__Quantitative Aptitude Practice Questions | IBPS 2017__**Dear Aspirants,**

Welcome to

**Mentor for Bank Exams**Quantitative Aptitude Quiz Section. The following covers Number Series (5 Questions), Data Sufficiency (5 Questions), Data Interpretation (5 Questions), Word Problems (5 Questions). All the best for upcoming**IBPS Exams 2017**.**Directions (1 – 5): In the following questions a number series is given. After the series, a number is given followed by (a), (b), (c), (d) and (e). You have to complete the series starting with the number given, following the sequence of the original series and answer the questions that follow the series.**

**1. What will come in place of (c) in**

2.5 |
22 |
170 |
1011 |
4032 |
8049 |

0.2 |
(a) |
(b) |
(c) |
(d) |
(e) |

a) 3

b) 12

c) -3

d) -93

e) None of these

**2. What will come in place of (d) in**

2.5 |
22 |
170 |
1011 |
4032 |
8049 |

0.2 |
(a) |
(b) |
(c) |
(d) |
(e) |

a) -90

b) 110

c) 10

d) 90

e) None of these

**3. What will come in place of (b) in**

5 |
293 |
546 |
766 |
955 |
1115 |

2 |
(a) |
(b) |
(c) |
(d) |
(e) |

a) 533

b) 553

c) 543

d) 540

e) None of these

**4. What will come in place of (c) in**

30 |
330 |
72 |
282 |
126 |
222 |

10 |
(a) |
(b) |
(c) |
(d) |
(e) |

a) 262

b) 362

c) 260

d) 282

e) None of these

**5. What will come in place of (c) in**

7 |
12 |
26 |
50 |
102 |
202 |

11 |
(a) |
(b) |
(c) |
(d) |
(e) |

a) 86

b) 84

c) 90

d) 94

e) 82

**Directions (6 – 10): Study the following information carefully to answer the questions that follow.**

**There are two trains A and B. Both trains have four different types of coaches viz. general coaches, sleeper coaches, first class coaches and AC coaches. In train A, there are total 700 passengers.**

**Train B has 30% more passengers than train A. 20% of the passengers of train A are in general conches. One-fourth of the total number of passengers of train A are in AC coaches. 23% of the passengers of train A are in sleeper class coaches.**

**Remaining passengers of train A are in first class coaches. Total number ofpassengers in AC coaches in both the trains together is 480.**

**30% of the number of passengers of train B is in sleeper class coaches. 10% of the total passengers of train B are in first class coaches.**

**6. If cost of per ticket of first class coach ticket is Rs. 550, what total amount will be generated from first class coaches of train A?**

a) Rs. 1000080

b) Rs. 1232000

c) Rs. 123200

d) Rs. 12320

e) None of the above

**7. Total number of passengers in general class coaches in both the trains together is approximately what percentage of total number of passengers in train A?**

a) 35%

b) 54%

c) 45%

d) 38%

e) 31%

**8. What is the difference between the number of passengers in the AC coaches of train A and total number of passengers in sleeper class coaches and first class coaches together of train B?**

a) 199

b) 178

c) 187

d) 179

e) None of these

**9. What is sum of the total number of passengers in the general coaches of train A and the AC coaches of train B together?**

a) 449

b) 459

c) 435

d) 445

e) None of these

**10. What is the ratio of the number of passengers in sleeper class coaches of train B to the number of passengers in first class coaches of train A?**

a) 13 : 7

b) 7 : 13

c) 39 : 32

d) Data inadequate

e) None of the above

**Directions (11 – 15): Given below is a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements is sufficient to answer the question. You should use the given data and your knowledge of Mathematics to choose between the possible answers. Give Answer**

a) The data in statement I alone is sufficient to answer the question,
while the data in statement II alone is not sufficient to answer the question.

b) The data in statement II alone is sufficient to answer the question,
while the data in statement I alone is not sufficient to answer the question.

c) The data in statement I alone or in statement II alone is sufficient
to answer the question.

d) The data in both the statements I and II is not sufficient to answer
the question.

e) The data in both the statements I and II together is necessary to
answer the question.

