Numerical Methods Exercise 5.2 Text Book Solutions

Dear Class 12 Samacheer Kalvi students, here are the Business Maths Chapter 5 Numerical Methods Exercise 5.2 Text Book Solutions for your study.

Numerical Methods Exercise 5.2 Text Book Solutions

1. Using graphic method, find the value of y when x = 48 from the following data:

x	40	50	60	70
y	6.2	7.2	9.1	12

Answer: Plot the points (40,6.2), (50,7.2), (60,9.1) and (70,12)

Scale: x axis: 1 cm = 10 units; y axis: 1 cm = 1 unit;

2. The following data relates to indirect labour expenses and the level of output.

Months	Jan	Feb	Mar	Apr	May	Jun
Units of output	200	300	400	640	540	580
Indirect labour expenses (Rs.)	2500	2800	3100	3820	3220	3640

Estimate the expenses at a level of output of 350 units, by using graphic method.

Answer: Plot the points (200,2500), (300,2800), (400,3100), (640, 3820), (540, 3220), (580,3640)

From the graph, when x=350, y=2900.

Expenses of 350 units = Rs. 2900

Scale: x axis: 1 cm = 100 units; y axis: 1 cm = 500 units

3. Using Newton’s forward interpolation formula find the cubic polynomial.

x	0	1	2	3
f(x)	1	2	1	10

Answer: Newton’s forward interpolation formula

$y=y_{0}+\dfrac{n}{1!}\: \Delta (\; y_{0})+\dfrac{n(n-1)}{2!}\: \Delta^{2} (\; y_{0})+\dfrac{n(n-1)(n-2)}{3!}\: \Delta^{3} (\; y_{0})+ \cdots$

$x=x;\; x_{0} =0; \; h=1\\$
$n=\frac{x-x_{0}}{h}\\$
$n=\frac{x-0}{1}\\$
$n=x$

The difference table is:

x	y	∆y	∆² y	∆³ y
0	1
1	2	1
2	1	-1	-2
3	10	9	10	12

$y=1+\dfrac{x}{1!} (1)+\dfrac{x(x-1)}{2!}\: (-2)+\dfrac{x(x-1)(x-2)}{3!}\: (12)\\$
$=1+\dfrac{x}{1}+\dfrac{x^{2}-x}{2} (-2)+ \dfrac{x(x^{2}-3x+2)}{6}(12)\\$
$=1+x-(x^{2}-x)+2 x(x^{2}-3x+2)\\$
$=1+x-x^{2}+x+2x^{3}-6x^{2}+4x\\$
$y= f(x)=2x^{3}-7x^{2}+6x+1$

4. The population of a city in a censes taken once in 10 years is given below. Estimate the population in the year 1955.

Year	1951	1961	1971	1981
Population in lakhs	35	42	58	84

$x=1955;\; x_{0} =1951; \; h=10\\$
$n=\dfrac{x-x_{0}}{h}\\$
$n=\dfrac{1955-1951}{10}\\$
$n=\dfrac{4}{10}$
$n=.4$

Since 1955 lies between 1951 and 1961, we can use Newton’s forward interpolation formula.

$y=y_{0}+\dfrac{n}{1!}\: \Delta (\; y_{0})+\dfrac{n(n-1)}{2!}\: \Delta^{2} (\; y_{0})+\dfrac{n(n-1)(n-2)}{3!}\: \Delta^{3} (\; y_{0})+ \cdots$

The difference table is:

x	y	∆y	∆² y	∆³ y
1951	35
1961	42	7
1971	58	16	9
1981	84	26	10	1

$y=35+\dfrac{.4}{1!} (7)+\dfrac{.4(.4-1)}{2!}\: (9)+\dfrac{.4(.4-1)(.4-2)}{3!}\: (1)\\$
$=35+\dfrac{2.8}{1}+\dfrac{.4(-.6)}{2} (9)+ \dfrac{.4(-.6)(-1.6)}{6}(6)\\$
$=35+2.8-1.08+0.064\\$
$y= 37.784\; lakhs$

5. In an examination the number of candidates who secured marks between certain interval were as follows:

Marks	0-19	20-39	40-59	60-79	80-99
No of candidates	41	62	65	50	17

Estimate the number of candidates whose marks are less than 70.

