SOLUTIONS
FOR
THE
EIT
SIMULATED
EXAM
ENGINEERING ECONOMICS
An automobile costs $5000.00 and has an estimated life of 5 years. Expenditure of extra money on certain additional maintenance items will extend this life to 10 years. Interest is 12.5% l.How much should be spent for this additional maintenance? b. $450 a. $500 d. $350 c. $400 e. $300 As= P(A/P) i = 12.5%, n = 5: 5000(.2809) = $1404.50 A 10 = P(A/P) i = 12.5%, n = 10: 5000(.1806) = $903 As- A 10 = 1404.50- 903 = 501.50 Answer (a) 2. What annual payment is required in order to have the renewal cost available in 5 yrs? a. $960 b. $840 c. $780 d. $690 e. $573 As= F(A/F) i = 12.5%, n = 5: 5000(.1559) = $779.50 Answer (c) 3. If the interest rate was reduced to 12% per annum, compounded monthly, what is the effective interest rate? b. 12.35% a. 12.25% d. 12.7% c. 12.5% e. none of the above i = 12%112 mo = 1%/mo (F/P) i = I%, n = 12 = 1.1268 Effective i = 1.1268- 1 = .1268 Answer(d) 4. If expenditures such as gasoline, oil, tires, batteries, etc., amount to $2300 per year, insurance costs $1000 per year, what is the annual cost of owning this car for the 5 year life if the salvage value is $1 000? a. $43,000.00 c. $4,548.60 e. $3,829.25
b. $4,147.50 d. $4,829.25
TAC = [(P- L)(A/P) + Lj] + AE = [(5000- 1000)(.2809) + (1000)(.125)] + 2300 + 1000 = 4548.60 Answer (c)
-
SESSION PM badoo.ge#EEEgaeff
13-45
13-46
ftp.adf#g_E--g_fo--Edf--f_dgdgE-fpaf-fg_----hgfghfffs_3dgf-hrE_ LS OF ENGINEERING EXAM REVIEW FUNDAMENTA
ELECTRICAL THEORY Problems 5-8 relate to the circuit shown below.
The circuit shown is a series R-L hook with a sinusoidal input equal to V. V = 141 sin wt; R=3Q;L=10.6 x 10·3h :f= 60Hz 5.
What is the effective steady state current? (include the relative phase angle. b) 141L.15._: a) 141L..2..:.. d) 100/73.2° c) 20/-53.11° e) 5/53.11° Effective means RMS which equals Peak/>12 (sinusoid) IRMS = V RMs/Z V = 141 sin wt = 141LOO VRMS = 141 sin wt/>12 = !4!LOOI>i2 =99.7LOO Z = R + j wL (complex)= >i(R2 + (wL)2) Ltan- (Polar) wL = 2rt:60 X ( 10.6 x 1{}3) = 4 Z = >1(32 + (4)2) Ltan-I (4/3) = 5L53.13° IRMS =99.7 LOot 5L53.13° =20L-53.13o Answer (c)
6.
Using the values of the components given compute the power factor. b. 0.6 a. 0.7 d. 0.8 c. 0.5 e. 0.78 PF =cos angle between V and I= -53.13° cos -53.13 ° = 0.6
7.
Answer (b)
Using the values given, what is the effective value of the voltage drop across the inductor? (include its phase angle.) b) 60 /-53 .1 ° a) 141/89.21° d) 100/73.2° c) 80/36.9° e) none of these IRMS = 20L-53.13°, ZL =j wL =j4 =4L90o VRMS (Inductor)= 20L-53.13o X 4L90o =80L36 .. 90
Answer (c)
WORKBOOK
SOLUTIONS
8.
FOR
t3oBtg_gf3g_fEa__#EgggE-fffff--EgEE--N-s_offTgfgfagepf-# PM THE EIT SIMULATED EXAM
What is the power consumed by the load? a. 60 watts b. 123 watts d. 319 watts c. 180 watts e. 1200 watts P = IRMs2R = 202 x 3 = 1200 watts Answer( e)
DYNAMICS Problems 9-11 relate to the following problem statement. The motion of a particle is defined by a= 200 + 6t2 ft!sec2. It has a velocity of 2 ft!sec and a distance s = 0, at time= 0 from a fixed point. 9.
Determine the average velocity during the first 10 sec. a. 8004 fps b. 4002 fps c. 4440 fps d. 8040 fps e. 6030 fps = so
Sf
Sf=
0 + 2(10) +
v
-so tt -to
Sf
vot + at 2
2
(200 + 6(10) 2)(10) 2 2
____0_ 10 - 0
=
=
40,020
4002 Answer(b)
10.
If a second particle has a motion defined by v = 160t + 12t2, which equation expresses the length of time required for it reach the first particle? S = 0, and t = 0 for the second particle. a. b. c. d. e.
(t3 - 8t2 + 40t + 4. t(t3 - 8t2 + 40t + 4. t(t4 - 8t3 + 40t2 + 4t. 2t(t3 - 8t2 + 40t + 4. 2(t3 - 8t2 + 40t + 4.
