Playing with Numbers Test

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Playing with Numbers Test - 7

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Question 1
1 / -0

A number is divisible by 3 and 4 both. By what other number(s) will it be always divisible?

A
2

B
12

C
24

D
All of these

Solution

If a number is divisible by 4, it is always divisible by 2 as well because 4 is a multiple of 2. Every number in the times table of 4 is also in the times table of 2.
The number will also be divisible by 12 and 24 because 12 and 24 are both common multiples of 3 and 4.
Question 2
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What least value should be given to 'a' so that the number 789a56 is divisible by 11?

A
2

B
5

C
7

D
9

Solution

Sum of odd places = 7 + 9 + 5 = 21
Sum of even places = 8 + a + 6 = 14 + a
If the number is divisible by 11, then either 21 - (14 + a) = 0 or 21 - (14 + a) = 11m, where m is any integer.
Therefore, a = 7
Question 3
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The numbers (49, 81) are ________ numbers.

A
prime

B
square

C
composite

D
Both 2 and 3

Solution

The numbers given are perfect square numbers.
Square numbers are numbers that we get by multiplying a number by itself once.
Square of 7 = 7 x 7 = 49
Square of 9 = 9 x 9 = 81
Question 4
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The biggest three-digit number when divided by 17, 29 and 39 will leave a remainder of _______, respectively.

A
13, 13 and 24

B
12, 13 and 24

C
13, 13 and 25

D
12, 14 and 25

Solution

Highest three-digit number = 999
999 ÷ 17 = 58 Quotient; Remainder = 13
999 ÷ 29 = 34 Quotient; Remainder = 13
999 ÷ 39 = 25 Quotient; Remainder = 24
Question 5
1 / -0

The LCM of two consecutive odd numbers is always ________.

A
even

B
the product of them

C
the difference between them

D
Cannot be determined

Solution

The LCM of two consecutive odd numbers is always their product.
Let 3 and 5 be two consecutive odd numbers.
LCM of 3 and 5 =

= 3 x 5 = 15, which is the product of 3 and 5.
Question 6
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The LCM of two numbers 120 and 52 is 1560. Find the HCF of these two numbers.

A
3

B
4

C
7

D
9

Solution

(First number) × (Second number) = (HCF) × (LCM)

120 × 52 = HCF × 1560

HCF == 4
Question 7
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Which of the following is not a composite number?

A
1

B
4

C
8

D
All are composite numbers.

Solution

A composite number is a number that has exactly more than two factors.
Factors of 1 = 1
Factors of 4 = 1, 2 and 4
Factors of 8 = 1, 2, 4 and 8
Therefore, 1 is not a composite number because it has only 1 factor.
Question 8
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Which of the following will be equal to the square of 13?

(A) 1 + 3 + 5 + ... + 21
(B) 1 + 3 + 5 + ... + 23
(C) 1 + 3 + 5 + ... + 25
(D) 1 + 3 + 5 + ... + 27

A
A

B
B

C
C

D
D

Solution

The first odd number is equal to the square of 1.
The sum of first two odd numbers is equal to the square of 2, and so on.

Therefore, the sum of first 13 odd numbers will be equal to the square of 13.
First 13 odd numbers go up to 25.

Therefore, 1 + 3 + 5 + ... + 25 will be equal to the square of 13.
13 x 13 = 169
Question 9
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The product of two different prime numbers below 100 always has ______ factors.

A
2

B
4

C
6

D
Cannot be determined

Solution

The product of two prime numbers always has 4 factors.
The product itself and 1, and the two prime numbers.
Question 10
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Which of the following numbers is divisible by 2, 3, 4, 5, 6 and 7?

A
15,540

B
15,690

C
15,760

D
15,790

Solution

15,540 is an even number, so it is divisible by 2.

A number is divisible by 3 if the sum of its digits is divisible by 3.
Sum of digits of 15,540 = 1+5+5+4+0= 15, which is divisible by 3, because 3x5 = 15
15,540 is divisible by 3 as well.

