Maths-
General
Easy

Question

A sports store sells a total of 70- Soccer balls in one month and collects a total of $2,400. Write and Solve a System of equations to determine how many of each type of soccer ball were sold.

Hint:

Given, Total income of the store for 70 soccer balls is $2,400.
There are two types of soccer balls and the cost of each type is different .
Frame equation considering no.of limited edition soccer balls sold be x
And no.of Pro NSL soccer ball sold be y and solve them to find  x and y.

The correct answer is: The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.


    Ans :- The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.
    Explanation :-
    Let no.of limited edition soccer balls sold be x ,no.of Pro NSL soccer ball sold be y.
    Step 1:- Frame equations
    Total no.of ball is 70
    I.e x + y = 70             —Eq1
    Total cost of balls is $2,400
    Cost of x  limited edition soccer balls is 65x (as per ball cost is given in diagram)
    And Cost of y Pro NSL soccer ball is 15x(as per ball cost is given in diagram)
    I.e 65x + 15y = 2,400            —Eq2
    Step 2:- Eliminate y to find x
    Do Eq2 -15(Eq1) to eliminate y
    65x + 15y - 15(x+y) = 2400 - 15(70)
    65x - 15x = 1350
    50x = 1350 ⇒ x = 27
    Step 3:- substitute value of x to find y
    x + y = 70 ⇒ 27 + y = 70
    ⇒ y = 70 - 27
    ∴y = 43
    ∴The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

    Related Questions to study

    General
    Maths-

    Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

    • Step by step explanation: 
      • Given:

    𝑚∠POQ = (x - 4) °  

    𝑚∠QOR = (5x - 20) °.

    • Step 1:
    • From the figure it is clear that,

    POR is right angle hence POR = 90o.
    and 

    POR = POQ + QOR 

    • Step 2:
    • Put values of ∠COD and ∠COP 

    ∠POR = ∠POQ + ∠QOR  

    90 = (x - 4) + (5x - 20)

    90 = 5x + x - (20 + 4)

    90 = 6x - 24

    6x = 90 - 24

    6x = 66

    x = 66 over 6

    x =  11

    • Final Answer:
    Hence, the x is 11.

    Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

    Maths-General
    • Step by step explanation: 
      • Given:

    𝑚∠POQ = (x - 4) °  

    𝑚∠QOR = (5x - 20) °.

    • Step 1:
    • From the figure it is clear that,

    POR is right angle hence POR = 90o.
    and 

    POR = POQ + QOR 

    • Step 2:
    • Put values of ∠COD and ∠COP 

    ∠POR = ∠POQ + ∠QOR  

    90 = (x - 4) + (5x - 20)

    90 = 5x + x - (20 + 4)

    90 = 6x - 24

    6x = 90 - 24

    6x = 66

    x = 66 over 6

    x =  11

    • Final Answer:
    Hence, the x is 11.
    General
    Maths-

    Identify all pairs of congruent angles and congruent segments.

    • Step by step explanation: 
    • Step 1:
    Find pair of congruent angles
    From the figure it is clear that

    ∠BAC = ∠QPR

    ∠BCA = ∠QRP

    ∠ABC = ∠PQR
    This are pair of congruent angles.

    • Step 2:
    Find pair of congruent segments
    From the figure it is clear that

    BA = QP                          
    This is pair of congruent segments.

    Identify all pairs of congruent angles and congruent segments.

    Maths-General
    • Step by step explanation: 
    • Step 1:
    Find pair of congruent angles
    From the figure it is clear that

    ∠BAC = ∠QPR

    ∠BCA = ∠QRP

    ∠ABC = ∠PQR
    This are pair of congruent angles.

    • Step 2:
    Find pair of congruent segments
    From the figure it is clear that

    BA = QP                          
    This is pair of congruent segments.

    General
    Maths-

    ∠𝐴 & ∠𝐵 are complementary.  ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅  ∠𝐶

    HINT – If sum of two angles is 90°, we say they are complementary angles
    SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
    rightwards double arrow ∠A + ∠B = 90°               ---- (1)
    Also, 𝐵 & ∠𝐶 are complementary
    rightwards double arrow ∠B + ∠C = 90°              ---- (2)
    From (1) and (2)
    We get, ∠A + ∠B = ∠B + ∠C
    rightwards double arrow ∠A = ∠C
    rightwards double arrow ∠𝐴 ≅  ∠𝐶
    Hence Proved.

