dat achiever prep test 1 answer (main)
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DAT Achiever Prep Test 1 Answers
Survey of the Natural Sciences
1. D
Overall elemental contents for all macromolecules presented:
Nucleotides – C, H, O, N, P
Carbohydrates – C, H, O
Lipids – C, H, O
Proteins – C, H, O, N, S
Cholesterols – C, H, O
2. B
Because the fetal lungs are the undeveloped and non‐functional, gas exchanges are phenomenally
carried out at the placenta. And as observed there, deoxygenate red blood cells (RBCs) in the umbilical
artery are adequately rejuvenated prior to giving rise to highly oxygenated RBCs flowing next to the
umbilical vein.
3. E
Additional comments to all of the statement presented:
a. Known samples are like streptococcus (spherically in chain), pseudomonas (rod‐shaped
bacillus), and spirillum (helically shaped).
b & d. Relatively small amount of protein is associated; smaller and simpler ones are called
plasmids
c. Generally, transformation occurs via uptake of naked genes from surrounding environment,
conjugation via one way passage of F plasmid, and transduction via the aid of viral carriers.
e. Peptidoglycans are polymers of modified sugars cross‐linked by short polypeptides. Both
gram positive and gram negative bacteria possess these components, with the cell walls of the
latter having lesser amount of peptidoglycans but are more structurally complex.
4. E
This question relates to the well‐known Hardy‐Weinberg theorem, asserting that population equilibrium
(same allele and genotype frequencies in successive generations) remains constant in the absence of
microevolution factors. Do not panic if you cannot find Hardy‐Weinberg equilibrium as one of the
answer choices. Question of this nature deliberately tests your ability to identify the closet alternative,
which is none other than E alone.
5. A
# of blue petals : # of white petal (299 + 302) : (103 + 98) 601:201 ~ 3:1 Parental genotype : Bb
x Bb (1)
# of long stems : # of short stems (299 + 103) : (302 + 98) 402 : 400 ~ 1:1 Parental genotype: Ll
x ll or ll x Ll (2)
Combining (1) and (2), the possible parental genotypes are: BbLl x Bbll or Bbll x BbLl
6. D
In bone remodeling, osteoblasts deposit Ca2+ in bones in the presence of calcitonins while release Ca2+
from bones in the presence of parathyroid hormones. The above processes are intimately linked to the
homeostatic control of blood calcium level.
7. B
The molecular structure of concern is characteristically represented by a metal/transition element
surrounded by a complex carbon ring, which in turn attaches itself to several other protein subunits.
Besides chlorophylls and hemoglobins, cytochromes found along the electron transport chains also
exhibit similar relationship discussed in this problem.
8. A
Cooperativity generally serves to amplify enzymatic responses to substrate bindings, as augmented by
information given in the question. Rest of the incorrect mechanisms is explained as follows:
Feedback inhibition – regulation of metabolic pathways via the threshold amount of end products
accumulated to inhibit front – end enzymes
Multienzye complex foration – an assembly of enzymes facilitating metabolic processes along the
pathway
Allosteric inhibition – inactive form of enzymes aggravated
Organelles specificity – membranes encapsulation of enzymes designated for specific metabolic
functions.
9. E
The birth of a type O child reveals that the father’s blood group genotype may only be heterozygous IAi.
Hence, crossing of the blood group genotypes from both parents again will produce either a type A (IAi)
or a type O (ii) child.
10. C
Distinguished for their affinities to exceptionally low pH environment, pepsins are a primary group of
enzymes designated to hydrolyze protein molecules to smaller polypeptides in the stomach.
11. B
Explanation for the INCORRECT descriptions:
I. A catalyst will never affect the position of a chemical equilibrium. It will nevertheless speed
up both the forward and reverse reactions as the system is tending toward an equilibrium
state.
III. Enzymatic activities are environment sensitive and optimally operate within tight
specifications of changing factors like temperature, pH, etc.
12. A
Formation of trophoblast preceds the development of a three‐layered embryo. The nerve cord,
belonging to the nervous system, develops from the ectoderm. While a major portion of notochord
develops from the mesoderm, the digestive tract has at least its epithelial layer develop from the
endoderm.
13. D
Myxomycota (plasmodial slime molds) falls under kingdom Protista. Other easily mistaken counterparts
are Acrasiomycota (cellular slime milds) and Oomycota (water molds), simple because of the similar
suffix (‐cota) appearing in these names. A to C are also commonly and respectively known as zygote, sac,
and club fungi.
14. D
Working backward, the minimum amount of energy mandated among the primary producers will be:
250/(0.05)2 = 100000 J
15. D
A, B, C , and E individually contribute to DNA replication: RNA primer initiates a polynucleotide strand,
which is subsequently elongated by DNA polymerase in a 5’ 3’ direction; helicase unwinds the
parental helix structure into two DNA strands, correspondingly stabilized by the single strand binding
protein. Nuclease, on the other hand, appears mostly in the digestive tract (small intestines) and
hydrolyzes nucleotides (RNA and DNA) to their respective sugars, phosphates, and bases.
16. D
Vitamin K, together with other initial clotting factors like calcium, broken vessels, etc…, jointly convert
prothrombin into the activated form of thrombin, from which fibrinogen is triggered to fibrin to
ultimately trap blood cells in the wounds.
17. A
Detoxification is generally carried out by enzymes in the smooth endoplasmic reticulum (SER). The liver
cells are especially loaded with SER when the body system takes in drugs and/or is poisoned.
18. E
Resting potential with an axon is ‐70mV resulted in imbalanced [K+] and [Na+] across the membrane.
Over the time, this ionic potential dissipates as K+ and Na+ diffuse down the respective concentration
gradients. Sodium‐potassium pumps therefore serve to maintain the desired potential value by acting
against these gradients at the expense of ATP molecules – a typical characteristic of active transport.
19. D
Under a repressible system, the associated proteins (end products) are usually produced continuously
until the amount of such proteins formed exceeds a desired (threshold) limit. Acting as co‐repressors
now, the end products accumulated will begin binding to the repressors to jointly stop the ongoing
transcription.
20. E
Blood pH level couple by subunit cooperativity basically determines O2 affinity of hemoglobin. At lower
pH value (higher concentration of CO2), hemoglobin will more readily unload O2 to surrounding cells and
vice versa.