**11. What day of week is today?**

**I.**

**Today is last day of month.**

**II.**

**There were five Sundays, Mondays and Tuesdays in this month.**

**12. When a dice is thrown, the outcome is an odd number Y. Which number is this?**

**I.**

**The probability of getting Y or a multiple of 3 in a throw of dice is ½.**

**II.**

**The probability of getting Y or a multiple of 3 in a throw of dice is 1/3.**

**13. A five digit number is taken. What will be the sum of its digits?**

**I.**

**The number lies between 12000 and 13000.**

**II.**

**There are exactly two unique digits in number.**

**14. How many zeroes will be there at the end of natural number P?**

**I.**

**The number of ways in which 12 persons can sit on 12 chairs is P.**

**II.**

**P is a multiple of both 25 and 32.**

**15. What is the compound interest obtained when Rs. 12000 are invested in compound interest?**

**I.**

**The rate of interest is 10% per annum.**

**II.**

**If same amount is invested in simple interest for 3 years, the interest obtained is Rs. 3600.**

**16. Blue and yellow colours were mixed in a plate A in ratio 7:4 and in other plate B it was mixed in ratio of 3:5. In what ratio should colours from these plates be mixed so as to form a mixture in ratio of 4:3.**

a) 91:70

b) 70:91

c) 40:121

d) 121:40

e) None of these

**17. Dane has some money in the form of 5 rupee and 10 rupee coins. The total number of coins he has is 54. However, while counting the money, he counted three of his 5 rupee coins as 10 rupee coins by mistake. He calculated that he has Rs. 355. How many coins of Rs. 5 do he has?**

a) 30

b) 36

c) 40

d) 45

e) 50

**18. Jack is warming up for his football match. He increases the intensity of warm up successively in three rounds. For first round of 10 minutes, he runs at a speed T km/hr. For next 20 minutes round, he doubles his speed. For final round of 40 minutes, he doubles his second round speed. Find the average speed of Jack in three rounds.**

a) 2T km/hr

b) 2.4T km/hr

c) 3T km/hr

d) 2.7 km/hr

e) 3.5 km/hr

**19. In a school, the student to teacher ratio is 40 : 1. However, as per rules, the ratio should be at most 25 : 1. If the minimum number of more teachers required to achieve desired ratio is 30, then how many students are there in the school?**

a) 1200

b) 1500

c) 1800

d) 2000

e) 2400

**20. After every 6 days of work, Allan takes a day off from work on Sunday. On this day, Paul works instead of Allan. The efficiency of Allan is one fifth of that of Paul. Due to this, a work that started on Monday gets finished in 41 days. In how many Allan alone would have finished this work?**

a) 50

b) 51

c) 56

d) 61

e) 66

**Solutions:**

**1. D)**The pattern of given series is:

→ 2.5,

→ 22 = (2.5 × 10) -3,

→ 170 = (22 × 8) -6,

→ 1011 = (170 × 6) -9,

→ 4032 = (1011 × 4) -12,

→ 8049 = (4032 × 2) -15,

Based on the above pattern we can create the
following series

a : -1 = (0.2 × 10) -3,

b : -14 = (-1 × 8) -6,

c : -93= (-14 × 6) -9,

Hence, (c) = -93

**2. B)**The pattern of given series is:

→ 30,

→ 5 = 30 × 0.5 - 10,

→ 40 = 30 × 1 + 10,

→ 35 = 30 × 1.5 - 10,

→ 70 = 30 × 2 + 10,

→ 65 = 30 × 2.5 - 10,

Based on the above pattern we can create the
following series

a : 15 = 50 × 0.5 - 10,

b : 60 = 50 × 1 + 10,

c : 65 = 50 × 1.5 - 10,

d : -110 = 50 × 2 + 10,

Hence, (d) = 110

**3. C)**The pattern of given series is:

→ 5,

→ 293 = 5 + (17)

^{2}- 1,
→ 546 = 293 + (16)

^{2}- 3,
→ 766 = 546 + (15)

^{2}- 5,
→ 955= 766 + (14)

^{2}- 7,
→ 1115 = 955 + (13)

^{2}- 9,
Based on the above pattern we can create the
following series

a : 290 = 2 + (17)

^{2}- 1,
b : 543 = 290 + (16)

^{2}- 3,
Hence, (b) = 543

**4. A)**The pattern of given series is:

→ 30,

→ 330 = 30 + 300,

→ 72 = 330 - 258, (300 – 42 = 258)

