NMTC Paper_Clas s IX And X_2014 (Stage 2)
JUNIOR_2014 1.
(a) p1, p2, ..............................p 2014 be an arbitrary arbitrary rearrangements of 1,2,3 .........n . Show that
1
1
+
p1 p2
p3
p2
1
+....................+
>
p2013 p2014
2013 2016
(a) We have t ofind the minimum value of 1 P1
1
+
P2
P2
P3
+......+
1 P2013
P2014
We have to make all term equal and denominator denominator should should be max so above exp in greater than 1 1 1 1 + + +......+ 1 2014 2 2013 3 2012 2 2013 1
= =
2015 2013 2015
1
+
2015
2013
and
1 P1 P2
+
2015
>
2015
+
1
1 P2 P3
+.....+
1 2015
2013 2016
+......+
1 P2013 P2014
>
2013 2016
n 1 +
n 1 is rational.
1.
(b) Fin Find pos posiitiv tive in intege tegerrs 'n 'n' su such that that
Sol. Sol.
Show Show that that the there re is no posit positiv ive e int intege egerr n for which which n 1 + If possible, let there be a positive integer n for which
n 1 is rational a n 1 is rational equal to b (say),
n 1 +
where a, b are positive integers. Then, a = b
n 1 + a = b
b a
=
..(1)
n 1
1 n 1 n 1 n1 n1 { n 1 n 1}{ n 1 n 1}
=
n
1
n 1
(n 1) (n 1)
=
n 1 n 1 2
2b = n 1 – n 1 a Adding (1) and (2) and subtracting subtracting (2) from (1), we get
..(2)
a 2b a 2b 2 n 1 = + and 2 n 1 = – b a b a
a2
n 1 =
n 1 and
2b 2
2ab
and
n 1 =
are rational n 1 ar
a2
2b 2
2ab ( a,b are integers
a2
2b 2
2ab
and
(n
+ 1 are rational ) (n + 1) and (n – 1) are perfect squares if poisitive integers This is not possible as any two perfect square differe at least by 3. Hence, there is no positive integer n for which ( n 1 +
n 1 ) is rational.
a2
2b 2
2ab
are rational )
2.
ABCD is is a quadr quadrililate ateral ral inscr inscrib ibed ed in a circl circle e of centr centre e O. Let BD BD bise bisect ct OC perpe perpend ndicu icula larl rly y. P is is a point on the diagonal AC such that PC = OC. BP cuuts AD at E and the circle ABCD at F. Prove that PF is the geometric mean of EF and BF.
Sol. Since
PC OC BC and CP
is similar to
AFP .
Next CB = CD = CP implies P is the incenter of ABD yielding yielding FAD FAD = ADF call it
.
We have PF = AF. AF AF..
ABD .
Then BF bisects
Alternatively we have FAD = PBD = FAD PB P BD
1
2 PCD
. Then
AFD 180º ACD 180º PCD
18 0º 2 PBD 18 0º 2
Hence, to
ADF ,
BFA .
So
Also we see AFE = BFA and
AF / EF BF / AF PF AF
3.
EAF ADF AB ABF
which implev
AFE is similar
. Then
EF BF BF
(a) The The Fi Fibon bonacci acci sequ equence ence is def defiined by F0 = 1, F1 = 1, Fn = Fn–1 + Fn–2 for natural numbers n 2.
3
3
3
Show that 7 F n 2 F n F n 1 is divisible by F n+3 Sol.
7F3n+2 – Fn3 – F3n+1 7 (Fn+3 – Fn+1)3 – Fn3 – F3n+1 7 (F3n+3 – F3n+1 – 3Fn+3Fn+1 (Fn+3 – Fn+1) – F3n – F3n+1 F3n+3 and 3Fn+3Fn+1(Fn+3 –Fn+1) are divisible by Fn+3 so we have the check only –7 F3n+1 – F3n – F3n+1 – 8 F3n+1 – F3n – [(2Fn+1)3 + F3n] It have factor (2Fn+1 + Fn) = Fn+1 + F n+1 + F n = Fn+1 + Fn+2 = Fn+3 so whole expression is divisible by Fn+3. (a)
x
4
4
2
(b) (b) If x, x, y, y, z are are eac each h grea greate terr th than 1, Show Show that that
Sol.
some correction z2 should replaced by z4
4.
ABCD is a square square E anf anf F are are point points s on BC and and CD res respect pectiv ivel ely y such such that that AE cuts cuts the the diago diagona nall BD at G and FG is perpendicular to AE. K is a point on FG such that AK = EF. Find the measure of the angle EKF EK F. A
a
45 45
B
G 1 2
E
8
7
Sol.. 4 45° D
6 F
5 C
( z 1) 2
+
z
3.
( y 1) 2
+
y
( x 1) 2
48.
By SAS ABG CBG AG = CG (CPCT) 4 = a (text of AGB) 1 = 45 + a 1 + 2 = 90 2 = 90 – (45 + a) = 45 – a GFC = 2 + 45° = 90 – a 5 = 90 – 4 = 90 – a In FAC 5 = 6 = 90 – a FG = GC By SAS AGK FGE GK = GE 7 = 8 = 45° EKF = 180 – 45 = 135 5. (a) Sol. Sol.
