# Quantitative Aptitude

(b) 52 : 115

(c) 53 : 115

(d) 54 : 115

(e) None of these

**Ans.c**

Ratio of milk in the containers are,

5 × ^{1}⁄_{6} : 4 × ^{3}⁄_{8} : 5 × ^{5}⁄_{12} = ^{5}⁄_{6} : ^{3}⁄_{2} : ^{25}⁄_{12}

and the ratio of water in the containers are,

5 × ^{5}⁄_{6} : 4 × ^{5}⁄_{8} : 5 × ^{7}⁄_{12} = ^{25}⁄_{6} : ^{5}⁄_{2} : ^{35}⁄_{12}

Ratio of mixture of milk and water in the containers

= (^{1}⁄_{6} × 5 + ^{3}⁄_{8} × 4 + ^{5}⁄_{12} × 5) : (^{5}⁄_{6} × 5 + ^{5}⁄_{8} × 4 + ^{7}⁄_{12} × 5)

= 106 : 230 = 53 : 115

^{1}⁄

_{9}%

(b) 11

^{1}⁄

_{11}%

(c) 11

^{9}⁄

_{10}%

(d) 9

^{1}⁄

_{10}%

(e) None of these

**Ans.a**

% change in rate =

For fixed expenditure, % change in consumption

=

=

(b) 5 < L < 5.5

(c) 4.5 < L < 5

(d) 4 < L < 4.5

(e) None of these

**Ans.b**

L × B × 2 = 48

⇒ L × B = 24

Now, 6 – 6 × 10% = 5.4,

5 – 5 × 10% = 4.5 and

Therefore, 5.4 × 4.5 = 24.3

Clearly, 5 < L < 5.5

(b) 4.5%

(c) 5%

(d) 3.46%

(e) None of these

**Ans.d**

Let the original rate be R%. Then,new rate= (2R)%.

∴

⇒ (2175 + 725)R = 33.50 × 100 × 3 = 10050

⇒ R = = 3.46%

(b) ₹ 26.90

(c) ₹ 31.61

(d) ₹ 32.40

(e) None of these

**Ans.a**

For first year,S.I. = C.I.

Now, ₹ 10 is S.I.on ₹ 100.

∴ 16 is S.I. on ₹ (^{100}⁄_{10} × 16) = ₹ 160.

So, S.I. on principal for 1 year at 10% is ₹ 160

∴ Principal = ₹ = ₹ 1600.

Amount for 2 years compounded half yearly

= ₹ = ₹ 1944.81

∴ C.I. = ₹(1944.81 – 1600) = ₹ 24.81.

S.I. = ₹ = ₹ 320.

∴ (C.I.) – (S.I.) = ₹ (344.81 – 320) = ₹ 24.81.

(b) 11%

(c) 12%

(d) 12

^{1}⁄

_{2}%

(e) None of these

**Ans.d**

Let the principal be ₹ P and rate of interest be R% per annum.

Difference of C.I. and S.I. for 3 years

=

=

Difference of C.I. and S.I.for 2 years = P(^{R}⁄_{100})^{2}

∴

⇒ = \frac{25}{8}

⇒ R = ^{100}⁄_{8} = 12^{1}⁄_{2}%

(b) 81

(c) 82

(d) 83

(e) None of these

**Ans.b**

Let x additional men employed.

117 men were supposed to finish the whole work in 46 × 8 = 368 hours.

But 117 men completed ^{4}⁄_{7} of the work in 33 × 8 = 264 hours

∴ 117 men could complete the work in 462 hours.

Now (117 + x) men are supposed to do ^{3}⁄_{7} of the work, working 9 hours a day, in 13 × 9 = 117 hours, so as to finish the work in time.

i.e. (117 + x) men are supposed to complete the whole work in 117 × ^{7}⁄_{3} = 273 hours.

∴ (117 + x) × 273 = 117 × 462

⇒ (117 + x) × 7 = 3 × 462

⇒ x + 117 = 3 × 66 = 198 ⇒ x = 81

∴ Required number of additional men to finish the work in time = 81.

^{2}⁄

_{3}min.

(b) 25 min.

(c) 40

^{2}⁄

_{3}min.

(d) 42

^{2}⁄

_{3}min.

