A) 500/3 | B) 700/3 |

C) 300 | D) 600 |

Explanation:

Let the initial amount be x (with gambler), then

$\left[\left[\left(x+100\right)\frac{1}{2}+100\right]\frac{1}{2}+100\right]\frac{1}{2}=\frac{x}{2}$

=> x = 700/3

A) 3 | B) 4 |

C) 2.5 | D) 5 |

Explanation:

Let p% of 50 is 2

=> p x 50/100 = 2

=> p = 4%

Hence, **2 is 4 percent of 50.**

A) 209 | B) 290 |

C) 309 | D) 390 |

Explanation:

Given **21% of 500 + 31% of 1100 - 30% of 790 = ?**

? = 21 x 5 + 31 x 11 - 3 x 79

? = 105 + 341 - 237

? = 446 - 237

**? = 209.**

** **

A) 76 | B) 60 |

C) 56 | D) 52 |

Explanation:

Total engineers = 4800

Number of female engineers = 3/8 x 4800 = 1800

Then, number of male engineers = 4800 - 1800 = 3000

Now, required is female engineers is what percentage of the number of male engineers

=> 1800 is what % of 3000

=> p x 3000/100 = 1800

=> 30p = 1800

**p = 60%**

Hence, female engineers is **60 percentage** of the number of male engineers.

A) 20% | B) 25% |

C) 30% | D) 15% |

Explanation:

Percentage of students passed in English **= 60%**

Percentage of students passed in Hindi **= 45%**

Percentage of students passed in both subjects **= 25%**

ATQ,

Percentage of students passed in only English **= 60 - 25 = 35%**

Percentage of students passed in only Hindi **= 45 - 25 = 20%**

Therefore, Percentage of students failed** = 100 - (35 + 25 + 20)**

**= 100 - 80 = 20%.**

A) 52.385 | B) 46.875 |

C) 42.155 | D) 40.555 |

Explanation:

75% of 450 + ? % of 560 = 600

$\frac{\mathbf{75}\mathbf{}\mathbf{x}\mathbf{}\mathbf{450}}{\mathbf{100}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{?}\mathbf{}\mathbf{x}\mathbf{}\mathbf{560}}{\mathbf{100}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{600}\phantom{\rule{0ex}{0ex}}\frac{75\mathrm{x}9}{2}+\frac{28\mathrm{p}}{5}=600\phantom{\rule{0ex}{0ex}}\frac{75\mathrm{x}9\mathrm{x}5+56\mathrm{p}}{10}=600\phantom{\rule{0ex}{0ex}}3375+56\mathrm{p}=6000\phantom{\rule{0ex}{0ex}}56\mathrm{p}=2625\phantom{\rule{0ex}{0ex}}\mathbf{p}\mathbf{}\mathbf{=}\mathbf{}\mathbf{46}\mathbf{.}\mathbf{875}$

Hence, in the given question **? = 46.875**

A) 1960 | B) 1954 |

C) 1947 | D) 1931 |

Explanation:

Given 82.5% of 2360

$\frac{\mathbf{82}\mathbf{.}\mathbf{5}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{2360}=0.825\mathrm{x}2360=\mathbf{}\mathbf{1947}$

A) 27 kgs | B) 25 kgs |

C) 24 kgs | D) 21 kgs |

Explanation:

Initial wheat per kg cost be Rs. 100

Then, monthly expenditure of the family = 30 x 100 = 3000

After increase in cost,

Let P be new monthly consumption of family.

**P x 132 = 110/100 x 3000**

**=> P = 25 kg.**

A) 8400 | B) 7200 |

C) 6000 | D) 5600 |

Explanation:

12% of 70,000

=> 12 x 70,000/100

=> 8400

Hence, **12% of 70,000 = 8,400**