5.(b)
If none of a, b, c, x, y, z, is zero and
a
3
=
prove that a3 + b3 + c3 + abc = 0 Giv Given equa equati tion on may may be be wri written tten as, as, x2 (y + z) = a3 ..(1) 2 3 y (z + x) = b ..(2) z2 (x + y) = c3 ..(3) xyz = abc From (1), (2), (3), x2y2z2(y + z)(z + x) (x + y) = a3b3c3 a2b2c2(y + z) (z + x)(x + y) = a 3b3c3 or abc = (y + z) (z + x) (x + y) = (yz + yx + z 2 + zx) (x + y) = xyz + x2y + xz2 + x2z + y2z + y2x + yz2 + xyz = 2xyz + x2(y + z) + y2(z + x) + z2(x + y) = 2abc + a3 + b3 + c3 Hence a3 + b3 + c3 + abc = 0 Solv Solve e the the equ equati ations ons x y
Sol.
x 2 (y z)
x y
+
+
Now
y z y z
x y
z
+
x z
+
x
=
= y
+
+
y x y x z
+
+
=
z y z y y
+
+
+
x z x z
z
= x + y + z = 3
= x + y + z = 3
+
x
y z x x z 2 2 2 or x z + y x + z y = y z + z x + x2y or x2(y – z) + y 2(z – x) + z 2(x – y) = 0 or (x – y) (y – z) (z – x) = 0 x = y = z 2
Now
2
x
y
x y
+ 1 +
y x
+
z
= 3 z x Putting Putting y = z, we get, y
+
= 3
y 2 (z x ) b
3
=
z 2 (x y) c
3
=
xyz abc
= 1 ,
or
x2
y2
yx
= 2
or x2 – 2xy + y2 = 0 Again x + y + z = 3 x+x+x=3 thus x = y = z = 1 7.
Sol.
and
or (x – y)2 = 0
A mechan mechantt bough boughtt a quan quanti tity ty of of cotton cotton : this this he exc exchan hanged ged for oil oil he he sold sold.. He obs observ erved ed that that the the number of kg of cotton, the number of litres of oil obtained for each kg and the number of rupees for which he sold formed a decreasing geometric progression. He calculated that if he had obtained 1 kg more of cotton, one litre more of oil for each kg and Rs. 1 more each litre, he would have obtained obtained Rs. 10169 more, whereas if he had obtained one kg less of cotton and one litre less of oil for each kg and Rs. 1 less for each litre, he would obtained Rs. 9673 less. How much he actually received..? let the the merchan merchantt bought bought x kg of cotton and change changed d each kg for y litr litres es of oil, oil, and sold sold each each llitr llitre e for z rs, then he obtained xyz rupees in the first case ATQ, we have (x + 1)( 1)(y y + 1)( 1)(z z + 1) = xyz xyz + 101 10169 69 ..... ....... .... .... .... .... .... .... ...( .(1) 1) and by the 2nd case c ase ATQ, ATQ, we have (x + 1) 1)(y – 1)( 1)(z – 1) 1) = xyz – 9673 9673 .... .......... ...... .... .... .... .... ......( ..(2) also x, y, z form a G.P so y 2 = xz now from eq e q (1) and (2) we have (yz + xy + zx) + (x + y + z) = 10168 ................. (3) (yz + xy + zx) – (x + y + z) = 9672 ................. (4) on adding and subtracting equation no. (3) and (4) yz + xy + zx = 9920 x+y+z = 248 now yz + xy xy + zx = 9920 yz + xy + y2 = 9920 (y2 = xz) y(z + x + y) = 9920 y=
9920 248
= 40 zx = 1600
z + x = 248 – 40 = 208 x = 200 , z = 8 thus he actually received = xyz = Rs. 64000
8.
There There are thr three ee town towns s A, B and and C a person personwal walkin king g from from A to B , dri drivi ving ng from from B to C and and rid ridin ing g from from C to A makes the journey journey in in 15
1 2
hours, By driving driving from A to B, riding riding from B to C and walking from C to
A, he could could make the the journey in 12 hours. hours. On foot he could could make the journey in 22 hours, hours, on horse horse back in 8 Sol.
1 4
hours and driving in 11 hours . To walk 1 KM, ride 1 KM and drive 1 KM , he takes
altogether half an hour, Find the rates at which he travels and the distance between the towns. let total total distanc distance e be x km and and sspeed sspeed of of walking, walking, riding riding & driv drivin ing g be w, w, r d respecti respectivel vely y.. ATQ ATQ..
22w =
and
2w =
1 r
So w =
1
w x
22
33 4
3 4
r =d
1
d =
r = 11d 11d = x
1 2
165 2 22
33
=
4 x
15 4
,r=
22 x
4 x 33
11 x
1 2
= 10 and and d =
33 88 44 4 x x
11
=
15 2
1 2
165 4 x
1 2
x =
165 2
km