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Practice Test: Number System- 2 - CAT MCQ


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20 Questions MCQ Test - Practice Test: Number System- 2

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Practice Test: Number System- 2 - Question 1

The HCF of two numbers is 11 and their LCM is 616. If one of the numbers is 77, find the other number.

Detailed Solution for Practice Test: Number System- 2 - Question 1

Calculation:

Let, 2nd number be m,

⇒ m × 77 = 11 × 616

⇒ m = 616/7

⇒ m = 88

∴ The 2nd number is 88.

Practice Test: Number System- 2 - Question 2

Let a, b, m and n be natural numbers such that a > 1 and b > 1. If amb= 144145, then the largest possible value of n - m is

Detailed Solution for Practice Test: Number System- 2 - Question 2

It is given that am. bn = 144145, where a > 1 and b > 1.

144 can be written as 144 = 24 x 32

Hence, am . bn = 144145 can be written as am . bn = (24 x 32) = 2580 x 3290 

We know that 3290 is a natural number, which implies it can be written as a1, where a > 1

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

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Practice Test: Number System- 2 - Question 3

A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

Detailed Solution for Practice Test: Number System- 2 - Question 3

After the first sale, the remaining quantity would be (x/2) - 0.5 and after the second sale, the remaining quantity
is 0.25x - 0.75

After the last sale, the remaining quantity is 0.125x - (7/8) which will be equal to 0 So 0.125x - (7/8) = 0 ⇒ x = 7

Practice Test: Number System- 2 - Question 4

Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?

Detailed Solution for Practice Test: Number System- 2 - Question 4

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5 * 7 * 9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

Practice Test: Number System- 2 - Question 5

How many divisors of 105 will have at least one zero at its end?

Detailed Solution for Practice Test: Number System- 2 - Question 5

In order for a divisor (or any number) to have a zero at its end, it must have a 10 as a factor, i.e., a 2-and-5 pair. Notice that
105 = 25 x 55 = (24 x 54) x (2 x 5)
Therefore, any divisors of 24 x 54 will have a zero at its end when multiplied by 2 x 5. Since 24 x 54 has (4 + 1)(4 + 1) = 25 divisors, 105 has 25 divisors that have at least one zero at its end.

Practice Test: Number System- 2 - Question 6

Find the remainder when 496 is divided by 6.

Detailed Solution for Practice Test: Number System- 2 - Question 6

 496/6, We can write it in this form
(6 - 2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6, It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4.
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.

Practice Test: Number System- 2 - Question 7

Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?

Detailed Solution for Practice Test: Number System- 2 - Question 7

Information Given:

  • The notebook has 96 leaves, which means it has 192 pages (since each leaf has two pages).
  • The pages are numbered from 1 to 192.
  • Tappo tore out the last 25 leaves of the notebook. Since each leaf has 2 pages, she tore out 50 pages.

Step 1: Determine the page numbers torn out

The last 25 leaves correspond to the last 50 pages in the notebook. Since the total number of pages is 192, the page numbers torn out would be from 143 to 192.

Step 2: Calculate the sum of the torn-out pages

The sum of an arithmetic series (in this case, the page numbers) is given by:

For the torn-out pages from 143 to 192:

  • First term (aaa) = 143
  • Last term (lll) = 192
  • Number of terms (nnn) = 50

So, the sum is:

Step 3: Analyze each option

  1. Option 1: She could have found the sum of pages as 1990.

    • To find if this is possible, subtract 1990 from the total sum of all pages (1 to 192):

    • Since the remaining sum does not match with any realistic remaining pages, this option is not possible.
  1. Option 2: She could have found the sum of pages as 1275.

    • Subtracting 1275 from the total sum: Remaining sum=18528−1275=17253
    • The sum is possible and reasonable, so this option is possible.
  2. Option 3: She could have found the sum of pages as 1375.

    • Subtracting 1375 from the total sum: Remaining sum=18528−1375=17153
    • The sum is possible and reasonable, so this option is possible.

Conclusion:

Option 1: She could have found the sum of pages as 1990 is not true because the sum 1990 cannot realistically be the sum of the pages torn out in this context.

Answer: Option 1

Practice Test: Number System- 2 - Question 8

There are 50 integers a1, a2,........, a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer is referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than:

Detailed Solution for Practice Test: Number System- 2 - Question 8

The smallest integer of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.

As this is a variable based question: the word "ANY" can be used.

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.

S1=1,2,3,4,.........24, S2=25,26,27,..........50.

Practice Test: Number System- 2 - Question 9

Srini wrote his class 10th board examination this year. When the result came out he searched for his hall ticket to see his roll number but could not trace it. He could remember only the first three digits of the 6 digit number as 267. His father, however, remembered that the number was divisible by 11. His mother gave the information that the number was also divisible by 13. They tried to recollect the number when all of a sudden Srini told that the number was a multiple of 7. What was the unit digits of the number?

Detailed Solution for Practice Test: Number System- 2 - Question 9

His roll number is divisible by 1001 and the first three digits are 267. Hence the last three digits will also be 267.

Practice Test: Number System- 2 - Question 10

When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?

