# Quantitative Aptitude

**DIRECTIONS (Qs. 1 - 5) :** What will come in place of the question mark (?) in the following questions ?

**1. 4003 × 77 – 21015 = ? × 116**

(b) 2478

(c) 2467

(d) 2476

(e) None of these

**Ans.d**

4003 × 77 – 21015 = ? × 116

⇒ 308 231 – 21015 = ? × 116 ⇒ 287216 = ? × 116

⇒ ? =

^{2}= ?

(b) 72√7

(c) 134

(d) 70√7

(e) None of these

**Ans.a**[(5√7 + √7) + (4√7 + 8√7)] - (19)

^{2}= ?

⇒ (6√7 × 12√7) - (361) = ?

⇒ 72 × √7 × √7 - 361 = ?

∴ ? = 504 – 361 = 143

(b) 290.4

(c) 295.4

(d) 285.4

(e) None of these

**Ans.b**

(4444 ÷ 40) + (645 ÷ 25) + (3991 ÷ 26) = ?

⇒ ? = (111.1) + (25.8) + (153.5) ⇒ ? = 290.4

(b) 33

(c) 34

(d) 28

(e) None of these

**Ans.e**- (83)

^{2}= (?)

^{2}+ (37)

^{2}

⇒ (?)^{2} = - (83)^{2} - (37)^{2}

⇒ (?)^{2} = 182 × 51 – 6889 – 1369

⇒ (?)^{2} = 9282 – 6889 – 1369

⇒ (?)^{2} = 1024

∴ ? = = 32

^{17}⁄

_{37}× 4

^{51}⁄

_{52}× 11

^{1}⁄

_{7}+ 2

^{3}⁄

_{4}= ?

(b) 305.75

(c) 303

^{3}⁄

_{4}

(d) 305

^{1}⁄

_{4}

(e) None of these

**Ans.b**

5^{17}⁄_{37} × 4^{51}⁄_{52} × 11^{1}⁄_{7} + 2^{3}⁄_{4} = ?

⇒

⇒ 303 + ^{11}⁄_{4} = ?

∴ ? = ^{1223}⁄_{4} = 305.75

**DIRECTIONS (Qs. 6 - 10) :** What approximate value should come in place of the question mark (?) in the following questions? (**Note :** You are not expected to calculate the exact value.)

(b) 140

(c) 180

(d) 100

(e) 280

**Ans.c**

8787 ÷ 343 × 50 = ?

⇒ 25 × 7 = ?

∴ ? = 175 ≈ 180

^{2}

(b) 38

(c) 28

(d) 18

(e) 58

**Ans.b**× (303 ÷ 8) = (?)

^{2}

⇒ 38 × 37.5 (?)^{2}

? =

? = 38

^{5}⁄

_{8}of 4011.33 +

^{7}⁄

_{10}of 3411.22 = ?

(b) 4980

(c) 4890

(d) 4930

(e) 4850

**Ans.c**

^{5}⁄_{8} of 4011.33 + ^{7}⁄_{10} of 3411.22 = ?

⇒ ^{5}⁄_{8} × 4010 + ^{7}⁄_{10} × 3410 ⇒ 2506 + 2387

⇒ 4893 ≈ 4890

(b) 6420

(c) 6320

(d) 6630

(e) 6360

**Ans.e**

23% of 6783 + 57% of 8431 = ?

⇒ ? = 1559 + 4805

∴ ? = 6364 ≈ 6360

(b) 1550

(c) 1420

(d) 1590

(e) 1400

**Ans.a**

335.01 × 244.99 ÷ 55

⇒ ? =

∴ ? = 1492 ≈ 1490

**DIRECTIONS (Qs. 11 - 15) :** In each of these questions a number series is given. In each series only one number is wrong. Find out the wrong number.

(b) 5425

(c) 4621

(d) 5135

(e) 5506

**Ans.a**

The given number series is based on the following pattern:

Hence, the number 5531 is wrong and it should be replaced by 5555.

