# FRACTIONS, DECIMALS, APPROXIMATION AND PERCENTAGES jamb questions

These solutions to jamb fraction, decimal, approximation and percentages jamb questions are for those who have completely exhausted the topic.

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## 1987-2: Find the length of the rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length.

### Solution

All you are asked to do here is to find the LCM of 40cm and 48cm

40=2\times 2\times 2\times 5 \\ 48=2\times 2\times 2\times 2\times 3 \\ LCM= 2\times 2\times 2\times 2\times 3\times 5=240cm

## 1987-3: A rectangular lawn has an area of 1815 square yards. If its length is 50metres, find its width in metres given that 1metre equals 1{\cdot}1yards

### Solution

1metre=1{\cdot}1yards \\ 1m^2=1{\cdot}1\times 1{\cdot}1=1{\cdot}21 square\hspace{1pt} yards \\ 1815 square\hspace{1pt} yards=\frac{1815}{1{\cdot}21}=1500m^2 \\ Area=L\times B \\ 1500m^2=50m\times width \\ width=30m

## 1987-4: Reduce each number to 2sf and then evaluate \frac{0{\cdot}021741\times 1{\cdot}2047}{0{\cdot}023789}

### Solution

In 2sf, \\ 0{\cdot}021741\approx 0{\cdot}022 \\ 1{\cdot}2047\approx 1{\cdot}2 \\ 0{\cdot}023789\approx 0(\cdot)024 \\ \therefore \frac{0{\cdot}022\times 1{\cdot}2}{0{\cdot}024} =1{\cdot}1

## 1987-6: If the length of a square is increased by 20\% while its width is decreased by 20\% to form a rectangle. What is the ratio of the area of the rectangle to the area of the square?

### Solution

Let the initial length and width be x

\therefore Area\hspace{1pt}of\hspace{1pt}square=x^2

Increasing the length by 20\% = x+20\% of x

which gives x+0{\cdot}2x=1{\cdot}2x

Reducing the width by 20\%=x-20\% of x

which gives x-0{\cdot}2x=0{\cdot}8x \\ Area\hspace{1pt}of\hspace{1pt}rectangle=1{\cdot}2x\times 0{\cdot}8x=0{\cdot}96x^2 \\

ratio of the area of the rectangle to area of the square is given by \frac{0.96x^2}{x^2}=0.96

In ratio form, rectangle:square=0{\cdot}96:1 \\ \therefore ratio=24:25

## 1989-6: What is the difference between 0{\cdot}007685 correct to 3sf and 0{\cdot}007685 correct to 4dp?

### Solution

correct to 3sf, 0{\cdot}007685\approx 0{\cdot}00769

correct to 4dp, 0{\cdot}007685\approx 0{\cdot}0077 \\ \therefore difference=0{\cdot}00769-0{\cdot}0077=0{\cdot}00001

## 1989-7: If a:b=5:8,x:y=25:16 evaluate \frac{a}{x}:\frac{b}{y}

### Solution

From the question,

a=5,b=8,x=25,y=16 \\ \frac{a}{x}:\frac{b}{y}=\frac{5}{25}:\frac{8}{16}=\frac{1}{5}:\frac{1}{2} \\ \therefore \frac{a}{x}:\frac{b}{y}=2:5

## 1989-8: Oke deposited \#800.00 in the bank at the rate of 12\frac{1}{2}\% simple interest. After some time, the total amount was one and a half times the principal. For how many years was the money left in the bank?

### Solution

P=\#800,R=12\frac{1}{2}\%,T=? \\ A=1\frac{1}{2}\times \#800=\frac{3}{2}\times \#800=\#1200 \\ A=I+P \\ I=A-P \\ I=\#1200-\#800=\#400 \\ I=\frac{PRT}{100} \\ 400=\frac{800\times \frac{25}{2}\times 2}{100} \\ T=\frac{100}{25} \\ T=4years

## 1989-9: If the surface area of a sphere is increased by 44\%, find the percentage increase in diameter

### Solution

Let the initial surface area be x

x=4\pi (\frac{d}{2})^2=x\hspace{10pt}eq(i)

New surface area=x+44\% of x=1{\cdot}44x \\ 1.44x=4\pi (\frac{\triangle d}{2})^2=1{\cdot}44x\hspace{10pt}eq(ii)

dividing eq(ii) by eq(i)

\frac{4\pi\triangle d^2}{4\pi d^2}=\frac{1.44x}{x} \\ \triangle d^2=1.44d^2 \\ \triangle d=1{\cdot}2d

increase in diameter=\frac{1{\cdot}2d-d}{d}\times 100\% \\ \frac{0{\cdot}2d}{d}\times 100\% \\ 20\%

## 1986-6: Three boys shared some oranges. The first received \frac{1}{3} of the oranges, the second received \frac{2}{3} of the remainder. If the third boy received the remaining 12 oranges, how many oranges did they share?

