, apply the algebraic identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\( a^2 + b^2 = 10^2 - 2(1) = 100 - 2 = 98 \)
In simple words: Simplify a to 5 + 2√6 and b to 5 - 2√6. Their sum is 10 and their product is 1. Using (a + b)² - 2ab gives 100 - 2 = 98.

Exam Tip: Calculating the sum (a + b) and product (ab) first makes solving symmetric expressions like a² + b² much simpler than squaring fractions directly.

Question 11. If \( a = 9 - 4\sqrt{5} \), find the value of \( \left[a - \frac{1}{a}\right]^2 \)
Answer: Given \( a = 9 - 4\sqrt{5} \).
Find \( \frac{1}{a} \) by rationalisation:
\( \frac{1}{a} = \frac{1}{9 - 4\sqrt{5}} = \frac{9 + 4\sqrt{5}}{(9 - 4\sqrt{5})(9 + 4\sqrt{5})} = \frac{9 + 4\sqrt{5}}{9^2 - (4\sqrt{5})^2} = \frac{9 + 4\sqrt{5}}{81 - 80} = 9 + 4\sqrt{5} \)
Next, compute the difference \( a - \frac{1}{a} \):
\( a - \frac{1}{a} = (9 - 4\sqrt{5}) - (9 + 4\sqrt{5}) = 9 - 4\sqrt{5} - 9 - 4\sqrt{5} = -8\sqrt{5} \)
Now, square this result:
\( \left[a - \frac{1}{a}\right]^2 = (-8\sqrt{5})^2 = (-8)^2 \times (\sqrt{5})^2 = 64 \times 5 = 320 \)
In simple words: The reciprocal of 9 - 4√5 is 9 + 4√5. Subtracting them yields -8√5, and squaring -8√5 gives 320.

Exam Tip: Remember that the square of a negative number is positive: (-8√5)² = +320.

Question 12. If \( x = 1 - \sqrt{2} \), find the value of \( \left[x - \frac{1}{x}\right]^3 \)
Answer: Given \( x = 1 - \sqrt{2} \).
First, find \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{1 - \sqrt{2}} = \frac{1 + \sqrt{2}}{(1 - \sqrt{2})(1 + \sqrt{2})} = \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{-1} = -1 - \sqrt{2} \)
Now, calculate \( x - \frac{1}{x} \):
\( x - \frac{1}{x} = (1 - \sqrt{2}) - (-1 - \sqrt{2}) = 1 - \sqrt{2} + 1 + \sqrt{2} = 2 \)
Finally, evaluate the cube:
\( \left[x - \frac{1}{x}\right]^3 = 2^3 = 8 \)
In simple words: Finding the reciprocal gives -1 - √2. Subtracting that from 1 - √2 leaves 2, and 2 cubed is 8.

Exam Tip: Watch the denominator (1 - 2 = -1) carefully; dividing by -1 flips both signs in the numerator to -1 - √2.

Question 13. If \( x = 3 + 2\sqrt{2} \), find the value of \( \left[\sqrt{x} - \frac{1}{\sqrt{x}}\right] \)
Answer: Given \( x = 3 + 2\sqrt{2} \).
First, determine \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \)
Now, consider the square of \( \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) \):
\( \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} - 2 \)
\( = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) - 2 = 6 - 2 = 4 \)
Taking the positive square root on both sides:
\( \sqrt{x} - \frac{1}{\sqrt{x}} = \sqrt{4} = 2 \)
Note: If the expression is cubed as in matching problem sets, \( \left[\sqrt{x} - \frac{1}{\sqrt{x}}\right]^3 = 2^3 = 8 \).
In simple words: Use the identity (√x - 1/√x)² = x + 1/x - 2. Since x + 1/x = 6, the square equals 4, which gives 2 (and its cube equals 8).

Exam Tip: Since x > 1, √x > 1/√x, meaning the value of (√x - 1/√x) must be positive 2, not -2.

Question 14. If \( x = 0.125 \), find the value of \( (1/x)^{1/3} \)
Answer: Convert the decimal \( x \) into fractional form:
\( x = 0.125 = \frac{125}{1000} = \frac{1}{8} \)
Therefore, the reciprocal is:
\( \frac{1}{x} = 8 \)
Now, evaluate \( (1/x)^{1/3} \):
\( \left(\frac{1}{x}\right)^{1/3} = 8^{1/3} = (2^3)^{1/3} = 2^{3 \times \frac{1}{3}} = 2^1 = 2 \)
In simple words: 0.125 is 1/8, so its reciprocal 1/x is 8. The cube root of 8 is 2.

Exam Tip: Expressing 0.125 directly as 1/8 simplifies roots and powers compared to calculating with decimals.

