I.Q. test 45
By jinuxnet
@jinuxnet (675)
Philippines
February 28, 2007 1:28am CST
45. Alice has seven times the amount of pens that Maurice has. Paul has two-thirds of the amount of pens as Alice and Suzy have combined. Dawn has a dozen more pens than Paul. Suzy has half the pens that Maurice has. If Suzy has 2 pens, how many does Dawn have?
2
20
28
32
40
3 responses
@josephperera (2906)
• Sri Lanka
28 Feb 07
Suzy has 2 pens. Suzy has half the pens that Maurice has.
Therefore Maurice has only 1 pen.
Alice has seven times the amount of pens that Maurice has.
Therefore Alice has 7 pens.
Alice and Susan together has 9 pens (7 + 2).
Paul has two-thirds of the amount of pens as Alice and Suzy have combined.
There for Paul has two-thirds of 9 pens which is 6.
Dawn has a dozen more pens than Paul.
Therefor Dawn has 18 Pens.
But the answer is not included in your discussion.
@josephperera (2906)
• Sri Lanka
28 Feb 07
Sorry! Made a big mistake with Maurice. He has 4 pens. Alice has 28 pens. Alice and Susan have 30 pens. Paul has 20 pens. Ultimately Dawn has 32 pens.
I messed up with poor Maurice. No harm done.
1 person likes this
@Modestah (11179)
• United States
28 Feb 07
Suzy =2 so Maurice =4 Alice =7x4= 28 Paul= 28+2 x .66 = aprx 20
Dawn= 20+12
I am not sure if I figured it out right for paul, if I did then the answer is aprx 32 (maybe one of the pens is a little shorter than the others? heh)
Thanks for these postings, make my old brain shake the dust a little!
1 person likes this
@mdchennai (2129)
• India
28 Feb 07
EQUATIONS:
A=7M, P=(2/3)(A+S), D=12+P, S=M/2
GIVEN:
S=2
TO FIND: D=?
SOLUTION:
IF S=2(given) then
2=M/2 = M=4
as we know A=7M, substituting the value of M we get A=28.
Substituting the value of A and S in the following equation:
P=(2/3)(28+2)
= P=20
Finally substituting the value of P we get D as
D=12+20
D=32
