L)-New-2016-Mathematics Exam

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Jun 14, 2017 - r=Anti-log{1.3222 βˆ’. 1 ..... Find the volume of the cylinder in terms of and a .... lmn=Nd3,N=lmn/d3 if l=m=n=1,d=1/2,N=1/(1/2)3 =1/(1/8)=8 cubic blocks of ..... (a)The first term of a given arithmetic progression is 3 and and the 11th term is 43 .... In the geometric series sum of six terms can be made to be 9.
Solutions to G.C.E(O/L)-New-2016-Mathematics Exam Paper questions Topic: Volume Question -1(G.C.E(O/L)-2016) Solid spherical glass ball of radius 21c.m is melted and 240 identical solid cylindrical glass discs are made. Assume that there is no change in the volume of glass in this process .If the radius of each disc is r centimeters and height is r/9 centimeters, show that r=21/201/3 and, using logarithm table ,find the value of r correct to two decimal places. Solution 4 4 The volume of the spherical glass ball is= 3 πœ‹π‘… 3 = 3 πœ‹213 c.m3 π‘Ÿ

1

The volume of each identical cylindrical glass disc= πœ‹π‘Ÿ 2 β„Ž = πœ‹π‘Ÿ 2 (9) = 9 πœ‹π‘Ÿ 3 c.m3 Since there is no change in the volume in the process of making 240 identical glass discs 240 4 πœ‹π‘Ÿ 3 = 3 πœ‹213 9 20r3=213 213

π‘Ÿ3 = π‘Ÿ=

21 3

√20

21

,π‘Ÿ =

20

1

203

c .m 1

π‘™π‘œπ‘”10 π‘Ÿ = π‘™π‘œπ‘”10 21 βˆ’ π‘™π‘œπ‘”10 203 1 =π‘™π‘œπ‘”10 21 βˆ’ 3 π‘™π‘œπ‘”10 20 {π‘™π‘œπ‘”

1

r=10 10 21βˆ’3π‘™π‘œπ‘”10 20} 1 r=Anti-log{π‘™π‘œπ‘”10 21 βˆ’ 3 π‘™π‘œπ‘”10 20} 1

r=Anti-log{1.3222 βˆ’ 3 1.3010} r=Anti-log{1.3222 βˆ’ 0.4336} r=Anti-log{0.8886}=100.8886=7.7374 c .m r=7.73c .m up to two decimals correctly π‘Ÿ 4 π‘Ÿ 𝑁 If β„Ž = 𝑛 π‘‘β„Žπ‘’π‘› 3 πœ‹π‘… 3 = π‘πœ‹π‘Ÿ 2 β„Ž = π‘πœ‹π‘Ÿ 2 𝑛 = 𝑛 πœ‹π‘Ÿ 3 4

𝑅3 = 3

𝑁

4𝑛

3

4𝑛

4𝑛 1

π‘Ÿ 3 , 3𝑁 𝑅 3 = π‘Ÿ 3 , π‘Ÿ = √3𝑁 𝑅 = (3𝑁)3 𝑅 𝑛

N=The number of discs π‘Ÿ β„Ž = = π‘œπ‘›π‘’ π‘›π‘‘β„Ž π‘œπ‘“ π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  in height 𝑛

If N=1000=103 and 24

1 3

1 3

8

23 1

n=6 then π‘Ÿ = (3000) 𝑅 = (1000) 𝑅 = (103 )3 𝑅 =

3

2 3 (10)

1

𝑅 = 5 𝑅=0.2R=one fifth of the radius of

the glass ball If N=4,n=3 then r=R=The radius of the glass ball are few prototype examples for the forth coming exams for new students. If N glass cones were made of height h=R of base radius r0 instead of making glass discs 4

1

1

then 3 πœ‹π‘… 3 = 𝑁 (3 πœ‹π‘Ÿ02 β„Ž) = 𝑁 3 πœ‹π‘Ÿ02 𝑅

2

4

22

4𝑅 2 = π‘π‘Ÿ02 ,π‘Ÿ02 = 𝑁 𝑅 2 , π‘Ÿ0 = √ 𝑁 𝑅 =

𝑅 βˆšπ‘

If N=10000=1002 =ten thousand then r0 =R/100=One hundredth of the radius of the ball is a far more mathematically realistic thought full problem than the one given. If glass ball was broken into discs of decreasing radius it is mathematically thinkable. Instead of melting glass if clay was given as the material problem is practical for students. Glass melting need high temperatures in industry not good mathematically may be good in science of physics . Another case if a water droplet of radius R is broken into N number of small water droplets of 4 4 𝑅 radius r0 then 3 πœ‹π‘… 3 = 𝑁 3 πœ‹π‘Ÿ03 , π‘Ÿ0 = 3 βˆšπ‘

6

3

𝑅

If N=1000000=10 =100 =ten lack then π‘Ÿ0 = 100 = 𝑂𝑛𝑒 β„Žπ‘’π‘›π‘‘π‘Ÿπ‘’π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘‘π‘Ÿπ‘œπ‘π‘™π‘’π‘‘ This type of a thing can happen in atmosphere while raining. Another problem is if Coca Cola company is planning to design Coke metal Can into a higher radius & smaller height with out changing Coca Cola liquid volume how high it will be if radius is n times the original cylindrical shape with preservation of cylindrical the shape . πœ‹π‘Ÿ 2 β„Ž = πœ‹(π‘›π‘Ÿ)2 𝐻 = 𝑉 = 𝑓𝑖π‘₯𝑒𝑑 π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ π‘œπ‘“ πΆπ‘œπ‘π‘Ž πΆπ‘œπ‘™π‘Ž π‘™π‘–π‘žπ‘’π‘–π‘‘ β„Ž

