1.A chocolate bar costs twice as an ice creams yash buys 2 chocolates bars and one ice cream and pays Rs.11.50 .Find the cost of and icecream bar three chocolate bars.

2.

3.

4.

Find three consecutive odd numbers whose sum is 141.

5.

 

 

 

A sum of Rs. 2100 is made of 500, 100 and 50 rupee notes. The number of 50 rupee

 

notes is 2 times the number of 100 rupee notes, and 4 times the number of 500 rupee

 

notes. Find the number of notes of each denomination.

The altitude of a triangle is 2/3 of its base. If the altitude is increased by 2cm, and the

 

base is decreased by 2cm, the area of the triangle would remain the same. Find the base and the altitude of the triangle.

Sum of two numbers is 2500. If 4% of one number is 6% of the other, find the numbers.

 ANS:  1. GIVEN: A chocolate bar costs twice as an ice creams yash buys 2 chocolates bars and one ice cream and pays Rs.11.50 .

Required: Find the cost of and icecream bar three chocolate bars.

Let the cost of ice cream = x

Therefore the CP of chocolate = 2x

2*2x +x = 11.50

5x = 11.50

x= 11.50/5 = Rs2.30

the CP of chocolate = 2x = 2*2.30 =Rs 4.60

Thus, the CP of Ice cream = Rs 2.30 and Chocolate = Rs4.60 respectively.

 nice

 ANS: Required : Find three consecutive odd numbers whose sum is 141.

Let the three consecutive odd numbers are x-2 , x and x+2

therefore x-2 +x + x+2 = 141

3x = 141

x= 141/3 = 47

The three consecutive odd numbers are :  x - 2 = 47 -2 = 45 

x= 47and x + 2 = 47 + 2 = 49

Thus, The three consecutive odd numbers are 45,47 &49 respectively.

 ANS: Given :A sum of Rs. 2100 is made of 500, 100 and 50 rupee notes. The number of 50 rupee notes is 2 times the number of 100 rupee notes, and 4 times the number of 500 rupee notes.

Required : Find the number of notes of each denomination.

 ANS: Let the number of Rs 50 notes = x

Then, the no of Rs 100 notes = x/2

and Rs 500 notes = x/4

50 x+ 100*x/2 + 500*x/4 =2100

50x +50x +125x =2100

225x = 2100

x=2100/225 = 9.333333333

The above question can not be solved by the above conditions.

thank you.it really helped me a lot

thanks a lot for helping

ANS:4:  GIVEN : The altitude of a triangle is 2/3 of its base. If the altitude is increased by 2cm, and the base is decreased by 2cm, the area of the triangle would remain the same. Find the base and the altitude of the triangle.

 Let the base of triangle be x

then altitude 2x/3

Area=  1/2 base *altitude

=> x/2*2x/3 = x²/3

On reduction of base by 2cm the base = x-2

and on increase of altitude by 2cm the altitude becomes = 2x/3 +2

therefore area 1(x-2) /2 * (2x/3 +2) = x²/3             [since both the areas are same]

=> (x-2)/2 * (2x+6)/3 = x²/3

=> (x-2)(2x+6) = 2x2 

=>2x²+ 6x - 4x - 12 = 2x²

=>2x - 12 = 0

=> 2x = 12

=> x = 6

therefore, base = 6cm

then, altitude = 2x/3 = 2*6/3 = 4cm

The base and altitude of the tringle are 6cm and 4cm respectively.

let the 3 consecutive odd numbers be x,x+1,x-1

3x=141

x=141/3 = 47

 47,47-1,47+1

46,47,48