Revision Test 3 : Ratio, proportion, powers, roots, indices and units
11.3 Practical problems involving simple equations
Solving simple equations 77
8. 2 a =3
8 9. 1
3n+ 1 4n = 7
24 10. x+3
4 =x−3 5 +2 11. 3t
20=6−t 12 +2t
15−3 2 12. y
5+ 7
20=5−y 4 13. v−2
2v−3 =1 3 14. 2
a−3 = 3 2a+1 15. x
4−x+6
5 =x+3 2 16. 3√
t=9 17. 2√
y=5 18. 4=
3 a
+3
19. 3√ x 1−√
x = −6 20. 10=5 x
2−1
21. 16=t2 9 22.
y+2 y−2
=1 2 23. 6
a =2a 3 24. 11
2 =5+ 8 x2
11.3 Practical problems involving
78 Basic Engineering Mathematics
Using a calculator,gas constant,R=4160 J/(kg K), correct to 4 significant figures.
Problem 20. A rectangular box with square ends has its length 15 cm greater than its breadth and the total length of its edges is 2.04 m. Find the width of the box and its volume
Letxcm=width=height of box. Then the length of the box is(x+15)cm, as shown in Figure 11.1.
(x115) x
x
Figure 11.1
The length of the edges of the box is 2(4x)+4(x+15) cm, which equals 2.04 m or 204 cm.
Hence, 204=2(4x)+4(x+15) 204=8x+4x+60 204−60=12x
i.e. 144=12x
and x=12 cm
Hence,the width of the box is 12 cm.
Volume of box=length×width×height
=(x+15)(x)(x)=(12+15)(12)(12)
=(27)(12)(12)
=3888 cm3
Problem 21. The temperature coefficient of resistanceαmay be calculated from the formula Rt=R0(1+αt). Findα, givenRt=0.928, R0=0.80 andt=40
SinceRt=R0(1+αt),then
0.928=0.80[1+α(40)]
0.928=0.80+(0.8)(α)(40) 0.928−0.80=32α
0.128=32α
Hence, α=0.128
32 =0.004 Problem 22. The distancesmetres travelled in timet seconds is given by the formula s=ut+1
2at2, whereuis the initial velocity in m/s andais the acceleration in m/s2. Find the
acceleration of the body if it travels 168 m in 6 s, with an initial velocity of 10 m/s
s=ut+1
2at2, ands=168,u=10 andt=6
Hence, 168=(10)(6)+1
2a(6)2 168=60+18a
168−60=18a 108=18a a=108
18 =6
Hence,the acceleration of the body is 6 m/s2. Problem 23. When three resistors in an electrical circuit are connected in parallel the total resistance RT is given by 1
RT = 1 R1+ 1
R2+ 1 R3
. Find the total resistance whenR1=5,R2=10and R3=30
1 RT =1
5+ 1 10+ 1
30 =6+3+1
30 =10
30=1 3 Taking the reciprocal of both sides givesRT =3 Alternatively, if 1
RT =1 5+ 1
10+ 1
30, the LCM of the denominators is 30RT.
Hence, 30RT
1 RT
=30RT
1 5
+30RT
1 10
+30RT
1 30
. Cancelling gives 30=6RT+3RT +RT
i.e. 30=10RT
and RT =30
10=3,as above.
Solving simple equations 79
Now try the following Practice Exercise Practice Exercise 44 Practical problems involving simple equations (answers on page 344)
1. A formula used for calculating resistance of a cable isR=ρL
a . GivenR=1.25,L=2500 anda=2×10−4, find the value ofρ.
2. ForceFnewtons is given byF=ma, where mis the mass in kilograms andais the accel-eration in metres per second squared. Find the acceleration when a force of 4 kN is applied to a mass of 500 kg.
3. PV=m RT is the characteristic gas equa-tion. Find the value ofmwhen
P=100×103,V=3.00,R=288 and T =300.
4. When three resistorsR1,R2andR3are con-nected in parallel, the total resistanceRT is determined from 1
RT = 1 R1+ 1
R2+ 1 R3
(a) Find the total resistance when R1=3,R2=6andR3=18. (b) Find the value ofR3given that
RT =3,R1=5andR2=10. 5. Six digital camera batteries and 3 camcorder
batteries cost £96. If a camcorder battery costs £5 more than a digital camera battery, find the cost of each.
6. Ohm’s law may be represented byI=V/R, where I is the current in amperes, V is the voltage in volts and R is the resistance in ohms. A soldering iron takes a current of 0.30 A from a 240 V supply. Find the resistance of the element.
7. The distance,s, travelled in timetseconds is given by the formulas=ut+1
2a t2 where u is the initial velocity in m/s and a is the acceleration in m/s2. Calculate the accelera-tion of the body if it travels 165 m in 3 s, with an initial velocity of 10 m/s.
