Practice Test: Number System- 1 - CAT MCQ

Let us assume that the number with which Anita has to perform the multiplication is 'x'.

Instead of finding 35x, she calculated 53x.

The difference = 53x - 35x = 18x = 540

Therefore, x = 540/18 = 30

So, the new product = 30 x 53 = 1590.

Step 1: Frequency of each light

• Red light: 3 flashes per minute → interval = 60 ÷ 3 = 20 seconds.

• Green light: 5 flashes in 2 minutes = 2.5 per minute → interval = 60 ÷ 2.5 = 24 seconds.

So red flashes every 20 seconds, green every 24 seconds.

Step 2: When do they meet?

They flash together every LCM of 20 and 24 seconds.

• 20 = 2 × 2 × 5

• 24 = 2 × 2 × 2 × 3

• LCM = 2 × 2 × 2 × 3 × 5 = 120 seconds.

So they coincide every 120 seconds = 2 minutes.

Step 3: Count in 1 hour

• 1 hour = 60 minutes.

• Coincide every 2 minutes.

• Number of coincidences = 60 ÷ 2 = 30.

Quick Check (shortcut)

In 2 minutes, red flashes 6 times, green flashes 5 times. They only meet once (at the start).
So in 60 minutes = 60 ÷ 2 = 30 times.

Step 1: What gives trailing zeroes?

A trailing zero comes from a factor of 10 = 2 × 5.
So we need to see how many pairs of (2,5) are present in the product.

Step 2: Look at the set S

S = {2, 3, 5, 7, 11, 13, …, 97} (all primes less than 100).

• Clearly, only one 2 is there (since 2 is prime and appears only once).

• Only one 5 is there (same logic).

Other primes are neither 2 nor 5, so they don’t contribute to factors of 2 or 5.

Step 3: Number of pairs (2,5)

• 2 contributes one factor of 2.

• 5 contributes one factor of 5.
  Together, they form only one pair of (2 × 5).

So the product has exactly one factor of 10.

Step 4: Answer

That means the product ends with exactly 1 trailing zero.

7⁴ = 2401 = 2400+1
So, any multiple of 7⁴ will always end in 01
Since 2008 is a multiple of 4, 7²⁰⁰⁸ will also end in 01

Step 1: Frame the problem

We want even integers n between 100 and 200 (inclusive) that are divisible by neither 7 nor 9.

So:

• First count total even numbers.

• Then remove those divisible by 7 or 9 (using Inclusion–Exclusion).

Step 2: Total even numbers between 100 and 200

Even numbers: 100, 102, …, 200.
This is an AP with first term 100, last term 200, common difference 2.

Number of terms = (200−100)/2 + 1 = 50 + 1 = 51.
So there are 51 even numbers.

Step 3: Count evens divisible by 7

We want numbers divisible by LCM(2,7) = 14.
Between 100 and 200:

Smallest multiple of 14 ≥ 100 is 14×8 = 112.
Largest multiple of 14 ≤ 200 is 14×14 = 196.
So terms are 112, 126, …, 196.

Count = (196−112)/14 + 1 = 84/14 + 1 = 6 + 1 = 7.
So 7 numbers.

Step 4: Count evens divisible by 9

We want multiples of LCM(2,9) = 18.
Between 100 and 200:

Smallest multiple ≥ 100 is 18×6 = 108.
Largest multiple ≤ 200 is 18×11 = 198.

So terms: 108, 126, …, 198.
Count = (198−108)/18 + 1 = 90/18 + 1 = 5 + 1 = 6.
So 6 numbers.

Step 5: Count evens divisible by both 7 and 9

LCM(2,7,9) = LCM(126) = 126.
Between 100 and 200: only 126.

So 1 number.

Step 6: Apply Inclusion–Exclusion

Numbers divisible by 7 or 9 = 7 + 6 − 1 = 12.

So numbers divisible by neither 7 nor 9 = 51 − 12 = 39.

