>> Elementary Mathematics >> Simplification

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- Elementary Mathematics
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- Indices and Surd
- LCM and HCF
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- Logarithms
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- Set theory
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- Simplification
- Statistics
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- Trigonometric ratio
- Trigonometry
- Volume and surface area

41). If \( \Large x + \frac{1}{x} = 6 \), then \( \Large x^{4} + \frac{1}{x^{4}} \) is
\( \Large x + \frac{1}{x} = 6 \) On squaring both sides, we get \( \Large \left(x + \frac{1}{x}\right)^{2} = \left(6\right)^{2} \) \( \Large x^{2} + \frac{1}{x^{2}} + 2 = 36 \) \( \Large x^{2} + \frac{1}{x^{2}} = 34 \) On squaring both sides, we get \( \Large \left(x^{2} + \frac{1}{x^{2}}\right)^{2} = \left(34\right)^{2} \) \( \Large x^{4} + \frac{1}{x^{4}} + 2 = 1156 \) \( \Large x^{4} + \frac{1}{x^{4}} = 1154 \) | ||||

42). If \( \Large x + \frac{1}{x} = 3 \), then \( \Large x^{5} + \frac{1}{x^{5}} \) is equal to
\( \Large x + \frac{1}{x} =3 \) ...(i) On squaring both sides, we get \( \Large \left(x + \frac{1}{x}\right)^{2} = \left(3\right)^{2} \) => \( \Large x^{2} + \frac{1}{x^{2}} + 2 = 9 \) =>\( \Large x^{2} + \frac{1}{x^{2}} = 7 \) ...(ii) Again squaring both sides, we get \( \Large \left(x^{2} + \frac{1}{x^{2}}\right)^{2} = \left(7\right)^{2} \) \( \Large x^{4} + \frac{1}{x^{4}} + 2 = 49 \) \( \Large x^{4} + \frac{1}{x^{4}} = 47 \) ...(iii) On cubing both sides, we get \( \Large \left(x + \frac{1}{x}\right)^{3} = \left(3\right)^{3} \) \( \Large x^{3} + \frac{1}{x^{3}}+ 3\left(x + \frac{1}{x}\right) = 27 \) \( \Large x^{3} + \frac{1}{x^{3}} + 9 = 27 \) [Because, \( \Large \left(x + \frac{1}{x}\right)=3 \) ] \( \Large x^{3} + \frac{1}{x^{3}} = 18 \) ...(iv) On multiplying Eqs. (i) and (iii), we get \( \Large \left(x^{4} + \frac{1}{x^{4}}\right) \left(x + \frac{1}{x}\right) = 47 \times 3 \) \( \Large x^{5} + \frac{1}{x^{5}} +x^{3} + \frac{1}{x^{3}} = 141 \) \( \Large x^{5} + \frac{1}{x^{5}} + 18 = 141 \) [from Eq. (iv)] \( \Large x^{5} + \frac{1}{x^{5}} = 123 \) | ||||

43). \( \Large 3x + 2y = 12 \) and \( \Large xy = 6 \), then the value of \( \Large 9x^{2} + 4y^{2} \) is
Given equations are \( \Large 3x + 2y = 12 \) ...(i) \( \Large XY = 6 \) ...(ii) On squaring Eq. (i) on both sides, we \( \Large \left(3x + 2y\right)^{2} = \left(12\right)^{2} \) => \( \Large 9x^{2} + 4y^{2} + 12xy = 144 \) => \( \Large 9x^{2} + 4y^{2} = 144 - 12xy \) = \( \Large 144 - 12 \left(6\right) = 144 - 72 = 72 \) | ||||

44). A 120 m long rope is cut into 3 equal parts. What is the length of each part?
Correct Answer: 40 m
Required length = \( \Large \frac{120}{3} \) = 40 m | ||||

45). When a number is multiplied 5 times with itself, it gives the value 1445. Find the number.
Let the number = x According to the question, \( \Large 5x^{2} = 1445 \) Therefore, \( \Large x^{2} = \frac{1445}{5} = 289 \) Therefore, \( \Large x = \sqrt{289} = 17 \) | ||||

46). Mani Ram divides Rs.17200 amongst his 5 sons, 4 daughters and 2 friends. If each daughter receives four times as much as each friend receives and each son receives five times as much as each friend receives, how much does each daughter receive?
Let share of each friend = x. Share of each daughter = 4 X 400 = 1600 | ||||

47). One-third of Rahul's marks in Mathematis exceeds one-half of his marks in Hindi by 30. If he got 480 marks in the two subjects together, how many marks did he get in Hindi?
Let Rahul's marks in Mathematics = x | ||||

48). Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtain 40 marks. The number of sums solved correctly is
Let the questions solved correctly Mohan = x According to the question, \( \Large 3 \times x - \left(30 - x\right) \times 2 = 40 \) => \( \Large 3x - 60 + 2x = 40 \) \( \Large 5x - 60 = 40 \) \( \Large 5x = 40 + 60 => x = \frac{100}{5} = 20 \) Therefore, Mohan attempted 20 questions correctly. | ||||

49). The number obtained by interchanging the digits of a two-digit number is more than the original number by 27 and the sum of the digits is 13. What is the original number?
Let the ten's-digit = x and unit's digit = Y | ||||

50). 0f three positive numbers the product of the first and second is 42, that of the second and third is 56 and that of third and first is 48. The third number is
Let first, second and third numbers be x, y and z, respectively. Then, xy = 42 ...(i) yz = 56 ...(ii) xz = 48 ...(iii) Multiplying Eqs. (i), (ii) and (iii), we get \( \Large \left(xyz\right)^{2} = 42 \times 56 \times 48 \) => \( \Large \left(xyz\right)^{2} = 112896 \) => \( \Large xyx = 336 \) Dividing Eq. (iv) by Eq. (i), we get \( \Large \frac{xyz}{xyz} = \frac{336}{42} => z = 8 \) |