# Quantitative Aptitude

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

1. 4003 × 77 – 21015 = ? × 116
(a) 2477
(b) 2478
(c) 2467
(d) 2476
(e) None of these

#### View Ans & Explanation

Ans.d

4003 × 77 – 21015 = ? × 116

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

⇒ ? = $\frac{287216}{116} = 2476$

2. [(5√7 + √7) + (4√7 + 8√7)] - (19)2 = ?
(a) 143
(b) 72√7
(c) 134
(d) 70√7
(e) None of these

#### View Ans & Explanation

Ans.a

[(5√7 + √7) + (4√7 + 8√7)] - (19)2 = ?

⇒ (6√7 × 12√7) - (361) = ?

⇒ 72 × √7 × √7 - 361 = ?

∴ ? = 504 – 361 = 143

3. (4444 ÷ 40) + (645 ÷ 25) + (3991 ÷ 26) = ?
(a) 280.4
(b) 290.4
(c) 295.4
(d) 285.4
(e) None of these

#### View Ans & Explanation

Ans.b

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

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

4. $\sqrt{33124} \times \sqrt{2601} - \left(83 \right)^{2} = \left(? \right)^{2} + \left(37 \right)^{2}$
(a) 37
(b) 33
(c) 34
(d) 28
(e) None of these

#### View Ans & Explanation

Ans.e

$\sqrt{33124} \times \sqrt{2601}$ - (83)2 = (?)2 + (37)2

⇒ (?)2 = $\sqrt{33124} \times \sqrt{2601}$ - (83)2 - (37)2

⇒ (?)2 = 182 × 51 – 6889 – 1369

⇒ (?)2 = 9282 – 6889 – 1369

⇒ (?)2 = 1024

∴ ? = $\sqrt{1024}$ = 32

5. 51737 × 45152 × 1117 + 234 = ?
(a) 303.75
(b) 305.75
(c) 30334
(d) 30514
(e) None of these

#### View Ans & Explanation

Ans.b

51737 × 45152 × 1117 + 234 = ?

$\left ( \frac{202}{37} \times \frac{259}{52} \times \frac{78}{7} \right ) + \left ( \frac{11}{4} \right ) = ?$

⇒ 303 + 114 = ?

∴ ? = 12234 = 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.)

6. 8787 ÷ 343 × √(50) = ?
(a) 250
(b) 140
(c) 180
(d) 100
(e) 280

#### View Ans & Explanation

Ans.c

8787 ÷ 343 × 50 = ?

⇒ 25 × 7 = ?

∴ ? = 175 ≈ 180

7. $\sqrt[3]{54821}$ × (303 ÷ 8) = (?)2
(a) 48
(b) 38
(c) 28
(d) 18
(e) 58

#### View Ans & Explanation

Ans.b

$\sqrt[3]{54821}$ × (303 ÷ 8) = (?)2

⇒ 38 × 37.5 (?)2

? = $\sqrt{38 \times 38}$

? = 38

8. 58 of 4011.33 + 710 of 3411.22 = ?
(a) 4810
(b) 4980
(c) 4890
(d) 4930
(e) 4850

#### View Ans & Explanation

Ans.c

58 of 4011.33 + 710 of 3411.22 = ?

58 × 4010 + 710 × 3410 ⇒ 2506 + 2387

⇒ 4893 ≈ 4890

9. 23% of 6783 + 57% of 8431 = ?
(a) 6460
(b) 6420
(c) 6320
(d) 6630
(e) 6360

#### View Ans & Explanation

Ans.e

23% of 6783 + 57% of 8431 = ?

⇒ ? = 1559 + 4805

∴ ? = 6364 ≈ 6360

10. 335.01 × 244.99 ÷ 55 = ?
(a) 1490
(b) 1550
(c) 1420
(d) 1590
(e) 1400

#### View Ans & Explanation

Ans.a

335.01 × 244.99 ÷ 55

⇒ ? = $\frac{335 \times 245}{55}$

∴ ? = 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.