(b) 26

(c) 69

(d) 37

(e) 9

**Ans.b**

The given number series is based on the following pattern :

Hence, the number 26 is wrong and it should replaced by 21.

(b) 36

(c) 4632

(d) 760

(e) 152

**Ans.d**

The given number series is based on the following pattern:

Hence, the number 760 is wrong and it should be replaced by 770.

(b) 5

(c) 9

(d) 20

(e) 51

**Ans.e**

The given series is

× 0.5 + 0.5, × 1 + 1, × 1.5 + 1.5,

× 2 + 2, × 2.5 + 2.5, × 3 + 3

(b) 2

(c) 6

(d) 157.5

(e) 45

**Ans.a**

The given number series is based on the following pattern :

Hence the number 1 is wrong and it should be replaced by 2.

**DIRECTIONS (Qs. 16 - 20) :** Study the following graph and table carefully and answer the questions given below :

(b) Vehicle C

(c) Vehicle F

(d) Vehicle B

(e) None of these

**Ans.d**

Vehicle B.

(b) 12 km/hr.

(c) 11 km/hr.

(d) 8 km/hr.

(e) None of these

**Ans.c**

Speed of vehicle A on day 1 = 52 km/hr

Speed of vehicle C on day 1 = 63 km/hr

Difference = 63 – 52 = 11 km/hr

(b) 12.8

(c) 11.5

(d) 13.8

(e) None of these

**Ans.e**

Speed of vehicle can day 2 = 45 km/hr

⇒ (45 × ^{5}⁄_{18}) m/sec = 12.5 m/sec

(b) 65

(c) 85

(d) 95

(e) 90

**Ans.e**

Percentage

= × 100

= ^{636}⁄_{703} × 100 ≈ ^{630}⁄_{700} × 100 ≈ 90%

(b) 17 : 13

(c) 13 : 11

(d) 17 : 14

(e) None of these

**Ans.b**

Speed of vehicle D on day 2 = 51

Speed of vehicle E on day 2 = 39

Required ratio = ^{51}⁄_{39} = ^{17}⁄_{13} or 17 : 13

(b) 7

(c) 5

(d) 3

(e) 6

**Ans.a**

Article purchased = ₹ 78350

Marked price = ₹ (78350 × ^{130}⁄_{100}) = ₹ 101855

After Discount Price of Article

= ₹ (101855 × ^{80}⁄_{100}) = ₹ 81484

Profit Percentage =

⇒

(b) 6

(c) 4

(d) 5

(e) None of these

**Ans.e**

9, 15, 27

9 – x, 15– x, 27 – x

⇒ (15 – x)^{2} = (27 – x)(9 – x)

⇒ 225 + x^{2} – 30x = 243 – 9x – 27x + x^{2}

⇒ –30x + 9x + 27x = 243 – 225

⇒ 6x = 18 ⇒ x = 3

(b) ₹ 26.28/-

(c) ₹ 31.41/-

(d) ₹ 23.22/-

(e) ₹ 21.34/-

**Ans.b**

Required difference = P(^{R}⁄_{100})^{2}

= 7300 × (^{6}⁄_{100})^{2} = ₹ 26.28/-

(b) 303.14

(c) 308.73

(d) 306.35

(e) 309.55

**Ans.e**

Let the numbers are x, x + 1, x + 2

sum of three consecutive numbers = 2262

x + x + 1 + x + 2 = 2262

3x + 3 = 2262

3x = 2259

x = 753

Number are 753, 754, 755

∴ 41% of 755 = 309.55

(b) 1440

(c) 5040

(d) 3600

(e) 4800

**Ans.d**

No. of vowels in the word THERAPY = 2 i.e. E and A

In such cases we treat the group of two vowels as one entity or one letter because they are supposed to always come together.Thus, the problem reduces to arranging 6 letters i.e. T, H, R, P,Y and EA in 6 vacant places.

No. of ways 6 letters can be arranged in 6 places = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

But the vowels can be arranged themselves in 2 different ways by interchanging their position. Hence, each of the above 720 arrangements can be written in 2 ways.

