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Linear Equations in One Variable Test - 34

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Linear Equations in One Variable Test - 34
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  • Question 1
    1 / -0
    Length of a rectangle is $$8$$ m less than twice its breadth. If the perimeter of the rectangle is $$56$$ m. Find its length and breadth.
    Solution
    Let the breadth of the rectangle be $$x$$ m. Then,
    length of the rectangle $$= (2x - 8)$$ m
    Given, perimeter $$= 56$$ m
    $$\Longrightarrow 2(2x-8+x)=56\\ \Longrightarrow 2(3x-8)=56\\ \Longrightarrow 6x-16=56\\ \Longrightarrow 6x=56+16\quad \quad \quad [Transposing\quad -16\quad to\quad RHS]\\ \Longrightarrow 6x=72\\ \Longrightarrow x=12$$  
    $$\text{Therefore, breadth of the rectangle}$$ $$= 12$$ $$\text{m and length of the rectangle}$$ $$= 2 \times 12 - 8 = 16$$ $$m$$
  • Question 2
    1 / -0
    $$70$$ coins of $$10$$ paise and $$50$$ paise are mixed in a purse. If the total value of the money in the purse is Rs. $$19$$, find the number of each type of coins in the purse.
    Solution
    Suppose the number of $$10$$ paise coins $$= x$$

    $$\therefore$$ the number of $$50$$ paise coins $$= 70 - x$$

    Now, value of $$10$$ paise coins $$=$$ Rs. $$\displaystyle \frac{x}{10}$$

    and value of $$50$$ paise coins $$=$$ Rs. $$\displaystyle \frac{70-x}{2}$$

    Total value of coins $$=$$ $$\displaystyle \frac{x}{10}+ \frac{70-x}{2}$$

    $$\therefore \displaystyle \frac{x}{10} + \frac{70-x}{2} = 19$$

    Multiplying both sides by $$10$$, (LCM of $$10,2 = 10$$), we get 

    $$x + 5 (70 - x) =19 \times 10$$    

    $$x + 350 - 5x = 190$$

    $$-4x = -160 $$

    $$\displaystyle \frac{-4x }{-4} = \frac{-160}{-4}$$

    $$\therefore x = 40$$

    $$\therefore$$ number of $$10$$ paise coins $$= 40$$ 

    Number of $$50$$ paise coins $$= 70 - 40 = 30$$.
  • Question 3
    1 / -0
    The digit in the units place of a $$2$$ digit number is $$4$$ times the digit in the tens place. The number obtained by reversing the digits exceeds the given number by $$54$$. Find the given number.
    Solution
    Let one's digit be $$4x$$. Then, ten's digit $$= x$$.
    Number $$=10x+4x=14x$$ ....(1)
    Number obtained by reversing the digits $$= 40x + x = 41x$$
    According to the given condition, 
    $$41x - 14x = 54$$
    $$27x =54 $$
    $$x=2$$
    Putting the value of $$x$$ in (1) , we get 
    Number $$=14 \times 2 = 28$$
    Hence, $$28$$ is the required number.
  • Question 4
    1 / -0
    The difference between a number and two third of itself is $$13$$. Find the number.
    Solution
    Let the number be $$x$$.
    According to the given condition,
    $$x-\cfrac { 2 }{ 3 } x=13$$
    $$ \Rightarrow \cfrac { 3x-2x }{ 3 } =13$$ ....(By taking L.C.M.)
    $$\Rightarrow \cfrac { x }{ 3 } =13$$
    Multiply both sides by $$3$$
    $$3 \times \dfrac {x}{3}=13 \times 3$$
    $$ \Rightarrow x=39$$ 
    Hence, the number is $$39$$.
  • Question 5
    1 / -0
    The distance between two towns is $$300$$ km. Car $$A$$ and Car $$B$$ starts simultaneously from these towns and move towards each other. The speed of Car $$A$$ is more than that of Car $$B$$ by $$7$$ km/h. If the distance between the cars after two hours is $$34$$ km. Find the speed of both cars.
    Solution
    Let the speed of car $$B$$ be $$x$$ km/h. 
    Then, speed of car $$A$$ $$= (x + 7)$$ km/h
    Now, distance covered by car $$B$$ in $$2$$ hours$$ = 2x  $$ km and
    Distance covered by car $$A$$ in $$2$$ hours $$= 2(x + 7)$$ km
    Given, distance between the cars after two hours is $$34$$ km and distance between both the towns is $$300$$ km.
    So, distance covered by both the cars in two hours moving in opposite direction $$= 300 - 34 = 266$$ km
    $$\Rightarrow 2(x+7)+2x=266$$
    $$ \Rightarrow 2x+14+2x=266$$
    $$\Rightarrow 4x=252$$
    $$ \Rightarrow x=63$$ km/h 

