Question:

Asim got thrice as many sums wrong as he got right. If he attempted 60 sums in all, how many sums did he solve correctly?

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In problems involving ratios and totals, set up an equation with the unknown and solve it algebraically.
Updated On: Apr 11, 2025
  • 25
  • 12
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  • 10
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The Correct Option is

Approach Solution - 1

  1. Define Variables:
    • Let \( R \) be the number of sums Asim got right.
    • Let \( W \) be the number of sums Asim got wrong.
  2. Formulate Equations Based on the Problem Statement:
    • "Asim got thrice as many sums wrong as he got right": This translates to the equation:

\[ W = 3R \quad \quad (1) \]

  • "If he attempted 60 sums in all": This means the total number of sums (right + wrong) is 60.

\[ R + W = 60 \quad \quad (2) \]

  1. Solve the System of Equations:
    • Substitute the expression for \( W \) from Equation (1) into Equation (2):

\[ R + (3R) = solved correctly (got right).\]

  • Let \(W\) be the number of sums Asim solved incorrectly (got wrong).
  1. Formulate Equations based on the Given Information:
    • "Asim got thrice as many sums wrong as he got right": This means the number of wrong sums is 3 times the number of right sums.

\[ W = 3R \quad \quad (1) \]

  • "If he attempted 60 sums in all": The total number of sums attempted is the sum of those solved correctly and those solved incorrectly.

\[ R + W = 60 \quad \quad (2) \]

  1. Solve the System of Equations:
    • We can substitute the expression for \(W\) from equation (1) into equation (2):

\[ R + (3R) = 60 \]

  • Combine the terms with \(R\):

\[ 4R = 60 \]

  • Solve for \(R\) by dividing both sides by 4:

\[ R = \frac{60}{4} = 15 \]

  1. Answer the Question:
    • The question asks for the number of sums Asim solved correctly, which is \(R\).
    • We found \(R = 15\).
    • (Optional check: If \(R=15\), then \(W = 3 \times 15 = 45\). Total sums = \(R+W = 15+45=60\). This matches the problem statement.)
  2. Compare with Options: The calculated value is 15.

The correct option is (e) 15.

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Approach Solution -2

Let the number of correct sums be $x$.  
Then the number of wrong sums is 3 times the number of correct sums, i.e., $3x$. 

Total number of sums attempted = 60
So, $$x + 3x = 60$$ $$4x = 60$$ $$x = \frac{60}{4} = 15$$ 
Answer: (e) 15

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Approach Solution -3

Let the number of sums Asim solved correctly be \( x \).
Then the number of sums he solved incorrectly is \( 3x \).
We are given that the total number of sums attempted is 60.
Thus: \[ x + 3x = 60 \] \[ 4x = 60 \quad \Rightarrow \quad x = 15 \] Therefore, Asim solved 15 sums correctly.
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