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 9** 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 9** will assist the students to know the expected questions from **Quantitative Aptitude**.

- 5 kmph

B. 6 kmph

C. 6.25 kmph

D. 7.5 kmph

**Answer**: Option A

**Explanation**:

Let Abhay’s speed be x km/hr.

Then = \(\frac{30}{x}\) – \(\frac{30}{2x}\) = 3

6x = 30

x = 5 km/hr.

**2. Robert is traveling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?**

- A. 8 kmph

B. 11 kmph

C. 12 kmph

D. 14 kmph

**Answer**: Option C

**Explanation**:

Let the distance travelled by x km.

Then, \(\frac {x}{10}\) – \(\frac {x}{15}\) = 2

3x – 2x = 60

x = 60 km.

Time is taken to travel 60 km at 10 km/hr = \(\frac {60}{10}\) hrs = 6 hrs

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed = \(\frac {60}{5}\) kmph. = 12 kmph

**3. It takes eight hours for a 600 km journey if 120 km is done by train and the rest by car. It takes 20 minutes more if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:**

- A. 2 : 3

B. 3 : 2

C. 3 : 4

D. 4 : 3

**Answer**: Option C

**Explanation**:

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then, \(\frac {120}{X}\) + \(\frac {480}{Y}\) = 8 ⇒ \(\frac {1}{X}\) + \(\frac {4}{Y}\) = \(\frac {1}{15}\)

And, \(\frac {20}{X}\) + \(\frac {400}{Y}\) = \(\frac {25}{3}\) ⇒ \(\frac {1}{X}\) + \(\frac {2}{Y}\) = \(\frac {1}{24}\)

Solving (i) and (ii), we get: x = 60 and y = 80.

Ratio of speeds = 60 : 80 = 3 : 4.

**4. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:**

- A. 14 km

B. 15 km

C. 16 km

D. 17 km

**Answer**: Option C

**Explanation**:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, \(\frac {X}{4}\) + \(\frac {(61 -x) }{9}\) = 9

⇒9x + 4(61 -x) = 9 x 36

⇒ 5x = 80

⇒ x = 16 km.

**5. The ratio between the speeds of two trains is 7: 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:**

- A. 70 km/hr

B. 75 km/hr

C. 84 km/hr

D. 87.5 km/hr

**Answer**: Option D

**Explanation**:

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x = \(\frac {400}{4}\) = 100

X = \(\frac {100}{8}\) = 12.5

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

I. The average mark obtained by Tarun in four subjects including English is 60.

II. The total marks obtained by him in English and Mathematics together are 170.

III. The total marks obtained by him in Mathematics and Science together are 180.

** **

- A. I and II only

B. II and III only

C. I and III only

D. None of these

**Answer**: Option D

**Explanation**:

I gives, total marks in 4 subjects = (60 x 4) = 240.

II gives, E + M = 170

III gives, M + S = 180.

Thus, none of (A), (B), (C), (D) is true.

**2. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?**

I. The captain is eleven years older than the youngest player.

II. The average age of 10 players, other than the captain is 27.3 years.

** III. Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively. **

- A. Any two of the three

B. All I, II and III

C. II only or I and III only

D. II and III only

**Answer**: Option C

**Explanation**:

Total age of 11 players = (28 x 11) years = 308 years.

I. C = Y + 11 C – Y = 11 …. (i)

II. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.

Age of captain = (308 – 273) years = 35 years.

Thus, C = 35. …. (ii)

From (i) and (ii), we get Y = 24

III. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.

C + Y = (308 – 249) = 59 …. (iii)

From (i) and (iii), we get C = 35.

Thus, II alone gives the answer.

Also, I and III together give the answer.

**3. The average age of P, Q, R, and S is 30 years. How old is R?**

I. The sum of ages of P and R is 60 years.

** II. S is 10 years younger than R.**

- A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

**Answer**: Option D

**Explanation**:

P + Q + R + S = (30 x 4) P + Q + R + S = 120 …. (i)

I. P + R = 60 …. (ii)

II. S = (R – 10) …. (iii)

From (i), (ii) and (iii), we cannot find R.

**4. How many candidates were interviewed every day by panel A out of the three panels A, B, and C?**

I. The three panels on average interview 15 candidates every day.

** II. Out of a total of 45 candidates interviewed every day by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B. **

- A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

**Answer**: Option B

**Explanation**:

I. Total candidates interviewed by 3 panels = (15 x 3) = 45.

II. Let x candidates be interviewed by C.

Number of candidates interviewed by A = (x + 2).

Number of candidates interviewed by B = (x + 1).

x + (x + 2) + (x + 1) = 45

3x = 42

x = 14

**5. What is the average age of children in the class?**

I. The age of the teacher is as many years as the number of children.

** II. The average age is increased by 1 year if the teacher’s age is also included. **

- A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

**Answer**: Option D

**Explanation**:

Let there be x children.

I give, age of teacher = x years.

II gives, average age of (x + 1) persons = (x + 1) years.

Teacher’s age = (x + 1) (x + 1) – \({x}^{2}\) = (\({x}^{2}\) + 1 + 2x) – \({x}^{2}\) = (1 + 2x)

Thus, the teacher’s age cannot be obtained.

- A. 2 : 3

B. 4 : 3

C. 6 : 7

D. 9 : 16

** Answer**: Option B

**Explanation**:

Let us name the trains as A and B. Then,

\(\sqrt{b}\)\(\sqrt{a}\) = \(\sqrt{16}\) = \(\sqrt{9}\) = 4 : 3.

**2. Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet? **

- A. 9 a.m.

B. 10 a.m.

C. 10.30 a.m.

D. 11 a.m.

** Answer**: Option B

**Explanation**:

Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x – 1) hours = 25(x – 1) km.

Therefore 20x + 25(x – 1) = 110

=> 45x = 135

=> x = 3.

**3. The last day of a century cannot be **

- A. Monday

B. Wednesday

C. Tuesday

D. Friday

** Answer**: Option C

**Explanation**:

100 years contain 5 odd days.

The last day of \({1}^{st}\) century is Friday.

200 years contain (5 x 2) 3 odd days.

The last day of \({2}^{nd}\) century is Wednesday.

300 years contain (5 x 3) = 15 1 odd day.

The last day of \({3}^{rd}\) century is Monday.

400 years contain 0 odd days.

The last day of \({4}^{th}\) century is Sunday.

This cycle is repeated.

Last day of a century cannot be Tuesday or Thursday or Saturday.

**4. A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?**

- A. 45

B. 60

C. 75

D. 90

** Answer**: Option D

**Explanation**:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

**5. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:**

- A. 20

B. 80

C. 100

D. 200

** Answer**: Option C

**Explanation**:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10 x – y = 20 …. (i)

and x + 20 = 2(y – 20) x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.