### Solution

Let the oranges shared be x

first boy received \frac{1}{3}x=\frac{x}{3}

The second boy received \frac{2}{3} of the remainder

remainder=x-\frac{x}{3}=\frac{2x}{3} \\ \therefore second boy received \frac{2}{3}\times \frac{2x}{3}=\frac{4x}{9}

third boy received the remaining 12 oranges

remainder=x-(\frac{4x}{9}+\frac{x}{3})=\frac{2x}{9} \\ \frac{2x}{9}=12 \\ 2x=108 \\ x=54

54 oranges were shared

## 1986-9: Udoh deposited \#150.00 in the bank. At the end of 5 years, the simple interest on the principal was \#55.00. At what rate per annum was the interest paid?

### Solution

SI=\frac{P\times R\times T}{100} \\P=\#150000\hspace{2pt}R=?\hspace{2pt}T=5years\hspace{2pt}SI=\#55{\cdot}00 \\ 55=\frac{150\times R\times 5}{100} \\ R=0{\cdot}7333 \\ R=73{\cdot}33\%

## 1986-10: A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2:3:5 respectively. If Bisi got 5, how many were shared out?

### Solution

Let the total No of pencils be x

Bisi share=\frac{2x}{10}=5 \\ x=25 \\ \therefore 25 pencils were shared out

## 1986-44: An open rectangular box externally measures 4m\times 3m\times 4m. Find the total cost of painting the box externally if it costs \#2{\cdot}00 to paint one square metre.

### Solution

Area of the shape=2(4\times 4)+2(3\times 4)+(3\times 4) [because the top is opened]

Area=68m^2

if it cost \#2{\cdot}00 to paint 1m^2,

total cost=68m^2\times \frac{\#2{\cdot}00}{1m^2}=\#136{\cdot}00

## 1987-7: Two brothers invested a total of \#5000{\cdot}00 on a farm project. The farm yield was sold for \#15000{\cdot}00 at the end of the season. If the profit was shared in the ratio 2:3, what is the difference in the amount of profit received by the brothers?

### Solution

profit=SP-CP \\ profit=\#15000-\#5000=\#10000

first brother share=\frac{2}{5}\times 10000=\#4000

second brother share=\frac{3}{5}\times 10000=\#6000

difference=\#6000-\#4000=\#2000

## 1987-9: A man invests a sum of money at 4\% per annum simple interest. After 3years, the principal amounts to \#7000{\cdot}00. Find the sum invested

### Solution

I=\frac{PRT}{100} \\ Amount= P+I \\ \therefore I=Amount-P \\ Amount-P=\frac{PRT}{100} \\ Amount=\frac{PRT}{100}+P \\ 7000=\frac{P\times 4\times 3}{100}+p \\ 7000=\frac{12P}{100}+P \\ 7000=\frac{112P}{100} \\ P=6250 \\ \therefore Sum invested=\#6250{\cdot}00

## 1987-10: By selling 20 oranges for \#1{\cdot}35, a trader makes a profit of 8\%. What is his percentage gain or loss if he sells the same 20 oranges for \#1{\cdot}10?

### Solution

SP=20\times 1.35=\#27 \\ profit=8\% \\ CP=? \\ \%profit=\frac{SP-CP}{CP}\times 100\% \\ 8=\frac{27-CP}{CP}\times 100 \\ \frac{27-CP}{CP}=0{\cdot}08 \\ 27-CP=0{\cdot}08CP \\ 1{\cdot}08CP=27 \\ CP=\#25

if he sells the same 20oranges for \#1{\cdot}10, \\ SP=20\times 1{\cdot}10=\#22

since SP is less than CP, a loss is made

\%loss=\frac{CP-SP}{CP}\times 100\% \\ \%loss=\frac{25-22}{25}\times 100\% \\ \%loss=\frac{3}{25}\times 100\% \\ \%loss=12\%

## 1987-13: Instead of writing \frac{35}{6} as a decimal correct to 3sf, a student wrote it correct to 3dp. Find the error in standard form

### Solution

\frac{35}{6}=5{\cdot}8333333333

correct to 3sf=5{\cdot}83

correct to 3dp=5{\cdot}833 \\ error=5{\cdot}833-5{\cdot}83=0{\cdot}003

in standard form, error=3\times 10^{-3}

## 1988-1: Simplify \frac{1\frac{1}{2}}{21\frac{1}{4}of\hspace{1pt} 32}

### Solution

\frac{1\frac{1}{2}}{21\frac{1}{4}of\hspace{1pt} 32} \\ \frac{\frac{3}{2}}{\frac{85}{4}\times 32} \\ \frac{\frac{3}{2}}{680} \\ \frac{3}{2\times 680}\rightarrow \frac{3}{1360}

