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1). A number consists of two digits. The sum of the digits is 10. On reversing the digits of the number, the number decreases by 36. What is the product of the two digits?
Let the unit's digit of the number be x and ten's digit be y. Therefore, Number = 10y + x According to the question, x + y = 10 ...(i) \( \Large 10x+y = \left(10x+x\right)-36 \) ...(ii) = 10x+y-10y-x=-36 = 9x - 9y = -36 = x - y = -4 On adding Eqs. (i) and (ii), we get x + y = 10 \( \Large \frac{x-y=-4}{2x=6} \) Therefore, x = 3 and y = 7 Therefore, Required product of two digits = \( \Large 3 \times 7 \) = 21 | ||||

2). If a number is multiplied by three-fourth of itself, the value thus obtained is 10800. What is that number?
Let the number be x. According to the question, \( \Large x \times \left(x \times \frac{3}{4}\right)=10800 \) \( \Large \frac{3x^{2}}{4} = 10800 \) \( \Large x = \sqrt{14400} = 120 \) | ||||

3). The sum of five consecutive numbers is 190. What is the sum of the largest and the smallest numbers?
Let the five consecutive numbers are x-2,x-1,x,x+ 1,x+ 2. Sum of numbers = 190 Therefore, x - 2 + x - 1 + x + x + 1 + x + 2 = 190 = 5x = 190 Therefore, x = 38 Sum of largest and smallest numbers = 190 = x + 2 + x - 2 = 2x = \( \Large 2 \times 38 \) = 76 | ||||

4). The product of two consecutive odd numbers is 19043. Which is the smaller one?
Let the two consecutive odd numbers are x and x + 2. According to the question, \( \Large x \left(x+2\right)=19043 \) \( \Large x^{2}+2x-19043=0 \) \( \Large x^{2}+139x-137x-19043=0 \) \( \Large x \left(x+139\right)-137 \left(x+139\right)=0 \) \( \Large \left(x+139\right) \left(x-137\right)=0 \) = x = 137, -139 x = 137 [ignore -ve sign] So, the smallest one is 137. | ||||

5). If the three-fourth of a number is subtracted from the number, the value so obtained is 163. What is that number?
Let the number be x. | ||||

6). The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers.
Let the numbers be x and y, Then, x + y = 10 ...(i) and xy = 20 ...(ii) Therefore, \( \Large \left(x-y\right)^{2}= \left(x+y\right)^{2}-4xy \) \( \Large \left(x-y\right)^{2}= \left(10\right)^{2}-4 \times 20=20 \) \( \Large x - y = \sqrt{20} = 2\sqrt{5} \) ...(iii) On adding Eqs. (i) and (iii), we get x + y = 10 \( \Large x - y = 2\sqrt{5} \) \( \Large 2x = 10 + 2\sqrt{5} \) Therefore, \( \Large x = 5 + \sqrt{5} \) On putting the value of x in Eq. (i), we x + y = 10 => \( \Large 5 + \sqrt{5} + y = 10 \) Therefore, \( \Large y = 5 - \sqrt{5} \) Sum of reciprocals of x and y = \( \Large \frac{1}{x} \) + \( \Large \frac{1}{y} \) \( \Large =\frac{1}{5+\sqrt{5}}+\frac{1}{5-\sqrt{5}}= \frac{5-\sqrt{5}+5+\sqrt{5}}{ \left(5+\sqrt{5}\right) \left(5-\sqrt{5}\right) } \) \( \Large = \frac{10}{ \left(5\right)^{2} - \left(\sqrt{5}\right)^{2} } = \frac{10}{20} = \frac{1}{2} \) | ||||

7). Five times of a positive integer is equal to 3 less than twice the square of that number. Find the number.
Let the number be x. | ||||

8). A man has given one-fourth part of his property to his daughter, half part to his sons and one-fifth part given as charity. How much part of his property he has given?
Total share given by a person | ||||

9). A chocolate has 12 equal pieces. Manju gave \( \Large \frac{1}{4} \)th of it to Anju, \( \Large \frac{1}{3} \)rd of it to Sujata and \( \Large \frac{1}{6} \)th of it to Fiza. The number of pieces of chocolate left with Manju is
The number of pieces of chocolate with Maniu = \( \Large 1 - \left(\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\right) \) = \( \Large 1 - \left(\frac{3 + 4 + 2}{12}\right) \) = \( \Large 1 - \frac{9}{12} = \frac{12 - 9}{12} = \frac{3}{12} \) Hence, number of pieces of chocolate left with Manju is 3. | ||||

10). A number consists of two digits whose sum is 10. If the digits of the number are reversed, then the number decreased by 36. Which of the following is/are correct? I. The number is divisible by a composite number. II. The number is a multiple of a prime number.
Let the two-digit number be 10x + y. |