11. 5531   5506   5425   5304   5135   4910   4621
(a) 5531
(b) 5425
(c) 4621
(d) 5135
(e) 5506

#### View Ans & Explanation

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.

12. 6   7   9   13   26   37   69
(a) 7
(b) 26
(c) 69
(d) 37
(e) 9

#### View Ans & Explanation

Ans.b

The given number series is based on the following pattern :

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

13. 1   3   10   36   152   760   4632
(a) 3
(b) 36
(c) 4632
(d) 760
(e) 152

#### View Ans & Explanation

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.

14. 7   4   5   9   20   51   160.5
(a) 4
(b) 5
(c) 9
(d) 20
(e) 51

#### View Ans & Explanation

Ans.e

The given series is

× 0.5 + 0.5, × 1 + 1, × 1.5 + 1.5,

× 2 + 2, × 2.5 + 2.5, × 3 + 3

15.  157.5   45   15   6   3   2   1
(a) 1
(b) 2
(c) 6
(d) 157.5
(e) 45

#### View Ans & Explanation

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 :

16. Which of the following vehicles travelled at the same speed on both the days ?
(a) Vehicle A
(b) Vehicle C
(c) Vehicle F
(d) Vehicle B
(e) None of these

#### View Ans & Explanation

Ans.d

Vehicle B.

17. What was the difference between the speed of vehicle A on day 1 and the speed of vehicle C on the same day ?
(a) 7 km/hr.
(b) 12 km/hr.
(c) 11 km/hr.
(d) 8 km/hr.
(e) None of these

#### View Ans & Explanation

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

18. What was the speed of vehicle C on day 2 in terms of meters per second?
(a) 15.3
(b) 12.8
(c) 11.5
(d) 13.8
(e) None of these

#### View Ans & Explanation

Ans.e

Speed of vehicle can day 2 = 45 km/hr

⇒ (45 × 518) m/sec = 12.5 m/sec

19. The distance travelled by vehicle F on day 2 was approximately what percent of the distance travelled by it on day 1 ?
(a) 80
(b) 65
(c) 85
(d) 95
(e) 90

#### View Ans & Explanation

Ans.e

Percentage

= $\frac{Distance \; travelled \; by \; vehicle \; F \; on \; day \; 2}{Distance \; travelled \; by \; vehicle \; F \; on \; day \; 1}$ × 100

= 636703 × 100 ≈ 630700 × 100 ≈ 90%

20. What is the respective ratio between the speeds of vehicle D and vehicle E on day 2 ?
(a) 15 : 13
(b) 17 : 13
(c) 13 : 11
(d) 17 : 14
(e) None of these

#### View Ans & Explanation

Ans.b

Speed of vehicle D on day 2 = 51

Speed of vehicle E on day 2 = 39

Required ratio = 5139 = 1713 or 17 : 13

21. An article was purchased for 78,350/-. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price ?
(a) 4
(b) 7
(c) 5
(d) 3
(e) 6

#### View Ans & Explanation

Ans.a

Article purchased = ₹ 78350

Marked price = ₹ (78350 × 130100) = ₹ 101855

After Discount Price of Article

= ₹ (101855 × 80100) = ₹ 81484

Profit Percentage = $\frac{Profit}{Cost \; Price} \times 100$

$\frac{81484 - 78350}{81484} \times 100 = 3.8 \approx 4\%$

22. When X is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of X ?
(a) 8
(b) 6
(c) 4
(d) 5
(e) None of these

#### View Ans & Explanation

Ans.e

9, 15, 27

9 – x, 15– x, 27 – x

$\frac{15 - x}{9 - x} = \frac{27 - x}{15 - x}$

⇒ (15 – x)2 = (27 – x)(9 – x)

