Test: Number System (January 9) - CAT MCQ

• 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 

The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66.

Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.

66 need not to be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide

429 and also divides 693. Hence, 33 is the correct answer for the size of each row.

For how many rows would be required we need to follow the following process:

Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.

Therefore, the correct answer is A

• There are several rational numbers between 4 and 5. The numbers are between 16/4 and 20/4. 
• Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.

Solve this question step by step:

1. Any number formed by this method is clearly divisible by 3.
2. Since it needs to be a square, it should be divisible by (3)^[2*k]. k varies over the natural numbers.
3. Now consider the original number. It has hundred 1’s, hundred 2’s and hundred 3’s. Sum of these digits is 600. This is not divisible by 9. Hence number is not divisible by 9.
4. If a number is divisible by (3)^[2*k], it is divisible by 3^k.
5. This number is not divisible by 3^k for any k > 1. 

Hence it is not a perfect square for any arrangement.

Given:

73 × 75 × 78 × 57 × 197 × 37 is divided by 34

Calculation:

73 × 75 × 78 × 57 × 197 × 3734

We have taken individual remainder like

When 73 is divided by 34 gives remainder is 5

Similarly

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours. So the correct option is B

Note: In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K.

Let k = 1; the number becomes 53

If it is divided by 84, the remainder is 53.

Therefore, the correct answer is Option D.

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.