# Arithmetical Reasoning Questions

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) 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) 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) 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.