Question 15. If \( x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \) then find the value of \( x^2 \)
Answer: First, rationalise the denominator of \( x \):
\( x = \frac{(\sqrt{3} + \sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 + 2 + 2\sqrt{6}}{3 - 2} = 5 + 2\sqrt{6} \)
Now, compute \( x^2 \) by squaring this expression:
\( x^2 = (5 + 2\sqrt{6})^2 = 5^2 + (2\sqrt{6})^2 + 2(5)(2\sqrt{6}) \)
\( = 25 + (4 \times 6) + 20\sqrt{6} = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6} \)
In simple words: Rationalise x to get 5 + 2√6, then square that using (a + b)² to get 49 + 20√6.

Exam Tip: Don't forget that (2√6)² equals 4 × 6 = 24, not 12.

Question 16. If \( x = \frac{1}{2 - \sqrt{3}} \), find the value of \( x^3 - 2x^2 - 7x + 5 \)
Answer: Rationalise \( x \):
\( x = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \)
Rearrange and eliminate the radical:
\( x - 2 = \sqrt{3} \)
Squaring both sides:
\( (x - 2)^2 = (\sqrt{3})^2 \)
\( \implies x^2 - 4x + 4 = 3 \)
\( \implies x^2 - 4x + 1 = 0 \)
Now, express the polynomial \( x^3 - 2x^2 - 7x + 5 \) in terms of \( (x^2 - 4x + 1) \):
\( x^3 - 2x^2 - 7x + 5 = x(x^2 - 4x + 1) + 2x^2 - 8x + 5 \)
\( = x(x^2 - 4x + 1) + 2(x^2 - 4x + 1) + 3 \)
Since \( x^2 - 4x + 1 = 0 \):
\( = x(0) + 2(0) + 3 = 3 \)
In simple words: Simplify x to 2 + √3, form the equation x² - 4x + 1 = 0, and divide the big polynomial by it. The quotient parts become 0, leaving just the remainder 3.

Exam Tip: Setting up the quadratic relation x² - 4x + 1 = 0 avoids having to calculate x³ and x² through long expansion.

Question 17. Find four rational numbers between 3/5 and 4/5
Answer: To find 4 rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \), multiply the numerator and denominator of both fractions by \( (4 + 1) = 5 \) or by 10:
\( \frac{3}{5} = \frac{3 \times 10}{5 \times 10} = \frac{30}{50} \)
\( \frac{4}{5} = \frac{4 \times 10}{5 \times 10} = \frac{40}{50} \)
Four rational numbers between \( \frac{30}{50} \) and \( \frac{40}{50} \) are:
\( \frac{31}{50}, \frac{32}{50}, \frac{33}{50}, \frac{34}{50} \)
In simplest form, these can be written as:
\( \frac{31}{50}, \frac{16}{25}, \frac{33}{50}, \frac{17}{25} \)
In simple words: Expand both fractions to have a larger common denominator like 50, then pick four fractions between 30/50 and 40/50.

Exam Tip: Multiplying by 10 is the safest and quickest choice to create plenty of space between numerators.

Question 18. Find two irrational numbers lying between \( \sqrt{2} \) and \( \sqrt{3} \)
Answer: The approximate decimal values are:
\( \sqrt{2} \approx 1.414213\ldots \)
\( \sqrt{3} \approx 1.732050\ldots \)
Any non-terminating and non-repeating decimal between these two values is an irrational number.
Two such irrational numbers are:
1. \( 1.5050050005\ldots \)
2. \( 1.6060060006\ldots \)
In simple words: Write decimals that lie between 1.414 and 1.732 with an ever-increasing pattern of zeros so they never end and never repeat.

Exam Tip: Show the approximate decimal values of √2 and √3 first to justify why your chosen numbers lie within the interval.

Question 19. Find two rational and irrational numbers between 0.3101 and 0.3222
Answer:
Two rational numbers (terminating decimals) between 0.3101 and 0.3222:
1. \( 0.315 = \frac{315}{1000} = \frac{63}{200} \)
2. \( 0.320 = \frac{32}{100} = \frac{8}{25} \)
Two irrational numbers (non-terminating and non-recurring decimals) between 0.3101 and 0.3222:
1. \( 0.312010010001\ldots \)
2. \( 0.315010010001\ldots \)
In simple words: Pick standard terminating decimals like 0.315 and 0.320 for the rational answers, and create non-repeating pattern decimals in the same range for the irrational ones.

Exam Tip: For irrational numbers, always use ellipsis (...) at the end to show that the decimal expansion continues indefinitely.

