SEBA Class 10 Maths Complete Revision Solutions – Full Exam Preparation Guide

SEBA Class 10 Maths Revision 1 Square and Square Roots Solutions

Q1. What will be the digits in the unit place of the squares of the following numbers?
(i) 272    (ii) 79    (iii) 400    (iv) 2637
Answer:
To find the unit digit of a square, we only need to look at the unit digit of the number.

(i) 272 → unit digit = 2 → \(2^2 = 4\)
(ii) 79 → unit digit = 9 → \(9^2 = 81\) → unit digit = 1
(iii) 400 → unit digit = 0 → \(0^2 = 0\)
(iv) 2637 → unit digit = 7 → \(7^2 = 49\) → unit digit = 9

Final Answer:
(i) 4    (ii) 1    (iii) 0    (iv) 9

Q2. Why do the following numbers are not perfect square?
(i) 1057    (ii) 7982    (iii) 2221    (iv) 640
Answer:
A perfect square can only have certain digits at the unit place: 0, 1, 4, 5, 6, or 9.
If a number ends with 2, 3, 7, or 8, it cannot be a perfect square.

(i) 1057 → unit digit = 7 → not a perfect square
(ii) 7982 → unit digit = 2 → not a perfect square
(iii) 2221 → unit digit = 1 → possible, but not a perfect square (no integer square root)
(iv) 640 → unit digit = 0 → but not a perfect square (not exact square)

Final Conclusion:
These numbers are not perfect squares because they do not satisfy the properties of perfect squares.

Q3. What are the squares of the following numbers?
(i) 19    (ii) 37    (iii) 53    (iv) 78
Answer:
To find the square of a number, we multiply the number by itself.

(i) \(19^2 = 19 \times 19 = 361\)
(ii) \(37^2 = 37 \times 37 = 1369\)
(iii) \(53^2 = 53 \times 53 = 2809\)
(iv) \(78^2 = 78 \times 78 = 6084\)

Final Answer:
(i) 361    (ii) 1369    (iii) 2809    (iv) 6084

Q4. Find the square roots of the following numbers.
(i) 1764    (ii) 9216    (iii) 7744    (iv) 9801
Answer:
To find square roots, we check which number when multiplied by itself gives the given number.

(i) \(42^2 = 1764\) → √1764 = 42
(ii) \(96^2 = 9216\) → √9216 = 96
(iii) \(88^2 = 7744\) → √7744 = 88
(iv) \(99^2 = 9801\) → √9801 = 99

Final Answer:
(i) 42    (ii) 96    (iii) 88    (iv) 99

Q5. Find the least numbers (integer) with which the following numbers are to be multiplied so that they become perfect squares.
(i) 125    (ii) 1008    (iii) 2028    (iv) 768
Answer:
We find prime factors and make all powers even.

(i) \(125 = 5^3\) → multiply by 5 → \(5^4\)
Answer = 5

(ii) \(1008 = 2^4 \times 3^2 \times 7\) → multiply by 7
Answer = 7

(iii) \(2028 = 2^2 \times 3 \times 13^2\) → multiply by 3
Answer = 3

(iv) \(768 = 2^8 \times 3\) → multiply by 3
Answer = 3

Final Answer:
(i) 5    (ii) 7    (iii) 3    (iv) 3

Q6. With what least numbers (integer) the following numbers are to be divided so that they become perfect squares.
(i) 468    (ii) 1584    (iii) 2645    (iv) 1620
Answer:
We divide by factors whose powers are odd.

(i) \(468 = 2^2 \times 3^2 \times 13\) → divide by 13
Answer = 13

(ii) \(1584 = 2^4 \times 3^2 \times 11\) → divide by 11
Answer = 11

(iii) \(2645 = 5 \times 23^2\) → divide by 5
Answer = 5

(iv) \(1620 = 2^2 \times 3^4 \times 5\) → divide by 5
Answer = 5

Final Answer:
(i) 13    (ii) 11    (iii) 5    (iv) 5

Q7. Find the square root of the following decimal numbers.
(i) 12.25    (ii) 24.01    (iii) 146.41    (iv) 102.01
Answer:
Convert decimals into fractions or recognize known squares.

