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Content text Advanced Series 8th Class Maths - Solutions.pdf

Class 8 Mathematics Table of Content 1. Number System 1 – 37 2. Polynomials 38 – 65 3. Pair of Linear Equations 66 – 94 4. Quadratic Equations 95 – 122 5. Direct and Inverse Proportions 123 – 144 6. Exponents and Powers, Square Roots, and Cube Roots 145 – 167 7. Progressions 168 – 198 8. Plane Geometry – Triangles 199 – 232 9. Plane Geometry – Quadrilaterals 233 – 245 10. Surds 246 – 262 11. Logarithms 263 – 281 12. Sets and Relations 282 – 309 13. Mensuration 310 – 376 14. Data Handling and Statistics 377 – 393 15. Coordinate Geometry 394 – 419 16. Time and Work, Time, and Distance 420 - 450

Class 8 – Maths | A-8 Number System 1 1. Number System Solutions LEVEL-I 1. Correct option: (C) Let the two numbers be a and b. Let a=917. We have to find b. We know that HCF(a, b) × LCM(a, b) = a × b ⇒ 131 × 8253 = 917 × b ⇒ 131 × 8253 917 = b ⇒ b = 1081143 917 . ∴ b = 1179. Therefore, the other number is 1179. 2. Correct option: (D) A number is divisible by 3 if the sum of the digits is evenly divisible by 3. Sum of digits divisible by 3 A) 5+6+1+1+1+0=14 B) 2+2+1+2+2+0+1= 10 C) 2+2+0+0+0+2+2= 8 D) 5+5+5+5+5+5= 30 Therefore, 30 is divisible by 3. Hence 555555 is divisible by 3. 3. Correct option: (C) List out all of the prime factors for each number: • The prime factors of 3 3 are 3, 3 and 3. • The prime factors of 3 is 3. • The prime factors of 4 2 are 2, 2, 2, and 2. • The prime factors of 4 are 2 and 2. The least common multiple is the product of all factors in the greatest number of their occurrence. Here the common and uncommon prime factors are 2, 2, 2, 2, 3, 3, 3. On Multiplying 2, 2, 2, 2, 3, 3, 3 to get the LCM of the given numbers. L.C.M of (3 3 , 3, 4 2 , 4) = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432. ∴ The required L.C.M least common multiple of 3 3 , 3, 4 2 , 4 is 432. 4. Correct option: (A) Observe that this number is a multiple of 3, since the sum of its digits is a multiple of 3. Specifically, the sum of its digits is 3 x 300 = 900, which is divisible by 3.
Class 8 – Maths | A-8 Number System 2 Therefore, the greatest number among the follo Which of these is a prime number 3123,219,573, 467? A) 3123 B) 219 C) 573 D) 467wing which divides 3333......3(300times) is 3. 5. Correct option: (B) The divisibility test of 6 states that a number is divisible by 6 if the number is even and the sum of the digits is divisible by 3. 54672 is divisible by 2 as unit's digit is 2, and It is also divisible by 3 because sum of digits = 5+4+6+7+2 =24 is a multiple of 3. Hence, it is divisible by 6. 6. Correct option: (A) The divisibility rule for 7 states that if a number is divisible by 7, then the result obtained by subtracting double the last digit from the remaining number should be 0 or divisible by 7. A) 343: • (3 x 2) = 6 and 34 – 6 = 28. • 28 is divisible by 7. • Hence 343 is divisible by 7. B) 3 7 = 2187. • 218 - (7 x 2) = 204 • Since 204 is not divisible by 7. • Hence 2187 is also not divisible by 7. C) 3 77 • Use Fermat's Little Theorem, which states that if p is a prime number and a is any integer not divisible by p, then a (p − 1) is congruent to 1 modulo p. In other words, a (p − 1) - 1 is divisible by p. • Since 7 is a prime number and 3 is not divisible by 7, we can use Fermat's Little Theorem to determine whether 3 6 - 1 is divisible by 7. We have: 3 6 - 1 = 729 - 1 = 728 • Since 728 is divisible by 7 (728 = 7 x 104), we can write: 3 6 - 1 = 7k (where k is an integer) • Multiplying both sides by 3 71 , we get: 3 77 - 3 71 = 7k x 3 71 • Since 3 6 - 1 is divisible by 7, we know that 3 6 is congruent to 1 modulo 7. Therefore, we have: 3 77 = (3 6 ) 12 x 3 5 • Using the fact that 3 6 is congruent to 1 modulo 7, we can simplify this expression as follows: 3 77 = (3 6 ) 12 x 3 5 = (1) 12 x 243 = 243 • Therefore, we have: 3 77 = 7 x 34 + 5

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