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11thKVS Maths Olympiad contest 2008

Solutions

Q.1.Find the value of S = 12 22+ 32 42 + .-982+ 992

Solution :Let

S = 12 22- 32 42+ - 982+ 992

= 12-22+32! -42+"2! #2+$2! .-

982

+ 992

= 12+ 3+2! 3-2! + 4+"! "-4!

+ # + $! $ - #! + + 99 - 98! 99 + 98!

= 1+ 2 + 3 + 4 + " + # +$ + + 98 + 99

Fu%the%&

S = 1 + 2 + 3 + 4 + " + # + $ + + 98+ 99

+S = 99+ 98 + 9$ + 9# + 9" + 94+ 93 + + 2 + 1

2S= 100+100+100 +100+100 + 100+ 100 + 100 +100

= 99 ' 100

S = "0 ' 99

= 49"0

Q.2. Find the s(allest (ulti)le of *1" su,h that ea,h diit of the (ulti)le is eithe% *0

o% *8.

Solution:

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S(allest (ulti)le of 1"& su,h that ea,h diit of the (ulti)le in eithe% 0 o% 8 a%e

/o h%ee diit nos Fou% diit and Five diit nos

80

880

808

800

8000

8008

8080

8800

8880

80888

80888

88088

So onl )ossiilit fo% (ulti)le of 1" i.e. divisile "

is last diit is 0 i.e.

i! 2 diits 80

ii! 3 diits 880& 800

iii! 4 diit s 8000& 8800& 8880& 8080

iv! " diit 88880 80000

88800 88000

88080

s 1" = "'3

So the nu(e% should e divisile 3 the su( of diit should e divisile 3.

en,e let us anal5e the su( of diits in i!& ii!& iii! and iv!&

i! 2 diit 6 not )ossile

ii! 3 diit 6 not )ossile

iii! 4 diit 6 /ith 888

su( in 8+8+8 = 24

that is divisile 3

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7ut last diit should e 0 and it should ,ontain th%ee nu(e%s of 8.

i. e.

Q. 3t the end of ea% 2002. a( /as half as old as his %andfathe%. he su( of

ea%s in /hi,h the /e%e o%n is 38"4. hat is the ae of a( at the end of ea%

2003:

Let ae of a( at the end of 2002 =x

so ae of his %and fathe% = 2x

so& a( /as o%n in 2002 -x!

a(s %and fathe% /as o%n in 2002-2x!

f%o( ;uestion 6 2002 x+ 2002 2x= 38"4

-3x= - 1"0

x = "0

So& the ae of a( at the end of 2003 in 1.

Q. !Find the a%ea of the la%est s;ua%e& /hi,h ,an e ins,%ied in a %iht anle

t%ianle /ith les *4 and *8 units.

Solution:

Let the side of s;ua%e is ' as in fiu%e.

F%o( )%o)e%t of si(ila% t%ianle

8880

8"#

!

8

!"#

#

#

#

#

4-' ' ' 8-'

=

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h

4-'! 8-'! = '2

32 4' - 8' + '2= '2

32 = 12' ' = = 2.#$

so %ea of la%est s;ua%e in s; units

= s; units.

Q.. = 4 unit

7> = '

> = 8 '

Let ?F = hand

>F = t

So %ea 7> = @ 4 @x = 2x

83

#4 9 1 9

$

12

7

?

F>

' t 8'-t

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%ea 7 = @ 9x @ 4 = 18'

Let ?F ise,t the a%ea of 7& and ?FA>

So&

a%ea ?F = 9x

= @8'-t! @ h

h= .1!

he a%ea of >F?

= @>+?F! @ t

= @ > + ?F! @ t

= 9x 2x! Bi.e. half a%ea a%ea 7> of t%ianleC

= $x

$'= @ 4+h! @ t

t = ....2!

)uttin t fo%( ii! in i!

18' = 8' - t!

= 8' h

18 = 8' h

12

12

18x8x- t

1212

12

14xh+4

14xh+4

14xh+4

8 h = 4! 14

h + 4

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18 = h

18 h + 4! = B8 h+4! - 14C h

18 h + $2 = B8h + 32 - 14C h

= B8 h + 18C h

= 8 h2+ 18 h

8 h2= $2

h2=

= 9

h = 3 so& heiht = 3 units ns.