→ 282 = 72 + 210, (258 – 48 = 210), (48 -42 = 6)

→ 126= 282 – 156, (210 – 54 = 156), (54 – 48 = 6)

→ 222 = 126 +96, (156 – 60 = 96), (60 – 54 = 6)

Based on the above pattern we can create the
following series

a : 310 = 10 + 300,

b : 52 = 310 - 258, (300 – 42 = 258)

c : 262 = 52 + 210, (258 – 48 = 210), (48 -42 = 6)

Hence, (c) = 262

**5. E)**The pattern of the given number series is as follows:

⇒ 7 × 2 – 2 = 12,

⇒ 12 × 2 + 2 = 26,

⇒ 26 × 2 – 2 = 50,

⇒ 50 × 2 + 2 = 102,

⇒ 102 × 2 – 2 = 202,

Now, the required number series starting from 11 on
the same pattern,

⇒ 11

⇒ A = 11 × 2 – 2 = 20,

⇒ B = 20 × 2 + 2 = 42,

⇒ C = 42 × 2 – 2 = 82,

⇒ D = 82 × 2 + 2 = 166,

⇒ E = 166 × 2 – 2 = 330

∴ the number in place of (c) is 82.

**6. C)**Total number of passengers in train A = 700

According to the information given in the question,
20% of the passengers of train A are in general conches, one-fourth (i.e. 25%)
of the total number of passengers are in AC coaches and 23% of the passengers
of train A are in sleeper class coaches.

∴ Number of passengers in first class coaches
of train A = [100 – (20 + 25 + 23)] % of 700 = 32% of 700 = 224

∴ Amount generated = 224 × 550 = Rs.
123200

**7. B)**Number of passengers in train A = 700

∵ Train B has 30% more passengers than train A,

Number of passenger in train B = 700 + 30% of 700 =
910

No. of passengers of train A in general coaches =
20% of 700 = 140

According to the information given in the problem,
one-fourth (i.e. 25%) of the total number of passengers in train A are in AC
coaches and total number ofpassengers in AC coaches in both the trains together
is 480.

∴ Number of passengers in the AC coaches of
train B = 480 – (25% of
700) = 480 – 175 = 305

∴ Number of passengers in general class coaches
in train B

= Total passengers – (passengers in sleeper +
passengers in 1

^{st}class + passengers in AC)
= 910 – [(30 % 0f 910) + (10% of 910) + 305] = 910 –
669 = 241

∴ Total number of passengers in general class
coaches in both the trains together = 140 + 241 = 381

∴ Required percentage = 381/700 ×
100 = 54.42% ≈ 54%

**8. E)**Number of passengers in train A = 700

∵ Train B has 30% more passengers than train A,

Number of passenger in train B = 700 + 30% of 700 =
910

According to the information given in the question,
one-fourth of the total number of passengers in train A are in AC coaches.

∴ Number of passengers in the AC coaches of
train A = 25% of 700 = 175

Also, Total number of passengers in sleeper class
coaches and first class coaches together of train B = 30% of 910 + 10% of 910 =
273 + 91 = 364

∴ difference between the two quantities = 364 – 175 = 189

**9. D)**Number of passengers in train A = 700

∵ Train B has 30% more passengers than train A,

Number of passenger in train B = 700 + 30% of 700 =
910

∵ In train A, 20% of the passengers are in
general conches,

Number of passengers in the general coaches of train
A = 20% of 700 = 140

According to the information given in the problem,
one-fourth (i.e. 25%) of the total number of passengers in train A are in AC
coaches and total number ofpassengers in AC coaches in both the trains together
is 480.

∴ Number of passengers in the AC coaches of
train B = 480 – (25% of
700) = 480 – 175 = 305

∴ Total number of passengers in the general
coaches of train A and AC coaches of train B together = 140 + 305 =445

**10. C)**Number of passengers in train A = 700

∵ Train B has 30% more passengers than train A,

Number of passenger in train B = 700 + (30% of 700)
= 910

∵30% of the number of passengers of train B is
in sleeper class coaches,

Number of passengers in sleeper class coaches of
train B = 30% of 910 = 273

According to the information given in the question,
in train A, 20% of the passengers are in general conches, one-fourth (i.e. 25%)
of the total number of passengers are in AC coaches and 23% of the passengers
of train A are in sleeper class coaches

∴ Number of passengers in first class coaches
of train A = [100 – (20 + 25 + 23)]% of 700 = 32% 0f 700 = 224

∴ required ratio = 273:224 = 39:32

**11. E)**

**From statement I:**

Today is last day of month. Last day of month can be
any day of week.