21. E
Proton pumps (being powered by exergonic reactions such as ATP depletions, electron transport chain
redox series, etc…) commonly function as intermediary to drive other useful processes via the generated
proton motive force (H+ gradient) across each cell membrane. A to C are part of the energy couplings
enabling the synthesis of ATP in mitochondria and chloroplasts. The presence of transport proteins is
required to facilitate both active transport and cotransport.
22. A
Vitelline membrane is an outer layer positioned closet to the plasma membrane of a fertilized zygote. It
is usually and fully detached from the zygote upon the first mitotic division. Extraembryonic membranes
consisting of B to E appear during and/ or past the gastrulation stage.
23. A
B and E are flagellated whereas parameciums are ciliated. Plasmodiums, the malaria‐causing parasites,
move by sliding along the bloodstreams of infected hosts.
24. A
Exponential growth, as characterized by decreased mortalities and inflated birthrates, has best been
represented by A. B and E are wrong as these exhibit zero growth in a population. Theoretically, an
exponential growth is unrestrained, rendering C and D inappropriate as well.
25. E
Promoter specifies which DNA strand to be used as template prior to determining the starting point of
transcription. Transcription factor facilitates binding of RNA polymerase to the promoter, and forming
the crucial transcription initiation complex that kicks off rest of the transcribing process.
26. B
Antibodies secreted by plasma cells in the humoral immune response agglutinate antigens upon binding
themselves at the matching receptors. This renders the antigens more susceptible to attack (e.g.
phagocytosis) by white blood cells.
27. A
In glycolysis, each glucose is degraded to two molecules of pyruvates via a series of steps involving many
others like D, B, and C. In the presence of oxygen, pyruvate molecules enter the pathway of aerobic
respiration entailing both the Krebs cycle and the electron transport chain. When devoid of oxygen,
fermentation takes place where ethanols, lactic acids, etc…, are formed depending on the type of
metabolic pathway involved.
28. C
“Dark reactions” simply means reactions may proceed without the need for sunlight. Whether or not
light in present, they will resume as long as ATP and NADPH molecules are continuously supplied to the
cycles.
29. A
Codon recognition: binding of a tRNA (amino acid bearing) anticodon to the mRNA codon positioned at
the A site of a ribosome.
Peptide bond formation: between the new amino acid and the growing polypeptide (catalyzed by the
ribosome).
Translocation: relative moving of the ribosome one codon forward in the 5’ 3’ direction along the
mRNA.
30. B
Both ammonia and cyanide are toxic to human body. Uric acid pertains to excretory systems needing
heavy water conservation such as those found in birds and certain reptiles. Protein will be too big a
molecules to be transported across the glomerulus at the Bowman’s capsule, and moreover, it may
never be the product of deamination. Not only is urea an acceptable form of dissolved waste, it is also
crucially needed to increase the osmolarity of interstitial fluid at the inner medulla.
31. E
Descriptions for all modes of cell communication:
Hormonal signaling – one form of distant communication between endocrine glands and target
cells via hormones traveling along the circulatory system
Synaptic signaling – one‐way local communication via neurotransmitters traveling across the
synaptic cleft at a neuromuscular junction or between two neurons
Signal transduction – cellular response to “signal” molecules via an accentuated chain of
molecular interactions
Cell‐cell recognition – direct communication between two adjacent cells via surface attached
molecules
32. C
In induction, specific genes in the target cells are turned on to differentiate these cells into designated
tissue type. Invagination occurs during gastrulation while neurulation takes place right after it.
Implantation describes how embryo attaches itself to the uterine wall, and organogenesis relates to the
organs formation during the first trimester of gestation.
33. D
Coenocytic hypha mechanically results from ongoing division of nuclei without cytoplasmic division.
Septa basically act as perforated cross walls, turning the otherwise unicellular hypha into multicellular
form. Haustoria are modified ends penetrating the host tissue for nutrients absorption whereas mycelia
are clusters of interwoven hyphae significantly enhancing this process.
34. D
Incorrect speciation modes are annotated as follows:
Adaptive radiation – diversely adapted from a common ancestor (related to geographical
isolation)
Ecological speciation – adaptation on the basis of location and activity
Allopatric speciation – isolation via geographical barrier
Morphological speciation – separation on the basis of physical measures and outward features
35. A
Non‐disjunction causes genetic disorder resulting in aneuploidy and/or polyploidy cells formation. Sex
linked diseases are usually X‐linked unless otherwise stated, where infected fathers are uniquely
observed as passing down defective genes to their daughters only. B and D are part of the Mendelian
laws.
36. E
Trophic hormones are targeted at other endocrine glands where secondary hormones are
correspondingly secreted to regulate designated functions in the body. Adrenocorticotrophic hormones
stimulate the adrenal cortex to produce and secrete steroid hormones, whereas thyroid stimulating
hormones stimulate the thyroid gland to produce and secrete triiodothyronines, thyroxines, and
calcitonins mostly via negative feedback mechanisms.
37. C
Metaphase plate is term adopted to depict the central plane of a dividing cell in metaphase stage.
Mitotic spindles are practically invisible at the end anaphase. B and E pertain to cytokinesis in plant cells.
38. E
2n = 10 {diploid representation}
n = 5 {haploid representation}
By using multiplication rule, 2n = 25 =32
39. D
High levels of estrogens and progesterones in the blood are usually sustained via preservation of corpus
luteum by human chorionic gonadotropins (HCGs) during the first trimester of pregnancy. These
hormones are particularly needed to maintain and strengthen the uterine walls. And understandably,
menstrual cycle ceases during this period of time. Elevated amount of progesterones also inhibits
release of luteinizing hormones (LHs) from the anterior pituitary, rendering ovulation through LHs surge
unlikely to happen.
40. D
Crossing over occurring in Prophase I of Meiosis I involves two non‐sister chromatids belonging to two
separate but homologous chromosomes.
41. B
Without t
required t
will there
42. B
Nitrogen,
a single b
atom. [No
Applying v
angularly
43. D
Reversible
temperat
are consta
44. E
Rate = k[A
Keeping [
.
.
Keeping [A
0.30 M:
.
.