⇒ 225 + x2 – 30x = 243 – 9x – 27x + x2

⇒ –30x + 9x + 27x = 243 – 225

⇒ 6x = 18 ⇒ x = 3

23. What is the difference between the simple and compound interest on ? 7,300/- at the rate of 6 p.c.p.a.in 2 years ?
(a) ₹ 29.37/-
(b) ₹ 26.28/-
(c) ₹ 31.41/-
(d) ₹ 23.22/-
(e) ₹ 21.34/-

#### View Ans & Explanation

Ans.b

Required difference = P(R100)2

= 7300 × (6100)2 = ₹ 26.28/-

24. Sum of three consecutive numbers is 2262. What is 41 % of the highest number ?
(a) 301.51
(b) 303.14
(c) 308.73
(d) 306.35
(e) 309.55

#### View Ans & Explanation

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

25. In how many different ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together ?
(a) 720
(b) 1440
(c) 5040
(d) 3600
(e) 4800

#### View Ans & Explanation

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

26. What is the respective ratio between the number of mobile phones sold of company B during July and those sold during December of the same company ?
(a) 119 : 145
(b) 116 : 135
(c) 119 : 135
(d) 119 : 130
(e) None of these

#### View Ans & Explanation

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

27. If 35% of the mobile phones sold by company A during November were sold at a discount, how many mobile phones of company A during that month were sold without a discount?
(a) 882
(b) 1635
(c) 1638
(d) 885
(e) None of these

#### View Ans & Explanation

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

28. If the shopkeeper earned a profit of 433/- on each mobile phone sold of company B during October, what was his total profit earned on the mobile phones of that company during the same month ?
(a) ₹ 6,49,900/-
(b) ₹ 6,45,900/-
(c) ₹ 6,49,400/-
(d) ₹ 6,49,500/-
(e) None of these

#### View Ans & Explanation

Ans.d

Mobile phones sold of company B during October = 1500

Total profit earned on the mobile phones

= (433 × 1500) = ₹ 6,49,500

29. The number of mobile phones sold of company A during July is approximately what percent of the number of mobile phones sold of company A during December ?
(a) 110
(b) 140
(c) 150
(d) 105
(e) 130

#### View Ans & Explanation

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 = $\frac{4080}{3150} \times 100 = 129.5 \approx 130\%$

30. What is the total number of mobile phones sold of company B during August and September together ?
(a) 10,000
(b) 15,000
(c) 10,500
(d) 9,500
(e) None of these

#### View Ans & Explanation

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.

31. What is the respective ratio of the total cost of wooden flooring to the total cost of marble flooring ?
(a) 1879 : 2527
(b) 1887 : 2386
(c) 1887 : 2527
(d) 1829 : 2527
(e) 1887 : 2351

#### View Ans & Explanation

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

32. If the four walls and ceiling of the branch managers room(The height of the room is 12 meters) are to be painted at the cost of 190/- per square meter, how much will be the total cost of renovation of the branch manager's room including the cost of flooring ?
(a) ₹ 1,36,800/-
(b) ₹ 2,16,660/-
(c) ₹ 1,78.790/-
(d) ₹ 2,11,940/-
(e) None of these

#### View Ans & Explanation

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

33. If the remaining area of the bank is to be carpeted at the rate of 110/- per square meter, how much will be the increment in the total cost of renovation of bank premises ?
(a) 5,820/-
(b) 4,848/-
(c) 3,689/-
(d) 6,890/-
(e) None of these

#### View Ans & Explanation

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

34. What is the percentage area of the bank that is not to be renovated ?
(a) 2.2
(b) 2.4
(c) 4.2
(d) 4.4
(e) None of these

#### View Ans & Explanation

Ans.b

percentage area of bank not to be renovated

$\frac{Area \; bank \; not \; be \; renovated}{Total \; area \; of \; bank} \Rightarrow \frac{48}{2000}$ × 100 = 2.4%

35. What is the total cost of renovation of the hall for customer transaction and the locker area ?
(a) ₹ 2,29,100/-
(b) ₹ 2,30,206/-
(c) ₹ 2,16,920/-
(d) ₹ 2,42,440/-
(e) None of these

