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Mathematics for construction

Table of ContentsTASK 1............................................................................................................................................3TASK 2............................................................................................................................................9Scenario 1.........................................................................................................................................9a. Producing histogram with frequency density.....................................................................9b. Producing cumulative frequency curve and extracting median and interquartile range. .11c. Extracting mean, range and standard deviation................................................................13Scenario 2.......................................................................................................................................13A.a Normal distribution test.................................................................................................13A.b Deciding interval allowed for replacement of not more than 10% prior to replacementType A..................................................................................................................................13A.c. Practical consideration for replacement policy.............................................................14A.d Replacement time affected Type B...............................................................................14A.e Increment of 25% cost is preferable or not....................................................................15A.f. Type C..........................................................................................................................15B Testing hypothesis with 5% level of significance with both two and one tailed test.......15TASK 3.........................................................................................................................................18TASK 4.........................................................................................................................................22

TASK 1Question – 1 A building services engineer is to design a water tank for a project. The tank has arectangular area of 26.5m2. With the design specifics of the width being 3.2m shorter than thelength, calculate the length and width to 3 significant figures for resource requirementsAns :width= 3.2 mlength = 3.2 marea = 26.5 meter squarearea of rectangle= l X b2225 = l x 3.2l= 26.5 / 3.2l = 8.28wherewidth = 3 marea = l x b26.5= lx 3l= 26.5 / 3l= 8.83 m

Question – 2 As an employee of company JR construction you have received a letter regardingthe project your company is working on. It has a penalty clause that states the contactor willforfeit a certain some of money each day for late completion. (i.e. the contractor gets paid thevalue of the original contract less any sum forfeit). If she is 5 days late she receives £4250 and ifshe is 12 days late she receives £2120. Calculate the daily forfeit and determine the originalcontract.Ans :Let the original contract amount be x and per day forfeit be y.When work is delayed by 5 days:x – 5y = 4250(1)Similarly when work is delayed by 12 days:x – 12y = 2120(2)On solving equations 1 and 2 we gety = 304.28 andx = 4250 +(5*304.28) = 5771.4ThusDaily forfeit = £304.28Original contract amount = £5771.4Question – 3 A car driving at an average speed 65 miles/hourAns :A)1). Average speed= 65 miles/ hour [1 mile = 1760 yards]65 miles per hour = 65 x 1.609 = 104.585 yard/ second65 miles per hour = 29.05746 meter per second2). Time = distance / speedTime = 100/ 29Time = 3.44 hour/ second

3). 30 miles / gallon(miles/100 km ) x(liters/gallon) / (miles/gallon)(3.785 * 62.14)/30 = 235.19/30 = 7.84 liters/100 km4). 100 km/ 7.84 liter = 12.755 km per litre= 7.84 l x 100 /100= 7. 84 litre for the journey.B). Formula → Lift =k x ρ x V^2 x Alift = meter2 x Kg/meter3 x in2Lift = Kg.m/s2 = newtons or slug1. An arithmetic sequence is given by b, , , 0....... 3 3 2b b Determine the sixth term Statethe kth term If the 20th term has value of 15 find the value of b and the sum of the first 20termsAns :a).a = a+(n-1)da = b(6-1)b/3a = bx 5b/3a = 8b/3 6thtermsb). The k term is given by :a(k) = b + (k-1)* (b/3)c). a(20)= b +(20-1)*(b/3)15 = (3b +19b)/3

45= 22bb= 2.04Sum of n terms of A.P. = (n/2)*[2a + (n-1)*d]n = 20 , a= b= 2.04, d =b/3 =0.68On substituting the values in equation we getS(20) = 10 * [4.08 + 12.92]= 170Thus the sum of first 20 terms of given AP is 1702. For the following geometric progression 1, ½, 1/4 ........ determinea) The 20th term of the progression = a(1- r^n) / 1-r= (1-(1/2)^20) / 1- ½= 39 x 2 / 40= 39 / 20b). Sn =∑ (n=0) ar^nSn = 1/2(1-r^∞) / 1-rwhen the value of r is infinity it means the entire system itSolve the following Equations for x :(a) 2Log (3x) + Log (18x) = 27Ans: log ((3x^2)) + log (18x) = 27log ((3x)^2 . 18X ) = 27log ((3x)^2 . 18X ) = log (10^27)(3x)^2.18x = 10^27

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