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.

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.

We have a + c = e so possible summation 6+4=10 or 4+2 = 6.
Also b = 2d so possible values  4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .

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

- We need to find even integers n between 100 and 200.
- The even integers in this range are: 100, 102, ..., 200. There are 51 even numbers.
- We check for divisibility:
- Even numbers divisible by 7: 104, 112, 120, 126, 140, 154, 168.
- Even numbers divisible by 9: 108, 126, 144, 162, 180, 198.
- Even numbers divisible by both (14): 126.
- Using inclusion-exclusion:
- Total divisible = 7 + 6 - 1 = 12.
- Thus, even integers not divisible by either = 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.

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 

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.

Each box contains at least 120 and at most 144 oranges.
So boxes may contain 25 different numbers of oranges among 120, 121, 122, .... 144.
Lets start counting. 1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.
Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.
Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.
Hence the number of boxes containing the same number of oranges is at least 6.

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4.
So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number.
Remaining 3 digits out of 4 can be selected in ways and further can be arranged in 3! ways .
So in total = 8*4*6 = 192

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.

let's say number is x.
  
x =1584