(e) None of these

**Ans.a**

Let A and B together work for x minutes than amount of water filled in the period = x(^{1}⁄_{30} + ^{1}⁄_{40}) = ^{7}⁄_{120}x

Remaining part = 1 – ^{7}⁄_{120}x =

Work done by A in (10 – x) minutes =

= 1 - ^{7}⁄_{120}x

^{7}⁄_{120}x + = 1 or 7x + 40 – 4x = 120

3x = 120 – 40 = 80

x = 26^{2}⁄_{3} min

(b) 7 : 56 am

(c) 8 : 36 am

(d) 8 : 56 am

(e) None of these

**Ans.b**

Let the distance between X and Y be x km. Then, the speed of A is ^{x}⁄_{4} km/h and that of B is ^{2x}⁄_{7} km/h.

Relative speeds of the trains

= (^{x}⁄_{4} + ^{2}⁄_{7}x) = ^{15}⁄_{28}x km/h

Therefore the distance between the trains at 7 a.m. = x - ^{x}⁄_{2} = ^{x}⁄_{2} km

Hence, time taken to cross each other

=

Thus, both of them meet at 7 : 56 a.m.

(b) 3 : 2

(c) 8 : 3

(d) Cannot be determined

(e) None of these

**Ans.c**

Let the man's upstream speed be S_{u} kmph and downstream speed be S_{d} kmph. Then, Distance covered upstream in 8 hrs 48 min.

d = Distance covered downstream in 4 hrs.

⇒ (S_{u} × 8^{4}⁄_{5}) = (S_{d} × 4) ⇒ ^{44}⁄_{5}S_{u} = 4S_{d} ⇒ S_{d} = ^{11}⁄_{5}S_{u}.

∴ Required Ratio :

=

= 8 : 3

**DIRECTIONS (Qs. 11 - 15) :** Study the pie charts given below and answer the following questions.

(b) ₹ 6000

(c) ₹ 3000

(d) ₹ 15,000

(e) None of these

**Ans.c**

Cost of gear box = 20 × = 3000

(b) 6.6%

(c) 6%

(d) 5.4%

(e) None of these

**Ans.c**

Cost of brake = = 6000

∴ Required percentage =

(b) ₹ 2250

(c) ₹ 3750

(d) ₹ 375

(e) None of these

**Ans.a**

Price of tyres = = 3000

Increased price of tyres = 3000 × ^{125}⁄_{100} = 3750

∴ Price should be increased = 3750 – 3000 = ₹ 750

(b) ₹ 4000

(c) ₹ 6000

(d) Cannot be determined

(e) None of these

**Ans.b**

Increased transmission cost = 20,000 × ^{120}⁄_{100} = 24000

∴ increase in transmission cost = 24000 – 20000 = ₹ 4000

Here, this increase will reduce the profit by 4000.

(b) 2%

(c) 3%

(d) Cannot be determined

(e) None of these

**Ans.a**

Price of clutch = 30% of 20,000 = 6,000

∴ Required percentage =

**DIRECTIONS (Qs. 16 - 20) :** Read the following information and answer the questions that follow.

In a huge Jewellery shop, the electric gadgets being used are 17 tubelights of 40 W each, 14 fans of 80 W each, 16 bulbs of 60 W each, 11 bulbs of 100 W each, 11 AC’s of 2100 W each, 9 laptops of 200 W each and 10 TV’s of 120 W each. In a day, tube lights and TV’s are used for 8 h but 60 W bulbs are used for 7 h and 100 W bulbs are used for 9 h whereas laptops and AC's are used for 5 h and 9 h respectively. However, fans are used for 11 h.(Note : 1000 W = 1 unit, 1 month = 30 days).

(b) 576

(c) 67.2

(d) 201.6

(e) None of these

**Ans.d**

Total electric energy consume by 60 W bulb in whole month = 16 × 60 × 7 × 30 W = 201.6 unit

(b) 136.88%

(c) 122.68

(d) 169.62%

(e) None of these

**Ans.b**

Electricity consumed by all fans = 14 × 80 × 11 × 30 W

Electricity consumed by all laptops = 9 × 200 × 5 × 30

Required% =

(b) Tubelights

(c) Laptops

(d) TV’s

(e) None of these

**Ans.a**

Electricity consumed by all fans

= 14 × 80 × 11 × 30 = 369600 W

Electricity consumed by all laptops

= 9 × 200 × 5 × 30 = 270000 W

Electricity consumed by all tubelights

= 17 × 40 × 8 × 30 = 163200 W

Electricity consumed by all TV’s

= 17 × 120 × 8 × 30 = 288000 W

Electricity consumed by 100 W bulb

= 11 × 100 × 9 × 30 = 297000 W

hence fans consumed more electricity.

