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Linear Equations in One Variable Test - 11

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Linear Equations in One Variable Test - 11
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  • Question 1
    1 / -0

    A rectangular box has a length (5P - 6)m and breadth is (P + 4)m. What is the breadth of the box, if its perimeter is 20 m

    Solution

    2 [(5P − 6) + (P + 4)] = 20 ⇒ 6P − 2 = 10  ⇒ p = 2 ∴ 5P − 6 = 4 & P + 4 = 6 = breadth 

  • Question 2
    1 / -0

    Reema bought x pens at Rs 2.60 each and y greeting cards at 80 paise each. If the pens costs Rs. 12 more than the cards, the equation involving x and y is

    Solution

    'x' pens 'y' greeting cards Cost of all greetings card be 0.8y Cost of all pen = 2.6x  ⇒ 2.6 × 0.8y + 12  ⇒ 26x − 8y = 120  ⇒ 13x − 4y = 60.

  • Question 3
    1 / -0

    If the length and breadth of a room are increased by 1 m each, its area is increased by 21 m2. If the length is increased by 1 m and breadth decreased by 1 m, the area is decreased by 5m2. Find the area of the room.

    Solution

     (length +1) (breadth +1) = A + 2 length (1 + 1)(b − 1)= A − 51b +  1 + b + 1 = A + 21         ....(I) 1b − 1 + b − 1 = A − 5          .....(II)   (I) − (II) ⇒ 21 + 2 = 26  ⇒ lenth = 12

  • Question 4
    1 / -0

    The sum of two numbers is 69 and their difference is 17. Find the numbers.

    Solution

    x  + y = 69 x − y = 17  ⇒ x = 43,y = 26.

  • Question 5
    1 / -0

    A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the original number. Find the original number

    Solution

    10a + b = 7(a + b) ⇒ 3a = 6b ⇒ a = 2b 10b + a or  10b + 2b = (10a + b) − 18 Or 12b = (20b +  b) − 18 Or 18 = 9b ⇒ b =  2;a = 4  ⇒ 7(a + b )  =42.

  • Question 6
    1 / -0

    Five years hence, a man's age will be three times his son's age and five years ago, he was seven times as old as his son. Find their present ages.

    Solution

    (x + 5) = 3(y + 5) and (x −  5) = 7(y − 5) ⇒ x = 40,y = 10

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