Marks	No of candidates	Cumulative frequency
-.5-19.5	41	41
19.5-39.5	62	103
39.5-59.5	65	168
59.5-79.5	50	218
79.5-99.5	17	235

x	y	∇y	∇² y	∇³ y	∇⁴ y
Less than 19.5	41
Less than 39.5	103	62
Less than 59.5	168	65	3
Less than 79.5	218	50	-15	-18
Less than 99.5	235	17	-33	-18	0

Since we need the number of candidates whose marks are less than 70, we can use Newton’s backward interpolation formula.

$y=y_{n}+\dfrac{n}{1!} \nabla y_{n}+\dfrac{n(n+1)}{2!} \nabla^{2} y_{n}+\dfrac{n(n+1)(n+2)}{3!} \nabla^{3} y_{n}+\dfrac{n(n+1)(n+2)(n+3)}{4!} \nabla^{4} y_{n}+\cdots\\$
$x_{n}=99; \; x=70; \; h=20\\$
$n=\dfrac{x-x_{n}}{h}\\$
$n=\dfrac{70-99}{20}\\$
$n=-1.45\\$
$y=235+\dfrac{-1.45}{1!} (17)+\dfrac{-1.45(-1.45+1)}{2!} (-33)+\dfrac{-1.45(-1.45+1)(-1.45+2)}{3!} (-18) \\$
$+\dfrac{-1.45(-1.45+1)(-1.45+2)(-1.45+3)}{4!} (0)\\$
$y =235+\dfrac{-1.45}{1} (17)+\dfrac{-1.45(-0.45)}{2} (-33) \\$
$+\dfrac{-1.45(-0.45)(0.55)}{6} (-18)+ 0\\$
$y=235 -24.65-10.76625-1.076625\\$
$y=198.507125\\$

6. Find the value of f (x) when x = 32 from the following table:

x	30	35	40	45	50
f(x)	15.9	14.9	14.1	13.3	12.5

Since x=32 lies between 30 and 35, we can use Newton’s forward interpolation formula.

x	y	∆y	∆² y	∆³ y	∆⁴ y
30	15.9
35	14.9	-1
40	14.1	-0.8	0.2
45	13.3	-0.8	0	-0.2
50	12.5	-0.8	0	0	0.2

$y=y_{0}+\dfrac{n}{1!}\: \Delta (\; y_{0})+\dfrac{n(n-1)}{2!}\: \Delta^{2} (\; y_{0})+\dfrac{n(n-1)(n-2)}{3!}\: \Delta^{3} (\; y_{0})+ \cdots$

$x_{n}=30; \; x=32; \; h=5\\$
$n=\dfrac{x-x_{n}}{h}\\$
$n=\dfrac{32-30}{5}\\$
$n=\dfrac{2}{5}\\$
$n=-0.4\\$

$y=15.9+\dfrac{.4}{1!}\: (-1)+\dfrac{.4(.4-1)}{2!}\: (.2)+\dfrac{.4(.4-1)(.4-2)}{3!}\: (-0.2)\\$
$+\dfrac{.4(.4-1)(.4-2)(.4-3)}{4!}\: (0.2)\\$
$y=15.9-\dfrac{.4}{1}\:+\dfrac{.4(-.6)}{2}\: (.2)+\dfrac{.4(-.6)(-1.6)}{6} \\$
$(-0.2)+\dfrac{.4(-.6)(-1.6)(-2.6)}{24}\: (0.2)\\$
$y=15.9-.4\:-0.024 -0.0.128 -0.00832\\$
$y=15.45$

7. The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy.

P	40	50	60	70	80	90
T	180	204	226	250	276	304

Find the melting point of the alloy containing 84 percent lead.

Since x=84 lies between 80 and 90, we can use Newton’s backward interpolation formula.

