Question 1: The product of two consecutive integer is 56. Find the integers.

$\displaystyle \text{Let the two consecutive integers be } x \text{ and } (x+1)$

$\displaystyle \text{Therefore } x(x+1) = 56$

$\displaystyle x^2 + x - 56 = 0$

$\displaystyle (x+8)(x-7) = 0$

$\displaystyle \Rightarrow x = 7 \text{ or }x = -8$

Therefore the two integers could be $\displaystyle (7 \text{ and } 8) \text{ or }(-8 \text{ and } -7)$

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Question 2: The sum of the square of two consecutive natural numbers is 41. Find the numbers.

$\displaystyle \text{Let the two consecutive integers be } x \text{ and } (x+1)$

$\displaystyle x^2 + (x+1)^2 = 41$

$\displaystyle x^2 + x^2 + 1 + 2x = 41$

$\displaystyle 2x^2 + 2x-40 = 0$

$\displaystyle \text{Or } x^2+x-20=0$

$\displaystyle \text{Or } (x+5)(x-4) = 0$

$\displaystyle \text{Or } x = -5 \text{ or }x=4$

Therefore the two integers could be $\displaystyle (4 \text{ and } 5) \text{ or }(-5 \text{ and } -4)$

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Question 3: Find the two natural numbers which differ by 5 and the sum of the square is 97.

Let the two numbers be $\displaystyle x \text{ and } (x+5)$

$\displaystyle x^2 + (x+5)^2 = 97$

$\displaystyle x^2 + x^2 + 10x + 25 = 97$

$\displaystyle 2x^2+10x -72 = 0$

$\displaystyle \text{Or } x^2+5x-36 = 0$

$\displaystyle \text{Or } (x-4)(x+9) = 0$

$\displaystyle \text{Therefore } x= 4$

Therefore the two natural numbers could be $\displaystyle (4 \text{ and } 9)$

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Question 4: The sum of a number and its reciprocal is 4.25. find the number.

$\displaystyle \text{Let the numbers be } x \text{ and } \frac{1}{x}$

$\displaystyle \therefore x + \frac{1}{x} = 4.25$

$\displaystyle \text{Or } x^2 +1 = 4.25x$

$\displaystyle \text{Or } x^2-4.25x +1= 0$

$\displaystyle \Rightarrow x = 4$

$\displaystyle \text{Let the numbers will be } 4 \text{ and } \frac{1}{4}$

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Question 5: Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.

$\displaystyle \text{Let the two natural numbers be } x \text{ and } (x+3).$

$\displaystyle \frac{1}{x} + \frac{1}{x+3} = \frac{7}{10}$

$\displaystyle 20x + 30 = 7x(x+3)$

$\displaystyle 20x + 30 = 7x^2+ 21x$

$\displaystyle 7x^2+x-30 = 0 \rightarrow x = 2$

Therefore the two natural numbers are $\displaystyle 2 \text{ and } 5$

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Question 6: Divide 15 into two parts such that the sum of reciprocal is 3/10.

$\displaystyle \text{Let the two parts be } x \text{ and } (15-x)$

$\displaystyle \frac{1}{x} + \frac{1}{15-x} = \frac{3}{10}$

$\displaystyle 15-x+x = \frac{3}{10}x(15-x)$

$\displaystyle 150 = 3x(15-x)$

$\displaystyle x^2-15x+50 = 0$

$\displaystyle \Rightarrow x = 10 \text{ or }5$

Therefore the two parts should be $\displaystyle 5 \text{ and } 10$

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Question 7: The sum of the square of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.

Let the two numbers be $\displaystyle x \text{ and } y$

$\displaystyle \text{Given } x^2+y^2=208$

and $\displaystyle y^2 = 18x$

$\displaystyle \text{Therefore } x^2+18x-208 = 0$

$\displaystyle \text{Or } (x-8)(x+26) = 0 \text{ or }x = 8 \text{ or }-26 \Rightarrow x= 8$

$\displaystyle \text{Therefore } y^2 = 18 \times 8 = 144 \Rightarrow y = 12$

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Question 8: The sum of the square of two consecutive positive even numbers is 52. Find the numbers.

Let the two consecutive numbers be $\displaystyle x \text{ and } (x+2)$

$\displaystyle \text{Therefore } x^2 + (x+2)^2 = 52$

$\displaystyle 2x^2 + 4x - 48 = 0 \Rightarrow x = 4$

Therefore the two numbers are $\displaystyle 4 \text{ and } 6$

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Question 9: Find two consecutive positive odd numbers. The sum of whose square is 74.

