Download New Questions for Junior High Number Sense (2011 and 2012)

New Questions for Junior High Number Sense
(2011 and 2012)
This document contains information on some of the new tricks that
will be appearing on the 2011 District and State Number Sense tests.
Some of these tricks are not new and students familiar with the Texas
Math and Science Coaches Association (TMSCA) contests will recognize many of the following topics.
The new tricks are sectioned in the order that they will appear on the test. Some of the
new tricks are given implicitly. For the others, the student is encouraged to search for an
easy mental math formula, procedure for working the problem, or information on the topic.
Multiplication of Two Numbers Whose Difference is Two [#20-40]
We can express two numbers whose difference is 2 by referencing the number in the middle
as N . The two numbers then become N − 1 and N + 1. From algebra, we know that
(N − 1) × (N + 1) = N 2 − 1. This is a special version of the Difference of Two Squares
formula. The shortcut to computing (N − 1) × (N + 1) is to square the number in the
“middle” and subtract 1.
For example, to find the product 21 × 23, the middle number is 22. We know that
222 = 484. Subtracting 1 gives 483.
Multiplication by 333 [#40-60]
I will leave the shortcut for you to discover. The trick you are searching for will involve
multiplying 333 by a 2- or 3-digit number that is a multiple of 3. Good luck!
Rhombuses [#40-60]
A rhombus is a parallelogram with four equal sides. As
D
C
such, the perimeter P of a rhombus is P = 4s, where s is
the side length. The area A of a rhombus can be found by
taking half of the product of its diagonals: A = 12 d1 d2 .
Because the rhombus is a parallelogram, adjacent angles are supplementary: the sum of the angles must be
A
B
180 degrees. If the angles are A and B, then we know
that A + B = 180◦ .
Finally, we can find the length of side s of the rhombus given its two diagonals as
√
s = 21 d21 + d22 . I will leave it to you to write out a proof for this formula.
Counting Passwords [#60-80]
When counting re-arrangements of passwords, we must take into account when letters repeat.
PSIA – 2009 – New Questions for Junior High Number Sense — Page 2
For example, how many distinct 3-letter passwords can be formed by the letter AAB? Since
there are 3 letters, there are 3! total re-arrangements. They are AAB, AAB, ABA, ABA,
BAA, BAA. Notice how since there are two A’s, you cannot distinguish between words where
just the A’s have been re-arranged. Since there are 2! ways to re-arrange just the A’s, we
must divide by this many. In total, there are 3!/2! = 3 ways to re-arrange AAB.
Similarly, how many distinct 6-letter passwords can be formed by the letters BANANA?
There are 6 letters in BANANA, so start with 6!. Since there are 2 N’s and 3 A’s, divide
6!
= 60.
out by (2! · 3!). This gives a total of
2! · 3!
Practice Questions – The following practice questions cover the above examples and
should be used to guide your inquiries into the new types of questions to be asked on the
number sense tests.
1. 16 × 18 =
. 13. The adjacent angles of a rhombus
measure 85◦ and
◦
.
2. 13 × 15 =
.
3. 29 × 27 =
.
4. 34 × 32 =
.
5. 51 × 53 =
. 15. How many distinct 4-letter passwords can
6. 333 × 36 =
.
7. 333 × 48 =
.
8. 51 × 333 =
.
9. 132 × 333 =
.
10. 258 × 333 =
.
11. The perimeter of a rhombus whose sides
equal 8 cm is
cm.
12. The area of a rhombus whose diagonals
are 7 inches and 12 inches is
2
in .
14. The diagonals of a rhombus are 6 mm and
8 mm. Find the length of each side.
mm.
be created using all the letters of PAPA?
.
16. How many distinct 5-letter passwords
can be created using all the letters of
YUMMY?
.
17. How many distinct 6-letter passwords
can be created using all the letters of
POSTER?
.
18. How many distinct 8-letter passwords can
be created using all the letters of
ABCAABCA?
.
This document was prepared by Doug Ray for PSIA, 2010-2012. If you have any questions about
the material presented, please email [email protected].