$y=y_{n}+\dfrac{n}{1!} \nabla y_{n}+\dfrac{n(n+1)}{2!} \nabla^{2} y_{n}+\dfrac{n(n+1)(n+2)}{3!} \nabla^{3} y_{n}+\dfrac{n(n+1)(n+2)(n+3)}{4!} \nabla^{4} y_{n}+\cdots\\$

x	y	∇y	∇² y	∇³ y	∇⁴ y	∇⁴ y
40	180
50	204	24
60	226	22	-2
70	250	24	2	4
80	276	26	2	0	-4
90	304	28	2	0	0	4

$x_{n}=90; \; x=84; \; h=10\\$
$n=\dfrac{x-x_{n}}{h}\\$
$n=\dfrac{84-90}{10}\\$
$n=\dfrac{-6}{10}\\$
$n=-0.6\\$

$y=304+\dfrac{-0.6}{1!} (28)+\dfrac{-.6(-.6+1)}{2!} (2)+0 +0 +\dfrac{-.6(-.6+1)(-.6+2)(-.6+3)(-.6+4)}{5!} (4)\\$
$y=304-16.8-0.24+0+0-0.91392\\$
$y=286.868\\$

8. Find f (2.8) from the following table:

x	0	1	2	3
f(x)	1	2	11	34

Since 2.8 lies between 2 and 3, we can use Newton’s backward interpolation formula.

$y=y_{n}+\dfrac{n}{1!} \nabla y_{n}+\dfrac{n(n+1)}{2!} \nabla^{2} y_{n}+\dfrac{n(n+1)(n+2)}{3!} \nabla^{3} y_{n}+\dfrac{n(n+1)(n+2)(n+3)}{4!} \nabla^{4} y_{n}+\cdots\\$

$x_{n}=3; \; x=2.8; \; h=1\\$
$n=\dfrac{x-x_{n}}{h}\\$
$n=\dfrac{2.8-3}{1}\\$
$n=-0.2\\$

x	y	∇y	∇² y	∇³ y
0	1
1	2	1
2	11	9	8
3	34	23	14	6

$y=34+\dfrac{-0.2}{1!} (23)+\dfrac{-0.2(-0.2+1)}{2!} (14)+\dfrac{-0.2(-0.2+1)(-0.2+2)}{3!} (6)\\$
$y=34-0.2(23)+\dfrac{-0.2(0.8)}{2} (14)+\dfrac{-0.2(0.8)(1.8)}{6} (6)\\$
$y=34-4.6-1.12-0.288\\$
$y=27.992\\$

9. Using interpolation estimate the output of a factory in 1986 from the following data.

Year	1974	1978	1982	1990
Output in 1000 tonnes	25	60	80	170

Answer: Since the intervals are unequal, we can use LaGrange’s formula.

$y=\dfrac{(x-x_{1})(x-x_{2})(x-x_{3})}{(x_{0}-x_{1})(x_{0}-x_{2})(x_{0}-x_{3})} y_{0}+ \\$
$\dfrac{(x-x_{0})(x-x_{2})(x-x_{3})}{(x_{1}-x_{0})(x_{1}-x_{2})(x_{1}-x_{3})} y_{1}+ \\$
$\dfrac{(x-x_{0})(x-x_{1})(x-x_{3})}{(x_{2}-x_{0})(x_{2}-x_{1})(x_{2}-x_{3})} y_{2}+ \\$
$\dfrac{(x-x_{0})(x-x_{1})(x-x_{2})}{(x_{3}-x_{0})(x_{3}-x_{1})(x_{3}-x_{2})} y_{3} \\$

$y=\dfrac{(1986-1978)(1986-1982)(1986-1990)}{(1974-1978)(1974-1982)(1974-1990)} 25+ \\$
$\dfrac{(1986-1974)(1986-1982)(1986-1990)}{(1978-1974)(1978-1982)(1978-1990)}60+ \\$
$\dfrac{(1986-1974)(1986-1978)(1986-1990)}{(1982-1974)(1982-1978)(1982-1990)} 80+ \\$
$\dfrac{(1986-1974)(1986-1978)(1986-1982)}{(1990-1974)(1990-1978)(1990-1982)} 170 \\$

$y=\dfrac{(8)(4)(-4)}{(-4)(-8)(-16)} 25+\dfrac{(12)(4)(-4)}{(4)(-4)(-12)}60+ \\$
$\dfrac{(12)(8)(-4)}{(8)(4)(-8)} 80+ \dfrac{(12)(8)(4)}{(16)(12)(8)} 170 \\$
$y=6.25-60+120+42.5 \\$
$y=108.75 \\$

10. Use Lagrange’s formula and estimate from the following data the number of workers getting income not exceeding Rs. 26 per month.

