ACCOUNTS D1

 

Prepare a profit and loss account for the year ended 31^ST May 2018 from the following balances:

Transport……………………..…………..sh. 59,000

Advertising …………..………………… sh. 70,000

General expenses…………..………. sh 34,000

Sundry expenses…………..……….. sh 5,000

Interest received …………..……….sh. 30,000

Carriage …………..……………………..sh 42,000

Gross profit …………..……………… sh 124,000

 

Solution

Dr               PROFIT AND LOSS ACCOUNT FOR THE YEAR ENDED ON 31TH, MAY-2018                   Cr
Particulars	Folio	Amount	Particulars	Folio	Amount
Transport		59,000	Gross profit	b/d	124,000
Advertising		70,000	Interest received		30,000
General expenses		34,000	Net Loss		56,000
Sundry expenses		5,000
Carriage		42,000
		210,000			210,000

 

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Prepare a profit and loss account for the year ended 31^ST November 2016 from the following balances:

Transport ----------------------sh. 77,000

Advertising --------------------sh. 57,000

General expenses------------- sh 38,000

Sundry expenses --------------sh 9,000

Interest received------------- sh. 34,000

Carriage -------------------------sh 24,000

Gross profit----------------------sh 104,000

ALGEBRA D1

 

The sum of four consecutive numbers is 606. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+1	n+2	n+3	606

 

Then, n + (n+1) + (n+2) + (n+3) = 606

 

4n + 1+2+3= 606

 

4n + 6 = 606

 

4n = 606 – 6

 

4n = 600 

 

4n = 600

 4        4

 

n = 150

 

3rd number = n +2

 

                   = 150 + 2

 

                   = 152

 

Hence the 3rd number is 152

 

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The sum of five consecutive numbers is 1035. Find the largest number.

FUNCTIONS D3

Given that F(x) = 3x  +  11. Find F(-2) – 2F(5)

 

Solution

 

CASE 1: We find F(-2)

 

 

F(x) = 3x  +  11

 

F(-2) = 3(-2)  +  11

 

F(-2) = -6  +  11

 

F(-2) = 5

 

 

Hence  F(-2) = 5

 

 

 

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CASE 2: We find F(5)

 

 

 

F(x) = 3x  +  11

 

F(5) = 3(5)  +  11

 

F(5) = 15  +  11

 

F(5) = 26

 

 

Hence  F(5) = 26

 

 

2F(5) = 2(26) = 52

 

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CASE 3: We find F(-2) – 2F(5)

 

 

= F(-2) – 2F(5)

= 5 – 52

= -47

 

 

TRY THIS………………..

 

 

 

Given that F(x) = 12x  +  78. Find F(-20) – 2F(10)

FUNCTIONS D2

 

LOGARITHMS D1

 

Change the following into Logarithmic form.

i)  4² = 16

ii)  5³ = 125

iii)  3⁰ = 1

Solution

 

 

4² = 16 ⇔ log₄ 16 = 2

5³ = 125 ⇔ log₅ 125 = 3

3⁰ = 1 ⇔ log₃ 1 = 0

TRY THIS………………….

Change the following into Logarithmic form.

i)  7² = 49

ii)  3⁴ = 81

iii)  8⁰ = 1 

FUNCTIONS D1

 

Given that F(x) = 7x  +  11. Find F(-2)

 

Solution

 

F(x) = 7x  +  11

 

F(-2) = 7(-2)  +  11

 

F(-2) = -14  +  11

 

F(-2) = -3

 

 

Hence  F(-2) = -3

 

 

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Given that F(x) = 16x  +  23. Find F(-3)

CAPACITY D1

 

Change 32.6 liters of water into centiliters

 

Solution

 

1L  =  100cl

32.6L = ?

___________________________  

 

 = 32.6 x  100

           1

 

= 32.6 x 100

 

= 3260cl. Answer

 

 

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Change 154.3 liters of milk into centiliters

PERFECT SQUARES C1

 

What must be added to x²  + 14x to make the expression a perfect square?

 

Solution

 

a=1, b=14,c=?

 

b² = 4ac

 

(14)² = 4 x 1 x c

 

196 = 4c

196 = 4c

 4       4

 

49 = c

 

Hence number to be added is 49

And the expression will be: x²  + 14x +49=0

 

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What must be added to x²  + 10x to make the expression a perfect square?   

VECTORS C1

 

FACTORIZE C1

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Factorize 1-25m²

 

Solution

 

We use difference of two squares a² – b² = (a - b)(a + b)

 

1-25m² = 1²-5²m²       

 

            = 1² - (5m)²       

 

            = (1 - 5m)(1 + 5m)

 

Hence 1-25m² = (1 - 5m)(1 + 5m)  answer

 

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Factorize 1 – 49y²

ALGEBRA C2

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If 12w² = 588; find w.

Solution

12w² = 588

 

12w² = 588               [Dividing by 12 both sides]

12          12

 

w² = 49

w = 7          Because the square root of 49 is 7.   

 

Hence w = 7 . 

 

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If 14u² = 224; find u.

LOGARITHMS C1

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If Log_ax = -20, find Log_a(¹/x⁶)

 

Solution

 

= Log_a(¹/x⁶)                        

 

= Log_ax^-6                   [since ¹/cⁿ =c^-n]

 

= -6 x Log_ax       [since Log_mxⁿ   = nLog_mx]               

 

= -6 x -20

 

= 120

 

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If Log_aW= -30, find Log_a(¹/W⁵)

ALGEBRA C1

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Express as a single fraction m - 3 + m+5

                                               6         2

Solution

 

= m - 3 + m+5

      6          2

 

= m - 3 + 3(m+5)              [Since LCM of 6 and 2 is 6]          

            6         

 

= m - 3 + 3m+15               [Since 3(m+5)  = 3m+15]               

            6         

 

= m + 3m + 15- 3

            6         

 

= 4m +  12

         6         

 

= 4(m +  3)

         6         

 

= 2(m +  3)

         3         

 

= 2m +  6       answer

         3         

  

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Express as a single fraction 4n - 1 + n+1

                                                3          2

EXPAND C1

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Expand (y – 10) (y – 16)

 

solution

 

= (y – 10) (y – 16)

 

= y (y – 16) – 10 (y – 16)

 

= y² – 16y – 10y + 160 

 

= y² – 26y + 160  since (– 16y – 10y) = – 26y

 

= y² – 26y + 160 answer

 

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Expand (w – 2) (w – 11)

METRIC UNITS C1

 

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Change 51260m into Hm.

  

Solution

 

 

1Hm = 100m

?  =  51260m

 

 

= 1  x  51260

       100

 

=     551260

       100

 

= 512.6Hm

 

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Change 3502m into Hm.

RATES AND VARIATIONS B1

 

RADICALS B1