Question

There are four consecutive odd numbers \(x_1, x_2, x_3\) and \(x_4\) and three consecutive even numbers \(y_1, y_2\) and \(y_3\). The average of odd numbers is 12 less than the average of even numbers. If the sum of three even numbers is 32 less than the sum of four odd numbers, then the average of \(x_1, x_2, x_3\) and \(x_4\) is

• A. \(60\)
• B. \(64\)
• C. \(62\)
• D. \(68\)

Hint:

For consecutive numbers: \[ \text{Average}=\text{Middle Term} \] For four consecutive odd numbers: \[ x,\;x+2,\;x+4,\;x+6 \] Average: \[ x+3 \]

Answer 1

The Correct Option is D

Step 1: Find the average of odd numbers.
Average of four odd numbers: \[ \frac{x+(x+2)+(x+4)+(x+6)}{4} \] \[ =\frac{4x+12}{4} \] \[ =x+3 \]

Step 2: Find the average of even numbers.
Average of three consecutive even numbers: \[ \frac{y+(y+2)+(y+4)}{3} \] \[ =\frac{3y+6}{3} \] \[ =y+2 \] Given: \[ \text{Average of odd numbers} = \text{Average of even numbers}-12 \] Therefore, \[ x+3=(y+2)-12 \] \[ x-y=-13 \] \[ y=x+13 \]

Step 3: Use the condition involving sums.
Sum of four odd numbers: \[ 4x+12 \] Sum of three even numbers: \[ 3y+6 \] Given: \[ 3y+6=(4x+12)-32 \] \[ 3y+6=4x-20 \] \[ 3y=4x-26 \] Substituting \(y=x+13\), \[ 3(x+13)=4x-26 \] \[ 3x+39=4x-26 \] \[ x=65 \]

Step 4: Calculate the required average.
Average of odd numbers: \[ x+3 \] \[ =65+3 \] \[ =68 \]

Step 5: Verify carefully.
Using the given conditions: \[ \text{Average of odd numbers}=68 \] \[ \text{Average of even numbers}=80 \] Difference: \[ 80-68=12 \] and \[ 4(68)-3(80)=272-240=32 \] Both conditions are satisfied. Therefore, \[ {68} \] Hence option (D) is correct.