# Arithmetical Reasoning Questions

A) Rs. 4, Rs. 23 | B) Rs. 13, Rs. 17 |

C) Rs. 15, Rs. 14 | D) Rs. 17, Rs. 13 |

Explanation:

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

A) 11 | B) 18 |

C) 20 | D) 21 |

Explanation:

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

A) 32 rolls | B) 54 rolls |

C) 108 rolls | D) 120 rolls |

Explanation:

Number of cuts made to cut a roll into 10 pieces = 9.

Therefore, Required number of rolls = (45 x 24)/9 = 120.

A) 20 years | B) 22 years |

C) 25 years | D) 27 years |

Explanation:

Let Varun's age today = x years.

Then, Varun's age after 1 year = (x + 1) years.

Therefore x + 1 = 2 (x - 12) => x + 1 = 2x - 24 => x = 25.

A) 33 | B) 24 |

C) 54 | D) 41 |

Explanation:

We can get this by any of two explanations.

**Answer 1**: 41

7 = 3 + 4 = 34

Similarly, 5 = 4 + 1 = 41

**Answer 2** : 41

7 - 3 = 4 => 34

Similarly, 5 - 4 = 1 => 41

A) 18 | B) 36 |

C) 45 | D) None of these |

Explanation:

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then,

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 => 6R = 108 => R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

A) 3, 3, 8 | B) 2, 6, 6 |

C) 1, 6, 12 | D) 1, 8, 9 |

Explanation:

Let’s break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 – sum(2, 2, 18) = 22

2, 4, 9 – sum(2, 4, 9) = 15

2, 6, 6 – sum(2, 6, 6) = 14

2, 3, 12 – sum(2, 3, 12) = 17

3, 4, 6 – sum(3, 4, 6) = 13

3, 3, 8 – sum(3, 3, 8 ) = 14

1, 8, 9 – sum(1,8,9) = 18

1, 3, 24 – sum(1, 3, 24) = 28

1, 4, 18 – sum(1, 4, 18) = 23

1, 2, 36 – sum(1, 2, 36) = 39

1, 6, 12 – sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 – sum(2, 6, 6) = 14

3, 3, 8 – sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

A) 6555 | B) 5685 |

C) 1705 | D) 870 |

Explanation:

Candidates passed in atleast four subjects

= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)

= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)

= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555