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 **IBPS RRB PO Quantitative Aptitude Quiz 9** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. **IBPS RRB** has released **IBPS RRB Officer 2019 **Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article **IBPS RRB PO Quantitative Aptitude Quiz 9** will assist the students to know the expected questions from **Quantitative Aptitude**.

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**Answer**: Option B

**Explanation**:

X = 37*40 = 1480

X = 38*41 = 1558

Teacher age = 1558 – 1480 = 78

Aftr 2 yrs = 78 + 2 = 80

**2. At present Priya is twice Sariga’s age. 8 yrs hence the respective ratio between Priya and Sariga’s age will be 22 :13, What is Sariga’s present age ?**

**Answer**: Option C

**Explanation**:

P = 2S

\(\frac{2s+8}{s+8}\) = \(\frac{22}{13}\)

26s + 104 = 22s + 176

4s = 72

S= 18yr

**3. The ratio between the present ages of X and Y is 4:9 respectively. 2yrs from now, Y’s age will become twice the age of X.What was the difference between their ages five yrs ago ?**

**Answer**: Option A

**Explanation**:

\(\frac{4x+2}{7x+2}\) = \(\frac{1}{2}\)

8x + 4 = 9x + 2

x = 2

x:y = 8:18

5 yrs ago = 3:13

D = 13-3 = 10

**4. The ratio of present age of Kaviya and Lakshmi is 6:7, the ratio of their ages after 8yrs would be 8:9, What is Kaviya’s present age?**

**Answer**: Option B

**Explanation**:

\(\frac{6x+8}{7x+8}\) = \(\frac{8}{9}\)

54x + 72 = 56x + 64

2x = 8

x = 4

K = 6*4 = 24yrs

**5. The ratio between the present age of Maha and Deepa is 5:X. Maha is 9yrs younger than Parveen. Parveen’s age after 9yrs will be 33yrs. The difference between the Deepa and Maha age is same as the present age of Parveen. Find X**

**Answer**: Option A

**Explanation**:

P present age = 33 – 9 = 24

M = 24 – 9 =15

D – M = 24

D – 15 = 24

D = 24 + 15 = 39

15:39 = 5:13

X = 13

**Answer**: Option D

**Explanation**:

A = 3N

A + 3 = N + 4 + 3

3N + 3 = N + 7

2N = 4

N = \(\frac{4}{2}\)

N = 2yrs

A = 3*2 = 6yrs

**2. The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is:**

**Answer**: Option A

**Explanation**:

Let the required mean score be a

Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100

1600 + 775 + 55a = 5200

55a = 2825

a = 51.4

**3. The average of a non-zero number and its square is 5 times the number. The number is:**

**Answer**: Option D

**Explanation**:

Let the number be x.

Then,

\(\frac{(x + x^2) }{2}\)= 5x

x^2 – 9x = 0

x (x – 9) = 0

x = 0 or x = 9.

**4. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one them weighing 65 kg. What might be the weight of the new person?**

**Answer**: Option C

**Explanation**:

Total weight increased = (8 x 2.5) kg = 20 kg

Weight of new person = (65 + 20) kg = 85 kg

**5. The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is:**

**Answer**: Option D

**Explanation**:

Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given

So, the data provided is inadequate

**Answer**: Option C

**Explanation**: a + b = 6

20 cm = 0.2 m

So number of boxes that can fit = 1.6 × 1 × \(\frac{0.6 }{0.2}\) × 0.2 × 0.2

**2. What is the volume of a cylinder whose curved surface area is 704 c\({m }^{2}\) and height is 14 cm? **

**Answer**: Option D

**Explanation**:

2ᴨrh = 704, h = 14

Solve both, so r = 8

Volume = ᴨ\({r}^{2\)h

**3. The total cost of painting the walls of a room is Rs 475. Find the cost of painting the walls of another room whose length, breadth and height each are double than the dimensions of the previous room. **

**Answer**: Option B

Area of first room = 2(l+b)*h

After all dimensions doubled, new area = 2(2l + 2b)*2h = 4[2(l+b)*h ] = 4 times previous area, so cost of painting = 4*475

**4. A circular wire of diameter 84 cm is bent into a rectangle with sides ratio 6 : 5. What are the respective sides of the rectangle? **

**Answer**: Option C

**Explanation**:

Length of wire = 2ᴨr = 2(\(\frac{22}{7}\))*42 = 264 cm

Perimeter of rectangle = 2(6x + 5x) = 264

Solve, x = 12

So dimensions – 12*6, 12*5

**5. The ratio of the outer and the inner perimeters of a circular path is 9 : 8. The path is 3 metres wide. What is the diameter of the outer circle? **

**Answer**: Option B

**Explanation**:

\(\frac{2ᴨr}{2ᴨR }\) = \(\frac{9}{8}\)

So \(\frac{r}{R}\) = \(\frac{9}{8}\), so r = (\(\frac{9}{8}\))R

r – R = 3, so (\(\frac{9}{8}\))R – R = 3

R = 24, so r = (\(\frac{9}{8}\))*24 = 27