## Filters

Sort by :
Clear All
Q

Q26.    A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

[Hint: If x is the length of the shortest board, then , and are the lengths of the second and third piece, respectively. Thus, and ].

Let  x is the length of the shortest board, then  and are the lengths of the second and third piece, respectively. The  man wants to cut three lengths from a single piece of board of length 91cm. Thus,                                    if the third piece is to be at least 5cm longer than the second, than                   We conclude that    and   . Thus , . Hence, the length of the...

Q25.    The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Let the length of the smallest side be x cm.  Then largest side = 3x cm. Third side = 3x-2  cm. Given: The perimeter of the triangle is at least 61 cm.                 Minimum length of the shortest side is 9 cm.

Q24.    Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Let x be smaller of two consecutive even positive integers. Then the other integer is x+2. Both integers are larger than 5.  Sum of both integers is less than 23.   We conclude    and    and x is even integer number.   x can be 6,8,10. The pairs of consecutive even positive integers are .

Q23.    Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Let x be smaller of two consecutive odd positive integers. Then the other integer is x+2. Both integers are smaller than 10.        Sum of both integers is more than 11.   We conclude    and    and x is odd integer number.   x can be 5,7. The two pairs of consecutive odd positive integers are .

Q22.    To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Sunita’s marks in the first four examinations are 87, 92, 94 and 95. Let x be marks obtained in the fifth examination. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations.                                                                                    Thus, Sunita must obtain 82 in the fifth examination to get grade ‘A’ in the course.

Q21.    Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Let x be marks obtained by Ravi in the third test. The student should have an average of at least 60 marks.                                             the student should have minimum marks of 35 to have an average of 60

Solve the inequality and show the graph of the solution on number line

Q20.

Given :             are  real numbers greater than  equal to  Hence, values of x can be  as   The graphical representation of solutions of the given inequality is as :

Solve the inequality and show the graph of the solution on number line

Q19.

Given :            are  real numbers greater than -1 Hence, values of x can be  as   The graphical representation of solutions of given inequality is as :

Solve the inequality and show the graph of the solution on number line

Q18.

Given :             are real numbers greater than equal to  -1. Hence, values of x can be  as   The graphical representation of solutions of the given inequality is as :

Solve the inequality and show the graph of the solution on number line

Q17.

Given :             are  real numbers less than 3 Hence, values of x can be  as   The graphical representation of solutions of the given inequality is as :

Solve the inequality for real

Q16.

Given :              are  real numbers less than equal 2. Hence, values of x can be  as  .

Solve the inequality for real x

Q15.

Given :              are  real numbers greater than 4. Hence, values of x can be  as

Solve the inequality for real

Q14.

Given :              are  real numbers less than equal to 2. Hence , values of x can be  as

Solve the inequality for real

Q13.

Given :              are  real numbers greater than 4 Hence , values of x can be  as

Solve the inequality for real

Q12.

Given :              are  real numbers less than equal to 120. Hence, values of x can be  as  .

Solve the inequality for real

Q11.

Given :              are  real numbers less than equal to 2. Hence, values of x can be  as

Solve the inequality for real .

Q10.

Given :              are  real numbers less than -6 Hence, values of x can be  as

Solve the inequality for real

Q9.

Given :              are  real numbers less than 6 Hence, values of x can be  as

Solve the inequality for real

Q8.

Given :              are  real numbers less than equal to 4 Hence, values of x can be  as

Solve the inequality for real  .

Q7.

Given :            are  real numbers less than equal to  -3 Hence , values of x can be as ,
Exams
Articles
Questions