Number System Test

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Number System Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

When the sum of two natural numbers is multiplied by each number separately, the products obtained are 2418 and 3666. What is the difference between the two numbers?

A
16

B
22

C
26

D
35

Solution

Suppose the numbers are x and y, then:
x(x + y) = 3666 … (1)
y(x + y) = 2418 … (2)

Adding 1 and 2, we get
(x + y)²=6084
Therefore, x + y = 78 … (3)

Subtracting (2) from (1), we get
x² - y² = 1248
Or, (x + y)(x - y) = 1248
Or, 78(x - y) = 1248
Or, x - y = 16
Question 2
1 / -0

If 'p' be a number between 0 and 1, which one of the following is true?

A
p > √q

B
1/p² > √q

C
p > 1/p

D
p³ > p²

Solution

Try with some natural numbers and get the answer.
Question 3
1 / -0

Dividing by 3/8 and then multiplying by 5/6 is equivalent to dividing by which number?

A
5/16

B
16/40

C
9/20

D
40/18

Solution

x / ( 3/8 ) * 5/6 = x* 8/3 *5/6 = x*40/18
equivalent to dividing by 18/40 = 9/20
Question 4
1 / -0

A number when divided by 238, leaves a remainder 79. What will be the remainder when that number is divided by 17?

A
8

B
9

C
10

D
11

Solution

N = 238a + 79 =17 × 14a + 17 × 4 + 11.
N = 17(14a + 4) +11

Hence, on dividing N by 17, the remainder is 11.
Question 5
1 / -0

What is the remainder when 17 to the power 23 is divided by 16?

A
0

B
1

C
2

D
3

Solution

17 when divided by 16 leaves 1 as remainder and we have 23 such 17s which when divided by 16 leave 1 as remainder.
Question 6
1 / -0

The product of two numbers is 16,200. If their LCM is 216, find their HCF.

A
75

B
70

C
80

D
Data inconsistent

Solution

The data is inconsistent as the LCM is always a multiple of HCF. The product of two numbers is always equal to the product of their LCM and HCF. Here, HCF = 75 and LCM = 216 do not show consistency, i.e. no such number is possible.
Question 7
1 / -0

The LCM of two numbers is 72 and their HCF is 12. If one of the numbers is 24, what is the other number?

A
38

B
26

C
36

D
42

Solution

72 * 12 = 24 * n ==> n = 36
Question 8
1 / -0

4a56 is a four digit number divisible by 33. What is the value of a?

A
3

B
4

C
5

D
6

Solution

By divisibility rule of 3, 4 + 5 + a + 6 = multiple of 3 ⇒ a =3 or 6 or 9 ⇒ by divisibility rule of 11 ⇒ (4 + 5) - (6 + a) = multiple of 11n or zero ⇒ A = 3.
Question 9
1 / -0

A number when divided by 5 leaves a remainder 3. What is the remainder when the square of this number is divided by 5?

A
4

B
3

C
9

D
Cannot be determined

Solution

A number leaving 3 as remainder when divide by 5 ends either 3 or 8 .The number when squared will always have the last digit 4 or 9, thus leaving a remainder of 4 when divisible by 5.
Question 10
1 / -0

A two digit number is such that cube of its 24^th part is same as the number obtained by interchanging the digits of the number. What is the number?

A
24

B
72

C
48

D
None of these

Solution

The best way is to check the options. (1) is not possible
(3) 1/24 x 48 = 2, cube of 2 = 8 which is not a 2 digit number
(2) 1/24 x 72 = 3, cube of 3 = 27 which is reverse of 72.
Question 11
1 / -0

What are the unit digits of 3⁶⁹, 6⁸⁶⁴and 4⁷²⁵, respectively?

A
9, 6 and 6

B
6, 6 and 6

C
3, 6 and 4

D
None of these

Solution

3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243, 3⁶ = 729, 3⁷ = 2167, 3⁸ = 6501.......

From the above expressions, we can observe that the unit digit of the result gets repeated after every four consecutive results. So, the unit digit of 3⁶⁸ will be 1.
Hence, the unit digit of 3⁶⁹ will be 3.

Similarly, the unit digit of 6⁸⁶⁴ will be unit digit of 2⁸⁶⁴ × 3⁸⁶⁴,which is 6 × 1 = 6.

And, unit digit of 4⁷²⁵will be 4 (as 725 = 2n + 1).
Question 12
1 / -0

If the last two digits of a four-digit number are interchanged, the new number obtained is greater than the original number by 54. What is the difference between the last two digits of the number?

A
9

B
12

C
6

D
Data inadequate

Solution

Suppose that the last two digits of the four-digit number are c and d, where c is at the tens place and d is at the units place.
We can ignore the first two digits.

This gives us,
(10c + d) - (10d + c) = 54

9(c - d) = 54
c - d = 6
Question 13
1 / -0

A number formed by writing any digit 6 times (say as 444444 or 999999) is always divisible by

A
1001

B
7

C
13

D
All of the above

Solution

aaaaaa = aaa * 1000 + aaa
aaa (1000 + 1) = 1001(aaa)

= 7 * 11 *13 (aaa)
This is obviously divisible by 7,11 and 13.
Question 14
1 / -0

A positive interger which is nearest to 1000 and divisible by 2, 3, 4, 5 and 6 is

A
1020

B
1040

C
960

D
1030

Solution

Only option 1 is divisible by three.
Question 15
1 / -0

There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and that of the last three is 1001. Find the first number.

A
5

B
7

C
11

D
17

Solution

The best way is to factorise the number i.e 385 = 5 x 7 x 11
1001 = 7 x 11 x 13

So, the four prime numbers are 5, 7, 11 and 13.
Hence, first prime number = 5

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Number System Test - 4

Result

Number System Test - 4

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Question 1

16

22

26

35

Solution

Question 2

p > √q

1/p2 > √q

p > 1/p

p3 > p2

Solution

Question 3

5/16

16/40

9/20

40/18

Solution

Question 4

8

9

10

11

Solution

Question 5

0

1

2

3

Solution

Question 6

75

70

80

Data inconsistent

Solution

Question 7

38

26

36

42

Solution

Question 8

3

4

5

6

Solution

Question 9

4

3

9

Cannot be determined

Solution

Question 10

24

72

48

None of these

Solution

Question 11

9, 6 and 6

6, 6 and 6

3, 6 and 4

None of these

Solution

Question 12

9

12

6

Data inadequate

Solution

1/p² > √q

p³ > p²