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IBPS RRB Quantitative Aptitude Quiz Day 5

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IBPS RRB Quantitative Aptitude Quiz Day 5

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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 IBPS RRB Quantitative Aptitude Quiz Day 5 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 Quantitative Aptitude Quiz Day 5 will assist the students to know the expected questions from Quantitative Aptitude.

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Directions(1-5): Arithmetic Word Problems

1. 20 men, 12 women and 18 boys were given a project of doing 3960 designs of a building in 5 days. The ratio of the number of designs made by them respectively in 1 day is 3 : 2 : 1. If on the \({1}^{st}\) day all of them worked, on the \({2}^{nd}\) day 4 women and 6 boys went absent and on the 3rd day, 6 men and 10 boys went absent but still the work got finished on the 3rd day. Then find the number of designs designed by them on the \({3}^{rd}\) day?

    A. 1021
    B. 1110
    C. 1621
    D. 1210


Answer : Option B

Explanation: Let the number of designed by men, women and boys in 1 day be 3x, 2x and x respectively.

Designs of building on the \({1}^{st}\) day

⇒ 20 × 3x + 12 × 2x + 18 × x

⇒ 102x

On the \({2}^{nd}\) day = 20 × 3x + 8 × 2x + 12 × x = 88x

On the \({3}^{rd}\) day = 14 × 3x + 12 × 2x + 8 × x = 74x

Now, 102x + 88x + 74x = 3960

⇒ 264x = \(\frac{3960}{264}\) x 74
⇒ 74x = 1110


2. Rohit can row a boat 65Km upstream and 130Km downstream in 23 hours, whereas he can swim 45Km upstream and 104Km downstream in 17 hours. Find the speed of boat in still water and the speed of stream.

    A. 4 kmph, 9 kmph
    B. 8 kmph, 5 kmph
    C. 9 kmph, 4 kmph
    D. 5 kmph, 8 kmph


Answer : Option C

Explanation: Upstream, U = Speed of boat – speed of stream

Downstream, D = Speed of boat + speed of the stream

\(\frac{65}{U}\) + \(\frac{130}{D}\) = 23

\(\frac{45}{U}\) + \(\frac{104}{D}\) = 17
On solving the above two equations, we will get

U = Speed of the boat – the speed of stream = 5

D = Speed of the boat + speed of the stream = 13

Thus, Speed of boat = 9 and speed of stream = 4


3. An exam was conducted in a state over 222 centers. The average number of applicants per centre was found to be 1560. However, it was later realized that in one centre, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants per centre (upto two decimals)?

    A. 1557.87
    B. 1558.20
    C. 1558.92
    D. 1559.51


Answer : Option D

Explanation: Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = \(\frac{110}{220}\) = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51


4. A chaiwala has 2 types of mixture of tea with him. In 56 kg of first mixture ratio of tea to impurity is 5 : 2 and in 44 kg of second mixture the ratio of tea to impurity is 3 : 1. If he mixes these two mixture with 17 kg of pure tea in a large container, then find the ratio of tea to impurity in the large container.

    A. 10 : 3
    B. 3 : 1
    C. 73 : 27
    D. 5 : 3


Answer : Option A

Explanation: In 56 kg of first mixture, Tea = 56 × \(\frac{5}{7}\)
= 40 kg and impurity = 56 – 40 = 16 kg
In 44 kg of second mixture, Tea = 44 × \(\frac{3}{4}\)
= 33 kg and impurity = 44 – 33 = 11 kg

In large container quantity of pure tea = 40 + 33 + 17 = 90 kg

In large container quantity of impurity = 16 + 11 = 27 kg

Required ratio = 90 : 27 = 10 : 3


5. In a 1500 m race, Chaitali beats Vrunali by 100 m and in 1200 m race, Vrunali beats Krutika by 75 m. If Chaitali and Krutika are compared, then for how much m Chaitali will beat Krutika in 900 m race?

    A. 115 m
    B. 112.5 m
    C. 110 m
    D. 120 m


Answer : Option B

Explanation: Chaitali can beat Vrunali by 100 m in 1500 m race.

Hence, when Chaitali covers 1500 m, Vrunali covers 1400 m.

So when Chaitali covers 900 m, Vrunali will cover 840 m.

Similarly, when Vrunali covers 1200 m, Krutika covers 1125 m.

So, when Vrunali covers 840 m, Krutika will cover 787.5 m.

∴ Chaitali will beat Krutika by 112.5 m.

Directions(1-5): Data Sufficiency
1. Who among A, B, C, D and E is the best painter?
Statement I: D is better painter than E, A and C but not as better as B.
Statement II: D is better painter than C but not as better as B who is better painter than E.