(b) ₹ 28683

(c) ₹ 78600

(d) ₹ 2900

(e) None of these

**Ans.a**

One unit cost = ₹ 2.70.

Power (used by AC’s) Unit cost = ₹ 3.70

Electricity consumed by all equipment (except AC’s)

= (201.6 + 369.6 + 270 + 163.2 + 288 + 297) unit

= (1292.4 + 297.00) unit = 1589.4 unit

Cost for these unit = 1589.4 × 2.7 = 4291.38

Electricity consumed by AC’s

= 11 × 2100 × 9 × 30 W = 623700 W = 6237 Unit

Cost for it = 6237 × 3.7 = 23076.9

Total cost = 23076.9 + 4291.38 = ₹ 27368

(b) 4 : 5

(c) 3 : 4

(d) 2 : 3

(e) None of these

**Ans.d**

Required ratio = ^{201.6}⁄_{297} = ^{2}⁄_{3}

**DIRECTIONS (Qs.21 – 25):** In each of the following questions two equations are given. You have to solve them and give answer accordingly.

(a) If x > y

(b) If x < y

(c) If x = y

(d) If x ≥ y

(e) If x ≤ y

I. 2x

^{2}+ 5x + 1 = x

^{2}+ 2x – 1

II. 2y

^{2}– 8y + 1 = – 1

**Ans.b**

I. 2x^{2} + 5x + 1 = x^{2} + 2x – 1

x^{2} + 3x + 2 = 0

x^{2} + 2x + x + 2 = 0

x(x + 2) + 1 (x – 2) = 0

(x + 2)(x + 1) = 0

x = –2 , –1

II. 2y^{2} – 8y + 1 = –1

2y^{2} – 8y + 2 = 0

y^{2} – 4y + 1 = 0

= 2 ± √12 = 2 ± 2√3

Hence, y > x

I.

^{x2}⁄

_{2}+ x -

^{1}⁄

_{2}= 1

II. 3y

^{2}– 10y + 8 = y

^{2}+ 2y – 10

**Ans.b**

I. x^{2} + 2x – 1 = 2

x^{2} + 2x – 3 = 0

x + 3x – x – 3 = 0

x (x + 3) – 1 (x + 3) = 0

(x + 3)(x – 1) = 0

x = –3 , 1

II. 2y^{2} – 12y + 18 = 0

y^{2} – 6y + 9 = 0

(y – 3)^{2} = 0

y = 3, 3

Hence, y > x

I. 4x

^{2}– 20x + 19 = 4x – 1

II. 2y

^{2}= 26y + 84

**Ans.b**

I. 4x^{2} – 24x + 20 = 0

x^{2} – 6x + 5 = 0

x^{2} – 5x – x + 5 = 0

x(x – 5) – 1(x – 5) = 0

(x – 5) (x – 1) = 0

x = 5, 1

II. y^{2} – 13y + 42 = 0

y^{2} – 7y – 6y + 42 = 0

y (y – 7) – 6 (y – 7) = 0

(y – 7) (y – 6) = 0

y = 7, 6

Hence, y > x.

I. y

^{2}+ y - 1 = 4 - 2y - y

^{2}

II.

^{x2}⁄

_{2}-

^{3}⁄

_{2}x = x - 3

**Ans.a**

I. 2y^{2} + 3y – 5 = 0

2y^{2} + 5y – 2y – 5 = 0

y (2y + 5) – 1 (2y + 5) = 0

(2y + 5) (y – 1) = 0

y = -^{5}⁄_{2}, 1

II. x^{2} – 3x = 2x – 6

x^{2} – 5x + 6 = 0

x^{2} – 3x – 2x + 6 = 0

x (x – 3) – 2 (x – 3) = 0

(x – 3) (x – 2) = 0

x = 3, 2

Hence, x > y

I. 6x

^{2}+ 13x = 12 – x

II. 1 + 2y

^{2}= 2y +

^{5y}⁄

_{6}

**Ans.e**

I. 6x^{2} + 14x = 12

3x^{2} + 7x – 6 = 0

(x + 3) (3x – 2) = 0

x = –3 , ^{2}⁄_{3}

II. 1 + 2y^{2} = 176y

12y^{2} – 17y + 6 = 0

12y^{2} – 8y – 9y + 6 = 0

4y (3y – 2) – 3 (3y – 2) = 0

(3y – 2) (4y – 3) = 0

y = ^{2}⁄_{3}, ^{3}⁄_{4}

Hence, x ≤ y

**DIRECTIONS (Qs. 26 - 30):** Study the following graph carefully to answer these questions.

(b) ₹ 9000

(c) ₹ 16000

(d) Cannot be determined

(e) None of these

**Ans.b**

Suppose Giridhar invested ₹ x in company A.

