Bits Pilani Goa Test - 6 Paper Expl Answers

BITS PILANI Test 6 – Answer Key & Explanatory Answers 1.A clock gains 15 minutes per day. If it is set right at 12 noon, the time it shows at 4 AM is 1 ) 4.20 AM 2 ) 4.30 AM 3 ) 4.02 AM 4 ) 4.10 AM 2.A home theatre set is Rs. 4950. If two successive discounts of 20% and 15% are given, then its selling price is 1 ) Rs.3366 2 ) Rs.6633 3 ) Rs.3636 4 ) Rs.6363 3.If 15% of x is three times of 10% of y, then x:y = 1 ) 1:2 2 ) 2:1

3 ) 3:2

4 ) 2:3

4.A bookseller bought 500 text books for ₹ 20,000. He wanted to sell them at a profit so that he get 50 books free. At what profit percent should he sell them? 1 ) 10 2 ) 20 3 ) 15 4 ) 10.5 5.20% of a man's salary is paid as rent, 60% are his living expenses and 10% are his savings. If he spends remaining ₹ 30 on the education of his children, find his salary 1 ) 300 2 ) 900 3 ) 3000 4 ) 9000 6.A gun is fired at a distance of 6.64 km away from Ram. He hears the sound 20 seconds later. Then the speed of sound is 1 ) 664 m/s 2 ) 664 km/s 3 ) 332 m/s 4 ) 332 km/s 7.The simple interest on Rs. 2000 for 2 years at Rs. 75 per thousand per annum will be 1 ) Rs.150 2 ) Rs.300 3 ) Rs.600 4 ) Rs.400 8.If x = 1 + √2 + √3, then the value of x2 - 2x - 4 is 1 ) √6 2 ) 2√3

3 ) 3√2

9.If ΔABC is an equilateral triangle of side 16 cm, then the length of altitude is 1 ) 2√3 cm 2 ) 4√3 cm 3 ) 8√3 cm

4 ) 2√6 4 ) 5√3 cm

10.O is the circumcentre of ΔABC. If AO = 8 cm, then the length of BO is 1 ) 12 cm 2 ) 3 cm 3 ) 6 cm 11. Find the range of all real values of x that satisfy the given inequality: |4x + 3| < 6|x| – 5 (1) (-∞, -4/5) U (1/5, ∞) (2) (-∞, -1) U (4, ∞) (4) (-∞, -4) U (1, ∞) (5) None of these

4 ) 8 cm

(3) (-∞, -1) U (1/5, ∞)

Solution: Given: |4x + 3| < 6|x| – 5 i) When |4x + 3| = 4x + 3 and 6|x| – 5 = 6x – 5. → 4x + 3 < 6x – 5 → 8 < 2x →x > 4. ii) When |4x + 3| = –4x – 3 and 6|x| – 5 = 6x – 5. → –4x – 3 < 6x – 5 → 2 < 10x →x > 1/5. iii) When |4x + 3| = 4x + 3 and 6|x| – 5 = –6x – 5. → 4x + 3 < –6x – 5 → 10x < –8 →x < –4/5. iv) When |4x + 3| = –4x – 3 and 6|x| – 5 = –6x – 5. → –4x – 3 < –6x – 5 → 2x < –2 →x < –1. Thus, x € (-∞, -4/5) U (1/5, ∞) Hence, the correct answer is option 1. 12. A Buy-Zone dealer sells laptops of three different brands – Sony, Dell and Acer. He has 30 laptops in all with him which cost him Rs 12.6 lacs. The cost price of one laptop of each brand is as follows: i) Sony – Rs 48000 (ii) Dell – Rs 42000 (iii) Acer – Rs 35000. Brand cost is defined as the total cost of all the laptops of that particular brand. If it is known that the dealer has atleast 5 laptops of each brand, then what is the maximum difference (in lacs of Rs) of the brand costs between 2 companies? (1) 3.78 (2) 4.20 (3) 5.04 (4) 8.40 (5) None of these © 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Solution: Let the number of laptops of Sony, Dell and Acer be denoted by s, d and a respectively. Then, by given data; s, d, a ≥ 5. Total number of laptops = s + d + a = 30 (1) Total cost of all the laptops = 1260000 = 48000s + 42000d + 35000a → 48s + 42d + 35a = 1260 (2) (1) × 42 – (2) gives 6s = 7a (3) (2) – (1) × 35 gives 13s + 7d = 210 (4) The set of integral values for s and d which satisfy the above equation are: i) (s, d) = (7, 17), using this in (3), a = 6. ii) (s, d) = (14, 4), using this in (3), a = 12. The set (ii) is not valid as d = 4 < 5. Hence, s = 7, d = 17, a = 6. Thus, brand cost of: Sony = 7 × 48000 = Rs 336000. Dell = 17 × 42000 = Rs 714000. Sony = 6 × 35000 = Rs 210000. Hence, the difference in the maximum and minimum brand costs = 714000 – 210000 = Rs 504000 = Rs 5.04 lacs. Hence, the correct answer is option 3. 2