Substitute
the need to e
to neutralize
fore be:
0.5
the central a
ond with oxy
ote: double/m
valence‐shell
resonance st
D
e reaction pre
ure and press
ant values inc
A]x[B]y ,where
B] constant, r
. 1
A] constant, r
. 9
e (2) and (3) i
xpress a full c
every mole o
.
atom, has five
gen atom, an
multiple bond
electron‐pai
tructure outli
esent in the q
sure, concent
corporated in
e x and y are e
reaction rate
2 x = 0
reaction rate
3 y = 2
nto (1) Ra
chemical equ
of NaOH. Usin
.20
e valence elec
nd the remain
is treated as
r repulsion (V
ned as follow
question is an
trations of pu
nto K.
experimental
is observed t
0
is observed t
2
te = k[A]0[B]2
ation, it is ob
ng simple pro
ctrons. Two o
ning two form
s one unit of b
VSEPR) rule, N
ws:
n example of h
ure solids {[Fe
ly determined
to stay consta
to increase by
2
bvious that on
oportion relat
of them form
m a double bo
bond]
NO2‐ will auto
heterogeneo
e] and [Fe3O4
d.
ant as [A] dou
y a factor of 9
nly 0.5 mole o
tion, amount
a lone pair, o
ond with the o
omatically ass
us equilibria.
4] in this case
(1)
ubles from 0.1
(2)
9 as [B] triples
(3)
of H2SO4 is
of H2SO4 req
one of them fo
other oxygen
sume the
At constant
} or pure liqu
10 M to 0.20
s from 0.10 M
uired
orms
uids
M:
M to
45. E
Applying
P1V1 = P2V
46. B
Ionic com
electrosta
which is a
dismantle
crystals.
47. C
Associate
48. D
Let y = ox
Note that
1:
y
y
49. E
Generally
bottom. A
portion of
50. C
From the
The amou
Boyle’s law,
V2
3.00
1.0
mpounds are g
atic attraction
a huge collect
es into smalle
d radioactive
D
idation numb
t Mn is part of
+ 4(‐2) = ‐1
= +7
y, within a gro
As exhibited b
f group VII) w
periodic tabl
unt of HCl req
10000
300
good conduct
ns among the
tion of unit ce
r pieces of un
e disintegratio
ber of Mn
f the anion, M
oup of elemen
by the periodi
will turn out to
e give, molec
quired is there
0
ors of electric
cations and a
ells stacked up
nit cells when
on series may
MnO4‐, where
nts (halogens
ic table give,
o be relatively
cular weight f
efore:
city only in m
anions result
p in systemat
n appreciably
y be outlined
the sum of a
in this case),
I2 (consisting
y largest whe
for HCl = 1 + 3
4 36.45
molten or aque
in high melti
tic order. The
knocked upo
as follows:
all elemental
, atomic radii
g of two heav
en compared
35.45 = 36.45
2 36.5
eous forms. S
ing point to th
e lattice is har
on. IV pertain
oxidation num
increase from
vier atoms fro
to the other
5 g
73.0
Strongly
he crystal latt
rd and brittle;
s more to me
mbers equals
m the top to t
om the bottom
four molecul
tice,
; and
etallic
s to ‐
the
m
es.
51. B
Using approximation where appropriate, divide each of the elemental percent compositions (numerical
values) by its respective relative atomic weight (as exhibited by the periodic table given) and arrange
these in simplified (division by smallest value) whole‐number ratios:
C H O
39.98/12/01 6.66/1 53.36/16
~ 3.33 6.66 ~ 3.34
1 2 1
The empirical formula derived for the compound is therefore CH2O.
52. C
From the Gibbs free energy, G, equation, ΔG = ΔH – TΔS
Spontaneity occurs if ΔG < 0. As such, reaction will assuredly be spontaneous when ΔH < 0 and ΔS > 0 at
any absolute temperature, T > 0K.
53. C
H3PO4(aq) + H2O(aq) H3O+(aq) + H2PO4
‐(aq)
According to the Bronsted‐Lowrt theory,
i. H2O is the conjugate base of H3O+
ii. H3O+ is the conjugate acid of H2O
iii. H3PO4 is the conjugate acid of H2PO4‐
iv. H2PO4‐ is the conjugate base of H3PO4
54. C
Atomic number, Z = 31 – 16 = 15
As exhibited by the periodic table given, only phosphorous, P, matches the specified condition.
55. D
A catalyst has no effect on the equilibrium position since it equally affects both the forward and the
reverse reactions. It will however speed up the process of reach chemical equilibrium, by lowering both
the forward and reverse activation energies.
56. B
According to Gay‐Lussac’s law of combining volumes and Avogadro’s principle 2 volumes of O2 are
required to completely burn every volume of CH4 in relation to the chemical equation given. As such,
volume of oxygen required: 2 x 10 = 20 liters.
57. A
By referring to any temperature‐pressure phase diagram, it can be visualized that boiling occurs when
pressures of both liquid and gas are in equilibrium along the vapor pressure curve for liquid.
58. D
For every mole of KMnO4 used, 5 moles of iron in the ore sample would have been reacted. In D, the
first two multiplications basically serve to calculate the number of moles of Fe2+ reacted. The
multiplication gives us the iron mass of the sample. And the rest is to quickly arrive at the mass percent
of iron present in the ore.
59. C
The following formula is based upon the fundamental principle of mass (in moles) conservation:
M1V1 = M2V2
6.00 30018.00
100
60. C
Using supplemented data exhibited by the periodic table, # of moles of CuSO4 in the sample:
6.463.55 32.07 4 16
6.4100
0.04
Sample weight for H2O: 10.0 – 6.4 = 3.6 g
# of moles of H2O in the sample: 3.6/18 = 0.2
Finally .
.5
61. B
Enthalpy change,
ΔH = ΣΔHf(products) – ΣΔHf(reactants)
ΔH = [ΔHf(H2O) +ΔHf(SO2)] – ΔHf(H2S)
‐562.6 = [‐285.9 + (‐296.9)] ‐ ΔHf(H2S)
ΔHf(H2S) = ‐20.2 kJ/mol
Note: the enthalpy change of an element (O2 in this case) in its standard state is zero.
62. A
Let PT = total pressure of gas mixture = 0.950 atm
PA = partial pressure of N2
XA = mole fraction of N2
Number of moles of O2: .
.
Number of moles of N2: .
.1
According to Dalton’s law of partial pressures,
PA = XAPT = 0.950 0.380
63. D
From the chemical equation given, it is evident that 2 moles of CaCO3 will be required to produce 2
moles of CO2.
Using supplemental data exhibited by the periodic table, molecular weight of CaCO3: 40.08 + 12.01 +
3(16) = 100.09 g ≈ 100 g
The amount of CaCO3 required is therefore: 2 x 100 = 200 g
64. A
V1N1 = V2N2
N1 = . .
