Quantitative Aptitude - SPLessons

IBPS RRB PO Quantitative Aptitude Quiz 3

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

IBPS RRB PO Quantitative Aptitude Quiz 3

shape Introduction

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.

The article IBPS RRB PO Quantitative Aptitude Quiz 3 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 3 will assist the students to know the expected questions from Quantitative Aptitude.

Click Here for Official IBPS Website


shape Quiz

Q1. Ramesh can finish a job in 20 days. He worked for 10 days alone and completed the remaining job working with Dinesh, in 2 days. How many days would both Dinesh and Ramesh together take to complete the entire job?

    A. 4
    B. 5
    C. 10
    D. 12


Answer: Option A

Explanation:
Ramesh alone finished \(\frac{1}{2}\) of the work in 10 days.
Remaining \(\frac{1}{2}\) was finished by Ramesh and Dinesh together in 2 days.
Therefore, they both together can finish the complete job in 4 days.


Q2. Ram decided to plough a farmland in 50 days. He employed 50 men in the beginning and 50 more after 35 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished?

    A. 5 days
    B. 6 days
    C. 8 Days
    D. 10 Days


Answer: Option A

Explanation:
Given that, 50 men employed for 35 days and 50 more men employed for 15 days so that the work could be finished in 50 days.
Thus, the total work = 50*35 + (50 + 50) 15 = 3250
Suppose, it takes x days to finish the whole work if additional men were not employed.
So, we have an equation here.
50*x = 3250
x = 65 days
Therefore, it takes 15 days more than the stipulated time.


Q3. 20 men can complete a piece of work in 15 days and 12 women can complete the same piece of work in 24 days. What is the ratio of amount of work done by 30 men in one day to the amount of work done by 16 women in 1 day ?

    A. 6:7
    B. 9:5
    C. 7:5
    D. 2:7


Answer: Option B

Explanation:
M : W = \(\frac{30}{20*15}\) : \(\frac{16}{12*24}\)
= \(\frac{1}{10}\) : \(\frac{1}{18}\) = \(\frac{1}{5}\):\(\frac{1}{9}\)
Ratio = 9:5


Q4. X alone can do a piece of work in 5 days. Y can do the same piece of work in 4 days. X and Y are assigned to do the work for Rs.5000.They complete the work in 2 days with the help of Z. How much is to be paid to Z ?

    A. Rs.750
    B. Rs.620
    C. Rs.700
    D. Rs.500


Answer: Option D

Explanation:
Z’s one day work = \(\frac{1}{2}\) – [ \(\frac{1}{5}\) + \(\frac{1}{4}\)] = 10 – 4 + \(\frac{5}{20}\) = \(\frac{1}{20}\)
Ratio = \(\frac{1}{5}\):\(\frac{1}{4}\):\(\frac{1}{20}\) = \(\frac{4}{20}\) : \(\frac{5}{20}\) : \(\frac{1}{20}\) = 4:5:1
Z = 5000*\(\frac{1}{10}\) = 500


Q5. 9 men and 12 women can complete the job in 12 days.In how many days can 3 men and 4 women finish the same job working together ?

    A. 36 days
    B. 42 Days
    C. 30 Days
    D. 28 Days


Answer: Option A

Explanation:
9m + 12w = \(\frac{1}{13}\)
3m + 4w = ?
3(3m + 4w) = \(\frac{1}{13}\)
3m + 4w = \(\frac{1}{36}\)

Q1. The average weight of 39 Students in a class is 23. Among them Sita is the heaviest while Tina is the lightest. If both of them are excluded from the class still the average remains same. The ratio of weight of Sita to Tina is 15:8.Then what is the weight of the Tina?

    A. 15
    B. 16
    C. 18
    D. 19


Answer: Option B

Explanation:

S+T = 23*(39-37) = 46
\(\frac{S}{T}\) = \(\frac{15}{8}\)
T = 16


Q2. The ages of Four members of a family are in the year 2010 are ‘X’,’X+12’,’X+24’ and ‘X+36’. After some years Oldest among them was dead then average reduced by 3. After how many years from his death, the average age will same as in 2010?

