# Splification - Quantitative Aptitude

Q41. How many times are the hands of a clock at right angle in a day ?

(a) 24

(b) 36

(c) 44

(d) 48

**Answer:** (c) 44

In 12 hours, no of right angles = 22 times

∴ In 24 hours, no of right angles = 2 X 22 = 44

Q42. At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other ?

(a) 16^{4}/_{11} min past 3

(b) 18^{4}/_{11} min past 3

(c) 19^{4}/_{11} min past 4

(d) 21^{4}/_{11} min past 5

**Answer:** (a) 16^{4}/_{11} min past 3

At 3 o'clock, the minute hand is 15 min spaces apart from the hour hand

To be coincident, it must gain 15 min spaces

55 min are gained in 60 min

15 min are gained in (^{60}/_{55}) X 15 min = 16 ^{4}/_{11} min

∴ The hands are coincident at 16^{4}/_{11} min past 3

Q43. Rohit was asked to find ^{7}/_{9} of a fraction. But he made a mistake of dividing the given fraction by ^{7}/_{9} and got an answer which exceeded the correct answer answer by ^{8}/_{9}. The correct answer is

(a) ^{43}/_{36}

(b) ^{49}/_{36}

(c) ^{51}/_{36}

(d) ^{53}/_{36}

**Answer:** (b) ^{49}/_{36}

Let the given fraction be x

∴ The correct answer be ^{7}/_{9}x

A/Q | x | - | 7 | x = | 8 |

^{7}/_{9} |
9 | 9 |

⇒ | 9x | - | 7 | x = | 8 |

7 | 9 | 9 |

⇒ | 81x - 49x | = | 8 |

63 | 9 |

⇒ | 32x | = | 8 |

63 | 9 |

⇒ x = | 8 | X | 63 | = | 7 |

9 | 32 | 4 |

∴ Correct answer = | 7 | X | 7 | = | 49 |

9 | 4 | 36 |

Q44. The highest score in an innings was ^{3}/_{11} of the total and the next highest was ^{3}/_{11} of the remainder. If the scores differed by 18, then the total score is

(a) 212

(b) 225

(c) 236

(d) 242

**Answer:** (d) 242

Let the total score be x

∴ The highest score = ^{3x}/_{11}

Remainder = x - | 3x | = | 8x |

11 | 11 |

Next highest score = | 3 | of | 8x | = | 3 | X | 8x | = | 24x |

11 | 11 | 11 | 11 | 121 |

A/Q | 3x | - | 24x | = 18 |

11 | 121 |

⇒ | 33x - 24x | = 18 |

121 |

⇒ | 9x | = 18 |

121 |

⇒ x = | 18 X 121 | = 242 |

9 |

Q45. A tin of oil was ^{4}/_{5} full. When 6 bottles of oil were taken out and 4 bottles of oil poured into it, it was ^{3}/_{4} full. How many bottles of oil can the tin contain ?

(a) 35 bottles

(b) 48 bottles

(c) 40 bottles

(d) 47 bottles

**Answer:** (c) 40 bottles

2 bottles = ( | 4 | - | 3 | ) |

5 | 4 |

= | 16 - 15 | = | 1 |

20 | 20 |

Now ^{1}/_{20} = 2 bottles

⇒ 1 = 2 X 20 = 40 bottles

∴ Tin can contain 40 bottles

Q46. If a+b+c = 3s, then [(s-a)^{2} + (s-b)^{2} + (s-c)^{2} + 3s^{2}] = ?

(a) (a+b+c)^{2}

(b) a^{2}+b^{2}+c^{2}

(c) 2(a+b+c)

(d) 2abc

**Answer:** (b) a^{2}+b^{2}+c^{2}

(s-a)^{2} + (s-b)^{2} + (s-c)^{2} + 3s^{2}

= s^{2} - 2as + a^{2} + s^{2} - 2bs + b^{2} + s^{2} - 2cs + c^{2} + 3s^{2}

= 6s^{2} + a^{2} + b^{2} + c^{2} - 2 (as + bs + cs)

= 6s^{2} + a^{2} + b^{2} + c^{2} - 2s (a + b + c)

= 6s^{2} + a^{2} + b^{2} + c^{2} - 2s X 3s

= 6s^{2} + a^{2} + b^{2} + c^{2} - 6s^{2}

= a^{2} + b^{2} + c^{2}

Q47. A tree increases annually by ^{1}/_{7} th of its height. What will be its height after 2 years, if it stands today 49 cm height ?

(a) 61 cm

(b) 64 cm

(c) 65 cm

(d) 68 cm

**Answer:** (b) 64 cm

After 1 year height = 49 + 49 X ^{1}/_{7} = 49 + 7 = 56

After 2 years height = 56 + 56 X ^{1}/_{7} = 56 + 8 = 64 cm

Q48. If a-b = 2, then (a^{3} - b^{3} + 3ab) = ?

(a) - 1

(b) - 3

(c) 1

(d) 3

**Answer:** (a) - 1

a - b = -1

⇒ (a - b)^{3} = (-1)^{3}

⇒ a^{3} - b^{3} - 3ab(a - b) = -1

⇒ a^{3} - b^{3} - 3ab(-1) = -1

⇒ a^{3} - b^{3} + 3ab = -1

Q49. A vessel full of water weights 80 kg. If it is half filled, its weight is 60 kg. The weight of the empty vessel is

(a) 28 kg

(b) 32 kg

(c) 35 kg

(d) 40 kg

**Answer:** (d) 40 kg

Let the weight of the empty vessel be x kg

Weight of water when the vessel is full = 80 - x kg

Weight of water when the vessel is half full = ½(80 - x) kg

A/Q ½ (80 - x) = 60 - x

⇒ 80 - x = 120 - 2x

⇒ x = 120 - 80 = 40 kg

Q50. A man purchased 130 stamps of 60 paise and 30 paise. The total amount he spent was Rs 60.30. How many 60 paise stamps were purchased by him ?

(a) 65

(b) 69

(c) 71

(d) 74

**Answer:** (c) 71

Let the no of 60 paise stamps be x

∴ No of 30 paise stamps = 130 - x

A/Q 60 X x + 30 (130 - x) = 6030

⇒ 2x + 130 - x = ^{6030}/_{30}

⇒ x + 130 = 201

⇒ x = 201 - 130 = 71

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