∴ Required no. of total arrangements when two vowels are together = 720 × 2 = 1440

Total no. of arrangements of THERAPY = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

No. of arrangement when vowels do not come together = 5040 – 1440 = 3600

**DIRECTIONS (Qs. 26 - 30):** Study the following pie-chart and table carefully and answer the questions given below :

PERCENTAGE WISE DISTRIBUTION OF THE NUMBER OF MOBILE PHONES SOLD BY A SHOPKEEPER DURING SIX MONTHS

Total number of mobile phones sold = 45,000

(b) 116 : 135

(c) 119 : 135

(d) 119 : 130

(e) None of these

**Ans.c**

Number of mobiles sold of company B in July = 3570

Number of mobiles sold of company B in December = 4050

Required Ratio = 3570 : 4050 = 119 : 135

(b) 1635

(c) 1638

(d) 885

(e) None of these

**Ans.c**

Total mobiles sold by company A during November = 2520

Total mobiles sold by this company at discount

= 35% of 2520 = 882

Total mobiles sold by company A without discount

= 2520 – 882 = 1638

(b) ₹ 6,45,900/-

(c) ₹ 6,49,400/-

(d) ₹ 6,49,500/-

(e) None of these

**Ans.d**

Mobile phones sold of company B during October = 1500

Total profit earned on the mobile phones

= (433 × 1500) = ₹ 6,49,500

(b) 140

(c) 150

(d) 105

(e) 130

**Ans.e**

Number of mobile phones sold of company A during July = 4080

Number of mobile phones sold by company A during December = 3150

Required percentage =

(b) 15,000

(c) 10,500

(d) 9,500

(e) None of these

**Ans.a**

Mobile phones sold of company B during August = 5500

Mobile phones sold of company B during September = 4500

Total number of mobile phones = 5500 + 4500 = 10,000

**DIRECTIONS (Qs. 31 - 35) :** Study the following information and answer the questions that follow :

The premises of a bank are to be renovated. The renovation is in terms of flooring. Certain areas are to be floored either with marble or wood. All rooms/halls and pantry are rectangular. The area to be renovated comprises of a hall for customer transaction measuring 23 m by 29 m, branch manager's room measuring 13 m by 17 m, a pantry measuring 14 m by 13 m, a record keeping cum server room measuring 21 m by 13 m and locker area measuring 29 m by 21 m. The total area of the bank is 2000 square meters. The cost of wooden flooring is 170/- per square meter and the cost of marble flooring is 190/- per square meter. The locker area, record keeping cum server room and pantry are to be floored with marble. The branch manager's room and the hall for customer transaction are to be floored with wood. No other area is to be renovated in terms of flooring.

(b) 1887 : 2386

(c) 1887 : 2527

(d) 1829 : 2527

(e) 1887 : 2351

**Ans.c**

Area of customer transaction room = 23m × 29m = 667 sq.m

Area of branch manager room = 13m × 17 m = 221 sq. m

Area of Pantry room = 14m × 13m = 182 sq.m

Area of Server room = 21m × 13m = 273 sq. m

Area of locker room = 29 m × 21 m = 609 sq. m

Total cost of wooden flooring = ₹ [(170 × (667 + 221)]

= ₹ (888 × 170)

Total cost of marble flooring

= ₹ [(190 × (182 + 273 + 609)] = ₹ (190 × 1064)

Required Ratio = 888 × 170 : 1064 × 190 = 1887 : 2527

(b) ₹ 2,16,660/-

(c) ₹ 1,78.790/-

(d) ₹ 2,11,940/-

(e) None of these

**Ans.c**

Area of 4 walls and ceiling of branch managers room

= 2(lh + bh) + lb = 2 [17 × 12 + 13 × 12] + 13 × 17

= 941 sq. m

Total cost of renovat in = ₹ 190 × 941 = ₹ 178790

(b) 4,848/-

(c) 3,689/-

(d) 6,890/-

(e) None of these

**Ans.e**

Total area of bank is 2000 sq. m

Total area of bank to be renovated = 1952 sq. m

Remaining Area = 2000 – 1952 = 48 sq. m

Total cost Remaining Area to be carpeted at the rate of ₹ 110/sq. meter = ₹ (48 × 110) = ₹ 5280