    A. 2 Years
    B. 3 Years
    C. 4 years
    D. 6 Years


Answer: Option B

Explanation:

In 2010: 4x+\(\frac{72}{4}\) = x + 18
After death : 3x + 36 + \(\frac{3N}{3}\) = x + 18 – 3
N = 3 years
3x + 36 + \(\frac{3N}{3}\) = (x + 18)
N = 6 years
6-3 = 3 years from his death


Q3. The average of Four numbers is 24.5. of the four numbers, the first is 1.5 times the second, the second is 1/3 rd of the third, and the third is 2 times the fourth number. Then what is smallest of all those numbers?

    A. 12
    B. 13
    C. 14
    D. 15


Answer: Option C

Explanation:
First = 1.5x Second = x Third = 3x Fourth = 1.5x
average = 24.5 = \(\frac{(1.5x + x + 3x + 1.5x)}{4}\)
x = 14


Q4. There are 459 students in a hostel. If the number of students increased by 36, the expenses of the mess increased by Rs .81 Per day while the average expenditure per head reduced by 1. Find the original expenditure of the mess?

    A. 7304
    B. 7314
    C. 7324
    D. 7344


Answer: Option D

Explanation:
Total expenditure = 459x
36 students joined then total expenditure = 459x +81
average = 459x + \(\frac{81}{495}\) = x-1
x = 16
original expenditure = 16*459 = 7344


Q5. The average cost 32 different Mobiles is Rs. 9000. Among them, Oppo which is the costliest is 70% higher price than the cheapest Mobile Lava. Excluding those both mobiles, the average of the Mobiles is Rs.8880. Then what is the cost of Oppo Mobile?

    A. Rs. 10000
    B. Rs. 11600
    C. Rs. 12400
    D. Rs.13600


Answer: Option D

Explanation:
L+O = 21600
O = L*\(\frac{170}{100}\)
O = 13600

1. When a number is subtracted from the number 8,12 and 20, the remainders are in continued proportion, Find the number ?

    A. 3
    B. 4
    C. 2
    D. 6


Answer: Option B

Explanation: \(\frac{8-x}{12 –x}\) = \(\frac{12-x}{20 –x}\)
(8-x)(20-x) = (12 – x)(12 –x )
160 – 8x – 20x + \({x}^{2}\) = 144 – 12x – 12x + \({x}^{2}\)
4x = 16
x = 4


2. The sum of square of 2 number is 1972 and the difference of their square is 620. Find the number ?

    A. 33,26
    B. 34,25
    C. 36,26
    D. 38,28


Answer: Option A

Explanation:
\({x}^{2}\) + \({y}^{2}\)= 1972
\({x}^{2}\) – \({y}^{2}\) = 620
Solve ths
2\({x}^{2}\) = 2592
\({x}^{2}\) = 1296
x = 36
\({y}^{2}\) = 1972 – 1296
\({y}^{2}\) = 676
y = 26


3. Product of one fourth of the number and 120% of another number is what % of the product of the original number ?

    A. 35%
    B. 23%
    C. 32%
    D. 30%


Answer: Option D

\(\frac{x}{4}\) = one fourth of the no
\(\frac{120y}{100}\) = another no = \(\frac{6y}{5}\)
% = [(\(\frac{x}{4}\) * \(\frac{6y}{5}\))*100]x \(\frac{1}{xy}\)
= \(\frac{600}{20}\) = 30%


4. If three numbers are added in pairs, the sums equal to 11,16 and 23.Find the three numbers

    A. 8, 6, 16
    B. 9, 2, 14
    C. 9, 5, 14
    D. 7, 2, 8


Answer: Option B

Explanation:
X + y = 11
Y + z = 16
Z + x = 23
X + y + z = 25
X = 25 – 16 = 9
y = 25 – 23 = 2
z = 25 – 11 = 14


5. What is the no of zeroes at the end of the product of the no from 1 to 100 ?

    A. 20
    B. 22
    C. 24
    D. 25


Answer: Option C

Explanation:
\(\frac{100}{5}\) + \(\frac{100}{25}\) = 20 + 4 = 24


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