p (x) satisfies the criteria

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

p (x) satisfies the criteria

by sanju09 » Sat May 08, 2010 6:12 am

A cubic polynomial p (x) satisfies the criteria that p (1) = 1, p (2) = 2, p (3) = 3 and p (4) = 5. Then the value of p (6) is
(A) 6
(B) 10
(C) 15
(D) 16
(E) 60

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Source: Beat The GMAT — Problem Solving | Sat May 08, 2010 6:12 am

shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Sat May 08, 2010 6:56 am

sanju09 wrote:A cubic polynomial p (x) satisfies the criteria that p (1) = 1, p (2) = 2, p (3) = 3 and p (4) = 5. Then the value of p (6) is
(A) 6
(B) 10
(C) 15
(D) 16
(E) 60

Let the polynomial be
ax3+bx2+cx+d
--> a+b+c+d = 1 .................................i
--> 8a+ 4b+ 2c+ d =2 ........................ii
--> 27a+ 9b +3c + d =3.......................iii
--> 64a + 16b + 4c+ d = 5......................iv

ii - i gives
7a + 3b + c =1......................v
iv-iii gives
37a + 7b + c =2.....................vi

from iii - ii gives
19a + 5b +c= 1.................vii
vi - v gives
30a + 4b = 1..................viii

vi-vii gives
18a + 2b =1 --> 36a+4b =2.........ix
ix-viii 6a =1--> a= 1/6

b = (1-18a)/2 = (1-3)/2 =-1

c= 1-3b-7a = 1-3(-1)-7/6 = 4-7/6 = 17/6

d=1-a-b-c = 1-1/6+1-17/6 = 2-18/6 = -1

so P(6)= 216a+ 36b+6c+d=216x(1/6) +36 (-1)+ 6x(17/6) -1 =[spoiler] 36 -36+17-1=16 Ans D[/spoiler]

Last edited by shashank.ism on Sat May 08, 2010 8:13 am, edited 1 time in total.

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Sat May 08, 2010 7:01 am

now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

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brijesh: Senior | Next Rank: 100 Posts; Posts: 84; Joined: Mon Apr 26, 2010 2:51 am; Thanked: 6 times; Followed by:1 members

by brijesh » Sat May 08, 2010 9:41 am

shashank.ism wrote:now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

I would like to know the final polynomial eq. P(x) which satifies the needed critera i.e. p (1) = 1, p (2) = 2, p (3) = 3 and p (4) = 5.

brijesh

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Sat May 08, 2010 6:47 pm

If we go by 2nd solution the final polynomial is
A(x-1)(x-2)(x-3) + x --> 1/6(x-1)(x-2)(x-3)+x

Also if we go by 1st soln. we get final polynomial as follows: Only difference is that it is in expanded form
1/6 X^3 - X^2 + 17/6 X -1

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ayushiiitm: Master | Next Rank: 500 Posts; Posts: 216; Joined: Wed Jul 23, 2008 2:35 am; Location: Pune, India; Thanked: 5 times; GMAT Score:700

by ayushiiitm » Sun May 09, 2010 2:24 am

Hey Shashank that seems to be a cool trick

But I am stuck at one place (may be I am doing wrong maths)

In the expression A(x-1)(x-2)(x-3)+X.........how do we get the X

Shouldn't it be A(x-1)(x-2)(x-3) + constant(k)?

anybody please reply soon

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ayushiiitm: Master | Next Rank: 500 Posts; Posts: 216; Joined: Wed Jul 23, 2008 2:35 am; Location: Pune, India; Thanked: 5 times; GMAT Score:700

by ayushiiitm » Thu Jul 15, 2010 8:17 am

shashank.ism wrote:now there is an alternate and easy solution which was given by my friend mausam
since P(1)=1, P(2)=2, P(3) =3
so we can easily write the polynomial as A(x-1)(x-2)(X-3) +X
now P(4) =5
A(3)(2)(1) + 4 = 5
--> A=1/6
so P(6) = 1/6x 5x4x3+ 6= 10+6 = 16 Ans D

This is a old post, but i had not got any replies.

Can anybody explain the method given above?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800