    ∠𝐴 & ∠𝐵 are complementary.  ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅  ∠𝐶

    Maths-General
    HINT – If sum of two angles is 90°, we say they are complementary angles
    SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
    rightwards double arrow ∠A + ∠B = 90°               ---- (1)
    Also, 𝐵 & ∠𝐶 are complementary
    rightwards double arrow ∠B + ∠C = 90°              ---- (2)
    From (1) and (2)
    We get, ∠A + ∠B = ∠B + ∠C
    rightwards double arrow ∠A = ∠C
    rightwards double arrow ∠𝐴 ≅  ∠𝐶
    Hence Proved.
    parallel
    General
    Maths-

    Given: Ray OR bisects ∠𝑃𝑂𝑆.
    Prove: 𝑚∠1 = 𝑚∠2

    SOL – It is given that ray OR bisects ∠𝑃𝑂𝑆 i.e. OR is an angle bisector.
    We know that an angle bisector divides an angle into two congruent angles.
    rightwards double arrow ∠ POR ≅ ∠ ROS
    rightwards double arrow 𝑚∠1 = 𝑚∠2
    Hence Proved.

    Given: Ray OR bisects ∠𝑃𝑂𝑆.
    Prove: 𝑚∠1 = 𝑚∠2

    Maths-General
    SOL – It is given that ray OR bisects ∠𝑃𝑂𝑆 i.e. OR is an angle bisector.
    We know that an angle bisector divides an angle into two congruent angles.
    rightwards double arrow ∠ POR ≅ ∠ ROS
    rightwards double arrow 𝑚∠1 = 𝑚∠2
    Hence Proved.
    General
    Maths-

    Solve the following by using the method of substitution
    Y = - 2X-3
    Y = - X-4

    Ans :- x = 1 ; y = -5
    Explanation :-
    ⇒  y = -2x - 3 — eq 1
    ⇒  y = -x - 4—- eq 2
    Step 1 :- find x by substituting y = 4x + 2 in eq 2.
    -2x - 3 =   -x – 4  ⇒  2x + 3x + 4
    ⇒  2x – x = 4 - 3 ⇒ x  = 4 - 3
    ⇒  x  = 1
    Step 2 :- substitute value of x and find y
    ⇒  y =  - x – 4  ⇒  y = -1 - 4
    ∴  y =  - 5
    ∴   x = 1 and y = - 5  is the solution of the given pair of equations.

    Solve the following by using the method of substitution
    Y = - 2X-3
    Y = - X-4

    Maths-General
    Ans :- x = 1 ; y = -5
    Explanation :-
    ⇒  y = -2x - 3 — eq 1
    ⇒  y = -x - 4—- eq 2
    Step 1 :- find x by substituting y = 4x + 2 in eq 2.
    -2x - 3 =   -x – 4  ⇒  2x + 3x + 4
    ⇒  2x – x = 4 - 3 ⇒ x  = 4 - 3
    ⇒  x  = 1
    Step 2 :- substitute value of x and find y
    ⇒  y =  - x – 4  ⇒  y = -1 - 4
    ∴  y =  - 5
    ∴   x = 1 and y = - 5  is the solution of the given pair of equations.
    General
    Maths-

    Solve the system of equations by elimination :
    X - 2Y = 1
    2X + 3Y= - 12

    SOLUTION:
    HINT: Perform any arithmetic operation and then find.
    Complete step by step solution:
    Let x - 2y = 1…(i)
    and 2x + 3y=-12….(ii)
    On multiplying (i) with 2, we get 2(x - 2y = 1)
    ⇒ 2x - 4y = 2…(iii)
    Now, we have the coefficients of x in (ii) and (iii) to be the same.
    On subtracting (ii) from (iii),
    we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y
    and RHS to be 2 - (- 12) = 14
    On equating LHS and RHS, we have - 7y = 14
    ⇒ y = - 2
    On substituting the value of y in (i), we get x - 2 × - 2 =1
    ⇒ x + 4 = 1
    ⇒ x = 1-4
    ⇒ x = - 3
    Hence we get x = - 3 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations

    Solve the system of equations by elimination :
    X - 2Y = 1
    2X + 3Y= - 12

    Maths-General
    SOLUTION:
    HINT: Perform any arithmetic operation and then find.
    Complete step by step solution:
    Let x - 2y = 1…(i)
    and 2x + 3y=-12….(ii)
    On multiplying (i) with 2, we get 2(x - 2y = 1)
    ⇒ 2x - 4y = 2…(iii)
    Now, we have the coefficients of x in (ii) and (iii) to be the same.
    On subtracting (ii) from (iii),
    we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y
    and RHS to be 2 - (- 12) = 14
    On equating LHS and RHS, we have - 7y = 14
    ⇒ y = - 2
    On substituting the value of y in (i), we get x - 2 × - 2 =1
    ⇒ x + 4 = 1
    ⇒ x = 1-4
    ⇒ x = - 3
    Hence we get x = - 3 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations
    parallel
    General
    Maths-

    Solve the equation. Write a reason for each step.
    𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

    HINT – Open the brackets
    SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
    Opening the brackets
    We get, x – 2 + 3x + 6 = 3x + 10
    rightwards double arrow 4x + 4 = 3x + 10      ( Adding similar terms )
    rightwards double arrow 4x – 3x = 10 – 4
    rightwards double arrow x = 6.