Question 20. Simplify the following: a) \( \left(\frac{576}{625}\right)^{-1/2} \) b) \( \left(\frac{343}{1000}\right)^{-1/3} \) c) \( (-1/27)^{-2/3} \) d) \( (0.008)^{4/3} \) e) \( (729)^{-1/6} \)
Answer:
(a) \( \left(\frac{576}{625}\right)^{-1/2} = \left(\frac{625}{576}\right)^{1/2} = \sqrt{\frac{25^2}{24^2}} = \frac{25}{24} \)
(b) \( \left(\frac{343}{1000}\right)^{-1/3} = \left(\frac{1000}{343}\right)^{1/3} = \left(\frac{10^3}{7^3}\right)^{1/3} = \frac{10}{7} \)
(c) \( \left(-\frac{1}{27}\right)^{-2/3} = (-27)^{2/3} = \left[(-3)^3\right]^{2/3} = (-3)^2 = 9 \)
(d) \( (0.008)^{4/3} = \left(\frac{8}{1000}\right)^{4/3} = \left[\left(\frac{2}{10}\right)^3\right]^{4/3} = \left(\frac{1}{5}\right)^4 = \frac{1}{625} \) (or 0.0016)
(e) \( (729)^{-1/6} = (3^6)^{-1/6} = 3^{-1} = \frac{1}{3} \)
In simple words: Negative powers invert the fraction. Then rewrite each base as a prime power so the fraction exponents cancel out nicely.

Exam Tip: In part (c), handle the negative sign carefully: taking the cube root of -27 gives -3, and squaring -3 yields positive 9.

Question 21. Simplify and express the result in the simplest form: \( \frac{(25)^{3/2} \times (243)^{2/5}}{(16)^{5/4} \times (8)^{4/3}} \)
Answer: Express each number in terms of prime factors:
Numerator:
\( (25)^{3/2} = (5^2)^{3/2} = 5^{2 \times \frac{3}{2}} = 5^3 = 125 \)
\( (243)^{2/5} = (3^5)^{2/5} = 3^{5 \times \frac{2}{5}} = 3^2 = 9 \)
Denominator:
\( (16)^{5/4} = (2^4)^{5/4} = 2^{4 \times \frac{5}{4}} = 2^5 = 32 \)
\( (8)^{4/3} = (2^3)^{4/3} = 2^{3 \times \frac{4}{3}} = 2^4 = 16 \)
Substitute these values into the expression:
\( \frac{125 \times 9}{32 \times 16} = \frac{1125}{512} \)
In simple words: Convert 25, 243, 16, and 8 to powers of 5, 3, and 2. Cancel the exponent fractions to get 125 × 9 on top and 32 × 16 on the bottom.

Exam Tip: Notice that the bases in the denominator are both powers of 2, so you can also combine them first: 2⁵ × 2⁴ = 2⁹ = 512.

Question 22. Find the value x, if \( 5^{x-3} \times 3^{2x-8} = 225 \)
Answer: Find the prime factorisation of 225:
\( 225 = 25 \times 9 = 5^2 \times 3^2 \)
The given equation becomes:
\( 5^{x-3} \times 3^{2x-8} = 5^2 \times 3^2 \)
Equating the exponents of corresponding bases on both sides:
For base 5:
\( x - 3 = 2 \implies x = 5 \)
For base 3:
\( 2x - 8 = 2 \implies 2x = 10 \implies x = 5 \)
Both equations yield the consistent result \( x = 5 \).
In simple words: Break 225 into 5² × 3². Compare the powers of 5 and powers of 3 to see that x must be 5.

Exam Tip: Always verify that both base equations yield the same value of x to confirm your answer is fully consistent.

Question 23. Solve: a) \( 49 \times 7^x = (343)^{1/3} \)
b) \( 2^x = (128)^{1/7} \times (\sqrt{2})^4 \)
c) If \( 3^x = \frac{9}{27^x} \), find x
d) \( (1/7)^{4-2x} = \sqrt{7} \)
Answer:
(a) Express all terms with base 7:
\( 7^2 \times 7^x = (7^3)^{1/3} \)
\( \implies 7^{2+x} = 7^1 \)
\( \implies 2 + x = 1 \implies x = -1 \)
(b) Express terms with base 2:
\( 2^x = (2^7)^{1/7} \times (2^{1/2})^4 \)
\( \implies 2^x = 2^1 \times 2^2 = 2^{1+2} = 2^3 \)
\( \implies x = 3 \)
(c) Cross-multiply and equate powers of 3:
\( 3^x \times 27^x = 9 \)
\( \implies 3^x \times (3^3)^x = 3^2 \)
\( \implies 3^{x + 3x} = 3^2 \implies 3^{4x} = 3^2 \)
\( \implies 4x = 2 \implies x = \frac{2}{4} = \frac{1}{2} \)
(d) Express both sides as powers of 7:
\( (7^{-1})^{4-2x} = 7^{1/2} \)
\( \implies 7^{2x-4} = 7^{1/2} \)
\( \implies 2x - 4 = \frac{1}{2} \)
\( \implies 2x = 4 + \frac{1}{2} = \frac{9}{2} \)
\( \implies x = \frac{9}{4} \)
In simple words: Convert every number in each equation into powers of the same base, add exponents using laws of indices, and set the powers equal to solve for x.