𝐻 = 𝑛2 & if new radius is twice the original then height is one fourth of the original Coke Can. H=h/4 but providing same amount of liquid to the customer in thirst is a practical problem & mathematically interesting. Another one if Coke Can shape becomes Conic like a Ice Cream but radius enlarged then how high the Coke Cone be if Coca Cola company planning to design with out changing the liquid capacity to extinguish the thirst in some seasons of humans . 1 πœ‹π‘Ÿ 2 β„Ž = 3 πœ‹(π‘›π‘Ÿ)2 𝐻 = 𝑉 3β„Ž

3

𝐻 = 𝑛2 if n=2 then H=4 β„Ž Similar things can be worked out if radius decreased as well then height get increases. If heat needed to increase one kilogram of glass to raise it’s temperature by one degree is S known as the specific heat capacity of the substance under consideration then the heat required to raise the glass ball to it’s melting temperature is m S(πœƒπ‘šπ‘’π‘™π‘‘π‘–π‘›π‘” βˆ’ πœƒπ‘Ÿπ‘œπ‘œπ‘š ) = 𝑄 is the physics behind & where m is mass of the glass ball. There is a prototype question not the above exam question of G.C.E(O/L) 2016 given. There are only two problems of this type given available from them. Question -2(G.C.E(O/L)-2016-Prototype) The height of a solid right circular cylindrical metal block of radius 10.5c.m is 20c.m .When this cylinder was heated and 25 identical solid metal spheres were made,230c.m3 of metal was left over . 22 (i.) Taking πœ‹ = 7 ,calculate the volume of the cylindrical metal block (ii.) What is the volume of one the spheres that was made? (iii.) If the radius of one of the spheres that was made denoted by r ,by assuming that πœ‹ = 3.14 and using logarithms table ,find the value of r3 to the nearest whole number

(iv.)By using the value obtained above for r3,find the radius of the sphere Solution Answers for (1.),(ii.),(iii),(iv) 22 21 22 21 21 The volume of the cylinder= πœ‹π‘… 2 β„Ž= 7 Γ—( 2 )2 Γ—20𝑐. π‘š3 = 7 Γ— 2 Γ— 2 Γ—20𝑐. π‘š3 = 11Γ—3Γ—

21 2

693

𝑐. π‘š3=

2

𝑐. π‘š3 =346.5c.m3

The metal left over =230 c.m3 The metal used=(346.5-230)c.m3=116.5c.m3=The volume of 25 identical metal 4πœ‹ 1 22 100 spheres=25( 3 π‘Ÿ 3 ) = 100Γ— 3 Γ— 7 r3= 3 Γ—3.14π‘Ÿ 3 = 1.0466Γ—100π‘Ÿ 3 = 104.66π‘Ÿ 3 𝑐. π‘š3 116.5

π‘Ÿ3 =

3

=1.1131,r= √1.1131 c. m=1.0363c.m

104.66

π‘™π‘œπ‘”10 π‘Ÿ 3 = 3π‘™π‘œπ‘”10 π‘Ÿ = π‘™π‘œπ‘”10 116.5 βˆ’ π‘™π‘œπ‘”10 104.66 = 2.0663 βˆ’ 2.0197 = 0.0466 0.0466 π‘™π‘œπ‘”10 π‘Ÿ = 3 = 0.015533 r=Anti-log0.015533=100.015533=1.0364c.m r=1.03 c. m up to two decimals correctly The given problems are good but has lot of labor work during the process students will have lot of hardship with logarithmic tables. Here everything was calculated using scientific calculator in the computer that is a advance equipment device but students are advised to use logarithmic tables only in the exam that will be supplied. No calculators will be provided in the exam. Topic: Surface Area Question-3 The radius of a right circular cylinder is 7c.m and it’ s altitude is 20c.m.Find the area of it’s 22 curved surface (take πœ‹ = ) 𝑆 = 2πœ‹π‘Ÿβ„Ž = 2Γ—

22 7

7

Γ—7Γ—20𝑐. π‘š2 = 880𝑐. π‘š2

Topic:

Perimeter,Area Question-4 The figure depicts a semi-circular plot of land of diameter 14 meters. Sand has been spread on a rectangular portion of this land ,of length 7 meters and breadth 3 meters .Grass is grown on the remaining portion (Take πœ‹ =

22 7

)

(i)what is the perimeter of the semi-circular plot of land ? (ii)what is the area of the portion of land on which grass is grown ? (iii)Find the ratio of land on which grass is grown to the area of the land in which sand is spread. (iv) It is required to join to this plot of land ,a rectangular plot of land of area equal to the area on which grass is grown .Illustrate by drawing on the given figure, a sketch of this with the measurements marked on it ,such that it lies outside the semi-circular plot and has AB as one of its boundaries. Solution 1 22 (i)The radius=7m, perimeter=2 (2πœ‹π‘Ÿ) + 𝑑 = πœ‹π‘Ÿ + 2π‘Ÿ = (πœ‹ + 2)π‘Ÿ = ( 7 +2)7=22+14=36m (ii)Area where sand has been spread=The rectangle area=7Γ—3π‘š2=21m2 1 1 22 The area of the semi circle=2 (πœ‹π‘Ÿ 2 ) = 2 Γ— 7 Γ—72 = 11Γ—7π‘š2 = 77π‘š2

