Quiz 2

Solution:

The sequence,

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4  

Number of oranges put in = Number of times 2 was ordered = 6

Number of oranges taken out = Number of times 4 was ordered = 4

Number of oranges at the end of the sequence  = 6 – 4 = 2

Hence, option 1.

Solution:

Number of fruits put in the basket = Number of times of orders 1, 2 or 3 = 19  

Number of fruits taken out = 2 fruits taken out for every order for 4 = 2 × 4 = 8

Number of fruits in the basket after the sequence  = 19 – 8 = 11

Hence, option 2 is the correct one.

Solution:

The sum of the first and last terms in T is 470.

Likewise, the sum of the second and second-last terms is also 470.

Since the sum of the nth term from the beginning and the nth term from the end is 470.  

Only one of each of these pairs of terms will be in S. (For instance only one of 3 and 467 can be in S)

The set S can have a maximum of half of the terms in T.

The terms in T are in A.P. with a common difference of 8.

Last Term = 467 = 3 + (n − 1) × 8 n = 59                       

Total number of terms in the set T = 59

There are 29 pairs of numbers in T that add up to 470 and the 59th number is 235, which occurs in the middle of the series.

• S will be a set with 30 terms, with 29 terms which are from the pairs adding up to 470, and 235.

Solution: 

Let a₁, a₂, a₃, …, a_n be n positive real numbers. Now, a₁ × a₂ × a₃ × … × a_n = 1 We know that, A.M. ≥ G.M.

a₁ + a₂ + a₃ + … + a_n ≥ n × (a₁ × a₂ × a₃ × … × a_n)^1/n

• a₁ + a₂ + a₃ + … + a_n ≥ n × (1)^1/n a₁ + a₂ + a₃ + … + a_n ≥ n

The first layer has 1 ball.

Second layer has 1 + 2 = 3 balls

Third layer has 1 + 2 + 3 = 6 balls

The n^th layer of the stack would have balls.

 

Total balls in all the layers

Only n = 36 satisfies the above equation.

Hence, option 3 is correct.

Solution:

Assuming that the first term of the progression is a and the common difference is d.

T₃ + T₁₅ = T₆ + T₁₁ + T₁₃

(a + 2d) + (a + 14d) = (a + 5d) + (a + 10d) + (a + 12d)   a + 11d = 0

But, T₁₂ = a + 11d

• The 12^th term of an arithmetic progression is 0.

Solution:

In the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e,...

the first letter of the alphabet is written once, the second is written twice, and the n^th letter is written n times.

∴ The number of letters written up to the n^th letter is equal to the sum of the first n natural numbers given by, n(n + 1)/2

 

For n = 23, n(n + 1)/2 = 276 and for n = 24, n(n + 1)/2 = 300

Hence the series contains 276 letters in all for the letter corresponding to n = 23 and 300 letters in all for the letter corresponding to n = 24.

∴ The letter corresponding to n = 24 will be the letter occupying the 277^th to the 300^th place in the series.

But, n = 24 corresponds to letter x.

∴ The 288^th letter in the series is x.

Hence, option 4 is the correct one.

Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first

term and 1000 as the last. Let d be the common difference. Then, 1000 = 1 + (n – 1) × d ∴ 999 = (n – 1) × d ... (i) 999 = 1 × 33 × 37 ∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999 Substituting in equation (i) If d = 1, n = 1000 If d = 3, n = 334 If d = 9, n = 112 If d = 27, n = 38 If d = 37, n = 28 If d = 111, n = 10 If d = 333, n = 4 If d = 999, n = 2, which is not possible as n > 2 ∴ 7 arithmetic progressions can be formed. Hence, option 4.

11/2 (2a + 10d) = 19/2 (2a + 18d)
22a + 110d = 38a + 342d ⇒ 16a + 232d = 0 ⇒ 2a + 29d = 0. Ans.(1)

Sum of the digits of n is 2.
For 10 < n < 102
total different possible values of n = 11, 20 i.e., 2
For 102
< n < 103
total different possible values of n = 101, 110, 200 i.e.,
3
.
.
.
For 1010 < n < 1011 total different possible values of n = 10000000001,
10000000010, 10000000100, 10000001000, 10000010000,
10000100000, 10001000000, 10010000000, 10100000000,
11000000000, 20000000000 i.e., 11. Ans.(1)