Arithmetical Problem : Different types of problem Formula and Tricks with Problem (Question) and Solutions

Example 1.Ashok’s mother was 3 times as old as Ashok 5 years ago. After 5 years she will be twice as old as Ashok. How old is Ashok today?

Ans-

Supose the present age of Ashok is x years and that of his mother is y years

5 years ago 

3 (x – 5) = (y – 5)

3x – 15 = y – 5

3x – y = 10 …..(i)

5 years hence,

2 (x + 5) = (y + 5)

 2x + 10 = y + 5

2x – y = – 5 …..(ii)

From equations (i) and (ii)

x = 15 years

Example 2.Two years ago a mother was 4 times as old as her son. Six years from now her age will become more than double her son’s age by 10 years. What is the present ratio of their ages?

Ans-

Suppose the present age of son = x years

Present age of mother= y years

According to question.

4 (x – 2) = y – 2

4x – 8 = y – 2

4x – y = 6 …..(i)

2 (x + 6) + 10 = y + 6

 2x + 12 + 10 = y + 6

 2x – y = – 16 …..(ii)

From equations (i) and (ii)

x = 11

y = 4 × 11 – 6 = 38

Required ratio 

= 38 : 1

Example 3.The weights of 4 boxes are 70, 100, 20 and 40 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once.

(a) 230

(b)190

(c) 160

(d) 200

Sol-

According to options,

1st option = 70 + 100 + 20+ 40= 230

2nd option= 70 + 100 + 20= 190

3rd option = 100 + 20 + 40= 160

But in option (d) the sum of any numbers given in question is not 200.

So, option (d) is different from other options.

Example 4.The weights of 4 boxes are 80, 60, 90 and 70 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes and in a combination a box can be used only once.

(a) 300

(b) 230

(c) 220

(d) 290

Sol-

According to options,

1st option = 80 + 60 + 90 + 70= 300

2nd option= 80 + 60 + 90 = 230

3rd option = 60 + 90 + 70 = 220

And in option (d) the sum of any numbers given in option is 290.

So, option (d) is different from other options.

Example 5.If 3 @ 3 * 3 = 3 and 48 @ 4 * 3 =36, then 91 @ 13 * 2 = ?

(a) 4

(b) 8

(c)10

(d) 14

Sol-

(3 ÷ 3) × 3 = 3

(48 ÷ 4) × 3 = 36

(91 ÷13) × 2 = 14

Example 6.If 75 $ 26 = 4, 69 $ 53 = 7 then what is the value of 82 $ 46 = ?

(a) 62

(b) 56

(c) 0

(d) 91

Sol-

(7 + 5) – (2 + 6) = 4

(6 + 9) – (5 + 3) = 7

(8 + 2) – (4 + 6) = 0

Example 7.In a row, 25 trees are planted at equal distance from each other. The distance between 1st and 25th tree is 48 m. What is the distance between 3rd and 15th tree?

(a) 8m

(b) 15m

(c) 16m

(d) 24m

Sol-

The distance between the two trees =48/24=2 metres

The distance between 3rd and 15th trees=2 x 12 = 24 meters