Area grass is grown=(77-21)m2=56m2=The area other than the rectangle with in the semi circle π‘”π‘Ÿπ‘Žπ‘ π‘  π‘Žπ‘Ÿπ‘’π‘Ž

56

7Γ—8

8

(iii) π‘ π‘Žπ‘›π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 21 = 7Γ—3 = 3

56

(iv)56=ABΓ—π‘₯ = 14Γ—π‘₯ = 14π‘₯, π‘₯ = 14 = 4π‘š=width of the new rectangle that should be attached to AB of length 14 meters & student has to draw accordingly by a pencil marker Question-5 A sketch of the floor of a theatre is shown in the figure. It consists of a semicircular part CED on which the stage is built and rectangular part ABCD where the auditorium is built. The length of DC is 14m. 22 In the following calculations, use 7 for the value of πœ‹ when required (i.) Find the arc length of the semicircle CED (ii.) Find the area of the floor on which stage is built (iii.) If the floor area of the auditorium is three times the area on which the stage is built, find the length of AD (iv.) Light bulbs have been fixed Ground the floor on which the stage is built ,with bulbs at C and D too. There is an equal gap of 1.4metres between adjacent bulbs on the line CD .The bulbs on the arc CED are also fixed with an equal gap .The number of bulbs on the line CD and on the arc CED are equal. Calculate the distance along the arc between two adjacent bulbs on the arc CED. Solution 1 22 (i.)2 (2πœ‹π‘Ÿ) = πœ‹π‘Ÿ = 7 Γ—7π‘š = 22π‘š=Arc length of the CED Semi circle 1

1

(ii.)2 (πœ‹π‘Ÿ 2 ) = 2 Γ—

22 7

Γ—72 = 77π‘š2

(iii.)3Γ—77 = 14×𝐴𝐷, 𝐴𝐷 = 14

33 2

π‘š = 16.5π‘š

(iv.)1.4 = 10 = π‘‡β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘”π‘Žπ‘π‘  π‘œπ‘› π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 𝐢𝐷 π‘Žπ‘šπ‘œπ‘›π‘” π‘‘β„Žπ‘’ 𝑏𝑒𝑙𝑏𝑠 𝑓𝑖π‘₯𝑒𝑑 The number of bulbs fixed on the line CD=10+1=11=The number of bulbs on the arc CED So that number of gaps among the bulbs fixed on the arc CED=10 22 Then the length of a gap among two bulbs fixed on arc CED= 10 = 2.2π‘š All together there are twenty bulbs fixed on the stage. The last three problems are easier than the first two problems. Question-6(A model question for 2017 ) Volume a.)A uniform cylindrical shaped glass with the solid hemispherical base is given in the figure. The radius of the base is a c. m and height of it is 4a c. m. (i.)Find the volume of the cylinder in terms of πœ‹ and a (ii.)Find the volume of the hemisphere in terms πœ‹ and a 10 (iii.)When the glass fills with water show that the volume of the water is given by 3 πœ‹π‘Ž3 Logarithms b.)Simplify decimal

9.51Γ—(0.89)2 √7.632

using the logarithm tables and give the answer to the nearest second

Solution Students have to sketch the figure by a pencil marker . (i.)Volume of the cylinder= πœ‹π‘Ÿ 2 β„Ž = πœ‹π‘Ž2 (4π‘Ž) = 4πœ‹π‘Ž3 𝑐. π‘š3 1 4 2 (ii.)Volume of hemisphere= 2 (3 πœ‹π‘Ž3 ) = 3 πœ‹π‘Ž3 𝑐. π‘š3 (iii.)Volume of the glass full of water=Volume of the cylinder-Volume of the hemispherical base 2

= (4πœ‹π‘Ž3 βˆ’ 3 πœ‹π‘Ž3 ) 𝑐. π‘š3 =

12βˆ’2 3

πœ‹π‘Ž3 =

10 3

πœ‹π‘Ž3 𝑐. π‘š3

Student s have to use logarithmic table for calculations. 9.51Γ—(0.89)2

π‘™π‘œπ‘”10

√7.632

1

= π‘™π‘œπ‘”10 9.51 + 2π‘™π‘œπ‘”10 0.89 βˆ’ 2 π‘™π‘œπ‘”10 7.632 1

= 0.9781 + 2(βˆ’0.0506) βˆ’ 2 (0.8826) =0.9781-0.0253-0.4413 =0.5115 Anti-log100.5115=100.5115=3.2471 =3.24 up to two decimals correctly Here for all calculation al purposes scientific calculator of the computer has been used but students get only logarithmic table in the exam. It should be noticed scientific calculator is far more accurate than the logarithmic table but up to two decimals logarithmic table is sufficient. Question-7 (A model question for 2017) (a.) A wet flour ball of radius R was made by mixing flour, very small pieces of coconut scrapers and water later using it such small flour ball of radius r were made & using them r0 radius coconut rotis were pressed and baked by heating a pan. Find the number of rotis which were baked and thickness of each of them. (b.)Huge Chocolate block of cubic shape of dimension b was used to make very small chocolate balls of radius r find the number of Chocolate balls manufactured. Solution 4 4 (a.)3 πœ‹π‘… 3 = 𝑛 3 πœ‹π‘Ÿ 3 = π‘›πœ‹π‘Ÿ02 𝑑 4 π‘Ÿ3