(i) \(3.5^2 = 12.25\) → √12.25 = 3.5
(ii) \(4.9^2 = 24.01\) → √24.01 = 4.9
(iii) \(12.1^2 = 146.41\) → √146.41 = 12.1
(iv) \(10.1^2 = 102.01\) → √102.01 = 10.1

Final Answer:
(i) 3.5    (ii) 4.9    (iii) 12.1    (iv) 10.1

Q8. Four options are given for each of the following. Find the correct option.

(a) Which of the following is a square of an odd natural number?

(a) 256
(b) 169
(c) 546
(d) 754

Answer: (b) 169

(b) Which of the following will have 1 (one) in the unit place?

(i) \(19^2\)
(ii) \(34^2\)
(iii) \(18^2\)
(iv) \(20^2\)

Answer: (i) \(19^2\)

(c) Between 18² and 19² how many natural numbers are there?

(a) 38
(b) 36
(c) 42
(d) 40

Answer: (b) 36

(d) Which of the following is not a perfect square?

(a) 441
(b) 572
(c) 576
(d) 729

Answer: (b) 572

(e) If \( \sqrt{2025} = 45 \), then \( \sqrt{20.25} \) is equal to

(a) 45
(b) 4.5
(c) 0.45
(d) 0.045

Answer: (b) 4.5

SEBA Class 10 Maths Revision 2 Cube and Cube Roots Solutions

Q1. Which of the following is not a perfect cube?
(i) 3757    (ii) 3375    (iii) 3332    (iv) 4096
Answer:
A perfect cube is a number which can be written as \(n^3\).

\(3375 = 15^3\), \(4096 = 16^3\)
But 3757 and 3332 are not perfect cubes.

Final Answer:
(i) 3757 and (iii) 3332

Q2. Find the cubes of the following numbers.
(i) 19    (ii) 21    (iii) 23    (iv) 27
Answer:
Cube means multiplying the number three times.

(i) \(19^3 = 6859\)
(ii) \(21^3 = 9261\)
(iii) \(23^3 = 12167\)
(iv) \(27^3 = 19683\)

Q3. Write the digit in the unit place of the cubes of the following numbers.
(i) 14    (ii) 18    (iii) 13    (iv) 27
Answer:
We only check the unit digit of the number.

(i) 14 → unit digit 4 → \(4^3 = 64\) → unit digit = 4
(ii) 18 → unit digit 8 → \(8^3 = 512\) → unit digit = 2
(iii) 13 → unit digit 3 → \(3^3 = 27\) → unit digit = 7
(iv) 27 → unit digit 7 → \(7^3 = 343\) → unit digit = 3

Q4. Find the smallest integers with which the following numbers are to be multiplied so that they become perfect cubes.
(i) 5324    (ii) 3087    (iii) 3125    (iv) 648
Answer:
For a perfect cube, powers of prime factors must be multiples of 3.

(i) \(5324 = 2^2 \times 11^3\) → multiply by 2 → \(2^3\)
Answer = 2

(ii) \(3087 = 3^2 \times 7^3\) → multiply by 3
Answer = 3

(iii) \(3125 = 5^5\) → multiply by \(5\) to make \(5^6\)
Answer = 5

(iv) \(648 = 2^3 \times 3^4\) → multiply by 3
Answer = 3

Q5. Find the smallest numbers with which the following numbers are to be divided so that they become perfect cubes.
(i) 10368    (ii) 2187    (iii) 5000    (iv) 8192
Answer:
We divide to make powers multiples of 3.

(i) \(10368 = 2^7 \times 3^4\) → divide by \(2 \times 3\)
Answer = 6

(ii) \(2187 = 3^7\) → divide by \(3\)
Answer = 3

(iii) \(5000 = 2^3 \times 5^4\) → divide by 5
Answer = 5

(iv) \(8192 = 2^{13}\) → divide by \(2\)
Answer = 2

Q6. Find the cube roots of the following numbers.
(i) 1331    (ii) 1728    (iii) 2197    (iv) 2744
Answer:
Cube root means finding a number whose cube gives the given number.

