Missing Data MCQ Quiz - Objective Question with Answer for Missing Data - Download Free PDF

Formula used:

1. Qualified candidates = ({Number of appeared candidates}) × ({Percentage of qualified candidates}) / 100

Calculations:

Step 1: Calculate qualified candidates in 2008:

Qualified candidates in 2008 = A × 70 / 100 = 0.7A .

Step 2: Calculate qualified candidates in 2009:

Qualified candidates in 2009 = 2A × 80 / 100 = 1.6A .

Step 3: Total qualified candidates:

0.7A + 1.6A = 552 .

⇒2.3A = 552 .

⇒ A = 552 / 2.3 = 240 .

Step 4: Calculate appeared candidates in 2009:

Number of appeared candidates in 2009 = 2A = 2 × 240 = 480 .

∴ The number of appeared candidates from State Z in 2009 is 480.

Given:

Number of female qualified candidates from State X in 2010 = 168

Ratio of male to female qualified candidates in 2010 = 5 : 7

Percentage of qualified candidates from State X in 2010 = 60%

Formula used:

Let the total number of qualified candidates from State X in 2010 be Q.

Since the ratio of male to female candidates is 5 : 7, the number of female qualified candidates = \(\dfrac{7}{12} \times Q\) and the number of male qualified candidates = \(\dfrac{5}{12} \times Q\).

Given that the number of female qualified candidates is 168, we can write: \(\dfrac{7}{12} \times Q = 168\).

Calculations:

\(\dfrac{7}{12} \times Q = 168\)

⇒ \(Q = 168 × 12 /7 \)

⇒ \(Q = 288\)

Now, the total number of qualified candidates from State X in 2010 is 288. Since 60% of the appeared candidates qualify, we can write:

\(\dfrac{60}{100} \times \text{Total appeared candidates} = 288\)

⇒ \(\text{Total appeared candidates} = \frac{288× 100}{60} \)

⇒ \(\text{Total appeared candidates} = 480\)

∴ The number of appeared candidates from State X in 2010 is 480.

Calculation

Let the number of students in IIT Madras be x.

Then, the number of students in IIT Bombay = x × (100 - 20)/100

= 4/5x

The number of students in IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

= 2080 - 1180

= 900

⇒ x + 4/5x = 900

⇒ 9/5x = 900

⇒ x = 500

⇒ 4/5x → 500 × 4/5 = 400

The total number of students who study in the Applied department in these two colleges = 400 × 32% + 500 × 35%

= 128 + 175 = 303

The total number of students who study in the Civil department in these two colleges = 400 × 18% + 500 × 25%

= 72 + 125 = 197

Difference ⇒ 303 - 197

= 106

The answer is 106.

Calculation

The number of students at IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

⇒ 2080 - 1180

⇒ 900

Let the number of students at IIT Madras be 5a.

The number of students at IIT Bombay = 5a - 5a × 20%

⇒  4a

⇒ 5a + 4a = 900

⇒ 9a = 900

⇒ a = 100

4a → 400

The number of students at IIT Bombay is 400.

Calculation

The number of students at IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

⇒ 2080 - 1180

⇒ 900

Let the number of students at IIT Madras be 5a.

The number of students at IIT Bombay = 5a - 5a × 20%

⇒  4a

⇒ 5a + 4a = 900

⇒ 9a = 900

⇒ a = 100

4a → 400

The number of students at IIT Bombay is 400.

Calculation

Let the number of students in IIT Madras be x.

Then, the number of students in IIT Bombay = x × (100 - 20)/100

= 4/5x

The number of students in IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

= 2080 - 1180

= 900

⇒ x + 4/5x = 900

⇒ 9/5x = 900

⇒ x = 500

⇒ 4/5x → 500 × 4/5 = 400

The total number of students who study in the Applied department in these two colleges = 400 × 32% + 500 × 35%

= 128 + 175 = 303

The total number of students who study in the Civil department in these two colleges = 400 × 18% + 500 × 25%

= 72 + 125 = 197

Difference ⇒ 303 - 197

= 106

The answer is 106.

Calculation

The number of students at IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

⇒ 2080 - 1180

⇒ 900

Let the number of students at IIT Madras be 5a.

The number of students at IIT Bombay = 5a - 5a × 20%

⇒  4a

⇒ 5a + 4a = 900

⇒ 9a = 900

⇒ a = 100

4a → 400

The number of students at IIT Bombay is 400.

Calculation

Let the number of students in IIT Madras be x.

Then, the number of students in IIT Bombay = x × (100 - 20)/100

= 4/5x

The number of students in IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)

= 2080 - 1180

= 900

⇒ x + 4/5x = 900

⇒ 9/5x = 900

⇒ x = 500

⇒ 4/5x → 500 × 4/5 = 400

The total number of students who study in the Applied department in these two colleges = 400 × 32% + 500 × 35%

= 128 + 175 = 303

The total number of students who study in the Civil department in these two colleges = 400 × 18% + 500 × 25%

= 72 + 125 = 197

Difference ⇒ 303 - 197

= 106

The answer is 106.

Given:

Number of female qualified candidates from State X in 2010 = 168

Ratio of male to female qualified candidates in 2010 = 5 : 7

Percentage of qualified candidates from State X in 2010 = 60%

Formula used:

Let the total number of qualified candidates from State X in 2010 be Q.

Since the ratio of male to female candidates is 5 : 7, the number of female qualified candidates = \(\dfrac{7}{12} \times Q\) and the number of male qualified candidates = \(\dfrac{5}{12} \times Q\).

Given that the number of female qualified candidates is 168, we can write: \(\dfrac{7}{12} \times Q = 168\).

Calculations:

\(\dfrac{7}{12} \times Q = 168\)

⇒ \(Q = 168 × 12 /7 \)

⇒ \(Q = 288\)

Now, the total number of qualified candidates from State X in 2010 is 288. Since 60% of the appeared candidates qualify, we can write:

\(\dfrac{60}{100} \times \text{Total appeared candidates} = 288\)

⇒ \(\text{Total appeared candidates} = \frac{288× 100}{60} \)

⇒ \(\text{Total appeared candidates} = 480\)

∴ The number of appeared candidates from State X in 2010 is 480.

Formula used:

1. Qualified candidates = ({Number of appeared candidates}) × ({Percentage of qualified candidates}) / 100

Calculations:

Step 1: Calculate qualified candidates in 2008:

Qualified candidates in 2008 = A × 70 / 100 = 0.7A .

Step 2: Calculate qualified candidates in 2009:

Qualified candidates in 2009 = 2A × 80 / 100 = 1.6A .

Step 3: Total qualified candidates:

0.7A + 1.6A = 552 .

⇒2.3A = 552 .

⇒ A = 552 / 2.3 = 240 .

Step 4: Calculate appeared candidates in 2009:

Number of appeared candidates in 2009 = 2A = 2 × 240 = 480 .

∴ The number of appeared candidates from State Z in 2009 is 480.