13. In the given equation: 3x + 5bx + 48 = 0, if b is an integer, such that one root of the equation is the cube root of the other root of the equation, what is the value of ‘b’? (1) 2 (2) -2 (3) 6 (4) 6 or -6 (5) 2 or -2 Solution: 2 Given equation: 3x + 5bx + 48 = 0 3 Let the two roots of the equation be denoted by p and p respectively (Since one root is the cube of the other root). 3 4 Product of the roots = 48/3 = 16 = p × p = p . Sum of the roots = -5b/3. 2 From above, p = √16 = ±4 2 2 Consider p = 4, then p = ±2; if p = -4, then p has no real roots. 3 3 i) If p = 2, then p = 8 and p + p = 10. 3 → p + p = 10 = -5b/3 → b = -6. 3 3 ii) If p = -2, then p = -8 and p + p = -10. 3 → p + p = -10 = -5b/3 → b = 6. Hence, b can take two values of ±6. Hence, the correct answer is option 4. 2

3

14. If the terms of the series T are defined as follows: 3, 5/x, 7/x , 9/x … ∞; where 0 < x ≤ 1, the sum of the terms of this series could be equal to which of the following? 2 2 2 (1) [x(5x – 3)/(x – 1) ] (2) [x(3x – 1)/(x – 1) ] (3) [(3x – 1)/ x(x – 1) ] 2 (4) [(5x – 3)/ x(x – 1) ] (5) None of these Solution: Let the sum of the terms of the given series: 2 3 T = 3 + (5/x) + (7/x ) + (9/x ) + … + ∞. 2 3 T/x = (3/x) + (5/x ) + (7/x ) + … + ∞. 2 3 Therefore, T – (T/x) = 3 + (2/x) + (2/x ) + (2/x ) + … → (T)[1 – (1/x)] – 3 = (2)/[1 – (1/x)] = 2x/ (x – 1) → (T)[(x – 1)/ x] = 3 + [2x/ (x – 1)] = (5x – 3)/ (x – 1) 2 → T = [x(5x – 3)/(x – 1) ] Hence, the correct answer is option 1. 2x

15. If f(x) = [(e + e (1) R

-2x

2x

-2x

)/(e – e ), then what is the domain of f(x)? (2) R – {0, 1/2} (3) R – {0}

2

(4) R – {e }

(5) None of these

© 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Solution: 2x -2x 2x -2x Given: f(x) = [(e + e )/(e – e )] Consider the numerator of f(x): 2x -2x 2x 2x 4x 2x (e + e ) = e + (1/e ) = (e + 1)/ e . 4x 4x Here, the numerator = (e + 1) cannot be equal to zero for real values of x, since e = -1 has no real roots. Since the numerator of f(x) cannot be equal to zero, f(x) is never equal to zero. Thus, the domain of f(x) would be all real values except zero. Hence, the domain of f(x) = R – {0}, where R denotes the set of real numbers. Hence, the correct answer is option 3. 16. In a circle with centre O, AB is the diameter, of length 12 cms, and P is a point on the circumference such that BP is 2 equal to the radius of the circle. What is the area (in cm ) of the sector bound by lines OP, OB and the minor arc BP? (1) 12π (2) 6√3π (3) 3π (4) 6π (5) 3√3π Solution: The given data can be represented as shown in the following diagram. Since AB = diameter = 12 cms, radius OA = OB = OP = 12/2 = 6 cms. PB = radius (given) = 6 cms. Method 1: Triangle APB is right angled at P since it has the diameter AB as one of its sides. 2 2 2 2 2 2 Thus, AP + PB = AB → AP + 6 = 12 2 2 → AP = √(12 – 6 ) = 6√3 cms. Tan (Angle PAB) = PB/AP = 6/6√3 = 1/√3 -1 → Angle PAB = tan (1/√3) = 30°. Angle POB = 2 × Angle PAB = 60° (Since angle subtended by a chord at the centre is twice the angle subtended at the circumference). Method 2: Alternately, in triangle OPB, OP = OB = PB = 6 cms, hence it is an equilateral triangle, thus Angle POB = 60°. 2