∴

or, ^{14}⁄_{100}x + 3250 - ^{13}⁄_{100}x = 3340

or, ^{x}⁄_{100} = 90 or, x = ₹ 9000

(b) ₹ 49000

(c) ₹ 48300

(d) ₹ 49563.50

(e) None of these

**Ans.a**

Amount of dividend received by Anuja in 2011 from company B

= = ₹ 6650

Total amount invested by Anuja in 2012 in Company A

= 35000 + 6650 = ₹ 41650

Reqd amount = 41650 × ^{120}⁄_{100} = ₹ 49980

(b) ₹ 6300

(c) ₹ 6480

(d) ₹ 6840

(e) None of these

**Ans.b**

Total dividend = 18000 × (^{20}⁄_{100} + ^{15}⁄_{100}) = ₹ 6300

(b) 5 : 6

(c) 3 : 4

(d) Cannot be determined

(e) None of these

**Ans.c**

Reqd ratio = = 3 : 4

(b) ₹ 1640 less

(c) ₹ 1860 less

(d) ₹ 1680

(e) None of these

**Ans.d**

From the graph it is obvious that Suraj will get less dividend in 2014 from company A than from B.

Reqd less amount = 3% of 56000 = ₹ 1680.

**DIRECTIONS (Qs. 31 - 35) :** Study the following table to answer the given questions.

(b) 110

(c) 136

(d) 118.25

(e) None of these

**Ans.b**

Production of company AVC in 2012 = 360 crore units

Average production of AVC over the given years = ^{1970}⁄_{6}

Hence, required percent =

(b) 2010

(c) 2014

(d) 2012

(e) None of these

**Ans.c**

Approximate per cent increase or decrease introduction from the previous year for SIO are as follows :

2010 = ^{2}⁄_{85} × 100 = 2.35%

2011 =

2012 =

2013 =

2014 =

**Quicker method :** See the difference of produced units between two consecutive years. The difference is maximum for 2013 to 2014, and production during all these years is almost same.Hence, in the year 2014, SIO registered maximum increase in production over the previous year.

(b) CTU

(c) ZIR

(d) TP

(e) None of these

**Ans.d**

Sum of the productions of companies in the first three years and the last three years is as follows :

Company | First three Years | Last three Years |
---|---|---|

TP | 358 | 349 |

ZIR | 238 | 267 |

AVC | 900 | 1070 |

CTU | 836 | 852 |

PEN | 90 | 127 |

SIO | 261 | 279 |

(b) 104.55

(c) 90.40

(d) 10.62

(e) None of these

**Ans.c**

Total production of the six companies in first two given years = 863 + 927 = 1790

Again, total production of the six companies in last two given years = 989 + 991 = 1980

Therefore, required per cent =

(b) 1,00,00,000

(c) 10,00,000

(d) 40,00,000

(e) None of these

**Ans.b**

The required difference = (91 – 92) crore units = 1 × 10000000 = 10000000 units

(b) 44% decrease

(c) 66% increase

(d) 75% increase

(e) None of these

**Ans.a**

Net effect = x + y + ^{xy}⁄_{100} = -20 + 80 +

= 60 – 16 = 44% increased

What is the area of the plot?

(b) 534

(c) 696.5

(d) 684

(e) None of these

**Ans.d**

(32 - y)^{2} + (24 - x)^{2} = 625 .....(1)

x^{2} + y^{2} = 625 .....(2)

⇒ (24)^{2} + (32)^{2} - 64y - 48x = 0 (From (1) & (2))

⇒ 64y + 48x = 576 + 1024

⇒ 4y + 3x = 36 + 64 = 100

or y =

∴ x^{2} + (From(2))

⇒ -600x + 16x^{2} + 10000 + 9x^{2} = 625 × 16

⇒ 25x^{2} - 600x + 10000 - 625 × 16 = 0

⇒ x = 24 and y = 7

∴ Area = (24 × 25) + ^{1}⁄_{2} × 24 × 7 = 684

(b) 480

(c) 720

(d) 5040

(e) None of these

**Ans.c**

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (120 × 6) = 720.

