# Free Algebra Miscellaneous - 02 Practice Test - CAT

### Question 1

Find the 10’s digit of 1!+2!.....30!.

#### SOLUTION

Solution :A

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

### Question 2

What is the average of all five digit numbers that can be formed using all the digits 1, 2, 3, 4, 5 exactly once?

#### SOLUTION

Solution :B

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

### Question 3

In the Lakme annual sale, Disha, in order to keep her expenses in control, decided to buy 20 beauty products from nail polishes, lipsticks, eye-liners and eye-shadows, selecting atleast one from each. In how many ways can she make this selection?

^{20}C

_{3}

^{19}C

_{4}

^{19}C

_{3}

^{20}C

_{4}

#### SOLUTION

Solution :C

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 =

^{19}C_{3}

### Question 4

How many solutions are possible for abcd = 5^{12,} where a, b, c and d are natural numbers?

#### SOLUTION

Solution :B

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 =^{15}C_{3}12 identical things to 4 distinct groups =

^{15}C_{3}= 455

### Question 5

A grid is set up as shown using 5 horizontal and 6 vertical lines. What are the no. of ways one can go from point P to point Q walking along the grids but not moving upwards or left or retracing a grid?

#### SOLUTION

Solution :C

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!) =

^{9}C_{4 = 9C5}Or, There are 4 rows and 5 columns. Using the one-zero method, number of paths will be

^{9}C_{4}= 126 ways

### Question 6

Abhay choses to bat and he scores exactly 50 runs in 12 balls. He makes only zeroes, fours and sixes in these 12 balls. In how many different ways can he do so?

^{12}C

_{11}

^{12}C

_{10}+

^{12}C

_{11}.

^{2}C

_{1}+

^{12}C

_{9}.

^{3}C

_{2}

#### SOLUTION

Solution :D

option (d)

50 runs can be scored from 12 balls as follows

Case 1:- 11 fours and 1 six =

^{12}C_{11}^{. 1}C1Case 2:- 1 zero + 8 fours + 3 sixes =

^{12}C_{8}.^{4}C_{3}Case 3:- 2 zeroes+ 5 fours+ 5 sixes =

^{12}C_{5}.^{7}C_{5}Case 4: - 3 zeroes + 7 sixes + 2 fours =

^{12}C_{3}.^{9}C_{2 .}^{7}C_{7}Answer = sum of all 3. option- d

### Question 7

In how many ways can the letters of the word “CONVENIENCE” be arranged?

#### SOLUTION

Solution :C

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)

### Question 8

In how many ways can the letters of the word “CONVENIENCE” be arranged so that they begin with 2Ns and end with 2Es?

#### SOLUTION

Solution :E

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)

### Question 9

Determine the number of ways in which 5 prizes can be distributed among 4 students.

^{5}

^{4}

#### SOLUTION

Solution :A

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 4

^{5}ways.

### Question 10

How many terms of the AP 12, 29, 46... are needed to have a sum of 720?

#### SOLUTION

Solution :D

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)

### Question 11

If 7 times the seventh term of an A.P. is equal to 11 times its eleventh term, the value of eighteenth term of the A.P. is:

#### SOLUTION

Solution :B

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)

### Question 12

If the p_{th} term of an AP is q and its q_{th} term is p then its m_{th} term is:

#### SOLUTION

Solution :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 – mHence option (b)

Alternatively, use assumption

Let the first term=2 and the 2

^{nd}term=1, therefore p=1 and q=2Let 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

### Question 13

None of the digits are repeated?

#### SOLUTION

Solution :B

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)

### Question 14

Digits can get repeated?

#### SOLUTION

Solution :D

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)

### Question 15

How many integral values of x satisfies the below inequality?

(x + 2)(x + 6) < 40

#### SOLUTION

Solution :D

-10, -9, ..... 0, 1, 2.

Total 13 values satisfies the given inequality.