Area of the park = (60 x 40) = 2400 sq.m  

Area of the lawn = 2109  sq.m 

Area of the crossroads = (2400 - 2109) = 291 sq.m   

Let the width of the road be x metres. Then,   

60x + 40x - (x * x) = 291  

=>(x * x) - 100x + 291 = 0    

=>(x - 97)(x - 3) = 0  

=>x=3

2(l+b)/b = 5/1 

=> 2l + 2b = 5b  

=> 3b = 2l  

=> b =(2/3) x l

 Then, Area = 216 sq.cm 

l x b = 216  

=> l x [(2/3)x l] =216  

=> l x l = 324 

=> l = 18 cm.

100 cm is read as 102 cm.

 A1 = (100 x 100)  and A2 =(102 x 102)   

(A2 - A1) = [(102 x 102)-(100 x 100) ]= 404    

Percentage error =[404/(100 x 100 )] x 100%= 4.04%

Perimeter = Distance covered in 8 min. = (12000/60) * 8 = 1600m 

Let length = 3x metres and breadth = 2x metres.  

Then, 2(3x + 2x) = 1600 or x = 160.  

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) = 153600 

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Face value of the bill = Rs. 6000.

Date on which the bill was drawn = July 14 at 5 months. Nominally due date =                  December 14.

Legally due date = December 17.

Date on which the bill was discounted = October 5.

Unexpired time  : Oct.               Nov.                Dec.

                                 26  +               30  +              17     = 73 days  =1/ 5Years

 B.D. = S.I. on Rs. 6000 for 1/5 year

= Rs.   (6000 x 10 x1/5 x1/100)= Rs. 120.

T.D. = Rs.[(6000 x 10 x1/5)/(100+(10*1/5))]

            =Rs.(12000/102)=Rs. 117.64.

B.G. = (B.D.) - (T.D.) = Rs. (120 - 117.64) = Rs. 2.36.

Money received by the holder of the bill = Rs. (6000 - 120) = Rs. 5880.

Sparse matrix,

Index generation

MainMenu component is a component that allows the display of Menus at runtime on a form.

Process of creating Menu using MainMenu Component:

a. Add MainMenu component on Windows Form.

b. Menu designer allows deciding the structure of the main menu by selecting the Type Here area and adding the Menu Items to be displayed on the menu.

c. Add functionality to Menu Items as required