For particle I s 1 = t4f2 + IOOt2 + 2t + c t = 0, s = c, thus c = 0 For particle 2 s2 = 4t3 + 80t2 + c t = 0, s = c, thus c = 0 SJ = Sz 4t3 + 80t2 = t4f2 + IOOt2 + 2t Combining: t(t3- 8t2 + 40t + 4) Answer(c)
SESSION
13-47
-ft#E-sgo#3Egffg3-f_#Eegg_sEMsgcs_--gfd_gEfs_f---#Eso_-daas_R--#3fLS OF ENGINEERING EXAM REVIEW WORKBOOK FUNDAMENTA
13-48
11. If the weight of the particle with v = 160t + 12t2 is 2 lbs, what constant force would be required to stop it if the force were applied at time t = 2.5 sec. Assume that the stopping distance is 1 ft. b. 5000 lbs d. 10,000 lbs
a. 500 lbs c. 7000lbs e. 14000 lbs F
F
X
X
D
= 1.
2 Vr=
1_ - 2-
2
2 32.2
-
M
v6)
o (160t + 12t 2 ) 2
F = 7007lbsr Answer (c)
MECHANICS OF MATERIALS Questions 12 and 13 refer to the following problem statement: A composite beam is made of wood and steel, as shown: Assume Est! =
20XEwood 6"
-8+12' I-I
3"
12. Determine the moment of inertia about the neutral axis. b. 3459 in4 a. 1688 in4 d. 1248 in4 c. 2586 in4 e. None of these y
IG
(72)(6) + (3
X
0.5
X
20 )(12. 25)
72 + 30 3 6 12 + 3 0 ( 4. 41) 2 + 72 12
(1. 83) 2
7.83" 1688 In 4
Answer(a)
SOLUTIONS
#Et____f-E__gffEd_g_fEEg_#-#-fg__eegf###Ep_f--Ee_###-g_ef__f---fg__of PM SESSION THE EIT SIMULATED EXAM FOR
13. The function of the connectors in a composite unit such as this is to: a. Help balance the loads in the beam b. Transfer moment from the plates to the beam beam c. Transfer horizontal shear from the plate to the d. Transfer vertical shear from the plate to the beam e. It has something to do with diagonal tension Answer (c) STATICS Problems 14 and 15 refer to the beam shown below which carries a moving load of 1500 lbs. Assume the beam to be supported at B by a rod BC as shown. The supports at A and C and the joint at B permit unrestricted rotation.
T.25
.25
x5l
X
4
.25 X 4
Beam Cross Section
Beam Data Modulus of Elasticity E = 30 x 1Q6 Moment of Inertia I = 20 in4 Cross Sectional Area A= 3.375 in2 Rod Data Modulus of Elasticity E = 30 x 1Q6 Cross Sectional Area A = 0.4418 in2 14. The maximum unit flexural stress in the beam is most nearly b.6000 psi a.3000 psi d.16000 psi c. 8000 psi e. 32000 psi R = 750, M = 750 750 f
X
6
X
X
Me I
6
20
X
12
12
X_ยง_
2
8100 psi
Answer (c)
13-49
--##--EEg_z-f___g___f#--Tcft-Eg__og_fsggg_g_f#--h-Estaaf----ffo_-EcffT-g__k#-Mgefgggf-gesBsf 13-50 FUNDAMENTALS OF ENGINEERING EXAM REVIEW WORKBOOK 15. The maximum unit tensile stress in the rod BC is most nearly b.6000 psi a.3000 psi psi d.16000 psi c.8000 psi e.32000 T = 15/9 x 1500 = 2500# f = T/A = 2500/.4418 = 5700#
Answer (b)
MATHEMATICS 16.
The sum of the circumferences of two circles is 36rt. The sum of the areas of the circles is 1807t. The radius of one of the circles is: b. 9 a. 10 d. 7 c. 8 e. 6 2rt (R 1 + R 2 ) = 36rt, R 1 + R 2 = 18 (R 12 + R 22) = 1807t, R 12 + R 22 = 180 R1 = 18- R 2 : (18- R 2 )2 = 180 2R 22- 36R2 + 144 = 0: R2 = 6
1t
Answer (e) 17.
Ground salt is poured in a pile on the ground at the rate of 4 cfm. it forms a conical pile whose height is 30% of the diameter of the base. How fast is the height of the pile increasing when the base is 5 ft in diameter? b.. 687 fpm d. 2.83 fpm
a . .458 fpm c .. 204 fpm e. 5.76 fpm
v v dV dt
7td2 l_Bh; B = - ,. h 4 3 7td3 40 37td2 dd 40 dt
dd dt
dV 40 dt3rtd 2
dh dt
. 3 dd_ . 3 dt
4 __1Q_ 3rt5 2 X
.68
.3d
.68 .204 Answer (c)
T-T_go_o_f-E_of-fdfgft-E-esgf-f.gg#ERsggfdgggggEEffgf--Rfgfggzf--sfsgpffSOLUTIONS FOR THE EIT SIMULATED EXAM PM SESSION 18.