.0.3750
Since each mole of diprotic H2SO4 is associated with 2 equivalents of acid, corresponding molarity is
therefore: .
0.1875
65. B
There is usually no question on the real test requiring you to draw any Lewis structure from scratch. Do
not be time trapped. Question of this nature deliberately tests your ability to identify one or two minute
details. A, C, and D may be easily eliminated due to the very fact that nitrogen, being an element
belonging to the second period that devoid of d orbital, is unlikely to violate the standard octet rule.
66. B
Ka =
4.0 100.40 0.20
4.0 10.
. {For very small value of y}
y = 2.0 x 10‐10 mol/liter
67. D
Behavior I pertains to ideal gases. According to the Van der Waals equation characteristics pertaining to
real gases may be mathematically represented as follows:
is added to P to compensate for intermolecular attractive forces, whereas nb is subtracted from V
due to an appreciably small amount of intrinsic volume occupied by the incompressible gas molecules.
68. A
The simple cubic unit cell has all 8 corners at 1/8 atom each, and contains the equivalent of 8 x (1/8) = 1
atom.
The body centered cubic unit cell has all 8 corners at 1/8 atom each plus 1 unshared atom in the center,
and contains the equivalent of [8 x (1/8)] + 1 = 2 atoms.
The face centered cubic unit cell has all 8 corners at 1/8 atom each plus 6 face‐centered atoms at ½
atom each, and contains the equivalent of [8 x(1/8)] + [6 x (1/2)] = 4 atoms
69. D
Let
According to Raoult’s law,
2.00
2.00 6.000.400
6.002.00 6.00
0.333
14
0.40034
13
0.100 0.250 0.350
70. A
The ionic
Since HCl
spectator
71. A
Phenol an
possess. S
H+ will mo
more elec
acidic by n
72. D
According
polyenes
ᴨ electron
73. A
Both carb
oxygen is
74. C
Generally
to 16; and
75. C
Due to rea
highest ab
76. D
Possible s
follows:
A
equation is s
and FeCl2 are
ions and are
A
nd ethanol are
Since electron
ore easily be d
ctronegative t
nature.
D
g to Huckel’s
having (4n +
ns possessed
A
bons in methy
sp2 hybridize
y, unbranched
d solids when
arrangement
bundance fol
D
sets of combin
upposed to a
e both water
therefore ca
e comparativ
ns withdrawn
donated to ot
than sp3 hybr
rule, aromati
2) ᴨ electron
is not solvab
yl groups are
ed.
d n‐alkanes (s
n is 17 or gre
t of the secon
lowed by B an
nation to yiel
ppear as follo
soluble, Cl‐ ap
nceled out fr
vely stronger a
n by phenolic
ther Bronsted
ridized carbon
city can only
s where n = 0
le through Hu
associated w
aturated hyd
eater at 25 °C
dary carboca
nd A in decre
d the desired
ows:
ppearing on b
rom the equat
acids because
oxygen can b
d base. Also,
ns in ethane,
be displayed
0, 1, 2, 3, 4, …
uckel’s equat
ith sp3 hybrid
drocarbons) a
C and 1 atm.
ation to a mor
easing amoun
d alkene via th
both sides of
tion.
e of the attac
be delocalized
sp hybridized
making the f
by planar, m
… D is non‐aro
ion, where (4
dization wher
re gases whe
re stable tert
t respectively
he Wittig rea
the equation
ched –OH gro
d into the aro
d carbons in a
former compa
monocyclic, fu
omatic becaus
4n + 2) isn’t eq
reas the one d
en n = 1 to 4;
iary structure
y.
ction may be
n become
up they both
omatic ᴨ syste
acetylene are
aratively mor
lly conjugate
se the numbe
qual to 8.
doubly bonde
liquids when
e, C is produc
outlined as
em,
re
d
er of
ed to
n = 5
ed in
77. B
Free radic
stabilized
1)
2)
78. E
Choose th
below:
79. E
Deshieldin
E. Only a l
reference
80. A
Isomers o
physical a
C
(stutural)
{The chair
of illustrat
81. D
The first r
Clemmen
cals, like carb
via mechanis
) Inductive e
cumulative
) Greater sta
conjugated
he longest cha
ng effects of
low level of ra
e.
A
of the same m
and chemical
onformationa
isomers
r half‐chai
tions for conf
D
reaction is Fri
sen reduction
ocations, pos
sms described
effects (electr
e and most st
ability is even
d allylic syste
ain and numb
iodine, an ele
adiation is ne
molecular form
properties ou
al isomers < e
r skew bo
formation iso
edel‐Crafts ac
n to form an a
ssess unfilled
d as follows:
ron donating
trongly obser
n achieved wh
m.
ber its carbon
ectronegative
eeded to chem
mula can gene
utlined as foll
enantiomers <
at flipped
omerism.}
cylation of be
alkylbenzene
2p orbital (m
) of methyl gr
ved in tertiar
hen the unpa
n from the en
e halide, are c
mically shift t
erally be arra
ows:
< diastereom
chair conform
enzene to for
e, illustrated a
may be visualiz
roups to fill u
ry radical.
aired electron
d nearest to t
cumulative an
he proton sig
nged with inc
mers and geom
mations of cy
m an acylben
as follows:
zed as half‐fil
up the unfilled
n can be deloc
the first bran
nd most inten
gnal downfiel
creasing orde
metric isomer
yclohexane ar
nzene, followe
lled) to be
d orbital are
calized into th
nch, as illustra
nsely exhibite
d from TMS
er of differenc
rs < constituti
re one typical
ed by
he
ated
ed in
ces in
ional
l set
82. B
Free radic
carbon to
83. A
The final p
84. D
D does no
with two
85. C
Ether, bei
an ideal s
is readily
polarized
86. D
Since the
aromatic
87. C
Glycine is
the form o
complete
apparentl
cal addition is
o produce the
A
product, amin
D
ot display geo
identical subs
ing functiona
olvent for rea
abstracted an
organometal
D
alkyl group a
substitution,
highly polar
of +H3NCH2CO
ly neutralized
ly possessing
s anti‐Markov
most stable
ne, is formed
ometric isome
stituents (‐CO
lly non‐reacti
actions involv
nd transferre
llic carbon, ul
ttached to th
major produ
and readily d
O2H, this poly
d with. At isoe
a zero net ch
vnikov where
form of 3° alk
via an interm
erism because
OOH).