(i) \(11^3 = 1331\) → ∛1331 = 11
(ii) \(12^3 = 1728\) → ∛1728 = 12
(iii) \(13^3 = 2197\) → ∛2197 = 13
(iv) \(14^3 = 2744\) → ∛2744 = 14

(a) The digit in the unit place in the cube of 23 is —

(i) 6
(ii) 7
(iii) 8
(iv) 9

Answer: (ii) 7

(b) Which of the following is a perfect cube?

(i) 652
(ii) 933
(iii) 343
(iv) 1002

Answer: (iii) 343

(c) The value of \( \sqrt[3]{1000} \) is —

(i) 30
(ii) 100
(iii) 10
(iv) 1000

Answer: (iii) 10

(d) If m is the cube root of n then the value of n is —

(i) \( \sqrt{m} \)
(ii) \( \sqrt[3]{m} \)
(iii) \( m^3 \)
(iv) \( m^2 \)

Answer: (iii) \(m^3\)

(e) The value of \( \sqrt[3]{8} + \sqrt[3]{27} + \sqrt[3]{64} \) is —

(i) 6
(ii) 7
(iii) 8
(iv) 9

Answer: (iv) 9

SEBA Class 10 Maths Revision 3 Solutions (Indices and Powers)

Q1. Find the value of:
(i) \(11^3\)    (ii) \(2 \times 10^4\)    (iii) \(\left(\frac{1}{2}\right)^5\)    (iv) \((-4)^2\)
Answer:
We will use laws of exponents to evaluate each expression.

(i) \(11^3 = 11 \times 11 \times 11 = 1331\)
(ii) \(2 \times 10^4 = 2 \times 10000 = 20000\)
(iii) \(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\)
(iv) \((-4)^2 = 16\)

Final Answer:
(i) 1331    (ii) 20000    (iii) \( \frac{1}{32} \)    (iv) 16

Q2. Express the following numbers in terms of powers of their prime factors:
(i) 720    (ii) 3125    (iii) 3600    (iv) 108 × 192
Answer:
(i) \(720 = 2^4 \times 3^2 \times 5\)
(ii) \(3125 = 5^5\)
(iii) \(3600 = 2^4 \times 3^2 \times 5^2\)
(iv) \(108 \times 192 = (2^2 \times 3^3)(2^6 \times 3) = 2^8 \times 3^4\)

Final Answer:
(i) \(2^4 \times 3^2 \times 5\)
(ii) \(5^5\)
(iii) \(2^4 \times 3^2 \times 5^2\)
(iv) \(2^8 \times 3^4\)

Q3. Simplify:
(i) \((-3)^4 \times (-5)^2\)    (ii) \((2^2 \times 2)^3\)    (iii) \(2^3 \times 3^2 \times 4^0\)    (iv) \(\left(\frac{5}{8}\right)^2 \times \left(\frac{8}{5}\right)^2\)
Answer:
(i) \((-3)^4 = 81,\; (-5)^2 = 25\)
\(81 \times 25 = 2025\)

(ii) \((2^2 \times 2)^3 = (2^3)^3 = 2^9 = 512\)

(iii) \(4^0 = 1\)
\(2^3 \times 3^2 \times 1 = 8 \times 9 = 72\)

(iv) \(\left(\frac{5}{8} \times \frac{8}{5}\right)^2 = 1^2 = 1\)

Final Answer:
(i) 2025    (ii) 512    (iii) 72    (iv) 1

Q4. Compare the following numbers:
(i) \(2^8\) and \(8^2\)    (ii) \(2.7 \times 10^5\) and \(1.5 \times 10^6\)
Answer:
(i) \(2^8 = 256,\; 8^2 = 64\)
So, \(2^8 > 8^2\)

(ii) \(2.7 \times 10^5 = 270000\)
\(1.5 \times 10^6 = 1500000\)
So, \(2.7 \times 10^5 < 1.5 \times 10^6\)

Final Answer:
(i) \(2^8 > 8^2\)
(ii) \(2.7 \times 10^5 < 1.5 \times 10^6\)

Q5. Express the following with the help of positive power:
(i) \(2^{-3} \times (-7)^{-3}\)    (ii) \((-3)^{-4} \times \left(\frac{5}{3}\right)^4\)
Answer:
(i) \(2^{-3} = \frac{1}{2^3},\; (-7)^{-3} = \frac{1}{(-7)^3}\)
\[ = \frac{1}{2^3 \times (-7)^3} \]

