í∫’®Ω’¢√®Ωç 2 -´÷-Ja 2017

Ñ-Ø√-úø’ While solving number series questions, first try to identify the numbers in the series (they may be squares, cubes, prime number etc.,). Then find the difference between the numbers. If unable to get any clue then see the relation between the numbers with multiplication division etc.

Quantitative Aptitude Directions (Q.1-5): In each of the following number series questions a number is wrong. You have to identify that number. 1. 7 10 24 88 432 2568 1) 24 2) 2568 3) 10 4) 432 5) 7 2. 4 6 12 52.5 236.25 1299.125 1) 52.5 2) 1299.125 3) 12 4) 6 5) 236.25 3. 12 13 30 99 410 2085 1) 410 2) 30 3) 99 4) 2085 5) 13 4. 153 159 177 207 247 303 1) 247 2) 303 3) 153 4) 177 5) 159 5. 781.5 312.5 125 50 20 8 1) 312.5 2) 781.5 3) 8 4) 125 5) 20 Directions (6-10): Study the table carefully to answer the questions that follow:

Candidates who appeared and passed in the test from four schools in six different years

Temperature on Thursday was..? 1) 2010 4) 2013

2) 2011 5) 2014

3) 2012

8. What was the respective ratio between the number of candidates who passed from School-C in the year 2012 and the number of candidates who appeared in the exam from School-B in the year 2014? 1) 11 : 13 2) 48 : 49 3) 25 : 26 4) 9 : 11 5) None of these 9. Number of candidates who passed in the exam from School B in the year 2013 was

A

C

Appeared Passed

D

Appeared Passed

Appeared

Passed

2010

124

78

338

219

454

343

546

345

2011

234

124

545

435

732

567

565

456

2012

456

235

664

454

693

576

345

212

2013

398

156

345

144

645

545

527

339

2014

546

346

588

354

354

258

628

396

2015

547

435

704

347

578

313

456

252

6. What was the total number of failed candidates from School-D in the year 2012, 2013 and 2014? 1) 537 2) 625 3) 553 4) 607 5) None of these 7. In which year was the difference between the number of candidates who appeared and passed in the exam from School-B second lowest?

EXPLANATIONS 1. 7 × 2 − 4 = 10 10 × 3 − 6 = 24 24 × 4 − 8 = 88 88 × 5 − 10 = 430 430 × 6 − 12 = 2568 2. 4 × 1.5 = 6 6 × 2.5 = 15 15 × 3.5 = 52.5 52.5 × 4.5 = 236.25 236.25 × 5.5 = 1299.375 3. 12 × 1 + 1 = 13 13 × 2 + 4 = 30 30 × 3 + 9 = 99 99 × 4 + 16 = 412 412 × 5 + 25 = 2085 4. 153 + 6 = 159 159 + 18 = 177 177 + 30 = 207 207 + 42 = 249 249 + 54 = 303 5. 781.25 ÷ 5 × 2 = 312.5 312.5 ÷ 5 × 2 = 125 125 ÷ 5 × 2 = 50 50 ÷ 5 × 2 = 20 20 ÷ 5 × 2 = 8 6. Required Total = (345 − 212) + (527 − 339) + 628 − 396) ⇒ 133 + 188 + 232 = 553 7. Failed candidates in School − B 2010 ⇒ 338 − 219 = 119 2011 ⇒ 545 − 435 = 110

1) 90

2) 75

4) 80

5) None of these

approximately what percent of number of candidates who passed from School-A in the year 2014? 1) 74% 2) 58% 3) 80% 4) 42% 5) 90% 10. What is the percent decrease in the number of candidates who failed in the exam from school-A in the year 2015 as compared to the previous year? 1) 44% 2) 64% 3) 88% 4) 22% 5) 38% 2012 ⇒ 664 − 454 = 210 2013 ⇒ 345 − 144 = 201 2014 ⇒ 584 − 354 = 230 2015 ⇒ 704 − 347 = 357 Lowest in 2011 8. Required Ratio = 576 : 588 = 48 : 49 144 9. Required percent = ⎯⎯ × 100 ≈ 42% 346 10. School A, failed in 2015 = 547 − 435 = 112 Failed in 2014 = 546 − 346 = 200 Percent decrease = 200 − 112 88 ⎯⎯⎯⎯⎯ × 100 = ⎯ × 100 = 44% 200 200 11. Let the distance travelled by car be 'x' km ∴ 14x + 5(220 − x) = 2270 ⇒ 14x + 1100 − 5x = 2270 ⇒ 9x = 1170 ⇒ x = 130 12. Surface area of a sphere = 4πr2 Total surface area of a cylinder = 2πr(h + r) ∴ 4πr2 = 2πr1 (h + r1) 22 22 ⇒ 4 × ⎯ × r2 = 2 × ⎯ × 4(4 + 4) 7 7 ⇒ r2 = 16 ⇒ r = 4 13. Let the second pipe be turned off after 'x' minutes 12 In 12 minutes First pipe completes ⎯ 16 part work and in 'x' minutes second pipe completes

x

⎯ part work 24 12 x ∴ ⎯+⎯=1 16 24

2) 6 cm

Writer

Dr. G.S. Giridhar Director, Race Institute

3) 130 what was the temperature on Thursday?