    Solve the equation. Write a reason for each step.
    𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

    Maths-General
    HINT – Open the brackets
    SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
    Opening the brackets
    We get, x – 2 + 3x + 6 = 3x + 10
    rightwards double arrow 4x + 4 = 3x + 10      ( Adding similar terms )
    rightwards double arrow 4x – 3x = 10 – 4
    rightwards double arrow x = 6.
    General
    Maths-

    Use Substitution to solve each system of equations :
    6X - 3Y = -6
    Y = 2X + 2

    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations  .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.
    Ans :- infinite no.of solutions .
    Explanation :-
    y = 2x + 2— eq 1
    6x - 3y = -6—- eq 2
    Step 1 :- find x by substituting y = 2x + 2 in eq 2.
    6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6
    -6 = -6
    Here we get -6 = -6 which is always true i.e always having a root .
    They coincide with each other and have infinite no.of solutions
    They have infinite no.of solutions for the given system of equations

    Use Substitution to solve each system of equations :
    6X - 3Y = -6
    Y = 2X + 2

    Maths-General
    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations  .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.
    Ans :- infinite no.of solutions .
    Explanation :-
    y = 2x + 2— eq 1
    6x - 3y = -6—- eq 2
    Step 1 :- find x by substituting y = 2x + 2 in eq 2.
    6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6
    -6 = -6
    Here we get -6 = -6 which is always true i.e always having a root .
    They coincide with each other and have infinite no.of solutions
    They have infinite no.of solutions for the given system of equations

    General
    Maths-

    Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅
    Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

    SOL – It is given that 𝑚∠𝑃 = 30° and  𝑚∠𝑄 = 30°

    rightwards double arrow 𝑚∠𝑃 = 𝑚∠𝑄           ---- (1)
    Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅    ---- (1)
    Using transitive property which states that if A = B and B = C then A = C
    We get from (1) and (2),

    𝑚∠P = 𝑚∠R = 30°
    Hence Proved

    Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅
    Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

    Maths-General
    SOL – It is given that 𝑚∠𝑃 = 30° and  𝑚∠𝑄 = 30°

    rightwards double arrow 𝑚∠𝑃 = 𝑚∠𝑄           ---- (1)
    Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅    ---- (1)
    Using transitive property which states that if A = B and B = C then A = C
    We get from (1) and (2),

    𝑚∠P = 𝑚∠R = 30°
    Hence Proved

    parallel
    General
    Maths-

    Use the given information and the diagram to prove the statement.
    Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°
    Prove: 𝑚∠𝑃𝑀𝐷 = 30°

    HINT – Use the information given.
    SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150°    ---- (1)
    𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°
    150° + 𝑚∠𝑃𝑀𝐷 = 180°             ( - From (1) )
    𝑚∠𝑃𝑀𝐷 = 180° - 150°
    = 30°
    Hence Proved.

    Use the given information and the diagram to prove the statement.
    Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°
    Prove: 𝑚∠𝑃𝑀𝐷 = 30°

    Maths-General
    HINT – Use the information given.
    SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150°    ---- (1)
    𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°
    150° + 𝑚∠𝑃𝑀𝐷 = 180°             ( - From (1) )
    𝑚∠𝑃𝑀𝐷 = 180° - 150°
    = 30°
    Hence Proved.
    General
    Maths-

    Solve the system of equations by elimination :
    3X + 2Y = 8
    X + 4Y = - 4

    Complete step by step solution:
    Let 3x + 2y = 8…(i)
    and x + 4y = - 4….(ii)
    On multiplying (ii) with 3, we get 3(x + 4y=-4)
    ⇒3x + 12y = - 12…(iii)
    Now, we have the coefficients of  in (i) and (iii) to be the same.
    On subtracting (i) from (iii),
    we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y
    and RHS to be - 12 - 8 = - 20
    On equating LHS and RHS, we have 10y = - 20
    ⇒y = - 2
    On substituting the value of y in (i), we get 3x + 2× - 2 = 8
    ⇒ 3x - 4 = 8
    ⇒ 3x = 8 + 4
    ⇒ 3x = 12
    ⇒x = 4
    Hence we get x = 4 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations

    Solve the system of equations by elimination :
    3X + 2Y = 8
    X + 4Y = - 4

    Maths-General
    Complete step by step solution:
    Let 3x + 2y = 8…(i)
    and x + 4y = - 4….(ii)
    On multiplying (ii) with 3, we get 3(x + 4y=-4)
    ⇒3x + 12y = - 12…(iii)
    Now, we have the coefficients of  in (i) and (iii) to be the same.
    On subtracting (i) from (iii),
    we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y
    and RHS to be - 12 - 8 = - 20
    On equating LHS and RHS, we have 10y = - 20
    ⇒y = - 2
    On substituting the value of y in (i), we get 3x + 2× - 2 = 8
    ⇒ 3x - 4 = 8
    ⇒ 3x = 8 + 4
    ⇒ 3x = 12
    ⇒x = 4
    Hence we get x = 4 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations
    General
    Maths-

    Give a two-column proof.
    Given:

    Prove: PR = 25 in

    HINT – Use Segment Addition Postulate
    SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
    rightwards double arrow PR = PQ + QR
    rightwards double arrow PR = 12 + 13
    rightwards double arrow PR = 25 in.
    Hence Proved.

    Give a two-column proof.
    Given:

    Prove: PR = 25 in

    Maths-General
    HINT – Use Segment Addition Postulate
    SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
    rightwards double arrow PR = PQ + QR
    rightwards double arrow PR = 12 + 13
    rightwards double arrow PR = 25 in.
    Hence Proved.
    parallel
    General
    Maths-

    Use Substitution to solve each system of equations :
    - 3X - Y = 7
    X + 2Y = 6

    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations.
    Ans :- x = -4; y = 5
    Explanation :-
    3x - y = 7 ⇒ - 3x - 7 = y                     — eq 1
    x + 2y = 6                                     —- eq 2
    x + 2 ( - 3x - 7) = 6 ⇒ x - 6x - 14 = 6
    ⇒ - 5x = 14 + 6 ⇒ - 5x = 20
    ⇒ x = -4
    Step 2 :- substitute value of x and find y
    ⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7
    ⇒ y = 12 - 7
    ∴ y = 5
    x = -4 and y = 5 is the solution of the given pair of equations.


    Use Substitution to solve each system of equations :
    - 3X - Y = 7
    X + 2Y = 6

    Maths-General
    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations.
    Ans :- x = -4; y = 5
    Explanation :-
    3x - y = 7 ⇒ - 3x - 7 = y                     — eq 1
    x + 2y = 6                                     —- eq 2
    x + 2 ( - 3x - 7) = 6 ⇒ x - 6x - 14 = 6
    ⇒ - 5x = 14 + 6 ⇒ - 5x = 20
    ⇒ x = -4
    Step 2 :- substitute value of x and find y
    ⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7
    ⇒ y = 12 - 7
    ∴ y = 5
    x = -4 and y = 5 is the solution of the given pair of equations.


    General
    Maths-

    If Q is the midpoint of PR, prove that PR = 2 PQ.
    Give a two-column proof.

    HINT – Use Segment Addition Postulate
    SOL – If Q is mid – point  Þ PQ = QR      ---- (1)
    Using Segment Addition Postulate,
    We get,  PR = PQ + QR
    rightwards double arrow PR = PQ + PQ
    rightwards double arrow PR = 2 PQ
    Hence Proved.

    If Q is the midpoint of PR, prove that PR = 2 PQ.
    Give a two-column proof.

    Maths-General
    HINT – Use Segment Addition Postulate
    SOL – If Q is mid – point  Þ PQ = QR      ---- (1)
    Using Segment Addition Postulate,
    We get,  PR = PQ + QR
    rightwards double arrow PR = PQ + PQ
    rightwards double arrow PR = 2 PQ
    Hence Proved.
    General
    Maths-

    Compare 𝑚∠𝑃𝑂𝑄𝑎𝑛𝑑∠𝑆𝑂𝑇.

    • Step by step explanation: 
      • Step 1:
      • Find POQ:
    • Align the protractor with the ray OP on 0o as shown above. 
    • Start reading the inner scale from the 0°.
    • Step 2:
    • From the figure we can see that ray OP is aligned on mark 0o. And ray OQ is aligned on mark 30o.
    Hence, the measure of POQ = (30o - 0o).
    POQ = 30o.
    • Step 3:
    • Find SOT:
      • Align the protractor with the ray OT on 0o as shown above. 
      • Start reading the outer scale from the 0°.
      • Step 4:
      • From the figure we can see that ray OT is aligned on mark 0o. And the ray OS is aligned on mark 20o.
    Hence, the measure of SOT = 20o.

    SOT = 20o.

    • Step 5:
    • Compare ∠POQ and ∠SOT
    Hence, ∠POQ is greater than ∠SOT by 10o.
    • Final Answer:
    Hence, ∠POQ is greater than ∠SOT by 10o.

    Compare 𝑚∠𝑃𝑂𝑄𝑎𝑛𝑑∠𝑆𝑂𝑇.