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This is one of our many quizzes on Variables. Click on this to access more such competitive quizzes on Algebra.

Variables Quiz 2

mathematics

 $10$ chocolates and $12$ wafers cost ₹$56$. $12$ chocolates and $10$ wafers cost ₹$54$. Find the cost of each chocolate and wafer.

 To solve such questions, form both the equations and then try to eliminate one of the variables.

Let the cost of each chocolate and each wafer be ₹ c and ₹ w respectively. $ \\ \Rightarrow 10c + 12w = 56 -----1 \\ \Rightarrow 12c + 10w = 54 -----2 \\ $ After adding both the equations, $ \\ 22c + 22w = 110 \\ \Rightarrow 22(c + w) = 110 \\ \Rightarrow c + w = 5 ----3 \\ $ Subtracting equation $2$ from $1$ $ \\ - 2c + 2w = 2 \\ \Rightarrow  - c + w = 1 ----4 \\ $ Adding equation $3$ and $4$, we get, $ \\ w = 3$ $ \\ $ By substituting the value in equation $3$, we get, $c = 3$

 Find the value of  $x$ by solving the following equations. $10x - 8y - 10 = 0$ and $6x - 10y + 20 = 0$

 To solve two equations with two variables simultaneously, try to eliminate one of the variables.

 Rewriting the equations, $ \\ 10x - 8y =  10 ----1 \\ \Rightarrow 6x - 10y =  - 20 ----2 \\ $ Multiplying equation $1$ by $3$ and equation $2$ by $5$ $ \\ 30x - 24y =  30 \\ \Rightarrow 30x - 50y =  - 100 \\ $ After solving these resultant equations, we’ll get $ \\ 26y = 130 \\ \Rightarrow y = 5 \\ $ Substituting the value of $y$ in equation $1$ we get, $x = 5$

 At present, the age of a mother is five years more than thrice her daughter’s age. Fifteen years later, the mother’s age will be twice her daughter’s age at that time. Find the present age of the daughter.

 To find one of the two variables, eliminate one of the  variables to obtain the other.

 Let the present age of mother and daughter be M and D respectively, $ \\ $ Given, $M = 3D + 5 \ldots  \ldots  \ldots .1$ $ \\ \Rightarrow M + 15 = 2(D + 15) \ldots  \ldots  \ldots 2 \\ $ Substituting the value of M in equation $2$ $ \\ \Rightarrow 3D + 5 + 15 = 2(D + 15) \\ \Rightarrow 3D + 20 = 2D + 30 \\ \Rightarrow 3D - 2D = 30 - 20 \\ \Rightarrow D = 10$

 Mr. Abhishek distributed 225 biscuits among his kids- P,Q,R and S. The biscuits he gave to P and S together is twice the number of biscuits he gave to Q and R. If Q received 15 more chocolates than R, find the number of biscuits received by R.

 To find out one of the variables out of four, we try to eliminate the other three variables.

 Let the number of chocolates with P,Q,R and S be p,q,r and s, respectively. $ \\ $ Given, $ p + q + r + s = 225 \ldots  \ldots  \ldots 1$ $ \\ \Rightarrow p + s = 2(q + r) \\ \Rightarrow q = 15 + r \\ $ Substituting the value of $p + s$ in equation 1 $ \\ 3(q + r) = 225 \\ \Rightarrow q + r = 75 \\ $ Substituting the value of $q$ obtained, in the above equation, $ \\ 15 + r + r = 75 \\ \Rightarrow 15 + 2r = 75 \\ \Rightarrow 2r = 60 \\ \Rightarrow r = 30 \\ $ Hence, number of biscuits with R are $30$

 Sohil took a test which had 120 questions. He gets 4 marks for every question he answers correctly, loses 2 marks for every question he answers incorrectly, and loses 1 mark for every unattempted question. If sohil scored 228 marks, then find the maximum number of questions he can answer incorrectly.

 To maximize the variable, simply minimize the other variable and vice-versa. 

 Let the number of correct answers, incorrect answers, unattempted questions be C,I,U, respectively. $ \\ $ Given, $C + I + U = 120 -----1$ $ \\ \Rightarrow 4C - 2I - U = 228 -----2 \\ $ To solve equation $1$ we’ll multiply it with $4$ $ \\ 4C + 4I + 4U = 480 \ldots  \ldots  \ldots 3 \\ $ Subtracting equation $2$ from equation $3$ $ \\ 6I + 5U = 252$ To maximize I, we need to minimize U. For, $U = 0,\;\;\;W = 42$ $ \\ $ Hence, the maximum number of questions he can answer incorrectly are $42$ .