12. The surface area of a sphere is same as the total surface area of a cylinder whose height is 4 cm and the diameter of the base is 8 cm. What will be the radius of the sphere? 4) 4 cm

B

Appeared Passed

11. Fares by bus and car are Rs.5 and Rs.14 per kilometer respectively. A man who travels 220 km spends Rs.2,270 in going a part of distance by bus and the remaining in car. How many kilometers did he travel in car?

1) 4.5 cm

School Year ↓

13

3) 8 cm

5) None of these

13. Two pipes A and B fill a cistern in 16 minutes and 24 minutes respectively. When should be the second pipe B turned off so that the cistern may take 12 minutes to fill? 1) 8 minutes

2) 6 minutes

3) 5 minutes

4) 7 minutes

5) None of these

14. Production of a company in 2012 was 468 tonnes. If it increases by 15% in the first year and decreases by 8% in the second year, find the production of the company after two years? 1) 500.76

2) 493.875

4) 495.144

5) None of these

1) 45°C

2) 36°C

4) 54°C

5) None of these

Directions (Q.16-20): In each of the questions below, two equations are provided. On the basis of these you have to find out the relation between P and Q. Give answer 1) if P = Q Give answer 2) if P > Q Give answer 3) if P < Q Give answer 4) if P ≥ Q Give answer 5) if P ≤ Q 16. P2 − 3P − 10 = 0 ; Q2 + 13Q + 40 = 0 17. 2P2 + 3P − 2 = 0 ; Q2 + Q − 2 = 0 18. P2 − P − 12 = 0 ; Q2 + 2Q − 3 = 0 19. 3P + Q = 1 ; 8P + 12Q = 5 20. P2 + 5P − 14 = 0 ; Q2 + 6Q − 7 = 0

3) 487.468

15. The mean temperature of Monday to Wednesday was 39°C and of Tuesday to Thursday was 36°C. If the temperature on 4 Thursday was ⎯ th that of Monday, then 5 36 + 2x ⇒ ⎯⎯⎯⎯ = 1 48 ⇒ 2x = 48 − 36 ⇒ x = 6 minutes Shortcut: 12 3 In 12 minutes first pipe completes ⎯ = ⎯ 16 4 part work 1 The remaining ⎯ part work second pipe 4 1 completes in ⎯ × 24 = 6 minutes 4 ∴ Second pipe should be turned off after 6 minutes 14. Required production = 115% of 92% of 468 115 92 ⇒ ⎯ × ⎯ × 132 = 495.144 100 100 15. Let the temperature of Monday be 'x' 4x Temperature of Thursday = ⎯ 5 Temperatures of Monday + Tuesday + Wednesday = 3 × 39 = 117° C Temperatures of Tuesday + Wednesday + Thursday = 3 × 36 = 108° C Difference between Monday and Thursday is 117 − 108 = 9° C 4x ∴ ⎯ = 9 ⇒ x = 45 5 4 × 45 ∴ Thursday's temperature = ⎯⎯ = 36° 5 16. P2 − 3P − 10 = 0 (P − 5) (P + 2) = 0 P = 5, −2

3) 36.5°C

Key 1-4

6-3

11-3

16-2

2-3

7-2

12-4

17-5

3-1

8-2

13-2

18-4

4-1

9-4

14-4

19-1

5-2

10-1

15-2

20-4.

Q2 + 13Q + 40 = 0 ⇒ (Q + 8)(Q + 5) = 0 ⇒ Q = −8, −5 ∴P>Q 17. 2P2 + 3P − 2 = 0 (2P − 1) (P + 2) = 0 1 P = ⎯, −2 2 Q2 + Q − 2 = 0 ⇒ (Q − 1) (Q + 2) = 0 ⇒ Q = 1, −2 ∴P≤Q 18. P2 − P − 12 = 0 (P + 3) (P − 4) = 0 P = −3, 4 Q2 + 2Q − 3 = 0 ⇒ (Q − 1) (Q + 3) = 0 ⇒ Q = 1, −3 ∴P ≥ Q 19. 3P + Q = 1 8P + 12 Q = 5 1 1 By solving P = ⎯ and Q = ⎯ 4 4 ∴P=Q 20. P2 + 5P − 14 = 0 ⇒ (P − 2) (P + 7) = 0 ⇒ P = 2, −7 Q2 + 6Q − 7 = 0 ⇒ (Q − 1) (Q + 7) = 0 Q = 1, −7 ∴P≥Q

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