(ii) \((-3)^{-4} = \frac{1}{(-3)^4}\)
\[ = \frac{(5/3)^4}{3^4} = \left(\frac{5}{9}\right)^4 \]

Final Answer:
(i) \( \frac{1}{2^3 \times (-7)^3} \)
(ii) \( \left(\frac{5}{9}\right)^4 \)

Q6. Express the following numbers in standard form:
(i) 3430.000    (ii) 70,40,000,000    (iii) 0.000000015    (iv) 0.00001436
Answer:
(i) \(3.43 \times 10^3\)
(ii) \(7.04 \times 10^9\)
(iii) \(1.5 \times 10^{-8}\)
(iv) \(1.436 \times 10^{-5}\)

Final Answer:
(i) \(3.43 \times 10^3\)
(ii) \(7.04 \times 10^9\)
(iii) \(1.5 \times 10^{-8}\)
(iv) \(1.436 \times 10^{-5}\)

Q7. Express the following in general form:
(i) \(1.0001 \times 10^9\)    (ii) \(3.02 \times 10^6\)
Answer:
(i) \(1.0001 \times 10^9 = 1000100000\)
(ii) \(3.02 \times 10^6 = 3020000\)

Final Answer:
(i) 1000100000
(ii) 3020000

Q8. Find the value of m such that \( (-3)^m \times (-3)^5 = (-3)^7 \)
Answer:
Using law: \(a^m \times a^n = a^{m+n}\)

\[ (-3)^{m+5} = (-3)^7 \]
So, \(m + 5 = 7\)
\(m = 2\)

Final Answer:
\(m = 2\)

Q9. (a) The value of \(3^{-3}\) is:

(i) 3
(ii) \( \frac{1}{3} \)
(iii) \( \frac{1}{27} \)
(iv) 3×3

Answer: (iii) \( \frac{1}{27} \)

Q9. (b) The value of \( \left(\frac{2}{3}\right)^2 \) is:

(i) \( \frac{1}{(2×3)^2} \)
(ii) \( (2×3)^2 \)
(iii) \( \left(\frac{3}{2}\right)^2 \)
(iv) \( \left(\frac{2}{3}\right)^2 \)

Answer: (iv) \( \left(\frac{2}{3}\right)^2 \)

(c) The value of \( \left(-\frac{2}{3}\right)^4 \) is:

(i) \( \frac{8}{12} \)
(ii) \( \frac{16}{81} \)
(iii) \( -\frac{16}{81} \)
(iv) \( \frac{8}{12} \)

Answer: (ii) \( \frac{16}{81} \)

(d) The standard form of 0.000064 is:

(i) \(64 \times 10^4\)
(ii) \(64 \times 10^{-4}\)
(iii) \(6.4 \times 10^5\)
(iv) \(6.4 \times 10^{-5}\)

Answer: (iv) \(6.4 \times 10^{-5}\)

(e) The value of \(2.03 \times 10^{-3}\) is:

(i) 0.203
(ii) 0.000203
(iii) 203000
(iv) 0.00203

Answer: (iv) 0.00203

SEBA Class 10 Maths Revision 4 Solutions (Factorisation)

Q1. Find common factors of the following:

(i) 14pq, 28p²q
Answer: Given: 14pq, 28p²q
We will find common numerical and variable factors.
Common number = 14
Common variables = p, q
So, common factor = \(14pq\)

(ii) 16x³ − 4x², 32x
Answer: Given: 16x³ − 4x², 32x
Factor first expression:
\(16x^3 - 4x^2 = 4x^2(4x - 1)\)
Common factor with 32x is \(4x\)
Final Answer: \(4x\)

(iii) 20pq, 30qr, 40rp
Answer: Given: 20pq, 30qr, 40rp
Common number = 10
No common variable in all three terms
Final Answer: 10

(iv) 3x²y³, 10x³y², 6x²y²z
Answer: Given: 3x²y³, 10x³y², 6x²y²z
Common number = 1
Common variables = \(x^2, y^2\)
Final Answer: \(x^2y^2\)

Q2. Factorise:

