Arithmetic Reasoning Arithmetic Reasoning - Expert Level

First ten odd numbers form an arithmetic progression of the form 1,3,5,7,9......

here a = first term = 1

d = common difference = 2

Average of first n numbers = (2a + (n - 1)d)/2

n th term of the AP = a + (n - 1)d

Substituting a = 1, d = 2 and n = 10 in the above formulas

average of first 10 numbers = 10

10 th term of the AP = 19

Therefore average of first 10 terms is 19 - 10 = 9 greater than the last term. Hence the average is greater than the sum by

9/19 X 100 = 900/19 %

15 girls are absent => b = 2(g-15)...... eqn...(1)

45 boys are absent => g - 15 = 5(b-45).....eqn...(2)

=> g = 5(b-45) + 15

=> b = 2((5(b-45)+15)-15)

=> b = 2(5b - 225 + 15 - 15)

=> b = 2(5b-225)

=> b = 10b - 450

=> 9b = 450

=> b = 50

so from equation(2)

g - 15 = 5(b - 45)

=> g - 15 = 5(50-45)

=> g - 15 = 5(5)

=> g = 25 + 15

=> g = 40

Therefore, the number of boys = 50 and girls = 40

Arithmetic mean = sum/members

=> (1x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7) / (1 + 2 + 3 + 4 + 5 + 6 + 7)

=> 140/28 = 5

Then, A:B = 2:3 and B:C = 5:8

= (5 x 3/5) : (8 x 3/5) = 3:24/5

A:B:C = 2:3:24/5

= 10:15:24 => B = 98 x 15/49 = 30.

It is given that the first digit is 1/3 of the second. There are 3 such possibilities

1)1 and 3

2)2 and 6

3)3 and 9

Now, the third digit is the sum of the first and second digits.

1)1 + 3 = 4

2)2 + 6 = 8

3)3 + 9 = 12

It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option.

Hence, the answer is 1349.

We know that in A.P series

18th term = a + 17d

29th term = a + 28d

But given

a + 17d = 29..........(i)

a + 28d = 18......... (ii)

Solving equation (i) and (ii), we get

d = -1

put d = -1 in any of the above equations and we get,

a = 46

Now, we know 49th term can be written as, a + 48d

putting the value of a and d in the above equation,

a + 48d = 46 + 48(-1)

= 46 - 48

= -2

Hence, the 49th term is -2.

x – y = 5

4x – 6y = 6

x = 12 y = 7

Let the number be x. Then,

- 4x = x + 50

- 5x - 50 = 0

(3x + 10)(x - 5) = 0

x = 5

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

Let Varun's age today = x years.

Then, Varun's age after 1 year = (x + 1) years.

Therefore x + 1 = 2 (x - 12) => x + 1 = 2x - 24 => x = 25.