1. The LCM of two no. 14 times their HCF. The sum of LCM & HCF is 600. If one no
is 280, the find the other no.
2. Show that one and only one out of n, n+2, n+4 is divisible by 3. Where n is any
positive integer.
3. Prove that v2 is an irrational number.
4. Explain, Why (7 x 11 x 13) + 13 and (7 X 6 X 5 X 4 X3 X2 X 1) +5 are
composite no.?
5. Can the no.61, n being no., end with the digit 5? Give reason
6. Prove that -3V2 is an irrational
7. Find the zeroes of the quadric polynomial V3x2 - 8x +4 03
8. Form a quadratic polynominal P(x) with 3 and as sum and product of its
zeros, respectively.
9. If x3 + 8x2 + Kx + 18 is completely divisible by x2 + 6x +9, then find the value
K.
10. Solve using cross - multiplication method 2x = 5y +4; 3x - 2y +16 =0
11. Find the two no. whose sum is 75 and difference is 15.
12. Solve for x and y: x+ 4y =27xy, x+2y =21xy.
13. Prove 3-5 is irrational.
14. The sums of digits of a two digits no. is 11. The number obtained by
interchanging the digits of the given no. exceeds that no. by 63 find the no.
Answers
Answer:
1: second number is 14×280/600= 196/30= 6.53
Answer:
Euclid's division Lemma any natural number can be written as: .
where r = 0, 1, 2,. and q is the quotient.
thus any number is in the form of 3q , 3q+1 or 3q+2.
case I: if n =3q
n = 3q = 3(q) is divisible by 3,
n + 2 = 3q + 2 is not divisible by 3.
n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.
case II: if n =3q + 1
n = 3q + 1 is not divisible by 3.
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.
case III: if n = 3q + 2
n =3q + 2 is not divisible by 3.
n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
thus one and only one out of n , n+2, n+4 is divisible by 3.
Hence, it is solved
THANKS