Income not exceeding(Rs.)	15	25	30	35
No. of workers	36	40	45	48

$y=\dfrac{(26-25)(26-30)(26-35)}{(15-25)(15-30)(15-35)} 36+ \\$
$\dfrac{(26-15)(26-30)(26-35)}{(25-15)(25-30)(25-35)}40+ \\$
$\dfrac{(26-15)(26-25)(26-35)}{(30-15)(30-25)(30-35)} 45+ \\$
$\dfrac{(26-15)(26-25)(26-30)}{(35-15)(35-25)(35-30)} 48\\$

$y=\dfrac{(1)(-4)(-9)}{(-10)(-15)(-20)} 36+ \dfrac{(11)(-4)(-9)}{(10)(-5)(-10)}40+ \\$
$\dfrac{(11)(1)(-9)}{(15)(5)(-5)} 45+ \dfrac{(11)(1)(-4)}{(20)(10)(5)} 4\\$

$y=-0.432+31.68+11.88-2.112\\$
$y= 41.016\\$
$y= 41 \; persons\\$

11. Using interpolation estimate the business done in 1985 from the following data.

Year	1982	1983	1984	1986
Business done (in lakhs)	150	235	365	525

Answer: Since the intervals are unequal, we can use LaGrange’s formula.

$y=\dfrac{(1985-1983)(1985-1984)(1985-1986)}{(1982-1983)(1982-1984)(1982-1986)} 150+ \\$
$\dfrac{(1985-1982)(1985-1984)(1985-1986)}{(1983-1982)(1983-1984)(1983-1986)}235+ \\$
$\dfrac{(1985-1982)(1985-1983)(1985-1986)}{(1984-1982)(1984-1983)(1984-1986)} 365+ \\$
$\dfrac{(1985-1982)(1985-1983)(1985-1984)}{(1986-1982)(1986-1983)(1986-1984)} 525\\$

$y=\dfrac{(2)(1)(-1)}{(-1)(-2)(-4)} 150+ \dfrac{(3)(1)(-1)}{(1)(-1)(-3)}235+ \\$
$\dfrac{(3)(2)(-1)}{(2)(1)(-2)} 365+ \dfrac{(3)(2)(1)}{(4)(3)(2)} 525\\$
$y=37.5-235+547.5+131.25 \\$
$y=481.25\; lakhs \\$

12. Using interpolation, find the value of f(x) when x = 15

x	3	7	11	19
f(x)	42	43	47	60

Answer: Since the intervals are unequal, we can use LaGrange’s formula.

$y=\dfrac{(15-7)(15-11)(15-19)}{(3-7)(3-11)(3-19)} 42+ \\$
$\dfrac{(15-3)(15-11)(15-19)}{(7-3)(7-11)(7-19)}43+ \\$
$\dfrac{(15-3)(15-7)(15-19)}{(11-3)(11-7)(11-19)} 47+ \\$
$\dfrac{(15-3)(15-7)(15-11)}{(19-3)(19-7)(19-11)} 60\\$

$y=\dfrac{(8)(4)(-4)}{(-4)(-8)(-16)} 42+ \dfrac{(12)(4)(-4)}{(4)(-4)(-12)}43+ \\$
$\dfrac{(12)(8)(-4)}{(8)(4)(-8)} 47+ \dfrac{(12)(8)(4)}{(16)(12)(8)} 60\\$
$y=10.5-43+70.5+15 \\$
$y=53\\$

Numerical Methods Exercise 5.2 Text Book Solutions Business Maths Class 12

Numerical Methods Exercise 5.2 Text Book Solutions

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