𝑅

𝑛 = ( π‘Ÿ )3 = π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘œπ‘‘π‘–π‘  π‘π‘Žπ‘˜π‘’π‘‘,𝑑 = 3 π‘Ÿ 2 = π‘‡β„Žπ‘–π‘π‘˜π‘›π‘’π‘ π‘  π‘œπ‘“ π‘’π‘Žπ‘β„Ž π‘Ÿπ‘œπ‘‘π‘– 0

3

(b.)𝑏 =

4

𝑁 3 πœ‹π‘Ÿ03,𝑁

3

= (4πœ‹) (

𝑏 π‘Ÿ0

3

)

Question-8(A model question for 2017) (i.)A part of a coconut tree of cylindrical shape of radius r and height h was used to cut into n number of rectangular wooden blocks(pol - parala) of dimensions a, b & c in width ,breadth and length respectively for a roof of a house find the volume of the coconut wood that has been wasted during the process. (ii.)A jar of milk of cylindrical shape of radius R and Height H was poured into N number of hemi-spherical cups of radius r find the value of the radius r. (iii.)A cake of dimensions l, m, n of rectangular shape was cut into cake cubes of d dimension find how many cubic cake blocks were cut. If l=m=n=1,d=1/2 find the number pieces of cake. (iv.) A pot of ice-cream of hemispherical shape of radius r was distributed among biscut

cups of conic shape of slant height l and height h find the number of conic-biscut cups(Assume they were filled only up to the brim and they are ice cream cones) . Solution (i.)πœ‹π‘Ÿ 2 β„Ž βˆ’ π‘›π‘Žπ‘π‘ = π‘‡β„Žπ‘’ π‘€π‘Žπ‘ π‘‘π‘’π‘‘ π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘œπ‘π‘œπ‘›π‘’π‘‘ π‘€π‘œπ‘œπ‘‘ 1 4

2

(ii.)πœ‹π‘… 2 𝐻 = 𝑁 2 (3 πœ‹π‘Ÿ 3 ) = 𝑁 3 πœ‹π‘Ÿ 3,π‘Ÿ 3 = 3

(iii.)lmn=Nd ,N=lmn/d 1 4

3

3𝑅 2 𝐻 2𝑁

,π‘Ÿ = (

3𝑅 2 𝐻 1 2𝑁

3

3𝑅 2 𝐻

)3 = √

2𝑁

3

if l=m=n=1,d=1/2,N=1/(1/2) =1/(1/8)=8 cubic blocks of cake

2

1

2

(iv.) 2 (3 πœ‹π‘Ÿ 3 ) = 3 πœ‹π‘Ÿ 3 = 𝑛 (3 πœ‹(𝑙 2 βˆ’ β„Ž2 )β„Ž) , 𝑛 = (𝑙2 βˆ’β„Ž2 )β„Ž,where 𝑙 2 = β„Ž2 + π‘Ÿ 2 , π‘Ÿ 2 = 𝑙 2 βˆ’ β„Ž2 , π‘Ÿ = βˆšπ‘™ 2 βˆ’ β„Ž2 =the radius of the conic biscut cup(ice-cream cone) Mathematics-1 1.π‘™π‘œπ‘”2 16 = 4 16=24,π‘™π‘œπ‘”2 24 = 4 2.Solve (x-1)(x-2)=0, x-1=0 or x-2=0 x=1 or x=2 3.Write all the positive integers that satisfy the inequality 2x+1≀ 5 2x+1-1≀ 5 βˆ’ 1 2x+0 ≀ 4,2π‘₯ ≀ 4 2π‘₯ 4 ≀2 2 π‘₯ ≀ 2,x=1,2 are the positive integers satisfying the derived inequality hence the given. 1 1 4.Simplyfy π‘₯ + 2π‘₯ 2

1

= 2π‘₯ + 2π‘₯ =

2+1 2π‘₯

3

= 2π‘₯

5.Find the least common multiple of the two algebraic expressions xy and x2 LCM of xy and x2 is x2y x2y=x(xy) 6.A bus which moves at a uniform speed, travels a distance of 48 meters in 3 seconds. Find the speed of the bus in meters per second . The speed is defined as the distance travel per second. 𝑣 = 𝑆𝑝𝑒𝑒𝑑 =

π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘‘π‘Ÿπ‘Žπ‘£π‘’π‘™π‘™π‘’π‘‘ π‘‘π‘–π‘šπ‘’ π‘‘π‘Žπ‘˜π‘’π‘›

=

π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘‘π‘–π‘šπ‘’

=

𝑑 𝑑

=

48 π‘šπ‘’π‘‘π‘’π‘Ÿπ‘  3 π‘ π‘’π‘π‘œπ‘›π‘‘π‘ 

=

3Γ—16π‘š 3𝑠

= 16

π‘š 𝑠

= 16π‘šπ‘  βˆ’1

Speed is a scalar and it has no specific direction but it has a value. When speed is considered with a specific direction it is known as the Velocity. 7.Using the information given in the table, find the first approximation of √90 x 9.3 9.4 9.5 9.6 2 x 86.49 88.36 90.25 92.16 88.36 < 90 < 90.25, √88.36 < √90 < √90.25,9.4 < √90 < 9.5 9.4+9.5

√90=

2

=

18.9 2

= 9.45

√90=9.4868 by a Scientific Calculator of a Computer so up to first decimal root 90 is 9.4 by the table given is correct.