^{21}⁄

_{46}

(b)

^{25}⁄

_{117}

(c)

^{1}⁄

_{50}

(d)

^{3}⁄

_{25}

(e) None of these

**Ans.a**

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= ^{25}C_{3}

=

n(E) = (^{10}C_{1} × ^{15}C_{2})

=

= 1050.

∴

(b) ₹ 15.70

(b) ₹ 19.70

(d) ₹ 20

(e) None of these

**Ans.c**

Let the amount taxable purchases be Rs. x.

Then, 6% of x = ^{30}⁄_{100}

⇒ x = (^{30}⁄_{100} × ^{100}⁄_{6}) = 5

∴ Cost of tax free items = Rs. [25 - (5 + 0.30)] = ₹ 19.70

**DIRECTIONS (Qs 41 - 45):** In this type of questions, one term in the number series is wrong. Find out the wrong term.

(b) 434

(c) 498

(d) 521

(e) None of these

**Ans.d**

The correct pattern is +6^{3}, +5^{3}, +4^{3}, +3^{3}, .....

So, 521 is wrong and must be replaced by (498 + 3^{3}) i.e.525.

(b) 48

(c) 24

(d) 2

(e) None of these

**Ans.c**

The correct pattern is ÷ 12, ÷ 10, ÷ 8,÷ 6,.....

So,24 is wrong and must be replaced by (48 + 6) i.e. 8.

(b) 61

(c) 122

(d) 509

(e) None of these

**Ans.a**

The terms of the series are (2^{3} – 3), (3^{3} – 3),(4^{3} – 3),(5^{3} – 3), (6^{3} – 3), (7^{3} – 3), (8^{3} – 3).

So, 27 is wrong and must be replaced (3^{3} – 3) i.e. 24.

(b) 20

(c) 40

(d) 26

(e) None of these

**Ans.c**

The given sequence is a combination of two series :

I. 11, 20, 40, 74 and

II. 5, 12, 26, 54

The correct pattern in I is + 9, + 18,+ 36,.....

So, 40 is wrong and must be replaced by (20 + 18) i.e. 38.

(b) 129

(c) 10

(d) 356

(e) None of these

**Ans.d**

The correct pattern is × 2 + 1, × 3 + 1, × 4 + 1, × 3 + 1 + 1,.....

So, 356 is wrong and must be replaced by (129 × 3 + 1)i.e. 388.

**DIRECTIONS (46 - 50):** What approximate value will come in place of question mark (?) in the following questions (You are not expected to calculate the exact value).

(b) 6000

(c) 6300

(d) 5700

(e) 5100

**Ans.d**

Having a glance at the given options one can find out that the two nearest values have a difference of 300.

So round off the numbers to the nearest ten's values.

9228.789 ≈ 9230; 5021.832 ≈ 5020 and 1496.989 ≈ 1500

Now the equation will become

9230 – 5020 + 1500 = ?

? = 5710

But the nearest value is 5700.

(b) 600

(c) 900

(d) 250

(e) 400

**Ans.a**

It can be rounded off to the nearest ten's places.

1002 ≈ 1000; 49 ≈ 50; 99 ≈ 100 and 1299 ≈ 1300

Now the equation will become

1000 ÷ 50 × 100 – 1300 = ?

20 × 100 – 1300 = ?

2000 – 1300 = ?

? = 700

(b) 320

(c) 210

(d) 280

(e) 350

**Ans.d**

The difference between two nearest values is 70 (210 and 280). So round off the numbers to the nearest integers. 29.8% of 260 ≈ 30% of 260; 60.01% of 510 ≈ 60% of 510 and 103.57 ≈ 104

Now the equation will become

30% of 260 + 60% of 510 – 104 = ?

30/100 × 260 + 60/100 × 510 – 104 = ?

78 + 306 – 104 = ?

? = 384 – 104 = 280

^{2}– (25.02)

^{2}+ (13.03)

^{2}= ?

(b) 120

(c) 10

(d) 65

(e) 140

**Ans.a**

(21.98)^{2} ≈ (22)^{2}

(25.02)^{2} ≈ (25)^{2}

and (13.03)^{2} ≈ (13)^{2}

The equation will becomes

22^{2} – 25^{2} + 13^{2} = ?

484 – 625 + 169 = ?

653 – 625 = ?

? = 28 so the nearest value is 25