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Test: Number System (January 8) - CAT MCQ


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10 Questions MCQ Test - Test: Number System (January 8)

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Test: Number System (January 8) - Question 1

Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

Detailed Solution for Test: Number System (January 8) - Question 1

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.

Test: Number System (January 8) - Question 2

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?

Detailed Solution for Test: Number System (January 8) - Question 2

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.

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Test: Number System (January 8) - Question 3

Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?

Detailed Solution for Test: Number System (January 8) - Question 3

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.

Test: Number System (January 8) - Question 4

In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:

I. a + c = e,
II. b – d = d and
III. e + a = b

Which of the following options are true?

Detailed Solution for Test: Number System (January 8) - Question 4

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 .

Test: Number System (January 8) - Question 5

The sum of the first 100 natural numbers, 1 to 100 is divisible by: 

Detailed Solution for Test: Number System (January 8) - Question 5

The sum of the first 100 natural numbers is:

=  (n * (n + 1)) / 2
=  (100 * 101) / 2
=  50 * 101

101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.

Test: Number System (January 8) - Question 6

In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?

Detailed Solution for Test: Number System (January 8) - Question 6

Let the 4 digit no. be xyzw.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.

Test: Number System (January 8) - Question 7

All the page numbers from a book are added, beginning at page 1. 

However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?

Detailed Solution for Test: Number System (January 8) - Question 7

The Correct Answer is C: 10

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Test: Number System (January 8) - Question 8

The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?

Detailed Solution for Test: Number System (January 8) - Question 8

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.

Test: Number System (January 8) - Question 9

What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?

Detailed Solution for Test: Number System (January 8) - Question 9

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

Test: Number System (January 8) - Question 10

How many factors of 1080 are perfect squares?

Detailed Solution for Test: Number System (January 8) - Question 10
  • 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.
     
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