Q.$ nu(e% *@ leaves the sa(e %e(ainde% /hile dividin "814& "430& "9"8.

hat is the la%est )ossile value of *@:

Solution:

Let )& ;& % and s e an nu(e% f%o( the ;uestion& if % in %e(ainde%.

"814 = ) @ + % . i!

"430 = ; @ + % . ii!"9"8 = s @ + % .. iii!

f%o( i! ii!

384 = )-;! @

f%o( ii! iii!

"430 "9"8 = ; - s! @

$28

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"28 = s - ;! @

f%o( iii! i!

"814 "9"8 = ) - s! @

144 = s - )! @

so /e et th%ee e;uation

384 = ) - ;! @

"28 = s - ;! @

144 = s - )! @

) - ;! @ = 2'2'2'2'2'2'2'3

s - ;! @ = 2'2'2'2' '3' 11

s - ;! @ = 2'2'2'2' '3' '3

So the F of these th%ee nu(e%s

= 2'2'2'2'3= 48

So the %e;ui%ed la%est nu(e% is 48

he,D6

48 ' 121 = "808 then + # = "814

48 ' 113 = "424 then + # = "430

48 ' 124 = "9"2 than + # = "9"8

Q.%. s)o%ts (eet /as o%ani5ed fo% fou% das. En ea,h da& half of e'istin total

(edals and one (o%e (edal /as a/a%ded. Find the nu(e% of (edals a/a%ded on

ea,h da.

Solution:

Let total (edals = (

edals dist%iuted on( ( + 2

2 2+ 1 =

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1stda =

e(ainin (edals fo% 2ndda

= ( B(edals dist%iuted on 1stdaC

= ( -

=

So& (edals dist%iuted on 2ndda

=

=

e(ainin (edals fo% 3%dda

= ( B(edals dist%iuted o% 1st+ 2nddaC

= (

=

So& (edals dist%iuted on 3%dda

( - 2

2

( - 2

2

( + 2

4+1

( + 2

4

( + 2 ( + 2

2 4+

( - #

4

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=

=

e(ainin (edals fo% 4thda

= ( B(edals dist%iuted on 1st+ 2nd+ 3%ddaC

= (

=

So& (edals dist%iuted on 4thda

=

=

So Final %e(ainin (edals afte% dist%iution on 4th

da= ( B(edals dist%iuted on 1st+2nd+ 3%d+ 4thdaC

++

++

++

+=

1#

2

8

2

4

2

2

2 mmmmm

1#

301" +

=

m

m

7ut s)o%t end hen,e (ethods %e(ained = 0

( - #

8+ 1

( + 2 8

( + 14

8

( - 14

1#+ 1

( + 2

1#

( + 2

2 ( + 2 4

( + 2

8+ +

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1# ( = 1"( + 30

( = 30

So da /ise (edal dist%iution

>a 1 = 1#

>a 2 = 8

>a 3 = 4

>a 4 = 2

1! otal = 30

Q.8. Let G7 e isos,eles /ith 7 = 7 = $80. Let > and ? e the )oints

on sides 7 and %es)e,tivel su,h that 7> = 240and 7? = "10. Find

the anle 7?> and Hustif ou% %esult.

Solution:

.

F%o( ;uestion 7 =

So 7> = 180 24+$8!

= $80

and >7 = $80

So 7 = >Fu%the%& 7? = 180 $8+"1!

7

>

$8 "1 24 $8

24

$8

"1

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= "1

So 7 = ? = >

Let >?7 =x&

So in isos,eles ?>&

>? = ?>

= "1+x

Let 7? and > (eet at F

So >F? = 10"0

en,e in G >F?

"1+x+ 10" +x= 1800

2' = 24

' = 120

Iustifi,ation

> ? = "1+12

= #3

en,e ?> = 180 $8+#3!

= 39

en,e & ?> = 180 24+39!

= 11$0

so 11$ + 12 + "1 = 1800

en,e Hustified