9

1

1

11

Further √90 = √81 + 9 = √{81 (1 + 81)} = √92 (1 + 9) = 9√1 + 9 = 9 (1 + 2 9) = 1

9

1

9 (1 + 18) = 9 + 18 = 9 + 2 = 9 + 0.5 = 9.5 not sufficient looks higher π‘₯

√1 + π‘₯=1+2 for x< 1 1

1

1 1

1

10

√90 = √100 βˆ’ 10 = √102 (1 βˆ’ 10) = 10√1 βˆ’ 10 = 10 (1 βˆ’ 2 10) = 10 (1 βˆ’ 20) = 10 βˆ’ 20 = 1

10 βˆ’ 2 = 10 βˆ’ 0.5 = 9.5 not sufficiently low higher in value π‘₯

√1 βˆ’ π‘₯ = 1 βˆ’ 2 π‘“π‘œπ‘Ÿ π‘₯ < 1 So that root of ninety is as 9.4 the extracted value from the table given well approximate up to the first decimal. 1 1 Further π‘™π‘œπ‘”10 √90 = 2 π‘™π‘œπ‘”10 90 = 2 (1.9542) = 0.9771 Anti-log10 0.9771=100.9771 =9.4863 is accurate enough but student has to use log table at this point if question comes in this specific log base style (fashion) (2016 Prototype) 1.Find the common ratio of the geometric progression given below 1,3,9,27,…… 27 9 3 = = = 3 = π‘‡β„Žπ‘’ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘‘β„Žπ‘’ π‘”π‘’π‘œπ‘šπ‘’π‘‘π‘Ÿπ‘–π‘ π‘ π‘’π‘Ÿπ‘–π‘’π‘  = π‘Ÿ ,r=3 9 3 1 a=1=The first term,𝑑𝑛 = π‘Žπ‘Ÿ π‘›βˆ’1 = (1)3π‘›βˆ’1 = 3π‘›βˆ’1=The π‘›π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š 𝑠𝑛 = 𝑠𝑛 =

π‘Ž(π‘Ÿ 𝑛 βˆ’1) (π‘Ÿβˆ’1) 1(3𝑛 βˆ’1) (3βˆ’1)

= π‘‡β„Žπ‘’ π‘ π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ 𝑛 π‘‘π‘’π‘Ÿπ‘šπ‘  π‘œπ‘“ π‘Ž π‘”π‘’π‘œπ‘šπ‘’π‘‘π‘Ÿπ‘–π‘ π‘ π‘’π‘Ÿπ‘–π‘’π‘  π‘“π‘œπ‘Ÿ π‘Ÿ > 1 1

= 2 (3𝑛 βˆ’ 1),n=1,2,3,……. 1

1

1

For n=4,𝑠4 = 2 (34 βˆ’ 1) = 2 (81 βˆ’ 1) = 2 (80) = 40 1

𝑠𝑛 = 2 (3𝑛 βˆ’ 1) β†’ ∞ π‘€β„Žπ‘’π‘› 𝑛 β†’ ∞ in the limit for very large n since 3𝑛 β†’ ∞ π‘“π‘œπ‘Ÿ 𝑛 β†’ ∞(π‘šπ‘Žπ‘‘β„Žπ‘’π‘šπ‘Žπ‘‘π‘–π‘π‘Žπ‘™ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑦) that is for very large n. 2.(2017 model) 1 1 1 1 Find the common ratio 1,3 , 9 , 27 , 81 … ….of the geometric progression 1 81 1 27

=

1 27 1 9

=

1 9 1 3

=

1 3

1

27

9

3

1

= 81 = 27 = 9 = 3 = π‘‡β„Žπ‘’ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘‘β„Žπ‘’ π‘”π‘’π‘œπ‘šπ‘’π‘‘π‘Ÿπ‘–π‘ π‘ π‘’π‘Ÿπ‘–π‘’π‘  1, 𝑠6 =

π‘Ž(26 βˆ’1) (2βˆ’1)

π‘Ž(π‘Ÿ 𝑛 βˆ’1) (π‘Ÿβˆ’1)

= π‘Ž + π‘Žπ‘Ÿ + π‘Žπ‘Ÿ 2 + 7

=a(26 βˆ’ 1) = 7,𝑑𝑛 = π‘Žπ‘Ÿ π‘›βˆ’1 ,𝑑5 = π‘Ž25βˆ’1 = π‘Ž24 =(26 βˆ’1) Γ—24 =

7Γ—16

= (64βˆ’1) 7

7Γ—16 63 7

=

7Γ—16 7Γ—9 7

=

16 9 7

4

= (3)2 1

1

1

1

1

a=26 βˆ’1 = 64βˆ’1 = 63 = 7Γ—9 = 9,𝑑𝑛 = 9 2π‘›βˆ’1 ,𝑑1 = 9 21βˆ’1 = 9 20 = 9 , 20 = 1 The geometric series is