(b) 2.4

(c) 4.2

(d) 4.4

(e) None of these

**Ans.b**

percentage area of bank not to be renovated

⇒ × 100 = 2.4%

(b) ₹ 2,30,206/-

(c) ₹ 2,16,920/-

(d) ₹ 2,42,440/-

(e) None of these

**Ans.a**

Total cost of hall of customer transaction = ₹ (170 × 667) = ₹ 113,390

Total cost of Locker area = ₹ (190 × 609) = ₹ 115710

Total cost of customer transaction hall + locker area

= ₹ (113390 + 115710) = ₹ 229100

(b) ₹ 270/-

(c) ₹ 230/-

(d) ₹ 280/-

(e) None of these

**Ans.a**

Let amount of B = ₹ x

B’s Share without error =

x = ^{3}⁄_{9} × Total Amount .......(1)

B's share after error =

x - 40 = ^{2}⁄_{14} × Total Amount .......(2)

From equation (1) and (2)

3x = 7(x – 40)

3x – 7x = –280

∴ x = 70

Total Amount = 7(70 – 40) = ₹ 210

(b) Either 11 or 12

(c) Either 10 or 11

(d) Either 9 or 11

(e) Either 9 or 10

**Ans.c**

By options

(a) Either 12 or 13

then ice-creams should not be given at least 9. This can be rejected.

(b) Either 11 or 12

Ice-cream should be at least 9. By this combination ice cream gets less than 9.

(c) Either 10 or 11

By giving cookies 10 or 11, we get all the possible condition fulfilled.

(d) and (e), the ice-cream distribution can be more than cookies which violates our condition.

∴ option (c) is the write answer.

(b) ₹ 39/-

(c) ₹ 36/-

(d) ₹ 31/-

(e) None of these

**Ans.c**

₹ [(x for first 5 km) + 13 × remaining kms] = Total pay

₹ x + ₹ 13 × 182 = ₹ 2402

x + 2366 = 2402

x = ₹ 36

(b) 100

(c) 60

(d) 70

(e) 90

**Ans.a**

Let the even consecutive numbers are 2n – 2, 2n, 2n + 2

(2n – 2) × (2n) × (2n + 2) = 4032 .....(1)

Product of 1st even number third even number = 252

Putting this in equation...(1)

252 × 2n = 4032 ⇒ n = 8

Numbers are 14, 16, 18

Five times of 2nd number is = 5 × 16 = 80

(b) 11 years

(c) 25 years

(d) 19 years

(e) 15 years

**Ans.a**

Let the 4 members are x_{1}, x_{2}, x_{3}, daughter

Sum of 4 members five years ago

= x_{1} + x_{2} + x_{3} + daughter = 94

After 5 years,

x_{1} + x_{2} + x_{3} + daughter = 114...(1)

daughter + daughter in law = 92

Daughter = 92 – daughter in law

Put this eqn. ...(1)

x_{1} + x_{2} + x_{3} + 92 – Daughter in law = 114

x_{1} + x_{2} + x_{3} = 22 + Daughter in law

So, the required difference is 22 years.

^{41}⁄

_{190}

(b)

^{21}⁄

_{190}

(c)

^{59}⁄

_{190}

(d)

^{99}⁄

_{190}

(e)

^{77}⁄

_{190}

**Ans.d**

No. of ways of getting 2 white balls = ^{13}C_{2}

No. of ways of getting 2 black balls = ^{7}C_{2}

Probability of getting 2 same colour ball

=

or

⇒

⇒

⇒

(b) 86

(c) 79

(d) 73

(e) None of these

**Ans.b**

Marks is subject B = 56% of 150 = 84

Total marks obtained = 54 % of Total marks

= ^{54}⁄_{100} × [ ∴ Maximum marks in each subject is 150]

= 243

Total marks obtained = A + B + C

243 = 73 + 84 + X

X = 86

(b) 1169.33 sq.mtr

(c) 1181.21 sq.mtr.