Let n can be a 2 digit or a 3 digit number.
First let n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y Now, Pn + Sn = n Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,.. ,99, so 9 cases .
Now if n is a 3 digit number.
Let n = 100x + 10y + z So Pn = xyz and Sn = x + y + z Now, for Pn + Sn = n ; xyz + x + y + z = 100x + 10y + z ; so. xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (single digit) value.
Hence option D.

Let the common remainder be x.

32506 – x is divisible by n.

34041 – x is divisible by n.

Difference of (32506 – x) and (34041 – x) = (32506 – x) – (34041 – x)

⇒ 32506 – x – 34041 + x

⇒ 32506 – 34041

⇒ 1535 

Factors of 1535 = 1 × 5 × 307 × 1535

3-digit number = 307

⇒ n = 307

∴ The value of n is 307.

Quick verification shortcut:
Difference = 1535. Among options, check divisibility:
1535 ÷ 307 = 5 exactly. So n = 307.

In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20:

⇒ 12 = 2 x 2 x 3
⇒ 15 = 3 x 5
⇒ 18 = 2 x 3 x 3
⇒ 20 = 2 x 2 x 5

∴ LCM = 2 x 2 x 3 x 5 x 3
Since the soldiers are in the form of a solid square. Hence, LCM must be a perfect square.
To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers: 

= 2 x 2 x 3 x 3 x 5 x 5
= 900

• The factors of 1080 which are perfect square:
• 1080 → 2³ × 3³ × 5
• For, a number to be a perfect square, all the powers of numbers should be even number.
• Power of 2 → 0 or 2
• Power of 3 → 0 or 2
• Power of 5 → 0 
• So, the factors which are perfect square are 1, 4, 9, 36.
• Hence, Option B is correct.
   

• Different possibilities for the number of pencils = 12 or 13.
• Since it cannot be divided into his 4 brothers and sisters, it has to be 13.
• The number of erasers should be less than the number of pencils and greater than or equal to 11. So the number of erasers can be 11 or 12.
• If the number of erasers is 12, then the number of pens = 38 - 13 - 12 = 13, which is not possible as the number of pens should be more than the number of pencils.
• So the number of erasers = 11 and therefore the number of pens = 14 

Step 1: What’s being asked?

We have 3 numbers of plants: 363, 429, 693.
We want to arrange one variety per row such that:

• Each row has the same number of plants.

• Total number of rows is minimum.

That means we need the GCD of (363, 429, 693).

Step 2: Find GCD step by step

• GCD(363, 429):
  429 − 363 = 66.
  GCD(363, 66) → 363 ÷ 66 = 5 remainder 33.
  GCD(66, 33) = 33.
  So GCD(363, 429) = 33.

• Now GCD(33, 693):
  693 ÷ 33 = 21 exactly.
  So GCD = 33.

Thus, row size = 33 plants per row.

Step 3: Find number of rows

• 363 ÷ 33 = 11 rows

• 429 ÷ 33 = 13 rows

• 693 ÷ 33 = 21 rows

Total rows = 11 + 13 + 21 = 45 rows.

Quick Trick:
When you see "equal rows" + "minimum rows" → always compute GCD of the given numbers. That gives row size. Then divide each number by GCD to get rows.

• To find three rational numbers between 4 and 5, first express 4 and 5 with the same denominator. Since 4 = 16/4 and 5 = 20/4, any fraction between 16/4 and 20/4 will be a rational number between 4 and 5.
• The numbers 17/4, 18/4, and 19/4 all lie between 16/4 (which is 4) and 20/4 (which is 5).
• Therefore, the three rational numbers between 4 and 5 are 17/4, 18/4, and 19/4.

Step 1: Possible values

Each box can contain between 120 and 144 oranges.
So there are 25 possible values (120, 121, …, 144).

Step 2: Distribute boxes

We have 128 boxes.

• If each of the 25 values appears in 5 boxes, that accounts for 25 × 5 = 125 boxes.

• After that, 3 boxes remain.

Step 3: Placement of remaining boxes

Those 3 boxes must also take values from 120 to 144.

• Already, each number is used 5 times.