1 9

1

1

1

, 9 2, 9 22 , … … … , 9 2π‘›βˆ’1

2. (2016 prototype question) Sumana cuts pieces of ribbon for a decoration according to a pattern ,such that the first piece is 20 c. m long ,second piece is 25 c. m long and the third piece is 30c.m long .The longest piece of ribbon requires for the decoration is of length 95 c. m (i.) Show with reasons that a roll of ribbon of length 10m is sufficient for the decoration . (ii.) For another decoration ,the length of the longest piece of ribbon cut as above ,is twice the length of the above longest pieces. Show by computing whether two rolls of ribbon of length 10m each are sufficient for this. Solution (i.)a =20c.m, a+ d=25 c. m, a+2d=30 c. m, .…a+(n-1)d=95 c. m, d=25-a=25-20=5 c. m, 20+(n-1)5=95=20-5+5n=15+5n=95,3+n=19,n=19-3=16 𝑛 16 𝑆𝑛 = 2 {π‘Ž + 𝑙} = 2 {20 + 95} = 8Γ—115 = 920𝑐. π‘š < 1000𝑐. π‘š = 10Γ—100𝑐. π‘š = 10π‘š,so that 10meters would be sufficient for the decoration & there will be a balance of (1000-920)c. m=80c.m once decoration is completed. (ii.)a+(n-1)d=2Γ—95𝑐. π‘š = 190𝑐. π‘š=20+(n-1)5=15+5n=190,3+n=38,n=38-3=35 35 35 𝑆𝑛 = 2 {20 + 190} = 2 Γ—210 = 35Γ—105 = 3675c.m> 2000𝑐. π‘š = 20Γ—100 𝑐. π‘š = 20π‘š = 2Γ—10π‘š = π‘‘π‘€π‘œ π‘Ÿπ‘œπ‘™π‘™π‘  π‘œπ‘“ π‘Ÿπ‘–π‘π‘π‘œπ‘› π‘œπ‘“ 10π‘š π‘™π‘œπ‘›π‘”, 𝑖𝑠 𝑖𝑛𝑠𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑑, {3675 βˆ’ 2000}𝑐. π‘š = 1675𝑐. π‘š = 16.75π‘š = 1.675 π‘œπ‘“ 10π‘š π‘Ÿπ‘–π‘π‘π‘œπ‘› π‘œπ‘“ 𝑒π‘₯𝑐𝑒𝑠𝑠 𝑖𝑠 𝑛𝑒𝑒𝑑𝑒𝑑 π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘‘π‘’π‘π‘œπ‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›. 3.(2017 model question) In a shop, bars of Soap are arranged on top of each other on a rack in such a way that the bottom raw has 33 bars ,the raw above that has 29 bars and the row above that has 25 bars and so on, i.) Find the number of bars in the 6th row from the bottom . ii.)If the top raw has one bar of soap, find the total number of rows. iii.)Find the total number of bars of soap. iv.)If the bars of soap are 5c.m wide ,find the minimum height of the rack that should be enable all the rows of soap to be placed in. Solution d=33-9=4=29-25=the common difference among the number of soap arranged in rows of one over the other in a rack of a shop ,a decreasing sequence so that must have the negative sign i.)𝑇𝑛 = π‘Ž + (𝑛 βˆ’ 1)𝑑, π‘Ž = 33, 𝑑 = βˆ’4, 𝑇6 = 33 + (6 βˆ’ 1)(βˆ’4) = 33 βˆ’ 5Γ—4 = 33 βˆ’ 20 = 13 ii.)1=33+(n-1)(-4)=33+4-4n=37-4n,4n=37-1=36=4Γ—9,n=9=The number of rows in the rack one over the other iii.)The sequence is 33,29,25,21,17,13,9,5,1 a descending sequence with nine terms 𝑛 𝑛 𝑆𝑛 = 2 {π‘Ž + 𝑙} = 2 {2π‘Ž βˆ’ (𝑛 βˆ’ 1)𝑑} π‘“π‘œπ‘Ÿ π‘Ž π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘” π‘ π‘’π‘žπ‘’π‘’π‘›π‘π‘’ , 𝑙 = π‘Ž + (𝑛 βˆ’ 1)(βˆ’π‘‘) =

9

9

9

9

π‘Ž βˆ’ (𝑛 βˆ’ 1)𝑑,𝑆9 = 2 {2Γ—33 βˆ’ (9 βˆ’ 1)4} = 2 {33 + 33 βˆ’ 32} = 2 {33 + 1} = 2 {34} = 9Γ—17 = 153 iv.)The minimum height of the rack=9Γ—5 = 45 𝑐. π‘š in height that enables all soaps to be kept over one another. But there can be instability and soaps can fall one over the other. The model question here is originated from a tution master. Exam question given in the year 2016 is a hard problem if they have dropped the removal of even terms and kept preserve for the sequence it could have been much easier for the student in the exam of 2016 but given fashion is decisive and it check the student but there is less time for such steps give uneasiness otherwise can remove odd positions in a next chance can give similar uneasiness or preserve for comfort. In the geometric series sum of six terms can be made to be 9 then first term becomes one seven slight change in the setting for future. Prototype question is the best among three problems it check the students computational ability. But it looks all are difficult for students even for teachers. (Grade-11-Mathematics text book Questions) 4.The first three terms of a geometric progression are respectively 4,(x+3) and (x+27) (i.)Find the value of x (ii.)Show that there are two such geometric progressions and find the first four terms of each progression Solution π‘₯+27 π‘₯+3 = 4 = π‘‡β„Žπ‘’ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘–π‘œ = π‘Ÿ ,4(π‘₯ + 27) = (π‘₯ + 3)2 = π‘₯ 2 + 6π‘₯ + 9 = 4π‘₯ + π‘₯+3 108, π‘₯ 2 + 2π‘₯ βˆ’ 99 = 0, (π‘₯ βˆ’ 9)(π‘₯ + 11) = 0, π‘₯ = 9 π‘œπ‘Ÿ π‘₯ = βˆ’11 9+3 12 4,12,36,108,……… The common ratio= 4 = 4 = 3 ,The nth term=4(3)n-1 4,-8,+16,-32,………The common ratio=