(i) 4a² + 8a³
Answer: Given: \(4a^2 + 8a^3\)
Take common factor \(4a^2\):
\[ 4a^2(1 + 2a) \]

(ii) 7x²y − 21xy²
Answer: Given: \(7x^2y - 21xy^2\)
Common factor = \(7xy\)
\[ 7xy(x - 3y) \]

(iii) a²bc + abc² + abc²
Answer: Given: \(a^2bc + abc^2 + abc^2\)
Common factor = \(abc\)
\[ abc(a + c + c) \] \[ = abc(a + 2c) \]

(iv) a³ − a²b
Answer: Given: \(a^3 - a^2b\)
Common factor = \(a^2\)
\[ a^2(a - b) \]

Q3. Factorise:

(i) x² + xy + 6x + 6y
Solution:
Given: \(x^2 + xy + 6x + 6y\)

Group terms:
\[ x^2 + xy + 6x + 6y = (x^2 + xy) + (6x + 6y) \] Take common factors:
\[ = x(x + y) + 6(x + y) \] \[ = (x + y)(x + 6) \]

(ii) xy + x + y + 1
Solution:
Given: \(xy + x + y + 1\)

Group terms:
\[ xy + x + y + 1 = (xy + x) + (y + 1) \] Take common factors:
\[ = x(y+1) + 1(y+1) \] \[ = (x+1)(y+1) \]

(iii) 24x²y + 12x² − 12xy − 6x
Solution:
Given: \(24x^2y + 12x^2 - 12xy - 6x\)

Group terms:
\[ = (24x^2y + 12x^2) - (12xy + 6x) \] Take common factors:
\[ = 12x^2(2y+1) - 6x(2y+1) \] Take common binomial factor:
\[ = (12x^2 - 6x)(2y+1) \] Factor further:
\[ = 6x(2x -1)(2y+1) \]

(iv) z − 7 + 7xy − xyz
Solution:
Given: \(z - 7 + 7xy - xyz\)

Rearrange terms:
\[ = (z - xyz) + (7xy - 7) \] Take common factors:
\[ = z(1 - xy) + 7(xy -1) \] \[ = z(1 - xy) - 7(1 - xy) \] Take common factor:
\[ = (z - 7)(1 - xy) \]

Q4. Express in factors:

(i) 4x² + 12x + 9
Solution:
Given: \(4x^2 + 12x + 9\)

We compare with identity:
\[ a^2 + 2ab + b^2 = (a+b)^2 \] Here,
\[ (2x)^2 = 4x^2,\; 2(2x)(3) = 12x,\; 3^2 = 9 \] So,
\[ 4x^2 + 12x + 9 = (2x + 3)^2 \]

(ii) 25m² + 30m + 9
Solution:
Given: \(25m^2 + 30m + 9\)

Compare with identity:
\[ a^2 + 2ab + b^2 \] \[ (5m)^2 = 25m^2,\; 2(5m)(3) = 30m,\; 3^2 = 9 \] So,
\[ 25m^2 + 30m + 9 = (5m + 3)^2 \]

(iii) x² − 10x + 25
Solution:
Given: \(x^2 - 10x + 25\)

Compare with identity:
\[ a^2 - 2ab + b^2 = (a-b)^2 \] \[ x^2 - 10x + 25 = x^2 - 2(x)(5) + 5^2 \] So,
\[ = (x - 5)^2 \]

(iv) 121b² − 88bc + 16c²
Solution:
Given: \(121b^2 - 88bc + 16c^2\)

Compare with identity:
\[ a^2 - 2ab + b^2 \] \[ (11b)^2 = 121b^2,\; 2(11b)(4c) = 88bc,\; (4c)^2 = 16c^2 \] So,
\[ = (11b - 4c)^2 \]

(v) 9p² − 16q²
Solution:
Given: \(9p^2 - 16q^2\)

This is a difference of squares:
\[ a^2 - b^2 = (a-b)(a+b) \] \[ 9p^2 = (3p)^2,\; 16q^2 = (4q)^2 \] So,
\[ = (3p - 4q)(3p + 4q) \]

(vi) (l + m)² − (l − m)²
Solution:
Given: \((l + m)^2 - (l - m)^2\)

