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Q1. Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 154. Find the integers.

Solution

Let the three consecutive positive integers be x, x + 1, x + 2. x² + (x + 1) (x + 2) = 154 x² + (x² + 3x + 2) = 154 2x² + 3x - 152 = 0 2x² - 16x + 19x - 152 = 0 2x (x - 8) + 19 (x - 8) = 0 (x - 8)(2x + 19) = 0 x = 8 or -19/2 But, x is a positive integer. So, x = 8. Thus, the numbers are 8, 9, 10.

Q2. A person can buy 15 books less for Rs. 900 when the price of book goes up by Rs. 3. Find the original price and the no. of copies he would buy at initial price.

Solution

 Therefore the number of copies that he can buy with the initial price 

Q3. The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution

Let base = x cm and height = (x - 7) cm Hypotenuse = 13 cm By Pythagoras theorem, x² + (x - 7)² = 13² x² + 49 + x² - 14x = 169 2x² - 14x - 120 = 0 x² - 12x + 5x - 60 = 0 x( x - 12) + 5 (x - 12) = 0 (x - 12) (x + 5) = 0 x - 12 = 0 or x + 5 = 0 x=12 or x=-5 (rejecting) The other two sides are 12 cm and 5 cm.

Q4. Solve the following equation and calculate the answer correct to two decimal places: x² - 5x - 10 = 0.

Solution

Q5. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller, find the numbers.

Solution

Let x and y be two positive integers. Since the sum of the squares of two positive integers is 208, we have x² + y² = 208…(1) Also given that the square of the larger number is 18 times the smaller. ⇒ x² = 18y…(2) Substituting the value of x² from equation (2) in equation (1), we get 18y + y² = 208 ⇒ y² + 18y = 208 ⇒ y² + 18y - 208 = 0 ⇒ y² + 26y - 8y - 208 = 0 ⇒ y(y + 26) - 8(y + 26)= 0 ⇒ (y - 8) (y + 26) = 0 ⇒ y - 8 = 0 or y + 26 = 0 ⇒ y = 8 or y = -26 Since the numbers are positive integers, y = 8 And hence, x² = 18y = 18 × 8 = 144 ⇒ x = 12 Thus, the positive integers are 12 and 8.  

Q6. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.

Solution

Sum of the areas of two squares = 468 m² Let a and b be the sides of the two squares. ⇒a² + b² = 468…(1) Also given that, the difference of their perimeters = 24m ⇒4a - 4b = 24 ⇒a - b = 6 ⇒a = b + 6…(2) We need to find the sides of the two squares. Substituting the value of a from equation (2) in equation (1), we get (b + 6)² + b² = 468 ⇒b² + 6² + 2 × b × 6 + b² = 468 ⇒2b² + 36 + 12b = 468 ⇒2b² + 36 + 12b - 468 = 0 ⇒2b² + 12b - 432 = 0 ⇒b² + 6b - 216 = 0 ⇒b² + 18b - 12b - 216 = 0 ⇒b(b + 18) - 12(b + 18) = 0 ⇒(b + 18)(b - 12) = 0 ⇒b + 18 = 0 or b - 12 = 0 ⇒b = -18 = 0 or b = 12 Side cannot be negative and hence b = 12 m. Therefore, a = b + 6 = 12 + 6 = 18 m. Therefore the sides of the two squares are 12m and 18m.

Q7. Solve for x and give your answer correct to two decimal places: 3x² - 5x = 1

Solution

Q8. Sum of the area of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution

Let the side of the squares be x and y meters. Areas of first square and second square are x² and y² respectively. Perimeters of first and second squares are 4x and 4y respectively. From the given information, we have: x² + y² = 468… (1) 4x - 4y = 24 x - y = 6 y = x - 6 Substituting the value of y in (1), we get x² + (x - 6)² = 468 x² + x² +36 - 12x = 468 2x² - 12x - 432 = 0 x² - 6x - 216 = 0 (x - 18)(x + 12) = 0 x = 18 or x = -12 As the side cannot be negative, x = 18 Hence, side of the first square is 18 m. Side of the second square is y = (18 - 6) m = 12 m

Q9. Find two consecutive positive integers, sum of whose squares is 25.

Solution

Let the required numbers be x and x+1. Given: x² + (x+1)² = 25 2x² + 2x - 24 = 0 Þ x² + x - 12 = 0 i.e. x² + 4x - 3x - 12 = 0 i.e. x(x + 4) - 3(x + 4) = 0 (x + 4) (x - 3) = 0 x = - 4 or x = 3 Rejecting x = -4, we get the given consecutive positive integers as x = 3 and x + 1 = 3 + 1 = 4

Q10. Solve the following quadratic equations by factorization : 4x² + 5x = 0

Solution

Consider the given quadratic equation, 4x² + 5x = 0 x (4x + 5) = 0 ⇒x = 0 or 4x + 5 = 0 ⇒x = 0 or 4x = -5