Mains exams are very important to clear every

**Answer**: Option B

**Explanation**:

Speed = [\(\frac{600}{5× 60}\) ]m/sec

= 2 m/sec.

Converting m/sec to km/hr

= 2 x\(\frac{18}{5}\) km/hr

= 7.2 km/hr.

**2. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hour, it must travel at a speed of:**

**Answer**: Option D

**Explanation**:

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = \(\frac{1200}{5/3}\) km/hr.

Required speed = 1200 x \(\frac{3}{5}\) km/hr= 720 km/hr

**3. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:**

**Answer**: Option A

**Explanation**:

Let the actual distance travelled be x km.

Then, \(\frac{×}{10}\) = \(\frac{× + 20 }{14}\)

⇒14x = 10x + 200

⇒4x = 200

⇒ x = 50 km.

**4. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:**

**Answer**: Option C

**Explanation**:

Let speed of the car be x kmph

Then, speed of the train = \(\frac{150}{100}\) x = \(\frac{3}{2}\)

∴ \(\frac{75}{×}\) – \(\frac{75}{3/2}\) x = \(\frac{125}{10 × 60 }\)

⇒\(\frac{75}{×}\) – \(\frac{50}{×}\) = \(\frac{5}{24}\)

⇒x = \(\frac{25 × 24}{5}\) = 120 kmph

**5. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?**

**Answer**: Option B

**Explanation**:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = [\(\frac{9}{54}\) x 60] min = 10 min

**Answer**: Option D

**Explanation**:

A number is divisible by 3 when sum of its digits is divisible by 3.

Here, sum of digits = 4 + 3 + 5 + 6 + 7 + X = 25 + X. So, X can be 2, 5, 8 which gives the sum 27, 30 and 33 respectively.

Therefore, X has 3 values here, for which the number is divisible by 3. So, the answer is option D.

**2. Simplify the expression using BODMAS rule (\(\frac{3}{7}\)) of (\(\frac{4}{5}\)) of 20 (\({25}^{2}\) – \({24}^{2}\))**

**Answer**: Option A

**Explanation**:

(\(\frac{3}{7}\)) × (\(\frac{4}{5}\)) of 20 (625 – 576)⇒ (\(\frac{3}{7}\)) × (\(\frac{4}{5}\)) × 20 × 49 =336

**3. Simplify the expression using BODMAS rule (105 + 206) – 550 ÷ \({5}^{2}\) + 10**

**Answer**: Option D

**Explanation**:

(105 + 206) – 550 ÷ 52 + 10

= 311 – 550 ÷ 25 + 10

= 311 – 22 + 10

= 289 + 10 = 299

**4. Simplify the expression using BODMAS rule:: (\(\frac{3}{3}\)) of (\(\frac{4}{7}\)) {(10 × 3) – (8 × 2)}**

**Answer**: Option B

**Explanation**:

Applying BODMAS rule = (\(\frac{3}{2}\)) of (\(\frac{4}{7}\)) {30 – 16} = (\(\frac{12}{14}\)) × 14 = 12

**5. The six-digit number 54321A is divisible by 9 where A is a single digit whole number. Find A.**

**Answer**: Option D

**Explanation**:

A number is divisible by 3 when sum of its digits is divisible by 3.

Here, sum of digits = 4 + 3 + 5 + 6 + 7 + X = 25 + X. So, X can be 2, 5, 8 which gives the sum 27, 30 and 33 respectively.

Therefore, X has 3 values here, for which the number is divisible by 3.

**Answer**: Option C

**Explanation**:

(A + B + C)’s 1 day’s work = \(\frac{1}{6}\)

(A + B)’s 1 day’s work = \(\frac{1}{8}\)

(B + C)’s 1 day’s work = \(\frac{1}{12}\)

∴(A + C)’s 1 day’s work = [2\(\frac{1}{6}\)] – [\(\frac{1}{8}\) + \(\frac{1}{12}\)]

= \(\frac{1}{3}\) – \(\frac{5}{24}\)

= \(\frac{3}{24}\)

= \(\frac{1}{8}\)

So, A and C together will do the work in 8 days.

**2. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:**

**Answer**: Option C

**Explanation**:

(B + C)’s 1 day’s work = [\(\frac{1}{9}\) + \(\frac{1}{12}\)] = \(\frac{7}{36}\)

Work done by B and C in 3 days =[ \(\frac{7}{36}\) x 3] = \(\frac{7}{12}\)

Remaining work = [1 – \(\frac{7}{12}\)] = \(\frac{5}{12}\)

Now, \(\frac{1}{24}\) work is done by A in 1 day

So, \(\frac{5}{12}\) work is done by A in [24 x \(\frac{5}{12}\)] = 10 days

**3. X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?**

**Answer**: Option A

**Explanation**:

Work done by X in 8 days = [\(\frac{1}{40}\) x 8] = \(\frac{1}{5}\)

Remaining work = [1 – \(\frac{1}{5}\)] = \(\frac{4}{5}\)

Now,\(\frac{4}{5}\) work is done by Y in 16 days.

Whole work will be done by Y in [16 x\(\frac{5}{4}\)] = 20 days

∴X’s 1 day’s work = \(\frac{1}{40}\), Y’s 1 day’s work = \(\frac{1}{20}\)

(X + Y)’s 1 day’s work = [\(\frac{1}{40}\) + \(\frac{1}{20}\)] = \(\frac{3}{40}\) = \(\frac{3}{40}\)

Hence, X and Y will together complete the work in \(\frac{40}{3}\) = 13\(\frac{1}{3}\)

**4. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:**

**Answer**: Option B

**Explanation**:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

∴[\(\frac{20 – ×}{×}\) x 100 = 25]

⇒2000 – 100x = 25x

⇒125x = 2000

⇒ x = 16.

**5. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?**

**Answer**: Option B

**Explanation**:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

∴Required percentage = [\(\frac{295}{420}\) x 100]% = \(\frac{1475}{21}\)% = 70%

Competitive Exams – Study Guide | ||
---|---|---|

Category |
||

Quantitative Aptitude |
Reasoning Ability |
General Awareness |

Computer Awareness |
English Knowledge |
Computer Awareness |

Descriptive Test |

Competitive Exams – College Entrance Exams | |||
---|---|---|---|

Category |
Notification |
||

Law |
JLEE 2019 |
MP PAT 2019 |
ASSAM PAT 2019 |

Engineering |
JEEDEC 2019 | AIAPGET 2019 |
SVNIRTAR PGET 2019 |

Diploma |
AUAT 2019 | AP PECET 2019 | Goa Diploma Admissions 2019 |

Graduate |
WBJEE JECA | NEST 2019 | TUET |

^{} Click Here For – All India Entrance Exam Notifications |

Competitive Exams – Practice Sets | |
---|---|

Category |
Quiz |

Quant Aptitude |
Probability |

Reasoning Ability |
Puzzles |

Current Affairs |
Current Affairs |

General Knowledge for Competitive Examinations | |
---|---|

Topic |
Name of the Article |

GK – World |
BCCI Awards List |

North America Countries Capitals Currencies | |

Nobel Prize Winners List | |

GK – India |
Atmosphere – Layers |

Revolt – 1857 | |

Vedic Literature | |

GK – Abbreviations |
Computer Aptitude Abbreviations |

Finance Abbreviations | |

Education Abbreviations | |

GK – Banking & Insurance |
First Banks in India |

Future of Insurance Sector | |

Money and Money Market |