Use identity:
\[ a^2 - b^2 = (a-b)(a+b) \] \[ = [(l+m) - (l-m)][(l+m) + (l-m)] \] Simplify:
\[ = (l + m - l + m)(l + m + l - m) \] \[ = (2m)(2l) \] \[ = 4lm \]

(vii) x² − 13x − 30
Solution:
Given: \(x^2 - 13x - 30\)

We find two numbers whose:
Sum = −13 and Product = −30

Numbers are −15 and 2

Split middle term:
\[ x^2 - 15x + 2x - 30 \] Group terms:
\[ = (x^2 - 15x) + (2x - 30) \] \[ = x(x - 15) + 2(x - 15) \] Take common factor:
\[ = (x - 15)(x + 2) \]

(viii) y² − 5y − 36
Solution:
Given: \(y^2 - 5y - 36\)

Find two numbers whose:
Sum = −5 and Product = −36

Numbers are −9 and 4

Split middle term:
\[ y^2 - 9y + 4y - 36 \] Group terms:
\[ = (y^2 - 9y) + (4y - 36) \] \[ = y(y - 9) + 4(y - 9) \] Take common factor:
\[ = (y - 9)(y + 4) \]

(ix) 4y² + 25y − 21
Solution:
Given: \(4y^2 + 25y - 21\)

Multiply 4 × (−21) = −84
Find two numbers whose sum = 25 and product = −84

Numbers are 28 and −3

Split middle term:
\[ 4y^2 + 28y - 3y - 21 \] Group terms:
\[ = (4y^2 + 28y) - (3y + 21) \] \[ = 4y(y + 7) - 3(y + 7) \] Take common factor:
\[ = (4y - 3)(y + 7) \]

(x) 3x⁶ − 6x²y − 45x²y²
Solution:
Given: \(3x^6 - 6x^2y - 45x^2y^2\)

Take common factor:
\[ = 3x^2(x^4 - 2y - 15y^2) \] Now factorise inside:
\[ x^4 - 2y - 15y^2 \] Find two numbers whose product = −15y² and sum = −2y
Numbers are −5y and 3y

Split middle term:
\[ x^4 - 5y + 3y - 15y^2 \] (Grouping may not simplify further neatly, so final form is:)
\[ = 3x^2(x^4 - 2y - 15y^2) \]

SEBA Class 10 Maths Revision 5 Solutions (Congruence of Triangles)

Q1. Two triangles are congruent if:

(a) Their areas are equal
(b) Their shapes are the same but sizes may differ
(c) All corresponding sides and angles are equal
(d) Only their perimeters are equal

Answer: (c) All corresponding sides and angles are equal

Q2. Which of the following is not a valid congruency rule?

(a) AAA
(b) SSA
(c) SAS
(d) ASS

Answer: (a) AAA

Q3. If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent by:

(a) ASA
(b) SAS
(c) AAS
(d) RHS

Answer: (b) SAS

Q4. Which congruence rule is used when two angles and the included side are equal?

(a) AAS
(b) ASA
(c) SAS
(d) RHS

Answer: (b) ASA

Q5. Right-angled triangles are congruent by RHS if they have equal:

(a) Hypotenuse and one acute angle
(b) Two acute angles
(c) Hypotenuse and one corresponding side
(d) All sides

Answer: (c) Hypotenuse and one corresponding side

Q6. AAS congruence rule means:

(a) Two angles and a side not included between them
(b) Two sides and an angle
(c) Three angles
(d) One side and one angle

Answer: (a) Two angles and a side not included between them

Q7. If ∠A = ∠D, ∠B = ∠E and AB = DE, then ΔABC ≅ ΔDEF by:

(a) SAS
(b) AAS
(c) ASA
(d) RHS

Answer: (c) ASA

Q8. In ΔABC and ΔPQR, if AB = PQ, BC = QR and AC = PR, triangles are congruent by:

(a) ASA
(b) SSS
(c) SAS
(d) RHS

Answer: (b) SSS

Q9. The symbol '≅' represents:

(a) Similarity
(b) Congruency
(c) Parallel
(d) Perpendicular

Answer: (b) Congruency

Q10. Congruent triangles have:

(a) Equal corresponding sides
(b) Equal corresponding angles
(c) Same shape and same size
(d) All of these

Answer: (d) All of these