## 1988-2: If x is the addition of prime numbers between 1 and 6; and y the HCF of 6,9,15. Find the product of x and y

### Solution

x=2+3+5=10\\ y=3

Product of x and y=10\times 3=30

## 1988-3: A 5{\cdot}0g of salt was weighed by Tunde as 5{\cdot}1g. What is the percentage error?

### Solution

error=5{\cdot}1-5{\cdot}0=0{\cdot}1g \\ \%error=\frac{error}{actual\hspace{2pt}measurement}\times 100\% \\ \%error=\frac{0{\cdot}1}{5{\cdot}0}\times 100\%\\ \%error=2\%

## 1988-5}: Two sisters; Taiwo and Kehinde own a store. The ratio of Taiwo’s share to Kehinde’s is 11:9. Later, Kehinde sells \frac{2}{3} of her share to Taiwo for \#720{\cdot}00. Find the value of the store

### Solution

Let the value of the store be

Taiwo's\hspace{2pt}share=\frac{11}{20}\times x \\ Kehinde's\hspace{2pt}share=\frac{9}{20}\times x \\ \frac{2}{3} of\hspace{2pt} Kehinde's\hspace{2pt} share=\#720{\cdot}00\\ \therefore\frac{2}{3}\times \frac{9}{20}x=720 \\ \frac{3}{10}x=720 \\ x=\#2400 \\ \therefore The value of the store equals \#2400

## 1988-6: A basket contains green, black and blue balls in the ratio 5:2:1. If there are 10 blue balls, find the corresponding ratio when 10 green and 10 black balls are removed from the basket

### Solution

Let the total number of balls be x

blue\hspace{2pt}balls=\frac{1}{8}x=10 \\ x=80balls\\ green\hspace{2pt}balls=\frac{5}{8}x=\frac{5}{8}\times 80=50balls \\ black\hspace{2pt}balls=\frac{2}{8}x=\frac{2}{8}\times 80=20balls

If 10 green and 10 black balls are removed from the basket,

blue\hspace{2pt}balls=10 \\ green\hspace{2pt}balls=40\\ black\hspace{2pt}balls=10

corresponding new ratio of black, green and blue balls will be 40:10:10=4:1:1

## 1988-7: A taxpayer is allowed \frac{1}{8} of his income tax-free and pays 20\% on the remainder. If he pays \#490{\cdot}00 tax, what is his income?

### Solution

Let his income be x

tax free income=\frac{1}{8}x

taxed income=x-\frac{1}{8}x=\frac{7}{8}x \\tax=20\% of \frac{7}{8}x=\frac{20}{100}\times \frac{7x}{8}=\#490{\cdot}00 \\ \frac{x}{40}=\#70{\cdot}00 \\ x=\#2800{\cdot}00 \\ \therefore His income is \#2800{\cdot}00

## 1988-11: The thickness of an 800 paged book is 18mm. Calculate the thickness of one leaf of the book giving your answers in metres and in standard form

### Solution

1 leaf=2 pages

\therefore 800\hspace{2pt} pages=400 leaves

If 400 leaves=18mm,

then, 1 leaf=\frac{400}{18}=22{\cdot}222mm \\ 1000mm=1m \\ \therefore 22{\cdot}222mm=\frac{22{\cdot}222}{1000}=0{\cdot}0222m

In standard form, the thickness=2{\cdot}22\times 10^{-2}m

## 1988-13}: If \frac{1}{p}=\frac{a^2+2ab+b^2}{a-b} and \frac{1}{q}=\frac{a+b}{a^2-2ab+b^2}. Find \frac{p}{q}

### Solution

From \frac{1}{p}, \\ a^2+2ab+b^2 \\ a^2+ab+ab+b^2 \\ a(a+b)+b(a+b) \\ (a+b)(a+b)\\ \therefore\frac{1}{p}=\frac{(a+b)(a+b)}{a-b}

Similarly; from\frac{1}{q} \\ a^2-2ab+b^2=(a-b)(a-b) \\ \frac{1}{q}=\frac{a+b}{(a-b)(a-b)} \\ \frac{p}{q}=\frac{a-b}{(a+b)+b(a+b}\times \frac{a+b}{(a-b)(a-b)} \\ \frac{p}{q}=\frac{1}{(a+b)(a-b)} \\ \frac{p}{q}=\frac{1}{a^2-b^2}

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