GMAT Official Guide Quantitative Review 2019_ Book

GMAT® Official Guide 2019 Quantitative Review

For !1RST, the length of any side is less than

the sum of the lengths of the other two sides as

shown below.

RS=4<13= 8 + 5 = TR+ TS

RT=8<9= 5 + 4 =ST+ SR

ST = 5 < 12 = 8 + 4 = TR+ RS

2 R

s

p^10 T

Note: Not drawn to scale.

To show that 10 can be the length of PT,

consider the figure above. For !1TQP, the length

of any side is less than the sum of the lengths of

the other two sides as shown below.

QT = 9 < 13 = 10 + 3 = PT+ PQ

PQ = 3 < 19 = 10+ 9 = PT+ TQ

PT = 10 < 12 = 3 + 9 = PQ + TQ

For !1RQT, the length of any side is less than

the sum of the lengths of the other two sides as

shown below.

RT = 8 <11 = 9 + 2 = QT+ QR

RQ=2<17= 9 + 8 =QT+RT

QT = 9 < 10 = 2 + 8 = QT+RT

For !1RST, the length of any side is less than

the sum of the lengths of the other two sides as

shown below.

RS= 4 <13 = 8 + 5 = TR+ TS

RT=8<9= 5 + 4 =ST+ SR

ST = 5 < 12 = 8 + 4 = TR+ RS

Therefore, 5 and 10 can be the length of PT, and

15 cannot be the length of PT.

The correct answer is C.

3, k, 2, 8, m, 3

PS07771

147. The arithmetic mean of the list of numbers above is 4.
     If k and m are integers and k * m what is the median
     of the list?

(A) 2

(Bl 2.5

(C) 3

(D) 3.5

(El 4

Arithmetic 1tistics

S. mce t e h ant. h met1c. mean= ------sum of values ,

number of values

th en -------3 +h:+ 2 + 8 + m + 3 = 4 , an d so

6

(^16) + k + m = 4, 16 + k + m = 24, k + m = 8. Since
6
k * m, then either k < 4 and m > 4 or k > 4 and
m < 4. Because k and m are integers, either k ::;; 3
and m :::: 5 or k :::: 5 and m ::;; 3.
Case (i): If k ::;; 2, then m :::: 6 and the six
integers in ascending order are k, 2,
3, 3, m, 8 or k, 2, 3, 3, 8, m. The two
middle integers are both 3 so the
d

.. 3 + 3
me ian 1s --= 3.
2
Case (ii): If k = 3, then m = 5 and the six
integers in ascending order are
2, k, 3, 3, m, 8. The two middle
integers are both 3 so the median is
3 + 3 = 3.
2
Case (iii): If k = 5, then m = 3 and the six
integers in ascending order are
2, m, 3, 3, k, 8. The two middle
integers are both 3 so the median is
3 + 3 =3.
2
Case (iv): If k:::: 6, then m ::;; 2 and the six
integers in ascending order are m, 2,
3, 3, k, 8 or m, 2, 3, 3, 8, k. The two
middle integers are both 3 so the
d
.. 3 + 3
me 1an 1s --= 3.
2

The corre,ct answer is C.