Playing with Numbers Test

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Playing with Numbers Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

A number is divisible by both 6 and 13. By which of the following other numbers will that number be always divisible?

A
65

B
72

C
78

D
81

Solution

A number is divisible by both 6 and 13.
Factors of 6 = 1, 2, 3, 6
Factors of 13 = 1, 13
As the common factors of these numbers is 1, the given two numbers are co-prime numbers and they will also be divisible by their products 13 x 6 = 78
Question 2
1 / -0

What least value should be given to *, so that the number 750*64 is divisible by 11?

A
1

B
2

C
4

D
7

Solution

Sum of odd places = 7 + 0 + 6 = 13
Sum of even numbers = * + 5 + 4 = 9 + *

If the number is divisible by 11, then either 13 - (9 + *) = 0
Or 13 - (9 + *) = 11 m (where m is any integer)

Now, for the least value of *, m should be zero
13 - 9 - * = 0
* = 4.
Hence, option 3 is correct.
Question 3
1 / -0

The number (17, 23) are _____ numbers.

A
co-prime

B
prime

C
both prime and co-prime

D
None of these

Solution

Factors of 17 = 1 and 17.
Factors of 23 = 1 and 23.
Therefore, 17 and 23 are prime numbers.
H.C.F of 17 and 23 = 1
Therefore, they are co-prime numbers also.
Question 4
1 / -0

The lowest number which when divided by 18, 24, 36 and 48 leaves a remainder of 15, 21, 33, 45, respectively is

A
147

B
144

C
151

D
156

Solution

The highest number which is exactly divisible by 18, 24, 36 and 48 is their LCM.
LCM = 6 x 2 x 3 x 2 x 2 = 144

18 - 15 = 3
24 - 21 = 3
36 - 33 = 3
48 - 45 = 3
The difference is 3, so there is a pattern.
Required number will be 3 more than LCM = 144 + 3 = 147
Question 5
1 / -0

The HCF of two consecutive even numbers is

A
0

B
1

C
2

D
4

Solution

Let two consecutive numbers be 2n and 2n + 2.
So, both have HCF 2.
So, HCF of two consecutive even numbers is 2.
Question 6
1 / -0

The HCF of two numbers is 36 and their LCM is 464. If one number is 116, then the other number is

A
132

B
144

C
156

D
164

Solution

We have one number = 116
Let the other number be x.
HCF = 36, LCM = 464
Now, the product of two numbers = HCF LCM
116 x = 36 464

x == 144
Question 7
1 / -0

If a number has exactly two factors, 1 and itself, then it is a/an

A
odd number

B
prime number

C
even number

D
composite number

Solution

If a number has exactly two factors, 1 and itself, then the number is a prime number.
Question 8
1 / -0

Directions: Observe the following patterns. Then find the sum of 1 + 3 + 5 + 7 + 9 + ....+ 21.

1 = 1 x 1 = 1
1 + 3 = 2 x 2 = 4
1 + 3 + 5 = 3 x 3 = 9
1 + 3 + 5 + 7 = 4 x 4 = 16
1 + 3 + 5 + 7 + 9 = 5 x 5 = 25
............................................................................................................
............................................................................................................

A
100

B
121

C
144

D
441

Solution

According to the given pattern the sum of 1st 'n' odd natural numbers is n².
Number of terms in 1 + 3 + 5 +.......+ 21 is 11.
So, the sum of odd numbers till 21 is equal to 11² = 121.
Question 9
1 / -0

The HCF of the denominator and the numerator of a fraction, which is in its lowest form,

A
is 1

B
is 0

C
is always an even number

D
cannot be determined

Solution

A fraction is in its lowest terms or in lowest form, if the HCF of its numerator and denominator, is 1.
Question 10
1 / -0

The two numbers nearest to 10,000, which are exactly divisible by each of 3, 5, 7, 8 and 9, are

A
7,560 and 10,080

B
8,270 and 11,420

C
9,360 and 11,880

D
9,960 and 12,340

Solution

The two numbers nearest to 10,000, which are exactly divisible by each of 3, 5, 7, 8 and 9, are the multiples of LCM of the given numbers.
LCM = 3 x 5 x 7 x 8 x 9 = 2,520
Now, dividing 10,000 by 2,520, we get remainder = 2,440
Number just less than 10,000 and exactly divisible by the given numbers = (10,000 - 2,440) = 7,560
Number just greater than 10,000 and exactly divisible by the given numbers = 10,000 + (2,520 - 2,440) = 10,080
Question 11
1 / -0

Which of the following pairs are twin prime numbers?