Area of the sector (A) bound by the lines OP and OB and the minor arc PB is given by: A = (π) × (OP) × Angle POB/360° 2 2 → A = (π) × (6) × 60°/360° = 6π cm . Hence, the correct answer is option 4. 2

17. A triangle XYZ has side lengths 9, 12 and 15 cms. Find the area (in cm ) of the triangle formed by joining the in-centre, centroid and the vertex containing the largest angle in the triangle XYZ? (1) 3.0 (2) 1.5 (3) 6.0 (4) 2.25 (5) 9.0 Solution: The sides of the given triangle are 9, 12 and 15 cms. 2 2 2 It can be observed that 9 + 12 = 81 + 144 = 225 = 15 , hence it is a right angled triangle. Let the co-ordinates of the triangle be A(0, 0), B(0, 9) and C(12, 0) Centroid D = (9/3, 12/3) = (3, 4). In-centre I = {[(9 × 12)/(9 + 12 + 15)], [(9 × 12)/(9 + 12 + 15)]} = (3, 3). The vertex with the right angle is A. By joining the vertex A, centroid D and in-centre I, the triangle ADI is obtained with the following sides: 2 2 a = Side AI = √(3 + 3 ) = 3√2 cms. 2 2 b = Side ID = √(0 + 1 ) = 1 cm. 2 2 c = Side AD = √(3 + 4 ) = 5 cms. s = (a + b + c)/2 = 3 + (1.5)√2 Area of the triangle ADI = A = √[(s)(s – a)(s – b)(s – c)]. → A = √{[3 + (1.5)√2][3 – (1.5)√2][(1.5)√2 + 2][(1.5)√2 – 2)]} → A = √{[9 – 4.5][4.5 – 4]} → A = √{[4.5][0.5]} = √(2.25) 2 → A = 1.5 cm . Hence, the correct answer is option 2. 18. The FOREX value (Number of INR for 1 USD) varies as follows during a certain period: st nd rd In the 1 quarter of a year, it increases by d%. In the 2 quarter of a year, it decreases by d%. In the 3 quarter of a th year, it increases by d%. In the 4 quarter of a year, it decreases by d%. © 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers It is also noted that the FOREX value decreases by 200 points in the first 2 quarters and by 192 units in the last 2 quarters of the year. If a person had 100 USD all throughout the year, what is the overall loss percentage of the value of USD, when exchanged with INR, from the start to the end of the year? (1) 1.99 (2) 3.84 (3) 17.19 (4) 7.96 (5) 7.84 Solution: Let the initial value be denoted by v. nd At the end of the 2 quarter, the value would be given by: 2 v × [(1 + (d/100)] × [(1 – (d/100)] = v × [(1 – (d /10000)] Decrease in the value in the first 2 quarters 2 2 = v – {v × [(1 – (d /10000)]} = v × (d /10000) = 200 (1) th

