Free Algebra Miscellaneous - 02 Practice Test - CAT 

OPTION (d)

for n≥10, n! is divisible by 100, hence the 10s digit will be zero for numbers greater than 10!

Hence we will have to find the 10's digit of numbers form 1! to 9!

we see that 1! ends with 1

2! ends with 2.... hence sum of 1! to 9! the last two digits will be 13 hence the option is a

Option(b)

To find the average of all numbers, we need to find the sum of all the possible numbers and divide it by the total such numbers possible.The sum of all possible numbers can be found using the formula

(n-1)!(sum of digits)(1111…n times)

here n=5

(5-1)!(15)(11111)= 4!(15)(11111)

Total number of such possible cases= 5!

Average= 3999960/5! = 33333

Short cut: - If we arrange given numbers, These numbers will be in A.P.

Ao, A.M. of these numbers = (First term + last term)/2

(12345+54321 )/2 = 33333

Option(c) 

This is S -> D type of grouping with a lower limit of 1

N+L+EL+ES= 20

N,L,EL,ES ≥ 1

Removing 4 from each side

N+L+EL+ES=16

Now, N, L, EL, ES ≥ 0

This is the arrangement of 16 zeroes and 3 ones. Answer = ¹⁹C₃

Similar to different grouping( permutation and combination)

a,b,c,d will all be some power of 5 from 0 to 12.
This question is same as a + b + c + d = 12
Solutions = ¹⁵C₃ 

12 identical things to 4 distinct groups = ¹⁵ C₃ = 455

Option (c)

If someone is going from P to Q. He has to travel 4 times vertically and 5 times horizontally. Ultimately, we are arranging 4 - V's and 5 - H's .

Number of ways to arrange 4 - V's and 5 - H's = 9!/(4! x 5!) = ⁹C_{4 = ⁹C5}

Or, There are 4 rows and 5 columns. Using the one-zero method, number of paths will be ⁹C₄ = 126 ways

option (d)

50 runs can be scored from 12 balls as follows 

Case 1:- 11 fours and 1 six = ¹²C₁₁^{. 1}C1

Case 2:- 1 zero + 8 fours + 3 sixes = ¹²C₈ . ⁴C₃

Case 3:- 2 zeroes+ 5 fours+ 5 sixes = ¹²C₅.⁷C₅

Case 4: - 3 zeroes + 7 sixes + 2 fours = ¹²C₃.⁹C_{2 .}⁷C₇

Answer = sum of all 3. option- d

We need to arrange 2 Cs, O, 3Ns, V, 3Es and I. This is permutation with repetition.

It can be done in = 20160

Hence option (c)

NN_ _ _ _ _ _ _EE

The 7 places have to be filled by 2 Cs, O, N, V, E and I. This can be done in 7!/(2!)

Hence option (e)

Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute k prizes among p students?".

 For each prize, there are p possibilities for the recipient of that prize. In total, that means there are p⋅p⋅.....⋅p (with k copies of p in the product) possible ways to assign the prizes. This is the p^k part.

or, we can say.  This is a type of “Different to Different” question.

5 prizes can be distributed among 4 students in 4⁵ ways.

As, we know sum of the n terms in an A.P.

(S_n)

In the given question S_n = 720 ;a = 12; d = 17; n =?

720 =

Solving we get n = 9. 9 terms are needed to get the same

Hence option (d)

Seventh term of the AP = a + 6d

Eleventh term of the AP = a + 10d

7(a + 6d) = 11(a + 10d)

-> 7a + 42d = 11a + 110d

-> 4a + 68d = 0

-> a + 17 d = 0

So, the eighteenth term is zero. Hence option (b)

p_th term = a + (p – 1) d = q ... (i)

q _th term = a + (q – 1) d = p ... (ii)

Subtracting (ii) from (i)

(p – 1) d – (q – 1) d = q – p

->pd – d – qd + d = q – p

->d (p – q) = q – p

-> d = -1a = q – (p – 1)d = q + (p – 1)

So, m_th term = q + p – 1 – (m – 1) = q + p – 1-m + 1 = p + q – m

Hence option (b)

Alternatively, use assumption

Let the first term=2 and the 2^nd term=1, therefore p=1 and q=2

Let m=3

Thus, 3^rd term=0. Substitute the values of p,q and m in the answer options and look for 0. Only option (b) gives this value

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.

Unit’s place can be filled in only 3 ways.

So, total number of nos. Possible = 4* 5*4*3 = 240

Hence option (b)

When the repetition is allowed:

All the places: Hundred’s, ten’s and unit’s can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 i.e. in 6 ways. Thousand’s place can be filled only in 4 ways. Hence no. numbers possible = 4*6*6*6 = 864.

Hence option (d)