A
41 and 43

B
47 and 61

C
71 and 79

D
83 and 97

Solution

A twin prime is a prime number, that is either 2 less or 2 more than another prime number.
(41, 43) are one of the twin prime numbers.
In other words, a twin prime is a prime that has a prime gap of two.
Question 12
1 / -0

Which of the following numbers has exactly 6 factors?

A
24

B
20

C
14

D
21

Solution

Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 14 = 1, 2, 7, 14
Factors of 21 = 1, 3, 7, 21
Question 13
1 / -0

LCM of two numbers is equal to

A
the greatest number which can be divided by both the numbers

B
the smallest number which can be divided by both the numbers

C
one of the smallest numbers, which can be divided by both the numbers

D
one of the greatest numbers, which can be divided by both the numbers

Solution

LCM of two numbers is the smallest number, which can be divided by both the numbers.
For example, LCM of 15 and 20 is 60 and LCM of 5 and 7 is 35.
Question 14
1 / -0

A number always divisible by 120 is

A
divisible by both 5 and 24

B
divisible by both 3 and 66

C
divisible by both 2 and 44

D
None of these

Solution

120 is divisible by both 5 and 24.
5 × 24 = 120
Question 15
1 / -0

What is the reciprocal of the prime number just before 100?

A

B

C

D

Solution

The prime number before 100 is 97.

Reciprocal of 97 is .
Question 16
1 / -0

Numbers that are either 2 less or 2 more than another number are called

A
co-primes

B
twin primes

C
composite numbers

D
even numbers

Solution

A twin prime is a prime number that is either 2 less or 2 more than another prime number. For example, either member of the twin prime pair (41, 43). In other words, a twin prime is a prime that has a prime gap of two.
Question 17
1 / -0

If 5529x is divisible by 9, then what can be the least value of x?

A
0

B
1

C
2

D
6

Solution

If 5529x is divisible by 9, then 5 + 5 + 2 + 9 + x = 21 + x is divisible by 9.
Therefore, x = 6
So, 21 + 6 = 27, which is divisible by 9.
Question 18
1 / -0

The least number which when increased by 5 is exactly divisible by 14, 21, 28 and 35, is

A
405

B
410

C
415

D
420

Solution

The least number, which is exactly divisible by 14, 21, 28, 35 is their LCM.
LCM = 7 x 2 x 3 x 2 x 5 = 420
Now, the required number is 5 less than the LCM = 420 - 5 = 415
Question 19
1 / -0

Find the least number, which will divide the least 3 digit number and the least 4 digit number exactly.

A
6

B
8

C
10

D
12

Solution

The least 3 digit number = 100
The least 4 digit number = 1,000
The least number which will divide both 100 and 1,000 exactly, is the HCF.
Question 20
1 / -0

The HCF and LCM of two numbers are 48 and 576, respectively. If one of the numbers is 512, then the other number is

A
50

B
52

C
54

D
56

Solution

We have one number = 512

Let the other number be x.

Product of numbers = HCF LCM

512 x = 48 576

x == 54
Question 21
1 / -0

There are 178 mangoes and 156 cherries. These fruits are to be arranged in stacks containing the same number of fruits. Then, the greatest number of fruits possible in each stack is

A
4

B
3

C
2

D
5

Solution

The greatest number of fruits in each stack is the HCF of 178 and 156.
Therefore, HCF (178, 156) = 2
Hence, greatest number of fruits possible in each stack is 2.
Question 22
1 / -0

The least number of square tiles that will be needed to cover a roof of 441 m by 42 m is

A
40

B
42

C
44

D
46

Solution

Largest length of side of square tile is the HCF 441 and 42.
HCF = 21
Number of tiles required = Area of roof/Area of 1 tile
== 42
Question 23
1 / -0

Five bells toll together at an instant. After that, they separately toll at intervals of 3, 5, 7, 8 and 10 seconds. After how long will they again toll together?