Detailed Solution for Practice Test: Number System- 2 - Question 10

When 7179 and 9699 are divided by another natural number N, remainder obtained is same.
 Let remainder is R, then (7179 — R) and (9699 — R) are multiples of N and {(9699 — R) — (7179 — R)} is multiple of N. Then 2520 is multiple of N or the largest value of N is 2520. Total factors of N which are multiples of 10 is 18. 

Practice Test: Number System- 2 - Question 11

At a bookstore, 'MODERN BOOK STORE' is flashed using neon lights. The words are individually flashed at the
intervals of 2.5 s, 4.25 s and 5.125 s respectively, and each word is put off after a second. The least time after
which the full name of the bookstore can be read again for a full second is

Detailed Solution for Practice Test: Number System- 2 - Question 11

In this problem, the lights are flashed at the intervals 2.5, 4.25 and 5.125 seconds and put off after one second
each.

The total duration of intervals of these lights are (2.5+1) = 3.5 s, (4.25+1) = 5.25 s and (5.125+1) = 6.125 s.

We have to find the minimum duration. It would be the LCM of thes three numbers.

Since each word is put after a second. So LCM [5/2 + 1) (17/4 + 1) (41/8 + 1)] = LCM of numerator / HCF of denominator = 49*3/2 = 73.5.

Hence they will glow for full one second after 73.5-1 =72.5 sec.

Practice Test: Number System- 2 - Question 12

Find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34.

Detailed Solution for Practice Test: Number System- 2 - Question 12

Remainder,
(73 * 75 * 78 * 57 * 197 * 37)/34 ===> (5 * 7 * 10 * 23 * 27 * 3)/34
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
(5 * 7 * 10 * 23 * 27 * 3)/34 ===> (35 * 30 * 23 * 27)/34 [Number Multiplied]
(35 * 30 * 23 * 27)/34 ===> (1 * -4 * -11 * -7)/34
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]

(1 * -4 * -11 * -7)/34 ===> (28 * -11)/34 ===> (-6 * -11)/34 ===> 66/34 ===R===> 32.
Required remainder = 32.

Practice Test: Number System- 2 - Question 13

Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?

Detailed Solution for Practice Test: Number System- 2 - Question 13

Prime numbers less than 10 = 2, 3, 5, 7.

If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.

The smallest and largest numbers are of form 2_7 and 7_2

Since it is given that the sum of the digits is >13, x will be 5.

Verifying, 752-257 = 495. Answer is option (b).

as 7*5*2 = 70

Practice Test: Number System- 2 - Question 14

Let N = 553 + 173 - 723. N is divisible by:

Detailed Solution for Practice Test: Number System- 2 - Question 14

553 + 173 - 723 = (55-72)k + 173. This is divisible by 17

Remainder when 553 is divided by 3 = 1

Remainder when 173 is divided by 3 = -1

Remainder when 723 is divided by 3 = 0

So, 553 + 173 - 723 is divisible by 3

So, the answer is d) 3 and 17

Practice Test: Number System- 2 - Question 15

Which of the following would always divide a six-digit number of the form ababab?

Detailed Solution for Practice Test: Number System- 2 - Question 15

Number = ababab

=ab×10000+ab×100+ab

=ab(10000+100+1)

=ab(10101)

Practice Test: Number System- 2 - Question 16

Find the unit digit: 
(76476756749)8754874878

Detailed Solution for Practice Test: Number System- 2 - Question 16

Correct Answer :- a

Explanation : The unit digit of the number will depend on the last digit.

As we know that 91 = 9

92  = 81

93 = 729

94  = 6561

The unit digit of the number is 1 and 9, from the options we can pick the answer

Hence option a) is correct

Practice Test: Number System- 2 - Question 17

Find the unit digit:
1719∧13

Detailed Solution for Practice Test: Number System- 2 - Question 17

17 is raised to the power of 19 and 19 is raised to the power of 13.
To find the last digit of the number of this kind we will start with the base, and the base here is 17.
To get the unit digit of a number our only concern is the digit at the unit place i.e.7.
The cyclicity of 7 is 4.
Dividing 1913  by 4.
Remainder will be 3.
7 raised to power 3 (73), the unit digit of this number will be 3.

Practice Test: Number System- 2 - Question 18

When a number is successively divided by 7,5 and 4, it leaves remainders of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8,5 and 6 ?

Detailed Solution for Practice Test: Number System- 2 - Question 18

The number would be in the form of (7X+4) as when this number is divide by 7, will give remainder 4.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied. 

Now, 123 divided by 8, quotient =15, remainder =3
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.

Practice Test: Number System- 2 - Question 19

Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?  

Detailed Solution for Practice Test: Number System- 2 - Question 19

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

No. of Times bell rings in 1 min (60 seconds) = 1 .

No. of times bell rings in 1 hr (60 minutes) = 1 x 60 = 60

No. of Times bell rings in 8 hr = 8 x 60 = 480 

No. of Times bell rings in 8 hr + the first time all bell rings also needs to be counted

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours.

Practice Test: Number System- 2 - Question 20

What would be the greatest number that divides 14, 20, and 32 leaving the same remainder?

Detailed Solution for Practice Test: Number System- 2 - Question 20

Here, the number which divides 14, 20, and 32 leaves the same remainder.
∴ We will be using HCF model 2
The required number will be the HCF of (20 - 14), (32 - 20), and (32 - 14).
i.e. HCF (6, 12, 18)
which will be 6.
Therefore, the required number is 6.

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