Number System- 1 - Free MCQ Practice Test with solutions, CAT Quant Aptitude

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.

• 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.

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.

• 1 hour = 60 minutes.

• Coincide every 2 minutes.

• Number of coincidences = 60 ÷ 2 = 30.

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.

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.

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.

• 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.

That means the product ends with exactly 1 trailing zero.

Calculation:

Let the second number be m.

Using the relationship between HCF, LCM, and the numbers, we have:

• m × 77 = 11 × 616
• Solving for m: m = 616 / 7
• m = 88

Therefore, the second number is 88.

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are the same. Again, in the same set, there are 10 numbers whose hundreds and tens digits are the same.

But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are the same as 111, 222, 333, 444, etc.
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further, there are 9 such sets of numbers.
Therefore such total numbers = 19 × 9 = 171

Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171

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.

A and B are playing a game with integers from 1 to 100, placing signs between them. A starts by putting a plus sign. The result is determined by whether the final sum is even or odd. If even, A wins; if odd, B wins.

The analysis of the game is as follows:

• All numbers from 1 to 100 include 50 odd numbers.
• Adding any two odd numbers results in an even number.
• The sum of 50 odd numbers will also be even.

Therefore, no matter how the signs are placed, the result will always be even. Thus, A will always win the game.

100a + 10b + c
c = 2b → b = c/2
c = 1.5a → a = c/1.5
c/1.5 + c/2 + c = 13
6.5c = 39
c = 6, b = 3, a = 4 ⇒ 436

Thus, the original number is 63.

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).

• 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.

• 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.

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

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.

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.

• 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.

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}.

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.

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.

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.

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

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

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

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.