ive and capab
ving orgnanom
d from the hy
ltimately form
he ring is activ
cts formed w
issolves in wa
yprotic (divale
electronic po
harge.
bromine rad
kyl radical, illu
mediate called
e one of its do
ble of dissolvi
metallic reage
ydroxyl end o
ming an alkan
vating and ort
will certainly b
ater through
ent) acid requ
int, glycine ap
ical selective
ustrated as fo
d imine, illust
oubly bonded
ing most non‐
ent, a Bronste
of any Bronste
ne.
tho/para dire
be I and III.
hydrogen bo
uires two equ
ppears as zwi
ly attacks the
ollows:
trated as follo
d carbon atom
‐polar substa
ed base. Acco
ed acid to the
ecting toward
ond formation
ivalents of ba
itterion, +H3N
e less substitu
ows:
ms is attache
nces, is no do
ordingly, a pro
e negatively
ds electrophil
ns. Appearing
ase to be
NCH2COO‐,
uted
d
oubt
oton
ic
g in
88. B
Oxidation
89. E
Rate of E1
the alkyl g
in D, its in
slowest ra
90. D
Carboxylic
possible p
91. D
You will u
carbonyl g
spectra is
92. B
Rotation p
sample is
amounts
93. E
Associate
n of alkynes us
1 elimination,
group associa
nductive effec
ate of E1.
D
c acid assume
prior to numb
D
sually be ask
group (~1700
also strongly
patterns for D
purest when
of mirror‐ima
d steps of rea
sually procee
, like SN1 subs
ated with CH3
ct is also least
es higher prec
bering the ass
ed to match a
0 cm‐1) on the
y recommend
D and E have
only a single
aged enantiom
actions are ou
eds via ozono
stitution, is p
CH2Br is prim
t exhibited to
cedence over
sociated carbo
absorption fr
e real test. Fam
ded.
simply been s
e type of enan
mers are race
utlined as foll
lysis, illustrat
roportional to
mary in structu
o stabilize the
r alcohol and
on atoms from
equencies (w
miliarization w
switched to c
ntiomers is pr
emically mixe
lows:
ted as follows
o the ease of
ure and comp
carbocation
the chain cho
m the carbox
wave numbers
with graphic
confuse you. O
resent; it reve
ed.
s:
f carbocation
paratively sho
formed, ultim
osen should b
xylic end, as il
s) to –OH gro
appearances
Optical activi
erts to zero w
formation. Si
orter than the
mately leadin
be the longes
lustrated bel
up (~3300 cm
of these IR
ty of any give
when two equ
ince
e one
g to
st
ow:
m‐1) or
en
ual
94. C
Said react
Note that
alkyl chain
95. A
The mech
96. C
Addition o
the less su
rearrange
bromide (
97. E
C and D a
meso form
tion is observ
t the carbony
n of amine on
A
hanisms and r
of HBr to (CH
ubstituted ca
ement via hyd
(Br‐) can be ca
re mirror ima
m), illustrated
ving Hoffman
l carbon is ab
ne carbon sho
reactions rela
3)2CHCH=CH2
rbon in alken
dride shift fav
aptured to fo
ages to each o
d as follows:
rearrangeme
bstracted from
ort.
ted to aldo co
observes Ma
ne to form a m
vors formation
rm (CH3)2CBr
other, both h
ent outlined a
m the amide a
ondensation
arkovnikov’s r
more stable se
n of an even
rCH2CH3 inste
aving a plane
as follows:
and turned in
are illustrate
rule where pr
econdary car
more stable t
ad of (CH3)2C
e of symmetry
nto a carbona
ed as follows:
roton from H
rbocation. Ho
tertiary carbo
CHCHBrCH3.
y that render
te, rendering
Br will first ad
wever,
ocation where
s them achira
g the
dd to
e
al (in
98. C
Phenyleth
phenylpro
{Note: Syn
secondary
99. E
Like aldo
reaction e
100.
Conjugate
(cycloadd
hyl cyanide (a
opanoic acid,
nthesis of car
y alkyl halides
condensation
equation can
B
e addition of
ition) reactio
a nitrile) is firs
illustrated as
rboxylic acids
s}
n, Claisen con
generally be
dienophile to
n, illustrated
st formed via
s follows:
via formatio
ndensation is
represented
o a diene to fo
as follows:
nucleophilic
n and hydroly
carried out v
as follows:
orm a 6‐C ring
substitution,
ysis of nitriles
ia similar nuc
g is character
followed by
s applies only
cleophilic add
ristic of Diel‐A
hydrolysis to
y to primary a
dition and its
Alder
and
Percept
1. D
Front:
Left:
2. D
Front:
Left:
3. B
Front:
tual Ability
D
D
y Test
Bac
Top
Bac
Top
Bac
ck:
p:
ck:
p:
ck:
Right:
Bottom:
Right:
Bottom:
Right:
Left:
4. B
Front:
Left:
5. B
Front:
Left:
Top
Bac
Top
Bac
p:
ck:
p:
ck:
Top:
Bottom:
Right:
Bottom:
Right:
Bottom:
6. E
Front:
Left:
7. C
Front:
Left:
8. B
Front:
Top:
Bac
Top
Bac
Bac
ck:
p:
ck:
ck:
Right:
Bottom:
Right:
Bottom:
Right:
Left:
9. C
Front:
Left:
10. B
Front:
Left:
Top:
Top:
Bac
Bac
Top
ck:
ck:
p:
Bott
om:
Right:
Bottom:
Right:
Bottom:
11. E
Front:
Left:
12. D
Front:
Left:
Top:
D
Bac
Bac
Top
ck:
ck:
p:
Right:
Bottom:
Right:
Bottom:
13. E
Front:
Left:
14. C
Front:
Left:
Bac
Top
Bac
Top
ck:
p:
ck:
p:
Right:
Bottom:
Right:
Bottom:
15. E
Front:
Left:
16. C
Bac
Top
ck:
p:
Right:
Bottom:
17. A
18. B
A
19. C
20. D
D
21. B
22. C
23. C
24. B
25. D
26. D
D
D
27. A
28. A
A
A
29. B
30. C
31. A
32. D
33. C
34. B
35. A
A
D
A
36. D
37. C
38. B
39. A
40. D
D
A
D
41. C
42. B
43. A
44. D
45. C
A
D
46. A
47. D
48. B
49. E
50. C
51. A
A
D
A
52. D
53. B
54. E
55. C
56. A
57. D
D
A
D
58. B
59. E
60. D
61. D
62. E
63. C
64. A
65. A
66. B
67. C
68. C
69. A
70. A
71. C
72. D
73. D
74. B
75. B
D
D
A
A
A
A
D
D
76. B
77. C
78. D
79. A
80. D
81. C
82. C
83. D
84. C
85. A
86. C
87. A
88. A
89. D
90. A
Reading Comprehension Test
1. B
Most of the passages on the real test are educational and/or instructive by default; and will usually be
associated with one or two questions of this kind. Candidates who have followed the instruction to read
the passage in full will find this always a bonus question given. B is most appropriate in view of the
enlightening information shared throughout the entire texts of the passage.