3βˆ’11 4

=

βˆ’8 4

= βˆ’2 ,The nth term=4(-2)n-1

5.The sum of the first n terms of the arithmetic progression -7,-5,-3,-1,….is 105 .Use your knowledge on progressions to the following (i.)Taking n to be the number of terms we are adding together set up a quadratic equation in n for the sum of the first n terms (ii.)Solve the above equation to find the number of terms n you have added Solution (i.)The common difference=-5-(-7)=7-5=-3-(-5)=5-3=-1-(-3)=3-1=2=d 𝑇𝑛 = π‘Ž + (𝑛 βˆ’ 1)𝑑 = βˆ’7 + (𝑛 βˆ’ 1)2=-9+2n=l 𝑛 𝑛 𝑛 𝑆𝑛 = 2 {π‘Ž + 𝑙} = 2 {βˆ’7 +-9+2n}= 2 {βˆ’16 + 2𝑛} = 𝑛(βˆ’8 + 𝑛) = 𝑛2 βˆ’ 8𝑛 = 105, 𝑛2 βˆ’ 8𝑛 βˆ’

105 = 0, 𝑛2 βˆ’ 8𝑛 βˆ’ 15Γ—7 = 0, 𝑛2 βˆ’ 15𝑛 + 7𝑛 βˆ’ 15Γ—7 = 𝑛(𝑛 βˆ’ 15) + 7(𝑛 βˆ’ 15) = (𝑛 βˆ’ 15)(𝑛 + 7) = 0, (ii. )𝑛 = 15 π‘œπ‘Ÿ 𝑛 = βˆ’7 positive 15 is the answer for number of terms since it cannot be negative Pythagoras Theorem 6.The hypotenuse of a right angled triangle is 2x+1 c. m .The lengths of the other two sides are x c. m and x+7 c. m respectively .Solve for x and find the lengths of all three sides. Solution (2π‘₯ + 1)2 = π‘₯ 2 + (π‘₯ + 7)2 , 4π‘₯ 2 + 4π‘₯ + 1 = π‘₯ 2 + π‘₯ 2 + 14π‘₯ + 49 = 2π‘₯ 2 + 14π‘₯ + 49,2π‘₯ 2 βˆ’ 10π‘₯ βˆ’ 48 = 0, π‘₯ 2 βˆ’ 5π‘₯ βˆ’ 24 = π‘₯ 2 βˆ’ 8π‘₯ + 3π‘₯ βˆ’ 8Γ—3 = π‘₯(π‘₯ βˆ’ 8) + 3(π‘₯ βˆ’ 8) = (π‘₯ + 3)(π‘₯ βˆ’

8) = 0, π‘₯ = βˆ’3 π‘œπ‘Ÿ π‘₯ = 8 x=-3,(2Γ—βˆ’3 + 1)2 = (βˆ’3)2 + (βˆ’3 + 7)2 ,52=32+42,(3,4,5) (2Γ—8 + 1)2 = 82 + (8 + 7)2 , 172 = 82 + 152 ,(8,15,17) are the sides of triangles Further (𝑛2 + 1)2 = (2𝑛)2 + (𝑛2 βˆ’ 1)2 n=2,(22+1)2=(2Γ—2)2 + (22 βˆ’ 1)2 , 52 = 42 + 32 n=4,(42 + 1)2 = (2Γ—4)2 + (42 βˆ’ 1)2 , 172 = 82 + 152 Pythagoras theorem states in a right angle triangle square of the hypotenuse that is the side in front of the right angle is equal to the sum of squares of two adjacent sides to the right angle of the triangle . Value of 𝝅 ( 2017 model question) 7.(a)Srinivasa Ramanujan had found ( (b)Srinivasa Ramanujan had found

2143 1

22 9801√2 4412

)4 gives eight decimals of πœ‹.Prove it by logarithms. gives six decimals of πœ‹.Prove it by logarithms.