    Therefore, speed of car $$B$$ $$= 63$$ km/h and speed of car $$A$$ $$= 63 + 7 = 70$$ km/h.
  • Question 6
    1 / -0
    The difference of two numbers is $$72$$ and the quotient obtained by dividing the one by the other is $$3$$. Find the numbers.
    Solution
    Let one number be $$x$$. 
    According to given condition, second number $$= 3x$$
    Given, difference of two numbers $$= 72$$
    $$\Rightarrow 3x-x=72\\ \Rightarrow 2x=72\\ \Rightarrow x=36$$
    So, one number $$= 36$$ and other number $$= 3 \times 36 = 108$$
  • Question 7
    1 / -0
    The sum of three consecutive odd numbers is $$63$$. Find them.
    Solution
    Let the three consecutive odd numbers be $$(2x+1), (2x+3)$$ and $$(2x+5)$$.
    According to the given condition, we have
    $$(2x+1) + (2x+3) + (2x+5) = 63$$
    $$6x + 9 = 63$$
    $$6x = 54 $$        ...[By transposing $$9$$ to RHS]
    $$x = 9$$
    So, three consecutive odd numbers are $$(2 \times 9 + 1), (2 \times 9 + 3)$$ and $$(2 \times 9 + 5)$$ i.e., $$19, 21$$ and $$23$$. 
  • Question 8
    1 / -0
    Nineteen boys turn out for baseball. Of these 11 are wearing baseball shirts and 14 are wearing baseball pants. There are no boys without one or the other. The number of boys wearing full uniforms is ________.
    Solution
    Let the no. of boys wearing full uniform be x


     
    No. of boys wearing baseball shirts only =11x=11−x

    and No. of boys wearing baseball pants only =14x=14−x

    11x+14x+x=19∴11−x+14−x+x=19

    or, x=191114=6

    x=6
  • Question 9
    1 / -0
    Sixteen years ago, Tanya's grandfather was $$8$$ times older to her. $$8$$ years from now he would be $$3$$ times for her age. Eight years ago. what was the ratio of Tanya's age to that of her grandfather?
    Solution
    Let, Tanya age $$16$$ years ago $$= x$$
    Grandfather's age $$16$$ years ago $$= 8x.$$
    $$8$$ years from now, $$3(x+16+8)=8x+16+8$$
    $$3x+48+24=8x+24$$
    $$\Rightarrow 3x-8x=24-72$$
    $$\Rightarrow -5x=-48$$
    $$\Rightarrow x=\cfrac{48}{5}$$
    Also given, $$8$$ years ago, ratio was
    $$\cfrac{x+8}{8x+8}\Rightarrow \cfrac{\frac{48}{5}+8}{8\times \cfrac{48}{5}+8}$$
    $$\Rightarrow\cfrac{ \cfrac{88}{5}}{\cfrac{424}{5}}$$
    $$\Rightarrow \cfrac{88}{424}$$
    $$\Rightarrow \cfrac{11}{53}$$
    The ratio of Tanya's age to that of her grandfather is $$11:53$$.
  • Question 10
    1 / -0
    A labourer was engaged for $$30$$ days on the condition that he will be paid Rs. $$35$$ for each day he works and will be fined Rs. $$10$$ for each day he is absent. He earned Rs. $$735$$ in total. Find how many days did he work?
    Solution
    Let the number of days he worked be x. Then
    $$x \times 35 -10(30 -x) = 735$$
    or $$35x -300 + 10x = 735$$
    or $$45x=735+300$$
    or $$x=\frac {1035}{45}=23$$
    $$\therefore x=23 days$$
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