Quiz 1

The given series is an AP with common difference of 7. So the terms in the
above series are in the form of 2013 + 7k.
We have to find the perfect squares in this format in the given series.
Given that $44^{2}$ = 1936.
Shortcut:

To find the next perfect square,
add 45th odd number to $44^{2}$
So $45^{2}$ = 1936 + (2 x 45 -1) = 2025
46^{2}[/latex] = 2025 + (2 x 46 - 1) = 2116
$47^{2}$ = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.

The given series is 1234, 12344, 123444, 1234444, .....
So the number of digits in each
term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms = 3n+n(n+1)/2
So 3n+n(n+1)/2 ≤200
For n = 16, We get 184 in the left
hand side. So after 16 terms
the number of digits equal to 184.
And 16 them contains 16 + 3 = 19 digits.
Now 17 term contains 20 digits and
123444......4 (17times). So the last digit is 4 and the last two digits are 44.

The last two digits in the 200th place is 55.

The last two digits in the 200th place is 55.

The number of letters in each term is in AP. 1, 2, 3, ...
So n(n+1)2≤120n(n+1)2≤120
For n = 15, we get LHS = 120. So 15th letter in the alphabet is O. So 15th term contains 15 O.

Consecutive squares are subtracted from the numbers.
70 - 54 = 16
54 - 45 = 9
45 - 41 = 4
So next we have to subtract 1.
So answer = 41 - 1 = 40

11,23,47,83,131
23–11 = 12
47–23 = 24
83–47 = 36
131–83 = 48
Therefore, 131+60=191

Problem-solving = 14 letters word.
So divide 2015 by 14 and find the remainder.
Here remainder is 13. so 13th letter in problem-solving is n.

Let the boys = b and girls = g
Given b/(g−12) = 2/1
Substitute b = g in the above equation. g = 24.
So total students = 24 + 24 = 48

As the number of children is in arithmetic progression starting with 2,
the five families have 2, 3, 4, 5, 6 kids respectively.
As each child has kept the pets equal to the number of kids in the
family, each family has n2 pets.
So total = 22+32+42 +52 +62 = 90