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SSC CPO Quantitative Aptitude Quiz 4

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SSC CPO Quantitative Aptitude Quiz 4

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What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.


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.


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1. If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum?

    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.

1. Which of the following statements is not correct?


    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.

1. The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?

    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


IBPS RRB PO – Related Information
IBPS RRB Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 2
Book for Quantitative Aptitude