At the end of the 4 quarter, the value would be given by: 2 v × [1 – (d /10000)] × [(1 + (d/100)] × [(1 – (d/100)] 2 2 = v × [1 – (d /10000)] Decrease in the value in the last 2 quarters 2 2 2 = {v × [1 – (d /10000)]} – {v × [1 – (d /10000)] } 2 2 = {v × (d /10000) × [1 – (d /10000)]} = 192 (2) Using (1) in (2), we get 2 {200 × [1 – (d /10000)]} = 192 2 → [1 – (d /10000)] = 192/200 = 0.96 2 → (d /10000) = 1 – 0.96 = 0.04. 2 → (d ) = 0.04 × 10000 = 400. → d = 20%. th Thus, the value at the end of the 4 quarter 2 2 2 2 = v × [(1 + (d/100)] × [(1 – (d/100)] = v × 1.2 × 0.8 = 0.9216v. Thus, decrease in the percentage value over the year or 4 quarters = 100 × (v – 0.9216v) /(v) = 7.84%. Since, the FOREX value has decreased by 7.84 % in the year; the loss percentage would also be equal to 7.84%. Hence, the correct answer is option 5. 19. The surface area of cube A is increased by 44% and the volume of the cube B is increased by 33.1%, while retaining the cubical form of both the objects during these changes. By what percentage is the percentage increase of the side of cube B less than the percentage increase of the side of cube A? (1) 25 (2) 50 (3) 33.3 (4) 100 (5) None of these Solution: i) Consider cube A: Let the original side and the final side of cube A be denoted by a and a’ respectively, then original surface area of cube 2 2 2 2 A = 6a . After the increase of 44%, the new surface area = 6a × 1.44 = 8.64a = 6(a’) . 2 → a’ = √(1.44a ) = 1.2a. Thus, side of cube A has increased by 20%. ii) Consider cube B: Let the original side and the final side of cube B be denoted by b and b’ respectively, then original volume of cube B = 3 2 2 3 b . After the increase of 33.1%, the new volume = b × 1.331 = 1.331b = (b’) . 3 1/3 → b’ = (1.331b ) = 1.1b. Thus, side of cube B has increased by 10%. Hence, the percentage by which the percentage increase of the side of cube B is less than the percentage increase of the side of cube A = 100 × (20 – 10)/20 = 50%. Hence, the correct answer is option 2. 20. An XBox dealer made a profit of 25% on selling one gaming console after a discount of Rs 4000. What was the marked price (in Rs) of the XBox, if it is known that it was 45% more than the original cost price of the XBox? (1) 25000 (2) 21000 (3) 20000 (4) 14500 (5) 29000 Solution: Let the cost price of one unit of XBox be denoted by c. Since, there is a profit of 25%, the selling price would be 25% higher than the cost price, thus selling price = s = 1.25 c. A discount of Rs 4000 was given on the marked price to get to the selling price, thus marked price = m = s + 4000 = 1.25c + 4000. © 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Also, it is known that the marked price is 45% higher than the cost price, thus marked price = m = 1.45c. Thus, m = 1.45c = 1.25c + 4000. → c = (4000)/(1.45 – 1.25) = Rs 20000. Hence, marked price = m = 1.45c = 1.45 × 20000 = Rs 29000. Hence, the correct answer is option 5. 21.Select the related word/letters/number from the given alternatives. Donkey : Brays :: Monkey : ? 1) Chatters 2) Trumpets 3) Bellows

4) Grunts

22.Select the related word/letters/number from the given alternatives. ABDE : FGIJ :: IJLM : ? 1) NOQR 2) NOPQ 3) NMOP

4) NPQR

23.Select the related word/letters/number from the given alternatives. 1 : 8 :: ? :64 1) 25 2) 36 3) 30

4) 27

24. For the following questions Find the odd word/letters/number pair from the given alternatives. 1) Swimming 2) Sailing 3) Diving

4) Driving

25. For the following questions Find the odd word/letters/number pair from the given alternatives. 1) RGTF 2) MLOK 3) CTES

4) VDZC

26. For the following questions Find the odd word/letters/number pair from the given alternatives. 1) 443 2) 633 3) 821

4) 245

27.Arrange the following words as per order in the dictionary 1. Dyke 2. Dwindle 3. Dwell 4. Dye 1) 3,2,4,1 2) 1,3,4,2 3) 2,1,3,4

4) 3,4,2,1

28.Choose the correct alternative that will replace the question mark in the given series: DHL, PTX, BFJ, ? 1) KOS 2) NRV 3) NPS 4) NRU 29.Find the wrong number in the series: 28, 33, 31, 36, 34, 29 1) 33 2) 36

3) 34

4) 29

30.The age of Rakhi is twelve times that of her daughter Anubha. If age of Anubha is 3 years now, then what was the age of Rakhi, 2 years earlier? 1) 20 years 2) 34 years 3) 30 years 4) 36 years 31 to 34 Directions for questions : Refer the following data to answer the questions given below. Six teams viz. Manchester United, Bayern Munich, Chelsea, Real Madrid, Arsenal and Liverpool participated in a tournament. None of the matches scheduled were abandoned. Each team has to play exactly one match with each of the other teams. If a team wins a match, it is awarded two points and gets no points for losing a match. In case a match is a tie, then each of the two teams playing the match would be awarded one point. It is known that there were no more than two ties in the tournament. After the tournament was over, it was known that the total sum of the points scored by (Manchester United & Chelsea), (Chelsea & Bayern Munich), (Real Madrid & Chelsea), (Chelsea & Liverpool) and (Arsenal & Chelsea) are respectively 16, 15, 14, 12 and 9 points.

