A) 1124 | B) 1420 |

C) 1345 | D) 1250 |

Explanation:

Let initially 'X' ticket has been sold.

So now in 2nd week 20% increases, so

=> X x 120/100

In 3rd week 16% increases, so

=> X x (120/100) x (116/100)

In 4th week 20% decrease, so

=> X x (120/100) x (116/100) x (80/100) = 1392

X = 1250.

A) 19500 | B) 21000 |

C) 23600 | D) 24000 |

Explanation:

6% of 350000

$\frac{6}{100}x3,50,000\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=6x3500\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=21,000.$

Hence, **6% of 3,50,000 = 21,000.**

A) 20 | B) 24 |

C) 32 | D) 36 |

Explanation:

20% of 120

20 x 120/100

= 4 x 6

= 24

Hence, **20% of 120 = 24.**

A) 32,000 | B) 64,000 |

C) 40,000 | D) 52,000 |

Explanation:

From the given data,

**20% of K = 32% of N**

**V = 4/5 N**

Given Vipin's salary = 3.84 lpa = 3.84/12 = 32,000 per month

Now, N = 5/4 x 32000 = 40000

Now K = 32/20 x 40000 = 64000

Hence, karthik's monthly salary = **Rs. 64000.**

A) 4 | B) 3 |

C) 2 | D) 1 |

A) 10 | B) 20 |

C) 30 | D) 35 |

Explanation:

From the given data,

$\frac{\mathbf{40}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{75}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{80}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{25}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{K}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{250}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}30+20=\frac{5\mathrm{K}}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}50\mathrm{x}2=5\mathrm{K}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{K}=\frac{100}{5}=20$

A) 1060 | B) 960 |

C) 860 | D) 760 |

Explanation:

From the given data,

**80 x 15/100 of 80 + 5 x 80 x 25/100**

= 960 + 100

= 1060.

A) Rs. 87400 | B) Rs. 68700 |

C) Rs. 74000 | D) Rs. 94060 |

Explanation:

Let the total amount he had with = Rs. p

From the given data,

p - 38460 - 24468 = 28p/100

p - 28p/100 = 24468 + 38460

72p/100 = 62928

72p = 62928 x 100

p = 62928 x 100/72

p = 87400

Hence, the total amount he had = **p = Rs. 87400.**

A) 14.5 lit | B) 12.8 lit |

C) 11.6 lit | D) 10.46 lit |

Explanation:

This can be solved as

$\mathbf{20}\mathbf{}{\left(\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{4}}{\mathbf{20}}\right)}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{20}\mathbf{}{\left(\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{1}}{\mathbf{5}}\right)}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=20{\left(\frac{4}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=20\left(\frac{16}{25}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{64}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=12.8$