#### View Ans & Explanation

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

36. A certain amount was to be distributed among A, B and C in the ratio 2 : 3 : 4 respectively, but was erroneously distributed in the ratio 7 : 2 : 5 respectively. As a result of this, B got ₹ 40 less. What is the amount ?
(a) ₹ 210/-
(b) ₹ 270/-
(c) ₹ 230/-
(d) ₹ 280/-
(e) None of these

#### View Ans & Explanation

Ans.a

Let amount of B = ₹ x

B’s Share without error = $\frac{B's \; \; ratio}{Total \;\; Ratio} \times Total \; \; Amount$

x = 39 × Total Amount .......(1)

B's share after error = $\frac{B's \; \; new \; \; ratio}{Total \;\; new \; \; Ratio} \times Total \; \; Amount$

x - 40 = 214 × Total Amount .......(2)

From equation (1) and (2)

3x = 7(x – 40)

3x – 7x = –280

∴ x = 70

Total Amount = 7(70 – 40) = ₹ 210

37. Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy?
(a) Either 12 or 13
(b) Either 11 or 12
(c) Either 10 or 11
(d) Either 9 or 11
(e) Either 9 or 10

#### View Ans & Explanation

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.

38. The fare of a bus is ₹ X for the first five kilometers and ₹ 13/- per kilometer thereafter. If a passenger pays ₹ 2402/- for a journey of 187 kilometers, what is the value of X?
(a) ₹ 29/-
(b) ₹ 39/-
(c) ₹ 36/-
(d) ₹ 31/-
(e) None of these

#### View Ans & Explanation

Ans.c

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

₹ x + ₹ 13 × 182 = ₹ 2402

x + 2366 = 2402

x = ₹ 36

39. The product of three consecutive even numbers is 4032. The product of the first and the third number is 252. What is five times the second number ?
(a) 80
(b) 100
(c) 60
(d) 70
(e) 90

#### View Ans & Explanation

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

40. The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law, the sum of their ages is 92. Assuming that there has been no other change in the family structure and all the people are alive, what is the difference in the age of the daughter and the daughter-in-law ?
(a) 22 years
(b) 11 years
(c) 25 years
(d) 19 years
(e) 15 years

#### View Ans & Explanation

Ans.a

Let the 4 members are x1, x2, x3, daughter

Sum of 4 members five years ago

= x1 + x2 + x3 + daughter = 94

After 5 years,

x1 + x2 + x3 + daughter = 114...(1)

daughter + daughter in law = 92

Daughter = 92 – daughter in law

Put this eqn. ...(1)

x1 + x2 + x3 + 92 – Daughter in law = 114

x1 + x2 + x3 = 22 + Daughter in law

So, the required difference is 22 years.

41. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?
(a) 41190
(b) 21190
(c) 59190
(d) 99190
(e) 77190

#### View Ans & Explanation

Ans.d

No. of ways of getting 2 white balls = 13C2

No. of ways of getting 2 black balls = 7C2

Probability of getting 2 same colour ball

= $\frac{Probability \; of \; 2 \; whiteballs}{Total \; number \; of \; balls \; drawn}$

or $\frac{Probability \; of \; 2 \; Black \; balls}{Total \; number \; of \; balls \; drawn}$

$\frac{13_{C_{2}} + 7_{C_{2}}}{20_{C_{2}}} \Rightarrow \frac{\frac{13!}{2! \times 11!} + \frac{7!}{2! \times 5!}}{\frac{20!}{18! \times 2!}}$

$\frac{\frac{13 \times 12 \times 11!}{2! \times 11!} + \frac{7 \times 6 \times 5!}{2! \times 5!}}{\frac{20 \times 19 \times 18!}{18! \times 2!}} = \frac{13 \times 12 + 7 \times 6}{20 \times 19}$

$\frac{198}{380} = \frac{99}{190}$

42. Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many marks did he score in subject C ?
(a) 84
(b) 86
(c) 79
(d) 73
(e) None of these