© 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Solutions to 31 to 34: Let us take the first letter of the name of each team to represent its name. Now, it is given that MU + C = 16 C + BM = 15 RM + C = 14 C + L = 12 A+C=9 Adding the above five equations, we get MU + 5C + BM + RM + L + A = 66 (1) The total number of games played is 5 + 4 + 3 + 2 + 1 = 15 Each game is worth of two points, hence total points is 15 × 2 = 30 Or MU + C + BM + RM + L + A = 30 (2) Comparing with equation (1) and (2), we get; 4C = 36 C = 9. The maximum point that C can score is 10 points. Now it is 9 points, so it has to be of (4 wins and 1 tie) MU= 7 (3 wins, 1 tie, 1 loss) BM = 6 (3 wins, 2 loss) L = 3 (1 win, 1 tie. 3 loss) RM = 5 (2 wins, 1 tie, 2 loss) A = 0. It is clearly seen that A lost all the five matches. Team L wins only 1 match and that is against A. As there are two ties they must be between MU, RM, C and L, but not BM & A. Now we can have the following possibilities of ties: Case I: C ties with MU then RM ties with L. Team MU BM C RM L A

Win (BM or RM), L, A (MU or RM), L, A BM, RM, L, A (MU or BM), A A -

Lose (BM or RM) (MU or RM), C (MU or BM), C MU, BM, C MU, BM, C, L, RM

Tie C MU L RM -

Case II: C ties with RM then MU ties with L. Case III: C ties with L then MU ties with RM. Team

Win

Lose

Tie

MU BM C

BM, RM, A RM, L, A MU, BM, L, A

C MU, C -

L RM

RM L

L, A A

MU, BM RM, BM, C

C MU

A

-

MU, BM, C, L, RM

-

Team

Win

Lose

Tie

© 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers MU BM C RM L A

BM, L, A RM, L, A MU, BM, RM, A L, A A -

C MU, C BM, C MU, BM, RM MU, BM, C, L, RM

RM L MU C -

31. How many total arrangements of wins and ties are possible in the tournament? (1) 8 (2) 6 (3) 5 (4) 3

(5) 4

Solution: The total number of arrangements 2 in case I, 1 in case II, 1 in case III. Hence, the correct answer is option 5. 32. Which of the following teams never tied with any other team in any of the five matches it played? (1) Manchester United (2) Arsenal (3) Bayern Munich (4) Chelsea

(5) Indeterminate

Solution: It can either be Bayern Munich or Arsenal. Hence, the correct answer is option 5. 33. Which of the following statements must be true? (1) Liverpool lost against Manchester United (2) Liverpool lost against Chelsea. (3) Liverpool lost against Real Madrid (4) Liverpool lost against Arsenal (5) Liverpool lost against Bayern Munich Solution: Only Bayern Munich must have won against Liverpool, as others have a chance of tying a match with Liverpool. And also we know that Liverpool won against Arsenal. Hence, the correct answer is option 5. 34. If Chelsea ties with Manchester United, then how many total arrangements of wins and ties are possible in the tournament? (1) 1 (2) 2 (3) 3 (4) 4 (5) None of these Solution: It happens in Case I. There are two such arrangements are possible. Hence, the correct answer is option 2.

© 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers 35 to 37 Directions for questions : Refer the data below to answer the following questions. The table below gives the defence expenditure of thirteen countries across five years. The expenditure figures in the table are crores of rupees. Country

