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udit pande
@upudit0
1,640

What is the maximum possible number of intersection points between three lines, five circles and seven triangles?

**OPTIONS**

1)430

2)433

3)436

4)333

5)336

how i approached when only lines =3c2 =3only circles=5c2*2 =20only triangles=7c2*6 =126line and circle =3c1*5c1*2=30line and triangle=3c1*7c1*2=42triangle and circle =7c1*5c1*6=2103+20+126+30+42+210=431

kahan galat kiya ??

- 5 Comments
- @Dexian 3 wala point maine bhi socha tha but kitna hoga 3.... 26 Apr '13.
- i think u have considered only wala cases and 2 waala cas.... 26 Apr '13.

Avishek Adhvaryu
@AvishekAdhvaryu
341

@Dexian 3 wala point maine bhi socha tha but kitna hoga 3 wala points?

2/3/5?

2/3/5?

Ashutosh Verma
@Dexian
3,866

i think u have considered only wala cases and 2 waala case... 3 wala cases bhi hoga jab all 3 meet.. IMO

udit pande
@upudit0
1,640

my cat 2012 question

(1+x+x^2)^10 =a0+a1x+a2x^2+.......+a20x^20

find

(a0+a2+a4........+a20 )/(a1+a3+a5+......+a19)

1)2^10+1/(2^10-1)

2) 2^10-1/(2^10+1)

3) 3^10+1/(3^10-1)

4) 3^10-1/(3^10+1)

- 7 Comments
- @anytomdickandhary Great solution Sir...... 24 Apr '13.
- put x=1 to get a0+a1+a2+......+a20 = (1+1+1)10 = 310.... 24 Apr '13.

put x=1 to get

a_{0}+a_{1}+a_{2}+......+a_{20} = (1+1+1)^{10} = 3^{10}........................ (i)

Put x = -1 to get

a_{0}-a_{1}+a_{2}-......-a_{19}+a_{20} = (1-1+1)^{10} = 1^{10} ...................... (ii)

Add (i) and (ii) to get

2*(a_{0}+a_{2}+a_{6}+......+a_{20}) = 3^{10} + 1.

Subtract (ii) from (i) to get

2*(a_{1}+a_{3}+a_{5}+......+a_{19}) = 3^{10} - 1.

Divide the results to get

a

Put x = -1 to get

a

Add (i) and (ii) to get

2*(a

Subtract (ii) from (i) to get

2*(a

Divide the results to get