(a) If a^x = b , b^y = c , c^z = a , then xyz = 0
(b) If p = a^x, q =a^y, (p^y * q^x)^z = a^z, then xyz = 1.
(c) If x^a = y^b = z^c and ab+bc+ca=0, then xyz=1.
Find the statements which are true !!
For how many prime numbers P, P+10 and P+14 will be prime numbers??
@kingsleyx said:For how many prime numbers P, P+10 and P+14 will be prime numbers??
P = 3 satisfies
In general for P > 3, P = 6k + 1 / 6k - 1 ( k >=1)
P + 10 = 6k + 11/ 6k + 9 (not prime)
P + 14 = 6k + 15(not prime)/ 6k + 13
Clearly for P > 3 , either P + 10 or P + 14 is not prime, so only one prime i.e. P = 3 ?
Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.
select all options which satisfy it !!
The balance of a trader weighs 10% less than what it should be. Still the trader marks up his goods to get an overall profit of 20%. What is the markup on the cost price.
@ananthchief said:The balance of a trader weighs 10% less than what it should be. Still the trader marks up his goods to get an overall profit of 20%. What is the markup on the cost price.
33.33% ?
@kingsleyx said:Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
all of them?
@ananthchief said:The balance of a trader weighs 10% less than what it should be. Still the trader marks up his goods to get an overall profit of 20%. What is the markup on the cost price.
1/0.9 * x = 1.2
x = 1.2 * 0.9 = 1.08
So markup on CP = 8% of CP
x = 1.2 * 0.9 = 1.08
So markup on CP = 8% of CP
@ananthchief said:The balance of a trader weighs 10% less than what it should be. Still the trader marks up his goods to get an overall profit of 20%. What is the markup on the cost price.
profit via balnc= 100/900 *100 = (100/9)
now let markup =x
x+ (100/9) + (x/9) =20
solve 😃 we get x=8
@kingsleyx said:Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
2?
@kingsleyx said:Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
Is it 2, 3, 12. ??
A result :
(a -b)(a - c)(a - d)(b -c)(b - d)(c - d) is divisible by 3 and not by 5
provided all a,b,c,d are integers.
Take an example: a = 1 ; b = 2 ; c = 3 ; d = 4;
@kingsleyx said:Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
2 , 3 & 12..?
took nos. as 1,2 3 & 4
@kingsleyx said:(a) If a^x = b , b^y = c , c^z = a , then xyz = 0(b) If p = a^x, q =a^y, (p^y * q^x)^z = a^z, then xyz = 1.(c) If x^a = y^b = z^c and ab+bc+ca=0, then xyz=1.Find the statements which are true !!
(c)?
(a) b = a^x
c = a^xy
a = a^xyz
xyz = 1
FALSE
(b) (p^y * q^x) = a
a^xy * a^xy = a
a^(2xy) = a
xy = 1/2
z may or may not be 2
FALSE
(c) ab + bc + ca = 0
1/a + 1/b + 1/c = 0
x^a = y^b = z^c = k
xyz = k^(1/a + 1/b + 1/c) = k^0 = 1
TRUE
(a) b = a^x
c = a^xy
a = a^xyz
xyz = 1
FALSE
(b) (p^y * q^x) = a
a^xy * a^xy = a
a^(2xy) = a
xy = 1/2
z may or may not be 2
FALSE
(c) ab + bc + ca = 0
1/a + 1/b + 1/c = 0
x^a = y^b = z^c = k
xyz = k^(1/a + 1/b + 1/c) = k^0 = 1
TRUE
@Aizen said:Is it 2, 3, 12. ??A result :(a -b)(a - c)(a - d)(b -c)(b - d)(c - d) is divisible by 3 and not by 5provided all a,b,c,d are integers.Take an example: a = 1 ; b = 2 ; c = 3 ; d = 4;
Thats the answer !!
@grkkrg said:(c)?(a) b = a^xc = a^xya = a^xyzxyz = 1FALSE(b) (p^y * q^x) = aa^xy * a^xy = aa^(2xy) = axy = 1/2z may or may not be 2FALSE(c) ab + bc + ca = 01/a + 1/b + 1/c = 0x^a = y^b = z^c = kxyz = k^(1/a + 1/b + 1/c) = k^0 = 1TRUE
This is what I am getting too !! but the book says that b) and c) are true!! Hence the doubt !!
@kingsleyx said:(b) If p = a^x, q =a^y, (p^y * q^x)^z = a^z, then xyz = 1.
@kingsleyx said:This is what I am getting too !! but the book says that b) and c) are true!! Hence the doubt !!
Then I think the question has a typo.
It would probably be
(p^y * q^x)^z = a^2
This makes sense. :)
It would probably be
(p^y * q^x)^z = a^2
This makes sense. :)
What is the approximate area of a hexagon with side lengths (in order) 5cm, 2cm, 3cm, 6cm, 1cm and 4 cm and all interior angles equal to 120 degrees ?
@vijay_chandola said:What is the approximate area of a hexagon with side lengths (in order) 5cm, 2cm, 3cm, 6cm, 1cm and 4 cm and all interior angles equal to 120 degrees ?
65root(3)/4?
Make a hexagon(rough one). Take any 3 sides none of which are consecutive. Extend these sides and You will see that they make equilateral triangles.
The side of the bigger equilateral triangle(which includes hexagon and the 3 smaller equilateral triangles) = 10 (It depends which sides you extended. Ultimately final answer will be same)
The side of the bigger equilateral triangle(which includes hexagon and the 3 smaller equilateral triangles) = 10 (It depends which sides you extended. Ultimately final answer will be same)
Area = root(3) / 4 * (10)^2= 100root(3)/4
Sum of areas of smaller triangles = 35 root(3)/4
Hence required area = (100-35)root(3)/4 = 65root(3)/4
I am in office so making a diagram would be very uncomfortable I am sorry for that...
Attached herewith is a horrible representation of my words..Inner one is a hexagon(somehow) Please mark the sides yourself
P.S. Please don't ponder too much over my drawing skills. (I didn't take mechanical//civil for this reason)