2006

2007 2008 2009 2010

A

334

773

947

1000 432

B

648

372

833

902

877

C

2845

1526 596

0

0

D

540

213

E

1125

1114 524

347

414

F

518

536

929

869

G

1382

1066 636

15

24

H

3709

4437 1410 1483 1197

I

363

384

561

285

J

2441

1931 2537 898

883

K

561

209

L

1264

2622 2414 1215 0

M

193

395

1175 745 419

641 43 28

621

1246 709 86

716

35. Which of following countries spends the highest for defence for maximum number of the given years? (1) J (2) H (3) L (4) C (5) A Solution: The country H’s spending is more than any other country’s spending in all the years except 2008. Hence, the correct answer is option 2. 36. Which of the following inferences about the table is true? (1) There was a consistent decrease in defence spending by country G. (2) The defence spending of country A is proportional to the defence spending of country H. (3) The country C has declared peace with all its neighbours by bringing down its defence spending to zero. (4) The average defence spending by the country ‘I’ is more than sixty percent of the average defence spending of the country B in the given period. (5) None of these Solution: Option 1: There is an increase from 2009 to 2010 in the spending of the country G. Option 2: Between 2007 and 2008, spending of country H decreased whereas that of A increased. Hence, option 2 is false. Option 3: Option 3 is a hypothetical one and it cannot be inferred. Option 4: Average spending of country B = (648 +372 + 833 + 902 + 877)/5 = 726.4 Average spending of I = (363 + 384 + 641 + 561 + 285)/5 = 446.8 Sixty percent of 726.4 is = 435.84. Hence, option 4 is true. Hence, the correct answer is option 4. 37. In the year 2009, the defence spending of which of the following countries recorded the maximum percentage of increase over that of the previous year? (1) M (2) B (3) H (4) K (5) D Solution: Option 1: County M It is (86 – 28)× 100/28 = 207.1% (By observation it is 200%) © 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Option 2: County B It is (902 – 833)× 100/833 = 8.28% (By observation it is close to 10%) Option 3: County H It is (1483 – 1410)× 100/1410 = 5.18% (By observation it is less than 10%) Option 4: County K It is (1246 – 43)× 100/43 = 2797.67% (By observation it is greater than 1000%) Option 5: Country D showed a decrease. Hence, the correct answer is option 4. Directions for questions 38 to 40: Refer the data below to answer the following questions.

A B C D E F G H I J K L M N O P Q R S

12.10 18.60 7 5.30 35 7.50 6.20 13.50 9.60 5.20 4.80 4.50 3 3.40 3.70 3.30 3 3.40 1.80

900 800 1800 1600 1100 800 750 750 600 300 230 250 140 100 160 140 90 150 120

175 200 250 300 420 290 170 250 100 300 200 140 120 100 80 30 50 50 60

12 15 14 8 5 10 12 8 6 15 10 10 35 10 10 25 10 15 20

Tyre Size (mm)

Fuel tank capacity (litres) Fuel consumption (km/litre)

Boot space (litres)

Price (in Rs. lakhs)

Model

The five different details of nineteen different model cars are given below.

110 120 160 150 140 90 100 120 120 110 110 90 80 80 100 100 80 90 90

38. If all cars start a rally which involves ten legs each of length 400 km with a full tank of the fuel, then what percentage of the cars given in the table cannot finish the rally, given that each car is allowed to refuel only once during the rally after the start? (1) 44% (2) 47% (3) 32% (4) 37% (5) 42% Solution: The total distance of 3600 km has to be covered using only one refueling. For example, for the car K, Total distance travelled on full tank = 200 × 10 = 2000 km. Car S can cover a distance of 4000 km on 1 additional fuel. The number of cars that can complete the rally is 11. The number of cars that cannot finish the rally is 8 (The cars I, L, N, O, P, Q, R and S). The percentage of cars that do not finish the rally = (8/19)*100 = 42.10% Hence, the correct answer is option 5. 39. How many cars here have a boot space to fuel tank capacity ratio of more than 1.50 and whose price is less than seven lakhs rupees? (1) 6 (2) 8 (3) 7 (4) 5 (5) 9 © 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus

BITS PILANI Test 6 – Answer Key & Explanatory Answers Solution: There are 12 cars that are less than seven lakhs. By inspection the number of cars satisfying the condition is 8 (The cars S, Q, P, R, O, L, D and G) Hence, the correct answer is option 2. 40. If all the cars are ranked in descending order on the basis of the number of revolutions of the tyre required to cover a distance of 100 km, then what is the average fuel consumption for the top seven ranked cars? (1) 17.36 km/litre (2) 16.33 km/litre (3) 14.25 km/litre (4) 15.70 km/litre (5) 13.54 km/litre Solution: The car that requires the highest number of revolutions to travel 10 km is based on its tyre size. Hence cars with the minimum tyre sizes are ranked from 1 to 5. The car that get this ranking are: Car-M (1), Car-Q (2) and car-N (3) are with tyre sizes 80 mm. Car-S (4), car-R (5), Car- L (7) and car-F (6) have equal tyre size and it is 90 mm and hence they come in the top seven. Average fuel consumption = (35 + 10 + 10 + 20 + 15 + 10 + 10) /7 = 15.71 km/litre Hence, the correct answer is option 4.

© 2017 ETHNUS

www.eguru.ooo

facebook.com/ethnus