Let the two consecutive numbers be $\displaystyle (2x+1) \text{ and } (2x+3)$

$\displaystyle \text{Therefore } (2x+1)^2+(2x+3)^2=74$

$\displaystyle 4x^2+1+4x + 4x^2+9+12x = 74$

$\displaystyle 8x^2+16x-64=0$

$\displaystyle x^2+2x-8=0 \Rightarrow x = 2$

Therefore the two numbers are $\displaystyle 5 \text{ and } 7$

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Question 10: The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.

$\displaystyle \text{Let the fraction be} \frac{x}{2x+1}$

$\displaystyle \frac{x}{2x+1} + \frac{2x+1}{x} = 2.9$

$\displaystyle 5x^2+4x+1 = 5.8x^2+2.9x$

$\displaystyle 0.8x^2-1.1x-1 \Rightarrow x = 2$

$\displaystyle \text{Therefore the fraction is } \frac{2}{5}$

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Question 11: Three positive numbers are in the ratio $\displaystyle \frac{1}{2}:\frac{1}{3}:\frac{1}{4}$ . find the numbers if the sum of their squares is 244.

$\displaystyle \text{Let the fractions be} \frac{1}{2x}:\frac{1}{3x}:\frac{1}{4x}$

$\displaystyle \text{Given } (\frac{1}{2x})^2+(\frac{1}{3x})^2+(\frac{1}{4x})^2=244$

$\displaystyle \frac{45+16}{16 \times 9 \times x^2} = 244$

$\displaystyle x^2 = \frac{1}{16 \times 9 \times 4}$

$\displaystyle \text{Or } x = \frac{1}{24}$

Therefore the numbers are $\displaystyle 12, 8 \text{ and } 6$

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Question 12: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.

$\displaystyle \text{Let the two parts be } x \text{ and } (20-x)$

$\displaystyle \text{Given } x^2 - (20-x) = 10$

$\displaystyle x^2+x-30 = 0 \Rightarrow x = 5$

Therefore the two parts are $\displaystyle 5 \text{ and } 15.$

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Question 13: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.

Let the three consecutive number be $\displaystyle x, (x+1) \text{ and } (x+2)$

$\displaystyle \text{Given } (x+1)^2 -[(x+2)^2-x^2] = 60$

$\displaystyle x^2+1+2x -[x^2+4+4x-x^2] = 60$

$\displaystyle x^2+1+2x-4-4x=60$

$\displaystyle x^2-2x-63=0 \Rightarrow x = 9, -7 \text{ or }x = 9$

Therefore the numbers are $\displaystyle 9, 10 \text{ and } 11.$

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Question 14: Out of three consecutive positive integers, the middle number is $\displaystyle p$. if three times the square of the larger is greater than the sum of the squares of the other two numbers by 67; calculate the value of $\displaystyle p$.

Let the three positive integers be $\displaystyle (p-1), p \text{ and } (p+1)$

$\displaystyle \text{Given } 3(p+1)^2 -[(p-1)^2+p^2] = 67$

$\displaystyle 3p^2+3+6p -[p^2+1-2p+p^2] = 67$

$\displaystyle p^2+8p-65=0 \Rightarrow p = 5$

Therefore the numbers are $\displaystyle 4, 5, \text{ and } 6.$

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Question 15: A can do a piece of work in $\displaystyle x$ days and B can do the same work in $\displaystyle (x+16)$ days. If both working together can do it in 15 days: calculate $\displaystyle x$ .

$\displaystyle \text{Given } \frac{1}{x} +\frac{1}{x+16} = \frac{1}{15}$

$\displaystyle x^2-14x-240 = 0$

$\displaystyle \Rightarrow x = 24 \text{ days }$

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Question 16: One pipe can fill a cistern in 3 hours less than the other. Two pipes together can fill cistern in 6 hours and 40 minutes. Find the time that each pipe will take to fill the cistern.

$\displaystyle \text{Given } \frac{1}{x} + \frac{1}{x-3} = \frac{1}{6\frac{2}{3}}$

$\displaystyle 20(x-3+x) = 3x(x-3)$

$\displaystyle 3x^2-49x+60 = 0 \Rightarrow x = 15 \text{ and } 1.333$ (ignore this as it is not possible)

Therefore the time taken by the first pipe to fill in the cistern is $\displaystyle 15$ hours and the second one $\displaystyle 12$ hours.

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Question 17: A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking $\displaystyle x$ as the smaller part of the two parts, find the number. [2010]

$\displaystyle \text{Let the two parts be } x \text{ and } y$

$\displaystyle \text{Given } x^2+y^2 = 20$

$\displaystyle \text{Also } y^2 = 8x$

Substituting it back $\displaystyle x^2+8x-20 = 0$

$\displaystyle \Rightarrow x = 2 \text{ or }-10$ (ignore this as the number is positive)

Therefore the larger part is $\displaystyle y^ = 16 \Rightarrow y = 4$

Hence the number is $\displaystyle 2+4 = 6$