Solution (a) π‘™π‘œπ‘”10 ( 1 4

2143 1 22

1

1

)4 = 4 {π‘™π‘œπ‘”10 2143 βˆ’ π‘™π‘œπ‘”10 22} = 4 {3.3310221710 βˆ’ 1.3424226808} =

(1.9885994901) = 0.4971498725

Anti-log100.4971498725=100.4971498725 =3.14159265 up to eight decimals correctly. (b)π‘™π‘œπ‘”10

9801√2 4412

1

1

= π‘™π‘œπ‘”10 9801 + 2 π‘™π‘œπ‘”10 2 βˆ’ π‘™π‘œπ‘”10 4412=3.9912703891+2 (0.3010299956)-

3.6446355037=3.9912703891+0.1505149978-3.6446355037=0.4971498832 Anti-log100.4971498832=100.4971498832=3.141592 up to six decimals correctly. For this calculation scientific calculator of the computer has been used but students have to use logarithm table it can do calculations up to four decimals only cannot confirm up to eight or six decimals accuracy of Srinivasa Ramanujan . 8.Mr.Kithsiri made a cash donation to a community centre 2 1 π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ β„Žπ‘’ π‘‘π‘œπ‘›π‘Žπ‘‘π‘’π‘‘ π‘€π‘Žπ‘  𝑒𝑠𝑒𝑑 π‘‘π‘œ 𝑏𝑒𝑦 π‘šπ‘’π‘ π‘–π‘π‘Žπ‘™ π‘–π‘›π‘ π‘‘π‘Ÿπ‘’π‘šπ‘’π‘›π‘‘π‘  π‘Žπ‘›π‘‘ 2 π‘‘π‘œ 𝑏𝑒𝑦 s 9 ports equipment. (i.) Find what fraction of the total amount was used to buy the musical instruments and sports equipment 1 of the remaining amount was used to buy books for the library . 5 (ii.)Find what fraction of the total amount was used to books The amount remaining after purchasing the books was used to renovate the community centre (iii.)Find what fraction of the total amount was used for the renovation (iv.)If the renovation cost was Rs.20,000,find the total amount Mr. Kithsiri donated. Solution: The total amount=X Rs. 2 The money used to buy musical instruments=9 𝑋 1

The money used to buy sports equipments=2 𝑋

2

1

4

9

13

The money used to buy both=9 𝑋 + 2 𝑋 = 18 𝑋 + 18 𝑋 = 18 𝑋 13

The remaining amount=X-18 𝑋 =

18βˆ’13 18 5

1

5

𝑋 = 18 𝑋 1

The money used to buy books =5 . 18 𝑋 = 18 𝑋 4

2

The money used to renovate=The balance=18 𝑋 = 9 𝑋 2

The renovation cost=20,000=9 𝑋, X=90,000Rupees=The amount Kithsiri donated. 9.(2016 Prototype) 3 2 From the total distance a man needed to travel ,he travelled 5 by train,3 of the remaining portion by bus ,and the rest of the distance he walked (i.)What fraction of the total distance remained after he had travelled by train? (ii.)What fraction of the total distance did he travel by bus? (iii.)Write down in the simplest form the ratio of the distance he travelled by train to the distance he walked ? (iv.)The total distance of the journey was 30 kilometers. The time travelled by train was 20 minutes .Find the average speed of the train in kilometers per hour ? Solution: The total distance=d 3 The distance he travelled by train=5 𝑑 3

2

(i.)The distance remained after he travelled by train=1 βˆ’ 5 𝑑 = 5 𝑑 2

2

4

(ii.)The distance he travelled by bus=3 Γ— 5 𝑑 = 15 𝑑

2

4

6

4

2

(iii.)The distance he walked=The balance distance=5 𝑑 βˆ’ 15 𝑑 = 15 𝑑 βˆ’ 15 𝑑 = 15 𝑑 π‘‡β„Žπ‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘‘π‘Ÿπ‘Žπ‘£π‘’π‘™ 𝑏𝑦 π‘‘π‘Ÿπ‘Žπ‘–π‘› π‘‡β„Žπ‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘€π‘Žπ‘™π‘˜π‘’π‘‘

=

3 𝑑 5 2 𝑑 15

3

= 5Γ—

15 2

9

= 2=9: 2

(iv.) d=30k.m,The time travelled by train=20minutes 3 The distance travel by train=5 Γ—30π‘˜. π‘š = 18π‘˜. π‘š The average speed of the train=

18π‘˜.π‘š 18π‘˜.π‘š 20 β„Ž 60

=

1 β„Ž 3

= 18Γ—3

π‘˜.π‘š β„Ž

= 54π‘˜. π‘šβ„Žβˆ’1 =

54π‘˜π‘–π‘™π‘œπ‘šπ‘’π‘‘π‘’π‘Ÿπ‘  π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ 10.The following describes how the students and teachers of a school are involved in a tree planting event organized by the school environmental society. A total of Rs.16,500 is collected, with each student contributing Rs.150 and each teacher contributing Rs.500 .Then 330 saplings are bought with this money ,and all of them are distributed to be planted ,by giving each student 5 saplings and each teacher two saplings. (i.) By constructing a pair of simultaneous equations and solving them ,find the number of students and number of teachers (ii.)If instead of the above method of distributing the saplings ,each student is given p saplings and each teacher q saplings, then some saplings will remain undistributed. Write an inequality in terms of p and q using this information.

Solution: (i.)Assume s is the number of students and t is the number of teachers then 150s+500t=16500, Γ· 50,3s+10t=330← (1. ) 5s+2t=330← (2. ) (2.)-(1.),2s-8t=0,s-4t=0,s=4t 330 30 By (1.) 3(4t)+10t=330,12t+10t=22t=330,𝑑 = 22 = 2 = 15 By (2.) 5s+2(15)=5s+30=330,5s=330-30=300,s=

300 5

= 60

t=The number of teachers=15,s=The number of students=60 (ii.)60p+15q< 330 4p+q