∴ Statement I alone is not sufficient to answer
the question.

**From statement II:**

There were five Sundays, Mondays and Tuesdays in
this month.

⇒ Remaining days will occur for four times.
This is only possible if there are 31 days in month.

⇒ Total days in month = 4 × 4 + 3 × 5 = 31

**From Statement I and II:**

Now, suppose T is first day of month. So, 29

^{th}day will be T as well. T, day next to T, and day next to next of T will occur five times. This is only possible if T is a Sunday.
⇒ Last day of month will be next to next of
Sunday, which is Tuesday.

∴ The data in both the statements I and II
together is necessary to answer the question.

**12. B)**

**From statement I:**

The probability of getting Y or a multiple of 3 in a
throw of dice is ½.

⇒ If Y is 1, probability of getting Y or a
multiple of 3 in a throw of dice = (1/6) + (2/6) = ½

If Y is 3, probability of getting Y or a multiple of
3 in a throw of dice = (2/6) = 1/3

If Y is 5, probability of getting Y or a multiple of
3 in a throw of dice = (1/6) + (2/6) = ½

So, Y can be 1 or 5.

∴ Statement I alone is not sufficient to answer
the question.

**From statement II:**

The probability of getting Y or a multiple of 5 in a
throw of dice is 1/3.

⇒ If Y is 1, probability of getting Y or a
multiple of 3 in a throw of dice = (1/6) + (2/6) = ½

If Y is 3, probability of getting Y or a multiple of
3 in a throw of dice = (2/6) = 1/3

If Y is 5, probability of getting Y or a multiple of
3 in a throw of dice = (1/6) + (2/6) = ½

So, Y must be 3.

∴ Statement II alone is sufficient to answer
the question.

**13. D)**

**From statement I:**

The number lies between 12000 and 13000.

Between 12000 and 13000, number can be 12111 or
12222 or 12232, and so on, and each of them has different sum of digits.

Sum of digits cannot be uniquely found.

∴ Statement I alone is sufficient to answer the
question.

**From statement II:**

There are exactly two unique digits in number.

Number can be 14111 or 32222 or 13232, and so on,
and each of them has different sum of digits.

Sum of digits cannot be uniquely found.

∴ Statement II alone is not sufficient to
answer the question.

**From statements I and II together:**

The number lies between 12000 and 13000. There are
exactly two unique digits in number.

Number can be 12111 or 12112 or 12222, and so on,
and each of them has different sum of digits.

Sum of digits cannot be uniquely found.

∴ Even using both the statements together, we
cannot answer the given question.

**14. A)**

**From statement I:**

The number of ways in which 12 persons can sit on 12
chairs is P.

⇒ P = 12!

Value of P can be found and hence number of zeroes
at end of P.

∴ Statement I alone is sufficient to answer the
question.

**From statement II:**

P is a multiple of both 25 and 32.

⇒ P is a multiple of LCM of 25 and 32, i.e.,
800.

Now, P can be 800 or 8000 or 80000, and so on.
Number of zeroes at end of P cannot be uniquely found.

∴ Statement II alone is not sufficient to
answer the question.

**15. D)**

**From statement I:**

The rate of interest and principal amount is known,
but period of investment is not known, and hence interest earned cannot be
found.

∴ Statement I alone is not sufficient to answer
the question.

**From statement II:**

If same amount is invested in simple interest for 3
years, the interest obtained is Rs. 3600.

We know, Simple interest = [Principal × rate × Time
]/100

⇒ 3600 = (12000 × rate × 3)/100

⇒ rate = 10 % per annum

We see that it provides same information as first
statement, and hence for the same reason, interest earned cannot be found.

∴ Statement II alone is not sufficient to
answer the question.

