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# Solutions to some algebraic questions

Question 1

A trader bought 100 oranges at 5 for #1.20, 20 oranges got spoilt and the remaining were sold at 4 for #1.50. find the percentage gain or loss.

1. 30% gain          B. 25% gain              C. 30% loss                     D. 25% loss

Solution

The cost price for the 100 oranges = 100/5 * 1.20 = #24

20 got spoilt

good oranges = 80

selling price for the 80 oranges = 80/4 * 1.50 = #30

since the selling price is more than the cost price, therefore there is a gain

percentage gain = 30 – 24 / 24 *100 = 25%

B is the correct option

Questio 2

If U = {1, 2, 3, 4, 5, 6}, P = {3, 4, 5}, Q= { 2, 4, 6} and R= {1, 2, 3, 4}, list the elements of (p u Q)’ n( R)

1. {1,2,3,4,5,6}   B. {1,2,3,4}    C. {1}    D null

solution

P U Q = {2,3,4,5,6}

(P U Q)’ = {1}

(P U Q’) n (R) = {1}

C is the correct option

Question 3

The sum of two numbers is twice their difference if the difference of the numbers is p, find the larger of the two numbers.

1. p/2       B. 3p/2         C. 5p/2        D. 3p

solution

let the two numbers be x and y

x + y = 2 (x – y)

x + y = 2p…i

x – y = p…ii

2x = 3p

x = 3p/2

substitute x = 3p/2 in equation ii

3p/2 – y = p

y = 3p/2 -p = p/2

The larger of the two numbers is 3p/2

Question 4

Three consecutive positive integers, k, l  and m are such that l2 = 3(k + m). Find the value of m

1. 4        B. 5      C. 6        D. 7

solution

k = m -2

l = m-1

the above is valid since they are consecutive numbers

substitute into the above equation

(m-1)2 = 3( m – 2 + m)

m2 -2m +1 = 6m -6

m2 – 8m +7 = 0

m2 – 7m -m + 7 = 0

m(m-7) -1(m-7) = 0

m = 1 0r m = 7

D is the correct option

Question 5

If the population of a town was 240000 in January 1998 and it increases by 2% each year, what would be the population of the town in January 2000

1. 480000  B. 249696  C. 249600  D. 244800

Solution

In 1999 the population will be = 2% of 240000 + 240000 = 244 800

In 2000 the population will be = 2% of 244800 + 2448000 = 249 696

Question 6

In a youth club with 94 memebers, 60 like modern music and 50 like traditional music. The number of members who like both traditional and modern music is three times those who do not like any type of music. How many members like only one type of music?

1. 8  B. 24   C. 62  D.  86

solution

number of members = 94

n(modern) = 60

n (m n t) is those that love both musics = x

n(m’ n t’) those that love neither = y

x = 3 * y

x + 60-x + 50-x + y = 94

substitute for x = 3y

3y + 60-3y + 50-3y + y = 94

110 – 2y = 94

2y = 110 – 94

2y = 16

y = 8

x = 3y = 24

number of members that love only one type of music = 50-24 + 60-24 = 62

C is the correct option

Question 7

Audu bought an article for #50000 and sold it to Femi at a loss of x%. Femi later sold the article to Oche at a profit of 40%. If Femi made a profit of #10000, find the value of x.

1. 60    B. 50   C. 40      D. 20

Solution

Femi bougth it at = 100-x/100 * 50000

Femi sold it at = 140/100 * (100-x/100 *50000)

profit made by femi = 10000 = 140/100(100-x/100 *50000) – 100-x/100 *50000

10000 = 100-x/100 *50000(1.4 – 1)

10000 = 100-x/100 * 50000(0.4)

10000/50000(0.4) = 100-x/100

0.5 = 100-x/100

cross multiply

50 = 100-x

x = 100-50

x = 50

B is the correct option

Question 8

A trader realises 10x -x^2 naira profit from the sale of x bags of corn. How many bags will give him the maximum profit

1. 4         B. 5     C. 6     D. 7

Solution

To have maximum profit dy/dx = 0

d/dx (10x – x2) = 10 – 2x

10-2x = 0

2x = 10

x = 5

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