Exam Tip: Remember that (1/7) = 7^(-1); distributing the -1 changes (4 - 2x) into (2x - 4).

Question 24. Evaluate: a) \( 125^{-1/3} \times 27^{1/3} (6^2 + 8^2)^{1/2} \)
b) \( (17^2 - 8^2)^{1/2} \)
c) \( 64^{1/3} (64^{1/3} - 64^{2/3}) \)
Answer:
(a) Evaluate each component:
\( 125^{-1/3} = (5^3)^{-1/3} = 5^{-1} = \frac{1}{5} \)
\( 27^{1/3} = (3^3)^{1/3} = 3 \)
\( (6^2 + 8^2)^{1/2} = (36 + 64)^{1/2} = (100)^{1/2} = 10 \)
Multiply the parts:
\( \frac{1}{5} \times 3 \times 10 = \frac{30}{5} = 6 \)
(b) Evaluate inside the bracket:
\( (17^2 - 8^2)^{1/2} = (289 - 64)^{1/2} = (225)^{1/2} = 15 \)
(c) Simplify the terms with base 64:
\( 64^{1/3} = (4^3)^{1/3} = 4 \)
\( 64^{2/3} = (4^3)^{2/3} = 4^2 = 16 \)
Substitute back:
\( 4 \times (4 - 16) = 4 \times (-12) = -48 \)
In simple words: Simplify roots and powers step-by-step: part (a) works out to 6, part (b) takes the square root of 225 to give 15, and part (c) multiplies 4 by -12 to give -48.

Exam Tip: In part (b), you can also factorize 17² - 8² as (17 - 8)(17 + 8) = 9 × 25 = 225 to find the root mentally without squaring 17.

Question 25. Simplify: a) \( \sqrt{45} + \sqrt{80} - 3\sqrt{20} \)
b) \( 7\sqrt{6} - \sqrt{252} - \sqrt{294} + 6\sqrt{7} \)
c) \( 4\sqrt{28} + 3\sqrt{7} \)
Answer:
(a) Factor out perfect squares:
\( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \)
\( \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \)
\( 3\sqrt{20} = 3\sqrt{4 \times 5} = 3(2\sqrt{5}) = 6\sqrt{5} \)
Combining like surds:
\( 3\sqrt{5} + 4\sqrt{5} - 6\sqrt{5} = (3 + 4 - 6)\sqrt{5} = \sqrt{5} \)
(b) Factor each radical:
\( \sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7} \)
\( \sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6} \)
Substitute into the expression:
\( 7\sqrt{6} - 6\sqrt{7} - 7\sqrt{6} + 6\sqrt{7} = (7\sqrt{6} - 7\sqrt{6}) + (-6\sqrt{7} + 6\sqrt{7}) = 0 \)
(c) Simplify \( \sqrt{28} \):
\( \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \)
\( 4\sqrt{28} = 4(2\sqrt{7}) = 8\sqrt{7} \)
Combining terms:
\( 8\sqrt{7} + 3\sqrt{7} = 11\sqrt{7} \)
In simple words: Pull square factors outside each root sign so like terms match, then add and subtract like surds.

Exam Tip: Group matching surds together (terms with √6 together and terms with √7 together) before simplifying to prevent arithmetic confusion.

Question 26. Give an example of two irrational numbers whose: (A) Sum is rational (B) product is rational (C) quotient is rational
Answer:
(A) Sum is rational:
Consider the two irrational numbers \( (5 + \sqrt{3}) \) and \( (5 - \sqrt{3}) \).
Their sum is \( (5 + \sqrt{3}) + (5 - \sqrt{3}) = 10 \), which is a rational number.
(B) Product is rational:
Consider the two irrational numbers \( \sqrt{8} \) and \( \sqrt{2} \).
Their product is \( \sqrt{8} \times \sqrt{2} = \sqrt{16} = 4 \), which is a rational number.
(C) Quotient is rational:
Consider the two irrational numbers \( \sqrt{75} \) and \( \sqrt{3} \).
Their quotient is \( \frac{\sqrt{75}}{\sqrt{3}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5 \), which is a rational number.
In simple words: Adding conjugate pairs removes the root, multiplying roots that make a perfect square gives a whole number, and dividing roots that leave a perfect square also gives a clean rational result.

Exam Tip: Using conjugates (a + √b) and (a - √b) is the most reliable way to create rational sums and products from irrational pairs.