2. D
As plainly portrayed in the last sentence of paragraph 14, the instrument is there to dislodge disc stuck
in front of the condyle (head of the jawbone as indicated in paragraph 5).
3. C
If you read paragraph 10 in full, you will notice that TMJ disk or ligaments do not show up in TMJ X‐rays
at all, making this essentially impractical for diagnosis purposes. Sonography might not be needed by a
well‐trained clinician but no additional remarks have been given to entail its impracticality.
4. E
Of the three primary factors introduced in paragraph 4, only trauma has been discussed more in details
as is evidenced in paragraph 7. Stress is implicity elaborated in paragraphs 8 and 9.
5. C
TMD treatments are termed basic (paragraphs 11 and 12), intermediate (paragraph 13), and surgical
(paragraphs 14‐16).
6. B
Notice in paragraph 7 that movement of jaw from side to side (bilateral) is undesirable when a person
has TMD, making B (up and down) the best inference here. C‐E are rejected as these would be easily be
imagined as entailing some lateral jaw movement.
7. A
“That there are many causes of headache and jaw pain, while TMD, which is also located in the head,
can hurt through many different ways” typically creates a difficult diagnostic dilemma for health care
practitioners. B to D, though true, have simply not been as crucial as A is.
8. A
Prescription and over‐the‐counter medications, if used correctly, help alleviate (lessen) the pain only;
whereas narcotics have problems of addiction and rebound pain. Psychological and psychosocial
approach, on the other hand, appears to be more maneuverable through the Minnesota Multiphasic
Personality Index. Hot and cold therapy is just a simple measure for treating mild pain, whereas
biofeedback has not been discussed in details in the passage.
9. E
A‐C are irreversible surgical treatments. Similarly, the removable splint can be irreversible therapy
depending on how it is made and used according to paragraph 11. Ultrasound treatment has only been
indicated in paragraph 13 to relieve soreness or improve mobility of the TMJ, making it the only non‐
invasive option here.
10. D
According the paragraphs 5 and 6, arthritis and joint degeneration are noted to be diseases that occur in
all (catch‐all) joints.
11. A
“Vascular disturbances” may implicitly be tied to the cold and warm therapy, but, it has literally
appeared once throughout the entire passage.
12. C
Answer A may easily be discarded through direct info given in paragraph 3. According to paragraph 4,
heredity serves only one of the three primary reasons for TMJ issue, rendering B invalid as well. And as
mentioned in paragraph 1, TMD can manifest many types of pain, making some of these manifestations
indistinguishable from other causes of headache and jaw pain. Also, it is the psychological problems
(instead of medications) that can end up the sole cause of pain. C is, on the other hand, supported by
the last sentence of paragraph 5.
13. B
You are supposed to look for the best diagnostic test, where only A and B fit in. Evidently, arthrography
with dye injected into the TMJ to clearly show the position of the disk would establish a concrete
verification here.
14. E
A‐D respectively relate on way or another with TMK inflammation in paragraphs 15, 12, 16, and 6.
15. D
Gum chewing causes only harm to the TMJ, making A, D, and E the possible answers. E is readily
discarded because a trauma is an accidental occurrence, whereas A is incorrect due to the presence of
the adverb, “often”, in the answer.
16. A
This is what the test would typically provide, although all other answers may indirectly be connected to
it.
17. E
E is wrong because of the disk is supposed to be held in place by ligaments and a capsule; the statement
would have been correct had the word “by” been replaced by “between”. The validity of D may be
inferred from the fact that the removable splint is also worn at night (when sleeping) to minimize
clenching or grinding of the teeth. A‐C may respectively be verified in paragraphs 7, 11, and 9.
18. B
Question of this nature will continue to appear on the test, underscoring the importance of reading the
passage once through with considerable understanding prior to answering all problems that follow.
Elaboration on human greed for gain, disregard for ecological impact, unscrupulous manner of disease
treatment, etc., as seen in the passage serve primarily to disclose scenes and ruling motives behind gene
amalgamation.
19. A
The first sentence of paragraph 1, that develops further to the last sentence of paragraph 2 and beyond,
explains it all for the answer of A. B to E might carry some traces of relation to problems arising from
genetic engineering, but they have not been considerable montioned or specifically detailed by the
author.
20. C
Although the first part of statement C is correct where Bt gene is known to kill corn borer, the second
half of it is contended by research in immunology from the University of Cincinnati College of Medicine.
21. D
As illustrated in paragraph 3 and subsequent writings, it is the popular evolutionary belief in existence
through accident that has pushed modern gene technocrats to fabricate in vitro recombinant DNA
without adequate caution for long‐term human and environmental outcomes. The unclear future
originating from such pre‐supposition is logically not an ultimate solution for genetic issue in this very
respect. The concept of evolution may have sparked the shooting of foreign genes to the host cells, but
it has not been told to be facilitating (making easy) the very work of gene splicing.
22. E
A and B are measures which appear well regulated and endorsed in England. C is invalid because the
Heritage team has not found any active bla in bacteria from either the Bt‐corn‐fed animals or the silage
runoff. D is an issue that concerns the US. As expounded in paragraphs 9 to 11, the incorporation of such
foreign plasmid to the bacterial DNA brings about new strains of pathogens that will be resistant to
antibiotic, making this of prime concern to the British authority.
23. C
As indicated in paragraphs 3 and 8, the potent Bt gene may be spliced into the corn DNA through
aggressive and strong viruses (vectors), with very poor aim though.
24. C
Basing on information supplied in the second half of paragraph 2, the ratio of Bt‐gene corns to
conventional (traditional) ones in the fields can quickly be computed as follows:
30% : (100% ‐ 30%) => 3 : 7
25. B
In the first sentence of paragraph 4, Bt gene (a toxin as explained/termed in paragraphs 4‐7) splicing
technique has been told spreading not only to corn but also to potatoes and rice. Similar technique
might have been adopted for cultivation of many other crops and plants. You should, nevertheless, limit
your answer pick to information provided by the passage, unless otherwise specified by the question
presented.
