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 4** 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 4** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. Rs.575

B. Rs.675

C. Rs.585

D. Rs.685

**Answer**: Option C

**Explanation**:

90 = latex]\frac {(450*4*R)}{100}[/latex]

R = 5%

I = \(\frac {(450*6*5)}{100}\) = 135

450 + 135 = 585

**2. The area of a base of a cone is 30 cm\(^ {2}\). If the height of the cone is 6cm, find its volume?**

- A. 120 cm\(^ {3}\)

B. 40 cm\(^ {3}\)

C. 50 cm\(^ {3}\)

D. 60 cm\(^ {3}\)

**Answer**: Option D

**Explanation**:

πr\(^ {2}\) = 30 h = 6

\(\frac {1}{3}\) * 30 * 6 = 60.

**3. Ratan takes turns to cycle around the circular garden and through the diameter of the garden on alternate days. His speed each day is 6m/min but when he cycles through the diameter, he takes 1 hr less to cross the park. Find the diameter of the park. **

- A. 1530 m

B. 1680 m

C. 1750 m

D. 3600 m

**Answer**: Option B

**Explanation**:

We know that, Time =\(\frac {Distance}{Speed}\)

Speed = 60m/min = \(\frac {60}{60}\) m/sec = 1 m/sec

We know that,

Time along the park (circumference) – Time along the diameter = 60 minutes

∴\(\frac {2πr}{1}\) – \(\frac {2r}{1}\) = 60 minutes

∴ 2r (\(\frac {22}{7}\)– 1) = 3600 seconds

∴ r = 840m

∴ Diameter = 2 x radius = 1680m.

**4. In what time a sum of money double itself at 3% per annum simple interest?**

- A. 29 years

B. 33 1/3 years

C. 23 1/3 years

D. 13 1/3 years

**Answer**: Option B

**Explanation**

P = (P*3*R)/100

R = 33 1/3%

**5. For a rectangular box, what is the product of the areas of the 3 adjacent faces that meet at a point?**

- A. The volume of the box

B. Twice the volume of the box

C. Half of the volume of the box

D. Square of the volume of the box

**Answer**: Option D

**Explanation**:

The area of the 3 adjacent faces of a rectangular box of length ‘l’, breadth ‘b’, and depth ‘h’ are ‘l*b’, ‘b*h’, and ‘l*h’.

Therefore, lb*bh*lh = l\(^ {2}\) b\(^ {2}\) h\(^ {2}\) = (lbh)\(^ {2}\) = square of the volume of the box.

- A. \(log_{10}\) 10 = 1

B. log (2 + 3) = log (2 x 3)

C. \(log_{10}\) 1 = 0

D. log (1 + 2 + 3) = log 1 + log 2 + log 3

**Answer**: Option B

**Answer**: Option

A. Since \(log_{a}\) a = 1, so \(log_{10}\) 10 = 1.

B. log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

log (2 + 3) ≠ log (2 x 3)

C. Since \(log_{a}\) 1 = 0, so \(log_{10}\) 1 = 0.

D. log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, B is incorrect.

**2. If in a game of 80, P can give 16 points to Q and R can give 20 points to P, then in a game of 150, how many points can R give to Q?**

- A. 48

B. 60

C. 54

D. 90

**Answer**: Option B

**Answer**: Option

When P scores 80, Q scores 64.

When R scores 80, P scores 60

Hence, when R scores 150, Q scores (60 * 64 * 150) / (80 * 80) = 90

Therefore, in a game of 150, R can give 60 points to Q.

**3. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for?**

- A. 160

B. 175

C. 180

D. 195

**Answer**: Option B

**Explanation**:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

**4. Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car.**

- A. 906 km

B. 960 m

C. 960 km

D. 966 km

**Answer**: Option C

**Explanation**

60(x + 1) = 64x

X = 15

60 * 16 = 960 km

**5. 5 mat-weavers can weave 5 mats in 5 days. At the same time, how many mats would be woven by 10 mat- weavers in 10 days?**

- A. 10 mats

B. 15 mats

C. 20 mats

D. 30 mats

**Answer**:

**Explanation**

More mats are weaved if more weavers work. Hence, this problem is related to a direct proportion.

Let the required number of mats be x.

Total 5 mats can be weaved in 5 days by 5 weavers.

\[5: x::

\begin{cases}

5 : 10 – – – (Weavers)\\

5 : 10 – – – (Days)

\end{cases}

\]

5 × 5 × x = 10 × 10 × 5

x = \(\frac {10 \times 10 \times 5}{5 \times 5}\) = 20

20 mats can be weaved in 10 days by 10 mat weavers.

- A. 720

B. 144

C. 120

D. 36

E. 204

**Answer**: Option B

**Explanation**

The word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144.

**2. Find the one which does not belong to that group ?**

- A. 27

B. 37

C. 47

D. 67

**Answer**: Option A

**Explanation**:

37, 47, 67 and 17 are prime numbers but not 27

**3. (1331)\(^ {– (2/3)}\) **

- A. –\(\frac {1}{11}\)

B. –\(\frac {11}{121}\)

C. \(\frac {1}{121}\)

D. \(\frac {121}{11}\)

**Answer**: Option C

**Explanation**

Cube root of 1331 is 11. Therefore,

(11)\(^ {(3)\times – (2/3)}\)

Remember the law of indices (x\(^ {(m){n}}\)) = x\(^ {mn}\)

(11)\(^ {– 3 \times (2/3)}\) = 11\(^ { –2}\)

x\(^ {–1 }\)=\(\frac {1}{x}\)

Hence, 11\(^ {–2}\) = \(\frac {1}{11^{2}}\) = \(\frac {1}{121}\)

**4. Two pipes can fill a tank in 10 and 14 minutes respectively and a waste pipe can empty 4 gallons per minute. If all the pipes working together can fill the tank in 6 minutes, what is the capacity of the tank?**

- A. 120 gallons

B. 240 gallons

C. 450 gallons

D. 840 gallons

**Answer**: Option D

**Explanation**

Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)

= 1/6 – [(\(\frac {1}{10}\) ) + (\(\frac {1}{14}\) )] = \(\frac {1}{6}\) – (\(\frac {24}{140}\) ) = \(\frac {-1}{210}\)

Here, negative (-) sign indicates emptying of tank.

To find the capacity, we need to determine the volume of 1/210 part.

Therefore, volume of \(\frac {1}{210}\) part = 4 gallons ———(given condition)

Hence, the capacity of tank = volume of whole = 4 x 210 = 840 gallons.

**5. A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is?**

- A. 10 kmph

B. 5 kmph

C. 20 kmph

D. 22 kmph

**Answer**: Option C

**Explanation**

M = 15

S = 5

DS = 15 + 5 = 20