# t & w q11

#### Saumyojit

##### Senior Member
Q.
Mini and Vinay are quiz masters preparing for a quiz. In 'x' minutes, Mini makes 'y' questions more than Vinay. If it were possible to reduce the time needed by each to make a question by 2 mins , then in 'x' minutes Mini would make '2y' questions more than Vinay. How many questions does Mini make in 'x' minutes?

1] 1/4[ 2 ( x+y) - ( 2 x^2 + 4 x y^2 )^1/2 ]
2] 1/4[ 2(x-y) - ( 2 x^2 + 4 y^2 )^1/2 ]
3] Either option 1 or 2
4] 1/4[ 2(x-y) - ( 2 x^2 - 4 y^2 )^1/2 ]

No of question Vinay per minute he makes = Q /x questions

No of question mini per minute he makes = (Q + y) /x questions

To make one question suppose vinay takes z minute and t minute by mini.

Case 2 : To make one question suppose vinay takes z minutes -2 and t minutes - 2 by mini.

Vinay In x minutes --> Q1 question
1 min --> Q1 / x questions

Mini
In x minutes --> Q1 + 2y questions
1 min --> ( Q1 + 2y) / x questions

How to solve further

#### lev888

##### Elite Member
Q.
Mini and Vinay are quiz masters preparing for a quiz. In 'x' minutes, Mini makes 'y' questions more than Vinay. If it were possible to reduce the time needed by each to make a question by 2 mins , then in 'x' minutes Mini would make '2y' questions more than Vinay. How many questions does Mini make in 'x' minutes?

1] 1/4[ 2 ( x+y) - ( 2 x^2 + 4 x y^2 )^1/2 ]
2] 1/4[ 2(x-y) - ( 2 x^2 + 4 y^2 )^1/2 ]
3] Either option 1 or 2
4] 1/4[ 2(x-y) - ( 2 x^2 - 4 y^2 )^1/2 ]

No of question Vinay per minute he makes = Q /x questions

No of question mini per minute he makes = (Q + y) /x questions

To make one question suppose vinay takes z minute and t minute by mini.

Case 2 : To make one question suppose vinay takes z minutes -2 and t minutes - 2 by mini.

Vinay In x minutes --> Q1 question
1 min --> Q1 / x questions

Mini
In x minutes --> Q1 + 2y questions
1 min --> ( Q1 + 2y) / x questions

How to solve further
I don't understand any of what you wrote.

The problem describes 2 relationships:
1. The number of questions Mini answers in x min EQUALS the number of questions Vinay answers PLUS y.
2. With timer per question reduced by 2min:
The number of questions Mini answers in x min EQUALS the number of questions Vinay answers PLUS 2y.

These 2 relationships allow you to set up 2 equations. Can you do it? What values are you picking as variables?

#### Saumyojit

##### Senior Member
The number of questions Mini answers in x min EQUALS the number of questions Vinay answers PLUS y.
2. With timer per question reduced by 2min:
The number of questions Mini answers in x min EQUALS the number of questions Vinay answers PLUS 2y
Q + y =Q --(i)

Q1= Q + k . ( Vinay will answer k questions more than Q as efficiency is increased )
Q1 + 2y =Q1 --(ii)
Q+k + 2y = Q + k

#### lev888

##### Elite Member
Q + y =Q --(i)

Q1= Q + k . ( Vinay will answer k questions more than Q as efficiency is increased )
Q1 + 2y =Q1 --(ii)
Q+k + 2y = Q + k

#### Saumyojit

##### Senior Member
Q + y =Q --(i)

Q= no of question solve by vinay .

Q + y= no of question solve by mini.

Q1= Q + k . ( Vinay will answer k questions more than last time Q as efficiency is increased )

#### lev888

##### Elite Member
Q + y =Q --(i)

Q= no of question solve by vinay .

Q + y= no of question solve by mini.

Q1= Q + k . ( Vinay will answer k questions more than last time Q as efficiency is increased )
This is not a good choice of variables. You now have 3 variables for 2 equations. Why introduce k, if the problem clearly states that the difference between Q1 and Q is y? You need to step back and review how to set up equations for word problems. Maybe go back to easier, single variable problems.

If you want to continue with this problem...
Let's take the first equation:
The number of questions Mini answers in x min EQUALS the number of questions Vinay answers PLUS y.

Your task here is to figure out how to make the expressions for the blue and green parts. These expressions should use the givens and variables. How do you do it? You find the relationships that tie various values mentioned in the problem.
What are the values?
1. Number of questions solved
2. Time needed by a particular student to solve one question
3. Time
What's the relationship?

Number of questions solved = Time / (Time needed for 1 question) ( e.g. 5 questions = 10min/(2min per q) )

Time is given as x.
Times per question for each student - not given, but we need them. NOW we introduce variables for needed values which are not given.
M min per q for Mini
V min per q for Vinay

The number of questions Mini answers in x min: x/M
The number of questions Vinay answers: x/V

Resulting equation: x/M = x/V + y

Can you set up the second equation?

#### Jomo

##### Elite Member
Q + y =Q --(i)

Q1= Q + k . ( Vinay will answer k questions more than Q as efficiency is increased )
Q1 + 2y =Q1 --(ii)
Q+k + 2y = Q + k
Isn't clear to you that if Q +y = Q, then y=0? After all, what would YOU add to a number to get back that same number?

#### Saumyojit

##### Senior Member
Q +y = Q, then y=0? After all, what would YOU add to a number to get back that same number?
oh yes .

### x/M-2 = x / V-2 + 2y​

Then?
The order of operations is extremely important in math. It's the difference between correct and incorrect result. Do you see what needs to be fixed in your equation?

#### Saumyojit

##### Senior Member
The order of operations is extremely important in math. It's the difference between correct and incorrect result. Do you see what needs to be fixed in your equation?

### x/M-2 = x / (V-2 ) + 2y (v-2 is in denominator)​

Yes, should be (v-2). Anything else?

#### Saumyojit

##### Senior Member
x/M = x/V + y

x/M-2 = x / V-2 + 2y

From first equation we get Vx = Mx + MVy

From 2nd equation we get
x/M-2 = x / V-2 + 2y

x/ M-2 = x + 2y (V- 2) / V-2

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?

#### lev888

##### Elite Member
From first equation we get Vx = Mx + MVy

From 2nd equation we get
x/M-2 = x / V-2 + 2y

x/ M-2 = x + 2y (V- 2) / V-2

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?
No comments until you fix all parentheses.

#### Saumyojit

##### Senior Member
No comments until you fix all parentheses.
From first equation we get Vx = Mx + MVy

From 2nd equation we get
x/M-2 = x / (V-2) + 2y (v-2 is in denominator)

x/ M-2 = x + 2y (V- 2) / (V-2)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?

#### Subhotosh Khan

##### Super Moderator
Staff member
Go back to response #11 and fix parentheses - and continue. Use pencil/paper & rewrite the problem.

#### Saumyojit

##### Senior Member
everything is right

x/ (M-2) = x / (V-2 ) + 2y (M-2 and v-2 is only in denominator)

#### lev888

##### Elite Member
everything is right

x/ (M-2) = x / (V-2 ) + 2y (M-2 and v-2 is only in denominator)
You now have 2 equations and 2 variables - M, V. Can you solve the system?

#### Subhotosh Khan

##### Super Moderator
Staff member
everything is right

x/ (M-2) = x / (V-2 ) + 2y (M-2 and v-2 is only in denominator)
What is the FIND of your OP?