26. D
Considering the root of all evils entailed in gene amalgamation arises from the rush for quick‐fix
pragmatism, the technocrats certainly need to stop, reflect, and think about the backfiring of their
“design by accident” philosophy, before more tragic and costly lessons will be needed later.
27. D
Effects of this toxin is delineated from there till it is finally termed as the Bt gene in paragraph 7,
although plasmid (a naked DNA associated by Dr. Heritage with the “bla” gene) may be related ehre
from a different perspective.
28. A
According to Heritage team, the saliva of cattle (where the mouth is) is the likeliest digestive site for
plasmid (containing “bla” that leads to antibiotic resistance) absorption by any bacterium thriving there.
29. C
In the field conditions simulated by Losey and his teams as described in paragraph 4, it is pollen from a
Bt hybrid that has killed and stunted the monarch caterpillars and no the moistened milkweed leaves.
Question of this nature deliberately tests your ability to abstract passage info with great speed and
accuracy, which may easily be achieved through exhibition of strong confidence and unobstructed
concentration during the test.
30. B
From information given in the second half of paragraph 4, the portion of monarch butterflies killed can
quickly be approximated as follows:
19/100 ~ 20/100 = 1/5
31. C
The author attempts to draw evidences (like B and E) to support his disapproval of churning out new
combinations of genes from almost anywhere—plant, animal, or man—to almost anywhere. D along
with A would all the more promote gene amalgamation. C, if found true, would somehow disrupt the
author’s tone in the passage.
32. A
The first statement is wrong because the trigger as traced in paragraph 13 refers to the irritation of
macrophages that sends off IL‐6 for the attack of the lung. The second statement may easily be verified
from the last sentence of paragraph 5.
33. E
Bt toxin was reported in the first half of paragraph 6 as mounting damages to mitochondria, and short
microvilli on the bowel surfaces of the lab mice. Answer A has not been discussed in the passage,
whereas the incorrectness of statements B and C has simply been due to the wrong input of descriptive
words. The sensible concern raised by the British government has certainly been in line with the central
theme of the passage.
34. C
Human spleen is larger in men and heaviest in young adults according to information given in the first
paragraph, making C the best estimate.
35. C
Sympathetic nerves have been known to shrink the spleen during an emergency via constriction of
associated veins, according paragraph 2. This renders C the incorrect statement.
36. B
Correct information is presented in paragraph 5, although rest of other answers might be connected to a
weakened spleen as observed here.
37. E
You are supposed to look for another befitting symptom associated with an enlarged spleen. As
mentioned in paragraph 14, splenomegaly can reduce the number of healthy cells in the bloodstream
including the platelets (cells that help the blood clot as indicated in paragraph 7), making E the likeliest
symptom among all answer choices here. In fact, this symptom is similar in nature to hemorrhage (as
noted in paragraph 14 again), but you need not possess this knowledge to land on the right answer.
38. C
As noted in paragraph 2, the spleen shrinks during an emergency just like the incident mentioned in C. It
enlarges during digestion or when the body is sick (paragraph 10), and understandably remains normal
when the body is resting.
39. D
As inferred from paragraph 13, contact sports like soccer, football, polo, hockey, etc render an enlarged
spleen more vulnerable to rupture just like a motorcycle (physical) accident would do.
40. E
According to data given in paragraph 1, the following calculations can quickly be carried out within a
couple of seconds:
Amount of white pulp: (1/4) x 400 = 100 mL
Amount of red pulp: 400 – 100 = 300 mL
41. D
In view of the many vital functions handled by the spleen (like catching cancer cells trying to metastasize
through the blood) as well as the consequences of post‐splenectomy, current practice among the
surgeons is to try keeping this organ (as statistically indicated in the Vanderbilt series) rather than
resorting to the old‐style hasty removal.
42. A
Note that the people of Ashkenazi Jewish ancestry are more prone to enlarging the spleen simple
because of their higher risk of contacting Gauchers or Niemann‐Pick disease. Logically, anyone having
one of these storage diseases would have an equal chance of developing splenomegaly (assuming all
other factors being equal).
43. A
Paragraph 1 and 10 collectively indicate that the spleen is a purple spongy organ located on the left
abdomen, and about the size of a fist. Based upon the roles and functions presented in the passage, the
spleen will more appropriately be associated as the primary or first line of defense to the body,
rendering D incorrect as well. Answer A may apparently be abstracted anywhere from paragraph 4 to
paragraph 10 in the passage.
44. B
According to paragraph 3, the nervous system and endocrine hormonal system are clearly noted as
directing the body’s immune system (associated with splenic infection‐fighting episodes).
45. A
According to paragraph 13, both delicate surgery and netting are the two options used in preserving a
ruptured spleen. Radiation and antibiotics, as pointed out in paragraph 17, respectively pertain to
conditions affected by Hodgkin’s disease and infection.
46. E
Correct answer can directly be traced in the second half of paragraph 14, where anemia or hemorrhage
is caused by the blood damaging process of a deranged and dysfunctional spleen. Answer A may be
discarded since measures like antibiotics and radiation will always be adopted prior to taking up the last
option—splenectomy. A dysfunctional spleen may have more white blood cells added to its content, but
nowhere in the passage would be found to support the very extent of having the white and red pulp
composition flipped to a ratio of 3:1.
47. B
In connection with the last paragraph of the passage, you are supposed to pick an answer that would
minimize the risk of post‐splenectomy infection. Obviously, confining yourself to a region free from any
endemic or epidemic would save your body the risky task of fighting the highly infectious disease.
48. C
Correct answer may be mapped out in paragraphs 4 and 9. Pitting and culling are characteristic of the
spleen, whereas the old iron from red blood cells gets recycled in the bone marrow.
49. A
Paragraph 10 is noted as saying “…The battle can become so hot that when losing it the spleen will even
resort to enlargement of itself by the addition of millions of extra soldiers (white blood cells as indicated
in paragraph 7)…” This explains for the answer of A. Answer B would shrink the spleen, while answer D is
simply a result (or a symptom) of splenomegaly where the enlarged spleen presses down on the upper
left abdomen, under (and against) the stomach.
