FRACTIONS, DECIMALS, APPROXIMATION AND PERCENTAGES jamb questions

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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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