(d) 1173.25 sq.mtr

(e) None of these

**Ans.d**

Area of square = 1444 sq. meters

Side of square =

Breadth of Rectangle = ^{1}⁄_{4} × side of square

⇒ ^{1}⁄_{4} × 38 = 9.5m

Length of Rectangle ⇒ 3 × breadth

⇒ 3 × 9.5 ⇒ 28.5 m

Area of Rectangle = 270.75 sq. m

Difference in area = 1444 – 270.75

⇒ 1173.25 sq. mtr

(b) ₹ 46,893/-

(c) ₹ 20.097/-

(d) ₹ 26,796/-

(e) ₹ 13.398/-

**Ans.e**

A and B ratio is 4 : 7

⇒ 4x + 7x = 73689

⇒ 11x = 73689

⇒ x = 6699

Share of A = ₹ 26796

Share of B = ₹ 46893

Difference = twice of share B – thrice of share A

= 2 × 46893 – 3 × 26796 = ₹ 13398

(b) 2 : 7

(c) 3 : 7

(d) 1 : 5

(e) 3 : 5

**Ans.d**

A + B 1 day’s work = ^{1}⁄_{20} .....(1)

B + C 1 day’s work = ^{1}⁄_{30} .....(2)

C + A 1 day’s work = ^{1}⁄_{40} .....(3)

Adding eqn. (1), (2) and (3)

2(A + B + C) = ^{1}⁄_{20} + ^{1}⁄_{30} + ^{1}⁄_{40}

2(A + B + C) =

⇒ A + B + C 1 day work together = ^{13}⁄_{240}

A’ Alone 1 day’s work = (A + B + C) 12 day’s work – (B + C) 1 day’s work

Number of days taken by A = ^{240}⁄_{5} days

C’ Alone 1 day’s work = (A + B + C) 12 day’s work – (A + B) 1’ day’s work

⇒

Number of days taken by C = ^{240}⁄_{1} days

Required Ratio ^{240}⁄_{5} : ^{240}⁄_{1}

⇒ 1 : 5

**DIRECTIONS (Qs. 46 - 50) :** Study the following information and answer the questions that follow :

The table given below represents the respective ratio of the production (in tonnes) of Company A to the production (in tonnes) of Company B, and the respective ratio of the sales(in tonnes) of Company A to the sales (in tonnes) of Company B.

Year | Production | Sales |
---|---|---|

2006 | 5 : 4 | 2 : 3 |

2007 | 8 : 7 | 11 : 12 |

2008 | 3 : 4 | 9 : 14 |

2009 | 11 : 12 | 4 : 5 |

2010 | 14 : 13 | 10 : 9 |

2011 | 13 : 14 | 1 : 1 |

(b) 38

(c) 23

(d) 27

(e) 32

**Ans.d**

Percentage increase =

=

(b) 73

(c) 79

(d) 83

(e) 69

**Ans.b**

Percent of production = ^{400}⁄_{550} \times 100 = 72.72 ≈ 73%

(b) 649

(c) 675

(d) 593

(e) 618

**Ans.c**

Year | Production of B |
---|---|

2006 | 600 |

2007 | 700 |

2008 | 800 |

2009 | 600 |

2010 | 650 |

2011 | 700 |

(b) 64 : 55

(c) 71 : 81

(d) 71 : 55

(e) 81 : 55

**Ans.e**

Total production of company A = 4050

Total sales of company A = 2750

Required ratio &rArrl 4050 : 2750 = 81 : 55

(b) 4 : 5

(c) 3 : 4

(d) 3 : 5

(e) 1 : 4

**Ans.c**

Required ratio = production of B in the year 2006 : Production of B in the year 2008

⇒ 600 : 800 ⇒ 3 : 4