A candidate with **quantitative aptitude** knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an **important** measure for a prospective business executive’s **abilities**.

The article **SSC CPO Quantitative Aptitude Quiz 24** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. The article **SSC CPO Quantitative Aptitude Quiz 24** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. 144

B. 169

C. 154

D. 178

**Answer**: Option B

**Explanation**:

In this question the pattern * represents a code.

=> 3² + 4² = 25

=> 6² + 8² = 100

=> 12² + 5² = 169

**2. Solve for the given equation**

**121 x 54 =?**

- A. 68225

B. 75625

C. 76569

D. 45854

**Answer**: Option B

**Explanation**:

This question can be solved by 2 approaches:

=> Direct Multiplication

=> Numerical Operations

Direct multiplication is quite time consuming and involves a lot of calculations. An easier approach is based on numerical operations.

=> 121 x \({10/2}^{4}\)(10/2)4 = 121 x \(\frac{10000}{16}\) = 75625 (Lesser calculations, Saves time)

**3. \(\frac{6}{7}\) of a certain number is 96. Find quarter of that number.**

- A. 112

B. 32

C. 56

D. 28

**Answer**: Option D

**Explanation**:

6/7x = 96

x = 112

y = \(\frac{x}{4}\) = \(\frac{112}{4}\) = 28

**4. 36 is divided in 2 parts such that 8 times the first part added to 3 times the second part makes 203. What is the first part?**

- A. 15

B. 19

C. 23

D. None of these

**Answer**: Option B

**Explanation**:

Let the first part be x.

8x + 3 (36 – x) = 203

8x + 108 – 3x = 203

5x + 108 = 203

5x = 95

x = 19

**5. The number 243 has been divided into three parts in such a way that one third of the first part, fourth of the second part and half of the third part are equal. Determine the largest part.**

- A. 108

B. 86

C. 74

D. 92

**Answer**: Option A

**Explanation**:

It is given that 243 has been divided into 3 numbers x y and z for instance.

Second condition mentioned in the question is:

=> \(\frac{X}{3}\) = \(\frac{Y}{4}\) = \(\frac{Z}{2}\)

Clearly the biggest number is y. So let us solve w.r.t. y.

=> x = 3\(\frac{Y}{4}\)

=> x =3\(\frac{Z}{2}\)

=> z = \(\frac{Y}{2}\)

=> x+y+z = 243

=> \(\frac{Y}{2}\) + 3\(\frac{Y}{4}\) + y = 243

9\(\frac{Y}{4}\) = 243

=> 243*\(\frac{4}{9}\)

=> 108

- A. 2400

B. 3600

C. 7200

D. 1600

**Answer**: Option B

**Explanation**:

3600. The number is perfect square of 60.

It is divisible by 36 24 and 20. Only two numbers 3600 and 7200 fit these conditions.

It is clear that 3600 is the smaller number.

**2. India has a run rate of 3.2 in the first 10 overs. What should be the run rate in remaining 40 overs to win a game of 282 runs?**

- A. 6.25

B. 5.5

C. 7.25

D. 8

**Answer**: Option A

**Explanation**:

Runs scored in first 10 overs:

=> Rate*Overs= 3.2*10 = 32 runs

=> Required runs = 282-32 = 250

=> Left overs= 40

Required rate = runs/overs

= \(\frac{250}{40}\)

= 6.25

=> 6.25 runs per over is needed to win the game.

**3. The difference between a number and its two-fifth is 45. What is the number?**

- A. 64

B. 75

C. 80

D. 89

**Answer**: Option B

**Explanation**:

Let the number be x.

x – (\(\frac{2}{5}\)) x = 45

(\(\frac{3}{5}\)) x = 45

x = 75

**4. 8900 ÷ 6 ÷ 4 = ? **

- A. 349

B. 541.75

C. 224.37

D. 370.833

**Answer**: Option D

**Explanation**:

Given Exp. 8900 * \(\frac{1}{6}\) * \(\frac{1}{4}\)

= 370.833

**5. Two-third of a positive number and \(\frac{16}{216}\) of its reciprocal are equal. Find the positive number.**

- A. \(\frac{ 9}{25}\)

B. \(\frac{ 14}{ 4}\)

C. \(\frac{ 4}{ 12}\)

D. \(\frac{ 144}{ 25}\)

**Answer**: Option C

**Explanation**:

Let the positive number be x.

Then, \(\frac{2}{3}\) x = \(\frac{16}{216}\) * \(\frac{ 1}{X}\)

x2 = \(\frac{16}{216}\) * \(\frac{ 3}{2}\)

= \(\frac{ 16}{144}\)

x = \(\frac{ √16}{144}\) = \(\frac{ 4}{13}\).

- A. 2

B. 5

C. 6

D. 8

** Answer**: Option D

**Explanation**:

Unit digit in the given product = Unit Digit in (4*8*3*3) = 8

**2. In an entrance exam, 3 marks is awarded for every correct answer and (-1) for every wrong answer. If a student gets 38 marks after attempting all questions, find the number of questions answered correctly if the total questions were 70.**

- A. 38

B. 40

C. 27

D. 43

** Answer**: Option C

**Explanation**:

Let x be the number of questions answered correctly. Hence, number of questions whose answer were wrong = (70 – x)

3*x + (70 – x)*(-1) = 38

4x = 38 + 70 = 108

x = 27

**3. If in a book a total of 3209 digits are used for numbering, then find out the total number of pages in the book. **

- A. 1079

B. 1000

C. 1100

D. 1075

** Answer**: Option A

**Explanation**:

Number of digits used in 1 digit numbers = 1*9 = 9

Number of digits used in 2 digit numbers = 2*90 = 180

Number of digits used in 3 digit numbers = 3*900 = 2700

Number of digits used in 1 digit numbers = 3209 – (9+180+2700) = 320

Hence, number of pages with 4 digit number = \(\frac{ 320}{4}\) = 80

Total number of pages = 1079

**4. How many numbers between 100 and 600 are divisible by 2, 3, and 7 together?**

- A. 11

B. 12

C. 14

D. cannot be determined

** Answer**: Option B

**Explanation**:

As the division is by 2, 3, 7 together, the numbers are to be divisible by: 2*3*7 = 42

The limits are 100 and 600

The first number divisible is 42*3 = 126

To find out the last number divisible by 42 within 600:

600/42 = 14.28

Hence, 42*14 = 588 is the last number divisible by 42 within 600

Hence, total numbers divisible by 2, 3, 7 together are (14 – 2) = 12

**5. What will come in place of the question mark (?) in the following question?**

**112.36 + 225.05 + ? = 815.30 **

- 474.46

B. 474.46

C. 524.41

D. 564.40

** Answer**: Option B

**Explanation**:

815.30 – ( 112.36 + 225.05) = ?

= 477.89