Question 35: The mean of the following frequency distribution is 18. The frequency f in the class interval 19-21 is missing. Determine f.

Class Interval | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
---|---|---|---|---|---|---|---|

Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |

**Answer:**

Class Interval | Frequency (fi) | Class Mark (xi) | fixi |
---|---|---|---|

11-13 | 3 | 12 | 36 |

13-15 | 6 | 14 | 84 |

15-17 | 9 | 16 | 144 |

17-19 | 13 | 18 | 234 |

19-21 | f | 20 | 20f |

21-23 | 5 | 22 | 110 |

23-25 | 4 | 24 | 96 |

Total | Σfi = 40 + f | Σfixi = 704 + 20f |

`x=(Σf_ix_i)/(Σf_i)`

Or, `(700+20f)/(40+f)=18`

Or, `700+20f=720+18f`

Or, `700+2f=720`

Or, `2f=7=20`

Or, `f=10`

OR

The following table gives production yield per hectare of wheat of 100 farms of a village:

Production Yield | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 |
---|---|---|---|---|---|---|

No. of farms | 4 | 6 | 16 | 20 | 30 | 24 |

Change the distribution to a ‘more than’ type distribution and draw its ogive.

**Answer:**

Production Yield | Cumulative Frequency |
---|---|

More than 40 | 100 |

More than 45 | 96 |

More than 50 | 90 |

More than 55 | 74 |

More than 60 | 54 |

More than 65 | 24 |

Question 36: From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

**Answer:** Let us assume the height of building = p = 20 m

Total height of building and transmission tower = p’

The angle of elevation to the top of the building = 45° and that to the top of transmission tower = 60°

Height of building subtracted from total height will give the height of transmission tower.

**1st case:**

`text(tan) 45°=p/b`

Or, `(20)/(b)=1`

Or, `b=20 m`

**2nd case:**

`text(tan) 60°=(p’)/(b)`

Or, `sqrt3=(p’)/(20)`

Or, `p’=20sqrt3`

Height of transmission tower can be calculated as follows:

`p'- p = 20sqrt3 - 20`

`= 20(sqrt3 - 1)`

`= 20(1.732 - 1)`

`= 20 x× 0.732 = 14.64 m`

Question 37: It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

**Answer:** 12A + 12B = 1 (Case 1)

Case 2: 4A + 9B = `1/2`

Or, 8A + 18B = 1

Or, 8A + 18B = 12A + 12B

Or, 12A – 8A = 18B – 12B

Or, 4A = 6B

Or, A = `3/2`B

Substituting the value of A in first equation, we get:

`12×3/2B+12B=1`

Or, 18B + 12B = 1

Or, 30B = 1

Or, B = `1/(30)`

B will fill the pool in 30 hours

Substituting the value of B in second equation, we get:

`8A+18×1/(30)=1`

Or, `8A+3/5=1`

Or, `(40A+3)/5=1`

Or, 40A + 3 = 5

Or, 40A = 3 – 5 = 2

Or, `A=2/(40)=1/(20)`

So, A will fill the pool in 20 hours

A takes 20 hours and B takes 30 hours to fill the swimming pool

Question 38: Prove that `sqrt5` is an irrational number.

**Answer:** Let us assume the contrary, i.e. `sqrt5` is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

`sqrt5=a/b`

Or, `bsqrt5=a`

Squaring on both sides, we get;

`5b^2=a^2`

This means that a^{2} is divisible by 5 and hence a is also divisible by 5.

This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.

This also contradicts our earlier assumption that `sqrt5` is irrational.

Hence, `sqrt5` is irrational.

Question 39: Draw a circle of radius 3.5 cm. From a point P, 6 cm from its center, draw two tangents to the circle.

**Answer:** Draw a circle with radius 3.5 cm and centre O.

Draw a line segment OP = 6 cm

Make perpendicular bisector of OP which intersects OP at point O’.

Take O’P as radius and draw another circle.

From point P, draw tangents to points of intersection between the two circles.

OR

Construct a Δ ABC with AB = 6 cm, BC = 5 cm and ∠B = 60°. Now construct another triangle whose sides are 2/3 times the corresponding sides of Δ ABC.

**Answer:** Making the triangle:

Draw a line segment BC = 5 cm.

Make a 60° angle from point B and draw AB = 6 cm.

Join A and C to get the triangle ABC.

Dividing the base:

Draw a ray at an acute angle from BC.

Plot 3 points on BA so that BB1 = B1B2 = B2B3

Join B3 to point A.

Draw a line from B2 parallel B3C so that it meets BC at point C’.

Draw A’C’ || AC.

Triangle A’C’B is the required triangle.

Question 40: A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. (Take `π=(22)/7`)

**Answer:** Volume of cone `=1/3πr^2h`

`=1/3×(22)/7×7^2×3.5`

`=1/3×22×7×3.5=(539)/3`

Volume of hemisphere `=2/3πr^3`

`=2/3×(22)/7×7^3`

`=(2156)/3`

Volume of solid `=(539+2156)/3=(2695)/3=898.33` sq cm

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