Class 10: Quadratic Equations – Exercise 8(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

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

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

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

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.

Answer:

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

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

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

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

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

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

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

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.

Answer:

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.

Answer:

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

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

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

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

$\displaystyle \Rightarrow x = 4$

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

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

Answer:

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

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

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

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

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

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.

Answer:

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

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

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

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

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

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

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

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

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

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

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

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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.

Answer:

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

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

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

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

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

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.

Answer:

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

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

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

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.

Answer:

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 $\displaystyle p$ .

Answer:

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 $\displaystyle x$ .

Answer:

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

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

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

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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.

Answer:

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

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

$\displaystyle 3x^2-49x+60 = 0 \Rightarrow x = 15 \text{ and } 1.333$ (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]

Answer:

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