Hey Junta!!! Good Morning!!! Hope u are all doing GREAT...am starting this section as a common thread for my queries in MATH...all others are invited to solve..to help around..to post thier own questions.... Hey Mods..if u think..this ma...

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A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

my solution- total vertices = 12*4=8*6=6*8=48

no. of edges=1/2(12*4 + 8*6 + 6*=72

no. of lines joining vertices through faces(excluding edges)=

=216

so pts lying in the interior of the polyhedron are 48c2-216-72=840 answer

DISCLAIMER-answer could be wrong as i m still a novice

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

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@chiya
309

Guys

what is 1 + 3/2 + 6/4 + 10/8 + 15/16 + .........infinite terms..

hey Vijay, lemme post the solution......

s=1+3/2+6/4+10/8+15/16.....

s/2= 1/2+3/4+6/8 +10/16....

(s-s/2=s/2)= 1+2/2+3/4+4/8+5/16.....

s/4= 1/2+2/4+3/8+4/16.....

subtracting again...

s/4=1+1/2+1/4+1/8+1/16

on the rhs we have a G.P.

s/4=2.

s=8.

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@vijay317
164

a_tired_soul Saysis the answer 8

Yep..was expecting u to answer it. Method please...

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Guys

what is 1 + 3/2 + 6/4 + 10/8 + 15/16 + .........infinite terms..

is the answer 8

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Guys

what is 1 + 3/2 + 6/4 + 10/8 + 15/16 + .........infinite terms..

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adding one to both sides of the given equation we get

(a+1)(b+1)=3*5^2*7; (b+1)(c+1)=3*7^2; (c+1)(d+1)=3*5*7

since the last 1st & 3rd eqns have one 7 each & middle eqn has 2 7s there

fore each c+1 & b+1 have one factor of seven in them.so c+1=21 or 7

if c+1=21 =>c=20=4*5 & d=4 b=6 a=74(not possible) as 8! does not have factor of 37.

so c+1=7=>c=6; b=20=4*5;d=14=2*7;a=24=8*3

clearly this satisfies all the condition. hence a-d=24-14=10

DISCLAIMER-accept the above solution at your own peril. i m still a novice.

Thanks,seems perfect..well done!

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Solve:

For positive integers a,b,c and d, have a product of 8! and satisfy

ab+a+b=524,

bc+b+c=146, and

cd+c+d=104

What is a-d?

adding one to both sides of the given equation we get

(a+1)(b+1)=3*5^2*7; (b+1)(c+1)=3*7^2; (c+1)(d+1)=3*5*7

since the last 1st & 3rd eqns have one 7 each & middle eqn has 2 7s there

fore each c+1 & b+1 have one factor of seven in them.so c+1=21 or 7

if c+1=21 =>c=20=4*5 & d=4 b=6 a=74(not possible) as 8! does not have factor of 37.

so c+1=7=>c=6; b=20=4*5;d=14=2*7;a=24=8*3

clearly this satisfies all the condition. hence a-d=24-14=10

DISCLAIMER-accept the above solution at your own peril. i m still a novice.

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For positive integers a,b,c and d, have a product of 8! and satisfy

ab+a+b=524,

bc+b+c=146, and

cd+c+d=104

What is a-d?

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Ths s an old question posted in the first page of the thread...

I did not check the answers by other puys...

Anyways..giving my logic..

Q-How many numbers are their between 1 n 10000 that r not divisible either by 2 or 5?

Soln.

consider from 1-10

one and 10000 are excluded(BETWEEN)..

so 2,4,6,8,10 and 5 are divisible either by 2 or 5.

Hence in 1-10, 6 numbers dnt come under the category.

3 numbers come under the category.

for 11-20 and for all..

4 numbers come under this category.

divide 10000/10=1000

1-10 have 3 nos.

rest all have 4 nos.

Implies 999*4=3996+3=3999 Exclude 10000..

Ans:3999

F course there r formulas for ths sort of Q(i am too bad in formulae)...but this simple logic takes a few sec..and No pen!

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