50. D
Paragraph 7 mentions that the spleen contains a third of all the platelets in the body, implying that [1 –
(1/3)] = 2/3 of the body platelets may be found elsewhere in the body. A, B, C, and E can all be rejected
via info given in paragraphs 1, 8, 9, and 6 respectively.
Quantitative Reasoning Test:
1. B
420
5 5
2. E
Present price per sq. ft.: $ , .
$316.25
3. E
Rearranging the given equation, 5x + 2y = ‐7
(1)
Comparing (1) to y = mx + c, slope of the equation:
Equation of the line passing through the origin is therefore,
5x + 2y = 0
4. E
5. D
Let x = median of the designated series
(x – 4) + (x + 4) = 34
2x = 34
x = 17
6. B
1
7. D
1602
8808
10
8. A
1
√45
7
513
45
1
3√5
7
5175
1
√5
1
√5
√5
√5
√55
9. C
√54 √96 √9 6 √16 6 3√6 4√6 7√6
10. A
Circle circumference: 2(5+2) = 14 in.
Using circle circumference formula,
2ᴨr = 14
/
2.227272… .
11. D
12. B
Among the 3 children born, there will be 2 daughters statistically alike. The number of permutations of
all children is therefore, !
!3
Since the total number of ways for having 3 children in a row is (2)3 = 8 {Multiplication Rule}, the
likelihood of having exactly 1 boy in a family planning for three children is finally obtained as 3/8.
13. B
cos6
cos6
√32
14. E
Account balance: $1000 x (1.02)2 = $1040.40
15. B
To avoid dealing with big numbers, mixed fractions may have their whole numbers first separated
(associative property) before adding them back to form an improper fraction. Hence,
√0.36 3712
129
256
0.6 3 1 2712
29
56
0.6 221 8 30
36
0.6 2136
610
72 136
7160
16. B
|2y + 5| < 3
2y + 5 < 3 or ‐(2y + 5) < 3
y < ‐1 or 2y + 5 > ‐3
y > ‐4
Combining both inequalities of y,
‐4 < y < ‐1
17. E
2 x 7 x 24 x 60 x60 = 1209600 seconds
18. B
Sphere area: 64
Using spherical surface are formula, 4 64
r = 4 cm
19. D
Express all fractions in unit prices. Apply knockout method starting from the top or bottom by
comparing numerators over each pair of common denominators outlined as follows:
A: 4/12 = 8/24 versus B: 7/24 {7 is smaller than 8; so A is out}
B: 7/24 = 196/672 versus C: 8/28 = 192/672 {B is out}
C: 8/28 = 256/896 versus D: 9/32 = 252/896 {C is out}
D: 9/32 = 360/1280 versus E: 12/40 = 384/1280 {D is the best buy}
Note: Do not bother calculating any single set of common denominators shown above. They are here
serving only to justify the numerator comparisons.
20. B
The probability of NOT seeing any Japanese car out of every three automobiles is (2/3)3 = 8/27.
{Multiplication Rule}
Using complementary rule, the probability of seeing at least one Japanese car out of every three
automobiles is therefore, 1 – (8/27) = 19/27
21. C
Substituting the value of x at both sides of equation,
tan ‐45° = ‐1 {4th quadrant}
tan [90 – (‐45)] = ‐1 {2nd quadrant}
22. C
Let x = Alex’s present age = 10
y = Mary’s present age
z = Craig’s present age
x = 2y
5 (1)
3 years from now,
y + 3 = 2(z + 3)
3 3 1 (2)
10 years from now, Craig will be 10 + 1 = 11 years old.
23. E
Let t1 = time taken by pump A to drain the pond {4 days}
t2 = time taken by pump B to drain the pond
T = time taken by both pumps to drain the pond {3 days}
Applying the work formula,
Multiply both sides of equation with 12t2,
3t2 + 12 = 4t2 t2 = 12 days
24. E
Note that the LCD of this equation is 4x2 – 9 or (2x + 3)(2x – 3). Hence,
9 2 3 2 2 3
9 2 3 4 6
4 8 12 0
2 3 0
3 1 0
x = ‐3 or x = 1
25. E
Let x = smallest integer of the designated series
y = second smallest integer of the designated series
z = largest integer of the designated series
z = x +3 (1)
y = x + 1 (2)
x + y = ‐17
x + x + 1 = ‐17
x = ‐9 (3)
Substitute (3) into (1),
z = ‐9 + 3 = ‐6
26. B
x + x = y
2x = y
x = y/2 (1)
y = x2/9 (2)
Substitute (1) into (2)
29
y2 – 36y = 0
y(y – 36) = 0
y = 0 (discarded) or y =36
27. C
28. A
Among the 5 tosses, there are 3 heads alike, and 2 tails alike. The number of permutations of all 5 tosses
is therefore, !
! !10
Since the total number of all possible 5 tosses is 25 = 32 {multiplication rule}, the probability of getting
three heads and two tails in a five consecutive tosses of a fair coin is finally obtained as 10/32 = 5/16.
29. B
Distance traveled by both trains: 90 x 8 = 720km
Trip tie consumed by cargo train: 16 hours
Average speed of cargo train: 720/16 = 45 km/hr
30. A
Putting both dimensions in ratios of similar unit,
7.5 cm x 5 cm : 15 m x 10 m
7.5 cm x 5 cm : 1500 cm x 1000 cm
1 : 40000
31. B
Applying circle circumference formula, 2ᴨr = 18ᴨ
r = 9 units from the origin (0,0) (0,9)
(0,9) is simply the one of the infinite number of coordinates sitting on the circle.
32. C
33. D
Original re
Total disc
34. C
35. D
6x + y = 8
3x + 4y =
Subtract (
3x
x
D
etail price: $2
ount: $
$
D
20
(2) from (1),
x – 3y = ‐12
– y = ‐4
200 + $100 =
. $1
(1)
(2)
$300
00%$
$
100% 550%
36. D
1
37. A
Temperat
5977 3
Note: Con
by heart a
38. D
Let l =
w
Applying p
2
l =
39. C
40. C
2
2
2 1
x = ½ o
(Rememb
D
1
A
ture in Celsius
3259
45
nversion form
and be able to
D
= length of re
w = width of re
perimeter for
(l + w) = 20
= 10 – w = 10
1
1 0
1 0
r x = ‐1
ber always to
2 1
s:
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mulas are rare
o apply them
ectangle
ectangle
rmula,
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express quad
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ely given on th
at finger tips
dratic equatio
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he real test. Y
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on such that f
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f(x) = 0 prior t
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gly advised to
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o remember t
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