Here are some further worked examples on solving simple equations in practical situations.
Problem 24. The extensionxm of an aluminium tie bar of lengthlm and cross-sectional area Am2 when carrying a load ofFNewtons is given by the modulus of elasticityE=Fl/Ax. Find the extension of the tie bar (in mm) if
E=70×109N/m2,F=20×106N,A=0.1m2 andl=1.4 m
E=Fl/Ax,hence 70×109 N
m2 =(20×106N)(1.4 m) (0.1m2)(x)
(the unit ofxis thus metres) 70×109×0.1×x=20×106×1.4
x=20×106×1.4 70×109×0.1 Cancelling gives x= 2×1.4
7×100m
= 2×1.4
7×100×1000 mm
=4 mm
Hence,the extension of the tie bar,x=4 mm.
Problem 25. Power in a d.c. circuit is given by P=V2
R whereV is the supply voltage andRis the circuit resistance. Find the supply voltage if the circuit resistance is 1.25and the power measured is 320 W
Since P=V2
R , then 320= V2 1.25 (320)(1.25)=V2
i.e. V2=400
Supply voltage, V=√
400= ±20 V Problem 26. A painter is paid £6.30 per hour for a basic 36 hour week and overtime is paid at one and a third times this rate. Determine how many hours the painter has to work in a week to earn £319.20 Basic rate per hour=£6.30 and overtime rate per hour
=11
3×£6.30=£8.40 Let the number of overtime hours worked=x Then, (36)(6.30)+(x)(8.40)=319.20
226.80+8.40x=319.20
80 Basic Engineering Mathematics
8.40x=319.20−226.80=92.40 x=92.40
8.40 =11
Thus, 11 hours overtime would have to be worked to earn £319.20 per week. Hence, the total number of hours workedis 36+11, i.e.47 hours.
Problem 27. A formula relating initial and final states of pressures,P1andP2, volumes,V1andV2, and absolute temperatures,T1andT2, of an ideal gas is P1V1
T1 = P2V2
T2
. Find the value ofP2given P1=100×103,V1=1.0,V2=0.266,T1=423 andT2=293
Since P1V1
T1 = P2V2
T2
then(100×103)(1.0)
423 = P2(0.266) 293 Cross-multiplying gives
(100×103)(1.0)(293)=P2(0.266)(423) P2=(100×103)(1.0)(293)
(0.266)(423) Hence, P2=260×103or2.6×105.
Problem 28. The stress, f, in a material of a thick cylinder can be obtained from
D d =
f +p f −p
. Calculate the stress, given that D=21.5,d=10.75 andp=1800
Since D d =
f+p f−p
then 21.5 10.75=
f +1800 f −1800
i.e. 2=
f +1800 f −1800
Squaring both sides gives 4= f+1800 f−1800 Cross-multiplying gives
4(f −1800)= f+1800 4f −7200= f+1800
4f− f =1800+7200 3f =9000
f =9000 3 =3000
Hence, stress,f=3000
Problem 29. 12 workmen employed on a
building site earn between them a total of £4035 per week. Labourers are paid £275 per week and craftsmen are paid £380 per week. How many craftsmen and how many labourers are employed?
Let the number of craftsmen be c. The number of labourers is therefore (12−c).
The wage bill equation is
380c+275(12−c)=4035 380c+3300−275c=4035
380c−275c=4035−3300 105c=735
c=735 105=7
Hence, there are 7 craftsmen and (12−7), i.e. 5 labourerson the site.
Now try the following Practice Exercise Practice Exercise 45 Practical problems involving simple equations (answers on page 344)
1. A rectangle has a length of 20 cm and a width bcm. When its width is reduced by 4 cm its area becomes 160 cm2. Find the original width and area of the rectangle.
2. GivenR2=R1(1+αt), findαgiven R1=5.0,R2=6.03 andt=51.5 3. Ifv2=u2+2as, findugivenv=24,
a= −40 ands=4.05
4. The relationship between the temperature on a Fahrenheit scale and that on a Celsius scale is given by F=9
5C+32. Express 113◦F in degrees Celsius.
5. Ift=2π w
Sg, find the value ofSgiven w=1.219,g=9.81 andt=0.3132
6. Two joiners and five mates earn £1824 between them for a particular job. If a joiner earns £72 more than a mate, calculate the earnings for a joiner and for a mate.
Solving simple equations 81
7. An alloy contains 60% by weight of copper, the remainder being zinc. How much copper must be mixed with 50 kg of this alloy to give an alloy containing 75% copper?
8. A rectangular laboratory has a length equal to one and a half times its width and a perimeter of 40 m. Find its length and width.
9. Applying the principle of moments to a beam results in the following equation:
F×3=(5−F)×7
where F is the force in newtons. Determine the value ofF.