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Simple Equations Test - 18

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Simple Equations Test - 18
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  • Question 1
    1 / -0
    The average of 11, 12, 13, 14, and x is 13. The value of x is 
    Solution
    11+12+13+14+x5=13\displaystyle \frac{11+12+13+14+x}{5}=13
    50+x5=13\displaystyle \frac{50+x}{5}=13
    50+x=6550+x = 65
    x=6550x = 65-50
    x=15x = 15
    Hence, the correct option is CC
  • Question 2
    1 / -0
    The average of four consecutive even numbers is 1515. The 22nd highest number is
    Solution
    Let the numbers be x2,x,x+2,x+4x - 2,x,x + 2,x + 4

    According to the question,

    x2+x+x+2+x+44=15\dfrac{x-2 + x + x + 2 + x + 4}{4} = 15

    4x+4=604x + 4 = 60

    4x=564x = 56 x=14\Rightarrow x = 14

    Hence, the second highest number i.e. x+2=14+2=16x + 2 = 14 + 2 = 16
  • Question 3
    1 / -0
    If the mean of x,x+2,x+4,x+6,x+8x, x+2, x+4, x+6, x+8 is 2020 then xx is
    Solution
    mean= x+x+2+x+4+x+6+x+85=20\dfrac{x+x+2+x+4+x+6+x+8}{5}=20
    5x=805x=80
    x=16x=16
  • Question 4
    1 / -0
    If the mean of 2,4,6,8,x,y2, 4, 6, 8, x, y is 55, then find the value of x+yx + y ?
    Solution
    Mean =Sum  of   the   termsTotal   number   of   terms=\cfrac{{Sum \; of\;  the\;  terms}}{{Total\;  number\;  of\;  terms}}

    2+4+6+8+x+y6=5\Rightarrow \cfrac{2+4+6+8+x+y}{6}=5

    20+x+y6=5\cfrac{20+x+y}{6}=5

    20+x+y=5(6)20+x+y=5(6)

    x+y=3020=10x+y=30-20=10
  • Question 5
    1 / -0
    The weight of Rahul is 5 kg5\ kg more than Sachin .The weight  of Samir is 12 kg12\ kg less than double the weight of Sachin .If the total weight of all the three is 9393 form the equation
    Solution
    Let weight of sachin =x kg=x \ kg

    Acc. to question.

    Weight of Rahul =(x+5) kg=(x+5)\ kg

    Weight of Samir =(2x12) kg=(2x-12)\ kg

    Total weight of all =93 kg=93\ kg

    That is ,

    x+(x+5)+(2x12)=93x+ (x+5)+ (2x-12)=93

    4x7=934x-7=93


  • Question 6
    1 / -0
    Convert this statement in mathematical form.
    The number is 88 greater than 80.80.
    Solution
    Let the number be xx. Here, the given statement is the number is 88 greater than 8080.

    The word 'greater than' suggests addition. So, the verbal expression the number is 88 greater than 8080 can be represented by the algebraic expression:

    x=80+8x8=80x=80+8\\ \Rightarrow x-8=80

    Hence, the statement the number is 88 greater than 8080 can be rewritten as x8=80x-8=80.
  • Question 7
    1 / -0
    66 more than twice the number is 22 22. Express the statement in mathematical form.
    Solution
    Let the number be x.x.

    Twice the number =2x=2x
    Six more than 2x2x will be 6+2x.6+2x.
    This is equal to 22.22.

    Hence, the answer is 6+2x=22.6+2x=22.
  • Question 8
    1 / -0
    If 66, pp, 121288 and  99 mean of the data is 99 then p=p= ?
    Solution
    Arithmetic mean,A=SN=6+p+12+8+95=9A=\dfrac { S }{ N } =\dfrac { 6+p+12+8+9 }{ 5 } =9

    35+p5=9\dfrac { 35+p }{ 5 } =9

    35+p=4535+p=45

    p=4535=10p=45-35=10
  • Question 9
    1 / -0
    If 16t4=016t-4=0 then find 3t3t
    Solution
    16t4=016t-4=0
    Add 44 both the sides

    \Rightarrow16t=416t=4
    Multiply both the sides by 1616
    \Rightarrowt=416t=\dfrac 4{16}

    \Rightarrowt=14t=\dfrac 14

    \Rightarrow3t=343t=\dfrac 34
  • Question 10
    1 / -0
    The value of xx if 23x+69=023x+69=0
    Solution
    23x+69=023x+69=0
    Subract both side by 69
    \Rightarrow23x=6923x=-69
    Divide both side by 23

    \Rightarrowx=6923x=\dfrac{-69}{23}

    \Rightarrowx=3x=-3
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