CUET LEVEL - 1 JEE ( PYQ ) TEST - 1

A letter lock consists of three rings, each marked with ten different letters,

In how many ways is it possible to make an unsuccessful attempt to open the lock ?

• 999

• 888

• 777

• 666

Answer explanation

Two rings may have the same letter at a time, but the same ring cannot have two letters at a time. Therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways.

Therefore, the total number of attempts = 10 × 10 × 10 = 1000.

But out of these 1000 attempts, only one attempt is successful.

Therefore, the required number of unsuccessful attempts = 1000 – 1 = 999.

The total number of positive integral solution for x, y, z such that x* y * z = 24, is ________.

• 30

• 40

• 50

• 60

Answer explanation

We have,

x* y * z = 24

x* y * z = 2³ × 3¹

The number of ways of distributing ‘n’ identical balls into ‘r’ different boxes is ^{(n + r − 1)}C_{(r − 1)}

Here we have to group 4 numbers into three groups

Number of integral positive solutions

= ^{(3 + 3 − 1)}C_{(3 − 1)} × ^{(1+ 3 − 1)}C_{(3 − 1)}

= ⁵C₂ × ³C₂

= 30

Find the total number of signals that can be made by five flags of a different colour when any number of them may be used in any signal.

• 325

• 425

• 564

• 879

Answer explanation

Case I: When only one flag is used. No. of signals made = ⁵P₁ = 5.

Case II: When only two flags are used. Number of signals made = ⁵P₂ = 5 * 4 = 20.

Case III: When only three flags are used. Number of signals is made = ⁵P₃ = 5 4 3 = 60.

Case IV : When only four flags are used. Number of signals made = ⁵P₄ = 5 4 3 * 2 = 120.

Case V : When five flags are used. Number of signals made = ⁵P₅ = 5! = 120.

Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.

Prove that if each of the ‘m’ points in one straight line is joined to each of the n points on the other straight line, excluding the points on the given two lines. The number of points of intersection of these lines is

• none of these

Answer explanation

To get one point of intersection, we need two points on the first line and two points on the second line. These can be selected from n-points in ⁿC₂ ways and m points in ^mC₂ ways.

Therefore, the required number = ^mC₂ × ⁿC₂ = (m(m-1))/2!  × (n(n-1))/2!

Two numbers are chosen from 1, 3, 5, 7,… 147, 149 and 151 and multiplied together in all possible ways. The number of ways which will give the product a multiple of 5 is ______.

• 1020

• 1039

• 1049

• 2031

Answer explanation

In the numbers 1, 3, 5, 7,…, 147, 149, 151, the numbers multiples of 5 are 5, 15, 25, 35, …, 145, which form an arithmetic sequence.

T_n = a + (n − 1) × d

145 = 5 + (n – 1) × 10

n = 15

and if the total number of terms in the given sequence is m, then

151 = 1 + (m – 1) × 2

m = 76

So, the number of ways in which a product is a multiple of 5 = (both two numbers from 5, 10, 15, 20, 25, 30, 35, … , 150) or (one number from 5, 10, 15, 20, 25, 30, 35, …, 150 and one from remaining numbers)

= ¹⁵C₂ + ¹⁵C₁ × ^{(76 – 15)}C₁

= 105 + 15 × 61

= 105 + 915

= 1020

If the letters of the word SACHIN are arranged in all possible ways, and these words are written in a dictionary, at what serial number does the word SACHIN appear?

• 902

• 349

• 601

• 602

Answer explanation

(1) Alphabetical order is A, C, H, I, N, S

No. of words starting with A – 5!

No. of words starting with C – 5!

No. of words starting with H – 5!

No. of words starting with I – 5!

No. of words starting with N – 5!

SACHIN-1

Hence, the required number = 5! + 5! + 5! + 5! + 5! + 1 = 120+ 120 + 120 + 120 + 120 + 1 = 601.

Assuming the balls to be identical except for the difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is ______.

• 987

• 879

• 576

• 986

Answer explanation

Number of white balls = p = 10

Number of green balls = q = 9

Number of black balls = r = 7

Total ways of selection = (p + 1) (q + 1) (r + 1) – 1

= [11 10 8] – 1

From 6 different novels and 3 different dictionaries, 4 novels and a dictionary are to be selected and arranged in a row on the shelf such that the dictionary is in the middle. What is the number of such arrangements?

• 70

• 67

• 89

• 90

Answer explanation

Since, 38808 = 8 × 4851

= 8 × 9 × 539

= 8 × 9 × 7 × 7 × 11

= 2³ × 3² × 7² × 11

So, the number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72.

This includes two divisors, 1 and 38808.

Hence, the number of divisors except 1 and n = 72 – 2 = 70.

An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is _______.

• 7

• 8

• 9

• 10

Answer explanation

Since any of the digits 2, 5 and 7 can be used at any place, the total number of such positive n-digit numbers are 3ⁿ.

Since we have to form 900 distinct numbers, 3ⁿ ≥ 900 ⇒ n = 7.

In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?

• 2890

• 2880

• 2467

• 2345

Answer explanation

Since the total number of ways in which boys can occupy any place is (5 − 1)! = 4!

5 girls can sit accordingly in 5! ways.

Hence the required number of ways = 4! × 5 !

= 24 × 120

= 2880