**From statements I and II together:**

We have found that both statements I and II provide
equivalent information. If answer cannot be found using any of them, it cannot
be found using both of them.

∴ Even using both the statements together, we
cannot answer the given question.

**16. D)**Let colors from these plates be mixed in ratio of x:y

∴ Blue color in resultant mixture due to plate
A = 7x/11

Blue color in resultant mixture due to plate b =
3y/8

Yellow color in resultant mixture due to plate A =
4x/11

Yellow color in resultant mixture due to plate B =
5y/8

Total quantity of mixture = (x + y)

As per requirement we require Blue and Yellow color
in ratio of 4:3

∴Total Blue color = 4(x + y)/7

∴ Blue color due to plate A + Blue color due to
plate B = Total blue color

⇒ 7x/11 + 3y/8 = 4(x + y)/7

⇒ (56x + 33y) / 88 = (4x + 4y)/7

⇒ 392x + 231y = 352x + 352y

⇒ 392x – 352x = 352y – 231y

⇒ 40x = 121y

⇒ x/y = 121/40

⇒ x:y = 121:40

**17. C)**Suppose Dane has M coins of 5 rupees and N coins of 10 rupees.

Total number of coins = 54

⇒ M + N = 54 ------- (i)

While counting, Dane counted three of his 5 rupee coins
as 10 rupee coins.

⇒ He counted (M-3) coins of 5 rupees and (N+3)
coins of 10 rupees.

Dane calculated that he has Rs. 355

⇒ 5 × (M – 3) + 10 × (N + 3) =
355

⇒ 5M – 15 + 10N + 30 = 355

⇒ 5M + 10N = 340

⇒ M + 2N = 68 ------- (ii)

Multiply (i) by 2, and subtract (ii) from it.

2(M + N) – (M + 2N) = 2 × 54 – 68

M = 40

∴ Number of 5 rupees coins are 40.

**18. C)**As per given information, there are three phases in which Jack runs.

For phase I), Speed = T km/hr,

Time taken = 10 minutes = (10/60) hours = 1/6 hours

Distance covered = T km/hr × 1/6 hr = (T/6) km

For phase II), in this phase speed is doubled, Speed
= 2T km/hr,

Time taken = 20 minutes = (20/60) hours = 1/3 hours

Distance covered = 2T km/hr × 1/3 hr = (2T/3) km

For phase III), in this phase speed is double of
speed in second round, Speed = 2 × 2T km/hr = 4T km/hr

Time taken = 40 minutes = (40/60) hours = 2/3 hours

Distance covered = 4T km/hr × 2/3 hr = (8T/3) km

From this,

Total distance covered = (T/6) km + (2T/3) km +
(8T/3) km = (7T/2) km

Total time taken = (1/6) hr + (1/3) hr + (2/3) hr =
(7/6) hr

Now, we know that, Average Speed = (Total Distance
Covered/Total Time Taken)

Average speed = (7T/2)km/(7/6)hr=3T km/hr

∴ Average speed for entire warm up = 3T km/hr

**19. D)**Suppose there are 40T students and T teachers.

As per given condition, if 30 more teachers are
added, ratio of students to teachers becomes 25 : 1.

⇒ 40T/(T+ 30) = 25/1

⇒ 40T = 25T + 750

⇒ T = 750/15 = 50

∴ Number of students in school = 40T = 40 × 50 = 2000

**20. B)**Let the hundred’s place digit, ten’s place digit and unit digit of the correct number are ‘a’, ‘b’ and ‘c’ respectively, so the number mistakenly written as ‘cba’ in the place of ‘abc’

We know that, formula for average-

A

_{E}= S_{E}/n_{E}or S_{E}= A_{E}× n_{E}
Where,

S

_{E}= sum of entities,
n

_{E}= number of entities,
A

_{E }= Average of entities.
Now, according to the question,

Average increased by 24.75, hence sum increased by
24.75 × 8 = 198.

The reserved no. is cba. Now,

⇒ cba – abc = 198

cba can be written as 100 c + 10b + a and

abc can be written as 100a + 10b + c, so

The difference = 100 c + 10b + a – 100 a – 10b - c =
198

⇒ 99 c – 99 a = 198

⇒ (c – a) 99 = 198

Therefore c – a = 198/99 = 2

Hence, the required difference is 2.