MATHEMATICS OBJ

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Mathz – Theory

Any number you don’t see kindly scroll down to see wher we solved it in paper.

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1 )

P =N 300 , 000

R =7 ^ 1 / 2 %

T =3 yrs

At the end of year 1

I =PRI / 100 =300000 * 15* 1 / 100 * 2

I =N 22500

2 nd Year

P =300000 + 22500+ 50000

=N 372500

I =372500 * 15* 1 / 200 =N 27937. 50

3 rd Year

P =372500 + 27937. 5 + 50000

=N 450437 . 50

I =450437 . 50 *15 * 1 / 200

I =N 33782 . 81

total saving of 3 years

=450437 . 5 + 33782. 81 + 50000

=N 534220 . 31

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2 a)

T =thickness

P =Pages

T *P

T =KP

T =3

P =900

3 =900 k

k =3 / 900 = 1 / 300

T =P / 300

WHERE T=45 CM

P =?

45 =P / 300

P =300 * 45

P =13500pages

2 b )

X – Y= 3 & x ^ 2 – Y^ 2 =15

( x + y ) ( x – y )= 15

x + y =15/ x – y =15/ 3 =5

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3 a)

n ( y )= 40

n ( c)= 35

n ( b )= 26

n ( CnB )= x

drawing

40 =35- x + x + 26 – x

40 =61- x

x =61 – 40 =21

21 student offer both

3 b )

U : { all positive < – 20 }

S : { all even number < 14 }

T : { all even nus < – 20 divisible by 3 }

U ={ 1 , 2 , 3 , , , 20 }

S ={ 2 ,4 , 6 , 8 , 10 , 12}

T ={ 6 ,12 , 18}

SUT={ 2 , 4 , 6 , 8 , 10 , 12, 18}

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4 a)

length of chord

=2 rsm ^θ / 2

=2 *4 sm^ 9 ^ 4 / 2

=85 m 47

=8 *0 . 7314

=5 . 85cm

4 b )

draw

7 cm >7 cm

area ( A )= θper ^ 2 / 360 – 1 / 2 r ^ 2 8 mθ

=r ^ 2 / 2 { 22/ 180 – 8 mθ }

=49 / 2 [ 90 / 180 * 22 / 7 – 8 m 90 ]

=49 / 2 { 1 / 2 * 22/ 77 – 1 }

=49 / 2 { 11 / 7 – 1 } =44 / 2 { 1 . 57 – 1 }

=49 / 2 { 0 . 57 } =13 . 97cm ^ 2

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5 A & B)

5 c)

mean ( x̅ ) = Ʃfx / Ʃf = 535 . 5 / 43

=12 . 5

median = 20 students from the histogram

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7 a)
y = x ^ 3 – 6 x ^ 2 + 9 x – 5 , the value of x when gradient dy / dx = 0

therefore :

dy / dx = 3 x ^ 2 – 12x + 9

3 x ^2 – 12 x + 9 = 0

3 x ^2 – 3 x – 9 x + 9 =0

3 x ( x – 3 ) – 9 ( x – 1 )= 0

( 3 x – 9 ) ( x – 1 )= 0

x =3 / 9 or x =1

x = 3 , 1

7 b )

y ^ 2 / 36 – 1 / 9 = 0

y ^ 2 / 36 – 1 / 9

therefore :

y ^ 2 / 6 ^2 = 1 / 3 ^ 2

y / 6 = 1 / 3

y = 6 / 3

=2

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9 a)

logbase 4 ( x ^ 2 + 7 x + 28 )= 2

logx ^2 + 7 x + 28 =4 ^ 2

logx ^2 + 7 x + 28 =log16

x ^ 2 + 7 x + 28=16

x ^ 2 + 7 x + 28- 16 =0

x ^ 2 + 7 x 12 =0

x ^ 2 + 3 x + 4 x + 12=0

x ( x + 3 ) + 4 ( x + 3 )= 0

( x + 3 ) ( x + 4 )= 0

x =- 3 or x =- 4

9 b )

y =3 x ^ 2 – 4

y + dy =3 ( x + dx )^2 – 4

y + dy =3 ( x ^ 2 + 2 xDx + Dx ^ 2 ) – 4

dy =3 ( x ^ 2 + 2 xDx + Dx + Dx ^ 2 ) – 4 – y

=3 x ^ 2 + 6 xDx+ 3 Dx ^ 2 – 4 – ( 3 x ^ 2 – 4 )

Dy =6 xDx + 3 Dx ^ 2

Dy / Dx =6 xDx+ 3 Dx ^ 2 / Dx

=6 x + 3 Dx

but as Dx =0

Dy / Dx =dy / dx

dy / dx=6 x + 3 ( 0 )

=6 x

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11 a)

Draw The diamgram

– Angular difference between P and Q = 48+ 36 = 84 °

– Angular difference betwwen Q and R = 42- 22 = 20 °

( a) distance PQ = θ / 360 * 2 πRcos ∝

∝ = 42°

PQ = 84 / 360 * 3 . 142 * 6400 cos42

=84 * 2 * 3 . 142 * 6400 * 0 . 7431 / 360

=2510398 . 68 / 360

=6

11 b )

Distance QR is along the great circle

QR = θ / 360 * 2 πR

=20 / 360 * 2 *3 . 142 * 6400

=2 *3 . 142 * 6400 / 18

=40217. 60/ 18 = 2234 . 31

=2230 km ( 3 s . f )

11 c)

Speed = Distance/ Time

Time = Distance/ Speed

Total distance = 6970 + 2230

= 9200 km

Speed = 600 km/ h

Time = 9200 / 600

=15 . 3 hrs

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