Self Studies

Algebraic Expressions Test - 17

Result Self Studies

Algebraic Expressions Test - 17
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If P=3x4y8zP=3x-4y-8z, Q=10y+7x+11zQ=-10y+7x+11z and R=19z6y+4xR=19z-6y+4x, then PQ+RP-Q+R is equal to
    Solution
    P=3x4y8zP=3x-4y-8z
    Q=10y+7x+11zQ=-10y+7x+11z
    R=19z6y+4xR=19z-6y+4x
    PQ+R=(3x4y8z)(10y+7x+11z)+(19z6y+4x)\therefore P-Q+R=(3x-4y-8z)-(-10y+7x+11z)+(19z-6y+4x)
    =3x4y8z+10y7x11z+19z6y+4x=3x-4y-8z+10y-7x-11z+19z-6y+4x
    =3x7x+4x4y+10y6y8z11z+19z=7x7x10y+10y19z+19z=0=3x-7x+4x-4y+10y-6y-8z-11z+19z=7x-7x-10y+10y-19z+19z=0
  • Question 2
    1 / -0
    What is the simplified result of following the steps in order?
    (1)(1) add 5y5y to 2x2x
    (2)(2) multiply the sum by 33
    (3)(3) subtract x++y from the product
    Solution

  • Question 3
    1 / -0
    The sum of three expressions is x2+y2+z2{ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }. If two of them are 4x25y2+3z24{ x }^{ 2 }-5{ y }^{ 2 }+3{ z }^{ 2 } and 3x2+4y2+2z2-3{ x }^{ 2 }+4{ y }^{ 2 }+2{ z }^{ 2 }, the third expression is
    Solution
    Add two expressions 

    (4x25y2+3z2)+(3x2+4y2+2z2)\Rightarrow \left ( 4x^{2}-5y^{2}+3z^{2} \right )+(-3x^{2}+4y^{2}+2z^{2})

    =(4x25y2+3z23x2+4y2+2z2)=\left ( 4x^{2}-5y^{2}+3z^{2}-3x^{2}+4y^{2}+2z^{2} \right )

    =4x23x25y2+4y2+3z2+2z2=4x^{2}-3x^{2}-5y^{2}+4y^{2}+3z^{2}+2z^{2}

    =x2y2+5z2=x^{2}-y^{2}+5z^{2}

    But x2y2+5z2+P(x)=x2+y2+z2x^{2}-y^{2}+5z^{2}+P(x)=x^{2}+y^{2}+z^{2}, where P(x)P(x) is the third expression

    P(x)=(x2+y2+z2)(x2y2+5z2)\therefore P(x)=\left ( x^{2}+y^{2}+z^{2} \right )-\left ( x^{2}-y^{2}+5z^{2} \right )

    =x2+y2+z2x2+y25z2=2y24z2=x^{2}+y^{2}+z^{2}-x^{2}+y^{2}-5z^{2}=2y^{2}-4z^{2}
  • Question 4
    1 / -0
    (4px2+5q2y9rz)(3q2y+7px2rz)=\left( 4p{ x }^{ 2 }+5{ q }^{ 2 }y-9rz \right) -\left( -3{ q }^{ 2 }y+7p{ x }^{ 2 }-rz \right) =
    Solution
    (4px2+5q2y9rz)(3q2y+7px2rz)(4px^{2}+5q^{2}y-9rz)-(-3q^{2}y+7px^{2}-rz)
    =4px2+5q2y9rz+3q2y7px2+rz=4px^{2}+5q^{2}y-9rz+3q^{2}y-7px^{2}+rz
    =4px27px2+5q2y+3q2y9rz+rz=4px^{2}-7px^{2}+5q^{2}y+3q^{2}y-9rz+rz
    =3px2+8q2y8rz=-3px^{2}+8q^{2}y-8rz
  • Question 5
    1 / -0
    The sum of three expressions is x2+y2+z2.x^{2}+y^{2}+z^{2}. If two of them are 4x25y2+z24x^{2}-5y^{2}+z^{2} and 3x2+4y2+2z2-3x^{2}+4y^{2}+2z^{2}, find the third expression.
    Solution
    Let the third expression be aa then,
    a+(4x25y2+z2)+(3x2+4y2+2z2)=x2+y2+z2 a+(4x^{2}-5y^{2}+z^{2})+(-3x^{2}+4y^{2}+2z^{2}) = x^{2}+y^{2}+z^{2}
    a+(x2y2+3z2)=(x2+y2+z2) \Rightarrow a+(x^{2}-y^{2}+3z^{2})=(x^{2}+y^{2}+z^{2})
    a=(x2+y2+z2)x2+y23z2 \Rightarrow a = (x^{2}+y^{2}+z^{2})-x^{2}+y^{2}-3z^{2}
     a=2y22z2 \Rightarrow  a = 2y^{2}-2z^{2}

    Hence, option DD is correct.
  • Question 6
    1 / -0
    Subtract 2x34x2+3x+52x^{3}-4x^{2}+3x+5 from 4x2+x2+x+64x^{2}+x^{2}+x+6, then the resultant value is 
    Solution

  • Question 7
    1 / -0
    Add: 2u+3v,  2v+3w,  3uv+2w2u+3v,\,\,-2v+3w,\,\,3u-v+2w
    Solution
    given, (2u+3v+3uv+2w)+(2v+3w)\left(2u+3v+3u-v+2w\right)+\left(-2v+3w\right)

    =5u+2v+2w2v+3w=5u+2v+2w-2v+3w

    =5u+5w=5u+5w
  • Question 8
    1 / -0
    Simplify: a[b{c(abc)}c]+aa-[b-\{c-(a-b-c)\}-c]+a
    Solution

  • Question 9
    1 / -0
    The sum of (6a+4bc+3),(2b3c+4),(11b7a+2c1)(6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1) and (2c5a6)(2c-5a-6) is
    Solution
    (6a+4bc+3)+(2b3c+4)+(11b7a+2c1)+(2c5a6)(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)

    \Rightarrow  6a+4bc+3+2b3c+4+11b7a+2c1+2c5a66a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6

    Now arranging and combining the like terms,

    \Rightarrow  (6a7a5a)+(4b+2b+11b)+(c3c+2c+2c)+(3+416)(6a-7a-5a)+(4b+2b+11b)+(-c-3c+2c+2c)+(3+4-1-6)

    \Rightarrow  6a+17b-6a+17b

    \therefore  The sum of (6a+4bc+3),(2b3c+4),(11b7a+2c1)(6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1) and (2c5a6)(2c−5a−6) is (6a+17b)(-6a+17b)
  • Question 10
    1 / -0
    The sum of x4xy+2y2{x}^{4}-xy+2{y}^{2} and x4+xy+2y2-{x}^{4}+xy+2{y}^{2} is
    Solution
    Required sum =x4xy+2y2+x4+xy+2y2={x}^{4}-xy+2{y}^{2}+-{x}^{4}+xy+2{y}^{2} 
    =x4xy+2y2x4+xy+2y2={x}^{4}-xy+2{y}^{2}-{x}^{4}+xy+2{y}^{2}
    =[x4+)x4)]+(xy+xy)+(2y2+2y2)=[{x}^{4}+)-{x}^{4})]+(-xy+xy)+(2{y}^{2}+2{y}^{2})
    =0+0+4y2=4y2=0+0+4{y}^{2}=4{y}^{2}
    4y24{y}^{2} is a monomial and polynomial in yy
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now