NMTC Paper_Clas s VII And VIII_2014 VIII_2014 stage 2 SUB_Junior_2014
1.
A hare hare purs pursue ued d by a grey grey hou hound nd,, is is 50 of her her own own leaps leaps ahead ahead of of him him . Whil While e the the hare hare takes takes 4 leap leapss 3
the grey hound takes 3 leaps. In one leap the hare goes 1 Sol. Sol.
4
3
metre and the grey hound 2
4
metre. In
how many leaps will the grey hound overtake the hare ? Let Greyh Greyhoun ound d requ requir ired ed x leaps leaps to catch catch the the Hare Hare
Distance covered by Greyhound = x 2
3 m 4
In 3 leap of Greyhound, Hare takes 4 leap
In 1 leap of Greyhound, Hare takes In x leap of Greyhound, Hare takes
4 leap 3
4 x leap 3
4x 11x 175 7 = + 3 4 2 4
11x 7 x 175 – = 3 4 2 2.
If
(a x ) +
33 x 28 x 175 = 12 2
(b x) +
5x 175 = 12 2
175
12
x = 2 5 =
210
(c x) = 0
show that (a + b + c + 3x) (a + b + c – x) = 4 (bc + ca + ab) Ans.
Given (a x ) + we have
(b x) + (c x) = 0
(a x ) +
(b x ) = –
(c x )
squaring both sides, we get, a – x + b – x + 2 {(a x )(b x )} = c – x or Again on squaring, (a + b – c – x)2 = 4(a – x) (b – x) 2 or a + b2 + c2 + x2 + 2ab – 2ac – 2ax – 2bc – 2bx + 2cx = 4(ab – ax – bx + x2) or a2 + b2 + c 2 + x2 + 2ab – 2ac + 2ax – 2bc + 2bx + 2cx 2c x = 0 2 2 or (a + b + c) + 2(a + b + c)x – 3x = 4 (ab + ac + bc) or (a + b + c – x) (a + b + c + 3x) + 4(ab + bc + ac) 3.
some some amou amount nt of work work has has to to be comple completed ted.. Anand Anand,, Bilal Bilal and and Charl Charles es offer offered ed to to do the the job. Anand Anand would alone take ‘a’ times as many days as Bilal and Charles working together. Bilal would alone take b times as many days as Anand and Charles together. Charles would alone take c times as many a b c + + = 2 a1 b 1 c 1 Anand, Anand, Bilal Bilal and Charles Charles complete complete the the work work in x, y, z days days respec respectiv tively ely.. ATQ ATQ 1 1 1 zy x = a 1 1 a = x a = x y z yz y z
days as Anand and Bilal togheter. Show that
Sol.
xz xy yz xz xy a = xz xy yz = xz xy yz a1 yz
..(1)
in the same way yz yx b = xz xy yz b 1 Add (1) + (2) + (3)
zx zy c = ..(3) xz xy yz c 1
..(2)
a b c + + = 2 a1 b 1 c 1 Let P be any any poi point nt on on the the diago diagoan anll BD of of a rectan rectangl gle e ABCD. ABCD. F is the the foot foot of the the perp perpen endi dicu cular lar from P to BC. H is a point on the side BC such that FB = Fh. PC cuuts AH in Q. show that Area of APQ = Area CHQ
we get
4.
C
D
a-2x
y
O
H
Q
Sol..
a F
p y
x B
A b
In BCD PF || DC By BPT
y
x = a
2
a b
2
ya a 2 b 2 = x
..(1)
In BCO BH 2x = HC a 2x BP = PO
..(2) y
a 2 b2 y 2
BP y = = 2 x PO ya a 2x y 2x From equation (2) and (3)
..(3)
BH BP = HC PO By conv. conv. of BPT PH || OC or PH || AC area APH = area PCH (two having same base and between same parallel pa rallel have equal area) subtract area of PHQ from both side area APQ = area CHQ 5.
Sol.
A number number of three digits in scale 7 when expressed expressed in scale 9 has has its its digits digits reversed reversed in order. order. Find the number in scale 7 and scale 10. a,b,c < 7 ATQ (abc)7 = (cba)9 49a + 7b + c = 81c + 9b + a b = 8(3a – 5c) as b < 7 3a – 5c = 0 3a = 5c a = 5, c = 3 so number in scale 7 and 9 are (503)7, (305)7 and in scale 10 is (248)10
6.
Sol.
a) Two Two regu regula larr polyg polygon ons s have have the the numm nummber ber of of thei theirr sides sides as 3 : 2 and and the the int interi erior or angl angles es as 10 : 9. find find the number of sides of the polygon.
(a)
(3 x 2)180 3x 10 ( 2x 2)180 = 9 2x
number of sides 3x = 3 4 = 12 6. Sol. Sol.
7.
3x 2
5
2x 2 = 3 = x = 4 and 4x = 4 4 = 16
b) Fin Find d two natu natural ral number numbers s such such tha thatt their their differ differen ence, ce, sum sum and and the the produ product ct is to one one anoth another er as 1, 1, 7 and 24. (b) (b) (x (x – y) : (x + y) y) : xy = 1 : 7 : 24 x–y=z ... (1) x + y = 7z .. (2) xy = 24z .. (3) By solving (1) & (2) x= 4z, y = 3z Put in (3) (4z) (3z) = 24z z=2 x = 8, y = 6 Find Find the the val value ue of a, a, b, b, c whi which ch will will make make each each of the the expr expres essi sion onss x4 + ax3 + bx2 + cx + 1 and x4 + 2ax3 2 + 2bx + 2cx + 1 a perfect square. 2
Ans.
2 ax 1 Suppose x4 + ax3 + bx2 + cx + 1 x 2 4 3 2 2 and x + 2ax + 2bx + 2cx + 1 = (x + ax + 1)2 Equating Equating like coefficien coeff icients, ts, from (1), (2), we get a2 + 2 and c = a 4 2b = a2 + 2, 2c = 2a b=
a 2 a2 2 2 + 4 a + 2 = 2 4 = 2 or 2a2 + 4 = a 2 + 8 or a 2 = 4 a = 2, c = 2, b = 3
..(1) ..(2)