#### View Ans & Explanation

Ans.b

Marks is subject B = 56% of 150 = 84

Total marks obtained = 54 % of Total marks

= 54100 × [ ∴ Maximum marks in each subject is 150]

= 243

Total marks obtained = A + B + C

243 = 73 + 84 + X

X = 86

43. The area of a square is 1444 square meters. The breadth of a rectangle is 1/4th the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle?
(a) 1152.38 sq.mtr.
(b) 1169.33 sq.mtr
(c) 1181.21 sq.mtr.
(d) 1173.25 sq.mtr
(e) None of these

#### View Ans & Explanation

Ans.d

Area of square = 1444 sq. meters

Side of square = $\sqrt{1444} = 38m$

Breadth of Rectangle = 14 × side of square

14 × 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

44. ₹ 73,689/- are divided between A and B in the ratio 4 : 7. What is the difference between thrice the share of A and twice the share of B ?
(a) ₹ 36,699/-
(b) ₹ 46,893/-
(c) ₹ 20.097/-
(d) ₹ 26,796/-
(e) ₹ 13.398/-

#### View Ans & Explanation

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

45. A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone ?
(a) 2 : 5
(b) 2 : 7
(c) 3 : 7
(d) 1 : 5
(e) 3 : 5

#### View Ans & Explanation

Ans.d

A + B 1 day’s work = 120 .....(1)

B + C 1 day’s work = 130 .....(2)

C + A 1 day’s work = 140 .....(3)

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

2(A + B + C) = 120 + 130 + 140

2(A + B + C) = $\frac{6 + 4 + 3}{120}$

⇒ A + B + C 1 day work together = 13240

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

$A = \frac{13}{240} - \frac{1}{30} \Rightarrow \frac{13 - 8}{240} = \frac{5}{240}$

Number of days taken by A = 2405 days

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

$A = \frac{13}{240} - \frac{1}{20} \Rightarrow \frac{13 - 12}{240} = \frac{1}{240}$

Number of days taken by C = 2401 days

Required Ratio 2405 : 2401

⇒ 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
46. What is the approximate percentage increase in the production of Company A (in tonnes) from the year 2009 to the production of Company A (in tonnes) in the year 2010 ?
(a) 18
(b) 38
(c) 23
(d) 27
(e) 32

#### View Ans & Explanation

Ans.d

Percentage increase = $\frac{2010 - 2009}{2009} \times 100$

= $\frac{700 - 550}{550} \times 100 = 27.2 \approx 27\%$

47. The sales of Company A in the year 2009 was approximately what percent of the production of Company A in the same year ?
(a) 65
(b) 73
(c) 79
(d) 83
(e) 69

#### View Ans & Explanation

Ans.b

Percent of production = 400550 \times 100 = 72.72 ≈ 73%

48. What is the average production of Company B (in tonnes)from the year 2006 to the year 2011 ?
(a) 574
(b) 649
(c) 675
(d) 593
(e) 618

#### View Ans & Explanation

Ans.c

Year Production of B
2006 600
2007 700
2008 800
2009 600
2010 650
2011 700
$\frac{600 + 700 + 800 + 600 + 650 + 700}{6} = 675$
49. What is the respective ratio of the total production(in tonnes) of Company A to the total sales (in tonnes) of Company A ?
(a) 81 : 64
(b) 64 : 55
(c) 71 : 81
(d) 71 : 55
(e) 81 : 55

#### View Ans & Explanation

Ans.e

Total production of company A = 4050

Total sales of company A = 2750

Required ratio &rArrl 4050 : 2750 = 81 : 55

50. What is the respective ratio of production of Company B(in tonnes) in the year 2006 to production of Company B (in tonnes) in the year 2008 ?
(a) 2 : 5
(b) 4 : 5
(c) 3 : 4
(d) 3 : 5
(e) 1 : 4

#### View Ans & Explanation

Ans.c

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

⇒ 600 : 800 ⇒ 3 : 4