A
42 minutes

B
28 minutes

C
14 minutes

D
56 minutes

Solution

Required interval is the least common factor of 3, 5, 7, 8 and 10.
LCM of 3, 5, 7, 8 and 10 = 840 seconds
Now, converting 840 seconds into minutes, we get: = 14 minutes
Hence, they will again toll together after 14 minutes.
Question 24
1 / -0

Two containers contain 540 litres and 320 litres of water, respectively. The maximum capacity of a container, which can measure the water out of both the containers is

A
34

B
12

C
20

D
22

Solution

Capacity of two containers is 540 litres and 320 litres.

Maximum capacity of container is the HCF of 540 and 320.
Therefore, HCF = 20
Question 25
1 / -0

The length, breadth and height of an auditorium are 505 m, 525 m and 625 m, respectively. Find the longest tape, which can measure the three dimensions of the auditorium exactly.

A
5 m

B
8 m

C
10 m

D
15 m

Solution

Dimensions of the room are 505 m, 525 m and 625 m.
The longest tape, which can measure the all dimensions of the room exactly is the HCF of 505, 525 and 625.

Length of the longest tape = 5 m
Question 26
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Directions: State 'T' for true and 'F' for false.

(i) The sum of three odd numbers is even.
(ii) Prime numbers do not have any factor.
(iii) The product of two even numbers is always even.
(iv) All prime numbers are odd.

(i) (ii) (iii) (iv)
(A) T F T F
(B) F T T F
(C) F F T F
(D) T T F T

A
A

B
B

C
C

D
D

Solution

1. The sum of three odd numbers is even. False. Sum of three odd numbers is always odd.

2. Prime numbers do not have any factor. False. A prime number has two factors, one and the number itself.

3. The product of two even numbers is always even. True

4. All prime numbers are odd. False. 2 is an even prime number.
Question 27
1 / -0

Directions: Read the following statements and mark the correct option accordingly.

Statement-1: A number for which the sum of its factors other than the number is equal to that number itself is called a perfect number.
Statement-2: The product of three consecutive numbers will always be a multiple of 6.

A
Statement-1 is true and Statement-2 is false.

B
Both Statement-1 and Statement-2 are true.

C
Both Statement-1 and Statement-2 are false.

D
Statement-1 is false and Statement-2 is true.

Solution

Statement-1 is true.
A number for which the sum of all its factors is equal to twice the number is called a perfect number.
Or in other words, a number for which the sum of its factors other than the number itself is equal to that number itself is called a perfect number.
For example,
Factors of 28: 1, 2, 4, 7, 14, 28
Sum of the factors other than 28 = 1 + 2 + 4 + 7 + 14 = 28

Statement-2 is true.
The product of three consecutive numbers will always be a multiple of 6.
For example: 7 x 8 x 9 = 504; and 504 is a multiple of 6.
Question 28
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Directions: Fill in the blanks.

(i) LCM of two co-prime numbers will be (A) of the numbers.
(ii) HCF of two consecutive even numbers (B) .
(iii) If a number is divisible by 6, then it must also be divisible by 2 and (C) .

A
(A) - product, (B) - 2, (C) - 3

B
(A) - product, (B) - 1, (C) - 3

C
(A) - sum, (B) - 2, (C) - 3

D
(A) - product, (B) - 1, (C) - 4

Solution

(i) In case the two numbers are co-prime, their LCM will be the product of the numbers.
(ii) Let two consecutive numbers be 2n and 2n + 2. So, both have highest common factor 2, so HCF of two consecutive even numbers is 2.
(iii) If a number is divisible by 6, then it must also be divisible by 2 and 3.
Question 29
1 / -0

Find the value of x + y + z, if 821x is divisible by 8, 992y is divisible by 12 and 446z is divisible by 6.

A
10

B
14

C
15

D
16

Solution

If 821x is divisible by 8,
then, the only possible value of x is 6, when last three digits of 821x is divisible by 8.
So, x = 6

992y is divisible by 12,
9 + 9 + 2 + y = 20 + y is divisible by 12
y = 4

446z is divisible by 6,
4 + 4 + 6 + z = 14 + z is divisible by 6
z = 4
x + y + z = 6 + 4 + 4 = 14