    A. The data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question
    B. The data even in both statements I and II together are not sufficient to answer the question
    C. The data either in statement I alone or in statement II alone are sufficient to answer the question
    D. The data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question


Answer : Option D

Explanation: From statement I:
D > E, A and C, B > D
B > D > E, A and C
Hence, B is the best painterFrom statement II:
D > C, B > D, B > E


2. Which of the following will indicate the color of clear sky in a coding system?
Statement I: Indigo’ means ‘Grey’, ‘Grey’ means ‘Black’, Black’ means ‘Blue’ in that system.
Statement II: ‘Black’ means ‘Blue’, ‘Blue’ means ‘Orange’; Orange’ means ‘Green’ in that system.

    A. The data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question
    B. The data even in both statements I and II together are not sufficient to answer the question
    C. The data either in statement I alone or in statement II alone are sufficient to answer the question
    D. The data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question


Answer : Option B

Explanation: ‘Indigo’ means ‘Grey’, ‘Grey’ means ‘Black’, Black’ means ‘Blue’
Here code for blue is not mentioned.
From statement 2:
‘Black’ means ‘Blue’, ‘Blue’ means ‘Orange’, ‘ Orange’ means ‘Green’
Clearly, it is given that ‘blue means orange’.
Hence, the color of the clear sky is orange.


3. How is ‘No’ coded in the code language ?
Statement I: ‘Ne Pa Sic Lo’ means ‘But No None And’ and ‘Pa Lo Le Ne’ means ‘If None And But’.
Statement II: ‘Le Se Ne Sic’ means ‘If No None Will’ and ‘Le Pi Se Be’ means ‘Not None If All’.

    A. I alone is sufficient while II alone is not sufficient
    B. II alone is sufficient while I alone is not sufficient

    C. Either I or II is sufficient
    D. Neither I nor II is sufficient


Answer : Option A

Explanation: In the two statements given in I, the common words are ‘But’, ‘None’, ‘And’ and the

common code words are ‘Ne’, ‘Pa’, ,’Lo’. So, ‘Ne’, ‘Pa’ and ‘Lo’ are codes for ‘But’, ‘None’ and ‘And’. Thus, in the first statement, ‘Sic’ is the code for ‘No’.


4. Who among P, Q, T, V and M is exactly in the middle when they are arranged in ascending order of their heights ?
Statement I: V is taller than Q but shorter than M.
Statement II: T is taller than Q and M but shorter than P.

    A. I alone is sufficient while II alone is not sufficient
    B. II alone is sufficient while I alone is not sufficient

    C. Either I or II is sufficient
    D. Both I and II are sufficient


Answer : Option D

Explanation: From I, we have: M > V > Q.

From II, we have: T > Q, T > M, P > T.

Combining the above two, we have: P>T>M>V>Q i.e. Q

Clearly, M is in the middle.


5. How many visitors saw the exhibition yesterday ?
Statement I: Each entry pass holder can take up to three persons with him/her.
Statement II: In all, 243 passes were sold yesterday.

    A. I alone is sufficient while II alone is not sufficient
    B. II alone is sufficient while I alone is not sufficient
    C. Either I or II is sufficient
    D. Neither I nor II is sufficient


Answer : Option D

Explanation: From I and II, we find that maximum (243 x 3) i.e. 729 visitors saw the exhibition.

But the exact number cannot be determined.

Directions(1-5): Missing Numbers
1. 4, 5, 6, 14, ?, 100.5


    A. 32.5
    B. 47.5
    C. 67.5
    D. 37.5


Answer : Option A

Explanation: 4 * 1 + 1 = 5
5 * 1.5 – 1.5= 6
6 * 2 + 2 = 14
14 * 2.5 – 2.5 = 32.5
32.5 * 3 + 3 = 100.5


2. 5, 6, 8, 30, 136, ?


    A. 645
    B. 680
    C. 650
    D. 690


Answer : Option A

Explanation: 5 * 1 + 1 = 6
6 * 2 – 2 = 10
10 * 3 + 3 = 33
33 * 4 – 4 = 128
128 * 5 + 5 = 645


3. 5, ?, 20, 34, 76, 142


    A. 4
    B. 5
    C. 7
    D. 8


Answer : Option D

Explanation: 5*2 – 2 = 8
8*2 + 4 = 20
20*2 – 6 = 34


4. 2, 9, 39, 161, ?, 2613


    A. 675
    B. 670
    C. 665

    D. 651


Answer : Option D

Explanation: 2 * 4 + 1 = 9
9 * 4 + 3 = 39
39 * 4 + 5 = 161
161 * 4 + 7 = 651
651 * 4 + 9 = 2613


5. 2, 5, 17, 50, 122, ?


    A. 252
    B. 258
    C. 257
    D. 225


Answer: Option C


Explanation: 2 + 1³ + 2 = 5
5 + 2³ + 4 = 17
17 + 3³ + 6 = 50
50 + 4³ + 8 = 122


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