Section-1

1

The number of possible rearrangements of the

number 1421 = 4!/2! = 12

Now, consider only the units place.

1 occurs 6 times in the unit’s place.

2 occurs 3 times in the unit’s place.

4 occurs 3 times in the unit’s place.

The total sum of digits of the units place in all the
12 arrangements combined

= 1 × 6 + 2 × 3 + 4 × 3 = 24

Similarly, the sum of digits in ten’s, hundred’s or
thousand’s place will also be 24.

So the sum of all the 12 rearrangements
= 24 × 1111

Hence, the average of all these rearrangements
= 24 × 1111/12 = 2222.

548

The common difference of the AP 38, 55, 72, …
is 17.

The smallest 3-digit number in the AP is 106 and
the largest 3-digit number in it is 990.

So 106 + 123 + … + 973 + 990

= 106 + (106 + 17) + (106 + 2 × 17) + … +

(106 + 52 × 17)

= 53 × 106 + 17(1 + 2 + 3 + … + 52)

= 53 × 106 + 17 × 52 × 53/2

= 53 × 106 + 17 × 26 × 53

Hence, average of all 3-digit numbers

= (53 × 106 + 17 × 26 × 53)/53 = 106 + 442 = 548.

4

3

less than the sum of the lengths of the other sides.

Let the fourth side be x.

**Case 1:** x is the largest side.

x < 1 + 2 + 4 or, x < 7

So x can take the values 4, 5, 6.

**Case 2:** 4 is the largest side and x is less than 4.

4 < x + 1 + 2 or, x > 1

So x can take the values of 2 or 3.

Hence, x can take 5 values i.e., 2, 3, 4, 5, 6.

1

150

Let the number of students in the school be N.

N < 5000

N leaves a remainder of 4 when divided by 9, 10,

12, or 25.

N leaves a remainder of 4 when divided by LCM (9,

10, 12, 25) = 900.

N leaves a remainder of 4 when divided by 900.

N = 900x + 4

Since N < 5000, x can take range from 0 to 5.

But 900x + 4 is a multiple of 11 only when x = 2.

So N = 900 × 2 + 4 = 1804

Since 1804 = 12 × 150 + 4

Hence, when we divide these 1804 students into
groups of 12, we get 150 groups.

1

Glass contains 500 cc of milk and cup contains
500 cc water.

After first transfer, glass contains 500 – 150
= 350 cc milk and the cup contains 150 cc milk
and 500 cc water.

Now, the ratio of milk and water in the cup
= 150 : 500 = 3 : 10

After second transfer, water in the glass
= 150 × 3/13 cc and milk in the cup will be same
i.e., 150 × 3/13 cc.

Hence, required ratio = 1 : 1.

2

3

9

14

12

2

1

63

1

For x = r,

f(r) = 2r – r = r

So f(f(r)) = f(r) = r

For x < r,

f(x) = r

f(f(x)) = f(r) = r

For x > r,

Let x = kr, where k is some number greater than 1.

f(x) = f(kr) = 2kr – r = (2k – 1)r

Since k > 0, 2k – 1 > k

So for any x > r, f(f(x)) ≠ f(f(r))

Hence, for f(f(x)) = f(x) to be satisfied, the necessary
condition is, x ≤ r.

1

Speed of slower car = 60 km/h and they take 1.5
hours if they travel towards each other, to meet
each other after starting at the same time.

In this case the distance covered by the slower
car = 60 × 1.5 = 90 km.

11

Since Bob can finish a job in 40 days.

Then, Alex and Cole can finish this job in 20 and
60 days respectively.

Total work = LCM (20, 40, 60) = 120 units
Alex, Bob and Cole can finish 6, 3 and 2 units of
the job per day.

In 3 days, they can finish 2(6 + 3 + 2) = 22 units
In 15 days, they can finish 5 × 22 = 110 units
Out of remaining 10, 9 units of work can finish by
Alex and Bob and 1 unit by Bob and Cole.

Hence, the total number of days Alex would have
worked when the job gets finished, is
2 × 5 + 1 = 11 days.

2

2

60

3

The sum of the two smallest numbers = 14 × 2
= 28

The sum of the two largest numbers = 28 × 2 = 56
To maximize the average of all the 6 numbers, we
must try to maximize the two numbers in between.

This is possible when the two largest numbers are
27 and 29.

The maximum average case is

a, b, 25, 26, 27, 29; where a + b = 28v
Hence, the average of these 6 numbers

= (a + b + 25 + 26 + 27 + 29)/6

= (28 + 25 + 26 + 27 + 29)/6 = 22.5.