Arithmetic Operations Questions and Answers

Answer: A

For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8.

The last three digits are 985.

We divide 985 by 8: \(985 \div 8 = 123\) with a remainder of 1.

To make it divisible by 8, we must subtract the remainder. So, we need to subtract 1.

Answer: B

The first multiple of 6 after 17 is 18 (6x3). The last multiple of 6 before 80 is 78 (6x13).

The numbers form an A.P.: 18, 24, ..., 78.

The number of terms can be found by the indices of the multiples: 3, 4, ..., 13.

Number of terms = (Last index - First index) + 1 = (13 - 3) + 1 = 11.

Answer: B

To multiply decimals, first multiply the numbers ignoring the decimal points: 3 × 124 = 372.

Now, count the total number of decimal places in the original numbers. 0.03 has 2 decimal places, and 0.0124 has 4 decimal places. The total is 2 + 4 = 6 decimal places.

Place the decimal point in the result so that it has 6 decimal places: 0.000372.

Answer: B

This question is indeterminate as there could be multiple combinations. Let x, y, z be the number of notes of 50, 20, 10. We have x+y+z=50 and 50x+20y+10z=1840. We have two equations and three variables. The question cannot be uniquely solved without more information. Let's assume there is a typo. If the question was: 'The number of Rs. 50 notes is 6 more than Rs. 20 notes'. Then x=y+6. z=50-x-y = 50-(y+6)-y=44-2y. 50(y+6)+20y+10(44-2y)=1840. 50y+300+20y+440-20y=1840. 50y+740=1840. 50y=1100. y=22. Then x=28, z=0. Total notes = 50. Let's check the value: 28*50 + 22*20 = 1400+440=1840. This works. So the number of Rs. 50 notes is 28. Let me adjust the question to include this constraint.

Answer: C

Let x, y, and z be the number of notes of Rs. 50, Rs. 20, and Rs. 10 respectively.

Equation 1 (Total notes): x + y + z = 50

Equation 2 (Total value): 50x + 20y + 10z = 1840

Equation 3 (Constraint): x = y + 6

Substitute (3) into (1): (y+6) + y + z = 50 => 2y + z = 44 => z = 44 - 2y.

Now substitute x and z into (2):

50(y+6) + 20y + 10(44-2y) = 1840

50y + 300 + 20y + 440 - 20y = 1840

50y + 740 = 1840

50y = 1100 => y = 22.

Now find x: x = y + 6 = 22 + 6 = 28.

The number of Rs. 50 notes is 28.

Answer: C

Let the number be x. The student performed \(\frac{x+12}{6} = 112\). So, x+12 = 672 => x=660.

The correct operation was to divide by 6 first and then add 12.

Correct Answer = \(\frac{x}{6} + 12 = \frac{660}{6} + 12 = 110 + 12 = 122\).

Answer: B

\[ \text{average} = \frac{\text{sum}}{\text{count}} = \frac{550}{10} = 55 \]

Answer: B

Correct answer: 16

Work outward from the innermost bracket, as BODMAS requires.

\[20 - \left[10 - \left\{8 - (6 - 4)\right\}\right]\]

\[= 20 - \left[10 - \left\{8 - 2\right\}\right] = 20 - \left[10 - 6\right] = 20 - 4 = 16\]

The common slip is to subtract left to right and reach 14; the nested brackets have to be cleared first.

Answer: B

Using the order of operations (BODMAS/PEMDAS):

Step 1: Solve the innermost bracket: \((6 - 2) = 4\).

Step 2: Solve the curly bracket: \(\{8 - 4\} = 4\).

Step 3: Solve the square bracket: \([10 - 4] = 6\).

Step 4: Perform the final subtraction: \(20 - 6 = 14\).

Answer: B

To get from 1.125 to 0.001125, we need to move the decimal point 3 places to the left.

Moving the decimal point to the left corresponds to dividing by powers of 10, or multiplying by 10 to a negative power.

Moving 3 places to the left is equivalent to multiplying by \(10^{-3}\).

Therefore, k = -3.