What is the general solution of the differential equation ydx - y2dy = dy? a. x = 3y2+ 2y + C b. X = y3/3 + y + C c. x =esc y + C d. x = y2/2 + In y + c e. x sin y cos y + In y + C ydx
dy + y 2 dy
==
dx = dy + ydy y
x
19.
==
y2 2
+ ln y + C
Answer(d)
A 3- ft long wire is cut into two pieces. One piece is used to form a square and the other piece used to form a circle. If the sum of the enclosed areas is to be a minimum, what length of the wire should be used to form the circle? a. 0.14 ft. b. 0.56 ft. c. 1.12 ft. d. 1.32 ft. e. None of the above L =2m+ 4x dL = 21t + 4 dx dr dr A= m2 + x2
0; dx
dr
dA = 27tr + 21t dx 0 dr dr x=2r L = 3 = 27tr + 4(2r); r = .21 Lcircle =2m= 27t (.21) = 1.32
= 27tr
1t 2 + 2x ( - 7t)
2
Answer(d) 20.
My pencil holder contains 3 red pencils, 4 green pencils, and 3 blue pencils. My secretary has a pencil holder nearby that contains 4 blue pencils, 4 red pencils and 2 green pencils. What is the probability of my reaching out and grasping a blue pencil? a. 7/1 b. 12/1 c. 7/10 d. 12/20 e. 7/20 Total red 3 + 4 = 7 Total green 4 + 2 = 6 Total blue 3+4=7 Total 10 + 10 = 20 Probability of a blue pencil is 7/20. Answer (e)
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#-fg_#Eg_f#f_sgE##3sg_#ffnfsfg___of#-dsss3-pdfg-#--ffgef---dsa_eaE-fFUNDAMEN TALS OF ENGINEER ING EXAM REVIEW WORKBOOK 13-52 Problems 21 and 22 refer to the figure below. Line CD is Perpendicular to line AB.
c
(1,5)
y
(9, 1)
B
X
A
21.
What is the slope of line BD? b .. 33333 a.-5 d. -3 c.-.33333 e. 3 M
22
(0, -2)
!:i.Y = 1 -
(-2)
]_
l
0
9
3
9 -
!:!.X
Answer (b)
What is the area of triangle ABC? b. 6'{10 a. s'(TO d.
c. 26 e. 60
30
We need point D Line AB m
= 1/3, B = -2 by inspection at A y=lx-2 3
Line CD perpendicular to line AB, thus m since y = mx + b: form point C 5 = -3(1) + B; b = 8 y = -3 X + 8
=3
At point D line AB = line CD
1_ X 3
AD BD AABc
=
-3 X + 8; x
l
AABc
CD
2
2
=
3 andy
-1
AD 路 CD + 1_ BD 路 CD
2
(-1)? Y!:!.x 2 + !:!.y 2 = V(3 -1)2 + (s h 2 +1 2 =ID y6 2 + 22 = 2ID = l 2ID 路 flO + l 2ID 路 2ID = 2
=
2
2ID
10 +
(2
Answer (d)
x
10)
30
E-dg_d_fd.cat#ffEfggfgff-EgssgggfEEcf3fg_f---fgffffgzgofEgE--Efg_-PM SESSION THE EIT SIMULATED EXAM FOR
SOLUTIONS
THERMODYNAMICS Problems 23 and 24 refer to the following problem statement Atmospheric air inlet to an air cooler has a specific humidity of .009 lbs water vaporflbs dry air at 80°F. Calculate the following. 23. Density of dry air a .. 0726 lb/ft3 c .. 0459 lb/ft3 e .. 0858 lb/ft3
b .. 0680 lb/ft3 d .. 0472 lb/ft3
See chart solution. Answer(a) 24. Enthalpy of the water vapor a. 41.5 Btullbs dry air b. 32.5 Btuflbs dry air c. 29.0 Btullbs dry air d. 39.6 Btuflbs dry air e. 24.8 Btullbs dry air See chart solution. Answer(a) FLUID MECHANICS Problems 25 and 26 refer to the following problem statement. A radial gate, 5 ft wide, is installed as shown.
Sea Water
16'
w = 65.0 p
25.
Neglecting the weight of the gate and any mechanical forces, such as friction, what force Pis required to open the gate? (Answers are rounded off. a. 100,000 lbs. c. 25,000 lbs. e. 0 lbs.
b. 50,000 lbs. d. 12,500 lbs.
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i--Efsff_fg__EE-fgfEEhsgT#nadf_gEEs_fMfggffEg__-gfa OF ENGINEERING EXAM REVIEW FUNDAMENTALS
13-54
Force of sea water acting down on the gate
&X 5
2
X
5
65 = 6400 lbs
X
4 Force of sea water acting horizontally against the gate
6 5 ( 16 + 21 ) 5
x
2 Force required to open gate is 0
5
=
3 0 , 0 6 0 lbs Answer(e)
26.
What force acts against the sea wall?
b. 40960 lb. a. 41600 lb. c. 79872lb. d. 39936lb. e. none of these p
=
5
65(16) 2 2
X
5
41,600 lbs Answer(a)
WORKBOOK