• Distributing the last 3 boxes means at least one number will appear a 6th time.

Thus, the maximum frequency of boxes with the same number of oranges is at least 6.
Therefore, the minimum value of X = 6.

Quick Verification

• If each value appears 5 times, that covers 125 boxes.

• 3 boxes are left → at least one value will appear the 6th time.
  So minimum X = 6.

Step 1: Divisibility rule for 4

A number is divisible by 4 if its last two digits form a number divisible by 4.
Example: 312 ends with 12, and 12 is divisible by 4, so 312 is divisible by 4.

So our task = find all valid last two digits from {1–6}.

Step 2: Which 2-digit endings (from digits 1–6) are divisible by 4?

Check systematically:

• Numbers ending with 2: 12, 32, 52 (divisible by 4).

• Numbers ending with 6: 16, 36, 56 (divisible by 4).

• Numbers ending with 4: 24, 64 (divisible by 4).

So valid endings = {12, 16, 24, 32, 36, 52, 56, 64}.
Total = 8 endings.

Step 3: Build the 5-digit number

We fix the last 2 digits first (e.g., take 12).
Now 2 digits are used → 4 digits left from the set.

We need only 3 of them (because total number length = 5).

Number of ways to arrange 3 digits in the first 3 places = 4 × 3 × 2 = 24.

So for each valid ending, there are 24 numbers.

Step 4: Multiply

Valid endings = 8.
Each gives 24 numbers.
So total = 8 × 24 = 192.

We are told:

• First divide the number by 3 → remainder 2.

• Then divide that quotient by 4 → remainder 1.

• Then divide that quotient by 7 → remainder 4.

We need the remainder when the original number is divided by 84.

Step 1: Start from the last division

Quotient after dividing by 3 and 4 is divided by 7 → remainder = 4.
So this quotient = 7k + 4.

Step 2: One step back

That quotient came from division by 4.
So, quotient after dividing by 3 = 4(7k + 4) + 1 = 28k + 17.

Step 3: Go to the original number

Original number = 3 × (28k + 17) + 2 = 84k + 53.

Step 4: Divide by 84

So the number always looks like 84k + 53.
That means remainder on dividing by 84 is 53.

To determine the base system used by the teacher, we will analyze the numbers provided in the context of different base systems. The numbers given are 24 boys, 32 girls, and a total of 100 students. We will convert these numbers from their respective base systems to decimal (base 10) and check if the sum equals 100.

We are provided with the equation (32) + (24) = (100). Let us assume our base be 'b'
Then,we can say:

⇒ 32 = 3 x b¹ + 2 x b⁰ = 3b+2
⇒ 24 = 2 x b¹+ 4 x b⁰ = 2b+4
⇒ 100 = 1 x b² + 0 x b¹ + 0 x b⁰ = b²

Now, according to our question:

⇒ 32 + 24=100
⇒ (3b + 2) + (2b + 4) = (b²)
⇒ 5b + 6 = b²
⇒ b² - 5b - 6 = 0
⇒ b² - 6b + b - 6 = 0
⇒ b(b - 6) + 1(b - 6) = 0
⇒ (b - 6) * (b + 1) = 0
⇒ b = 6,- 1

Base can't be negative. Hence b = 6.
∴ Base assumed in the asked question must be 6.

The correct option is A

16
'abc' has 2 factors.
This means 'abc' is a prime number (Only a prime number can have exactly 2 factors).
Now, 'abcabc' = 'abc'×1001
'abcabc' = 'abc' × 7 × 11 × 13
Since 'abc' is prime we can write 'abcabc' as - p¹×7¹×11¹×13¹
No. of factors = (1+1) (1+1) (1+1) (1+1) = 16 factors.

The given number = 8, 9, 12, 15

Concept used: LCM: The smallest number which is divisible by two or more given numbers.

Calculation: LCM of 8, 9, 12 and 15 is 360

Now, according to the question, adding the remainder in LCM, we get ⇒ (360 + 1) = 361

∴ The required least number is 361.