A number is divisible by 4 if the number formed by the last two digits of the number is divisible by 4.
Number formed by the last two digits of 15,540 = 40, which is divisible by 4, because 4x10 = 40
15,540 is divisible by 4 as well.

15,540 ends with '0', so it is divisible by 5.

A number is divisible by 6 if it is divisible by both 2 and 3.
15,540 is divisible by 6 as well.

To check if a number is divisible by 7, remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7.
15,540 = 1554 - 2(0) = 1554
1554 = 155 - 2(4) = 147
147 = 14 - 2(7) = 0
Therefore, 15,540 is divisible by 7 as well.
Question 11
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How many pairs of co-prime numbers can be made from the following numbers?

78, 66, 77, 96, 110, 45

A
One

B
Two

C
Three

D
More than three

Solution

Co-prime numbers are numbers whose HCF is 1.
Therefore, all the even numbers 78, 66, 96 and 110 can't make a pair of co-prime numbers, because 2 is a factor of all these numbers.
66, 77 and 110 also can't make a pair of co-prime numbers because they are all in the times table of 11 and have a common factor 11.
110 and 45 also can't make a pair of co-prime numbers because they are both in the times table of 5 and have a common factor 5.
78, 96, 66 and 45 also can't make a pair of co-prime numbers because they are all in the times table of 3 and have a common factor 3.

Numbers that can make a pair of co-prime numbers = 78 and 77; 77 and 96; 77 and 45
HCF of 78 and 77 =
= 1

HCF of 77 and 96 =
= 1

HCF of 77 and 45 =
= 1
Question 12
1 / -0

Which of the following numbers has the least number of factors?

A
64

B
80

C
84

D
96

Solution

Factors of 64 = 1 x 64, 2 x 32,4 x 16, 8 x 8
= 1, 2, 4, 8, 16, 32, 64
= 7 factors

Factors of 80 = 1 x 80, 2 x 40, 4 x 20, 5 x 16, 8 x 10
= 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
= 10 factors

Factors of 84 = 1 x 84, 2 x 42, 3 x 28, 4 x 21, 6 x 14, 7 x 12
= 1,2,3,4,6,7,12,14,21,28,42,84
= 12 factors

Factors of 96 = 1 x 96, 2 x 48, 3 x 32, 4 x 24, 6 x 16, 8 x 12,
= 12 factors
Question 13
1 / -0

What is the HCF of the biggest two-digit number and the smallest two-digit number?

A
1

B
2

C
3

D
4

Solution

Biggest two-digit number = 99
Smallest two-digit number = 10

HCF of 10 and 99 =
HCF of 10 and 99 = 1
Question 14
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A number is always divisible by 6 if it is divisible by ________.

A
2 and 3

B
3 and 4

C
12

D
All of the above

Solution

2 and 3 are the prime factors of 6, so a number divisible by both of them will also be divisible be 6.
3 and 4 are the factors of 12, which is a multiple of 6. If a number is divisible by a multiple of 6, it will also be divisible by 6.
12 itself is a multiple of 6, so every number in the times table of 12 is also in the times table of 6. Therefore, if a number is divisible by 12, it will also be divisible by 6.
Question 15
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What is the reciprocal of the smallest two-digit composite number?

A

B

C

D

Solution

A composite number is a number which has more than two factors.
Smallest composite number = 10
Factors of 10 = 1, 2, 5, 10

Reciprocal of 10 =
Question 16
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The numbers which have three or more factors are called __________ numbers.

A
prime

B
co-prime

C
composite

D
twin prime

Solution

Any number that has more than two factors is called a composite number.
For example, 4 is a composite number because it has three factors 1, 2 and 4.
Question 17
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If 457N8 is divisible by 4, what could be the least possible value of N?

A
0

B
1

C
2

D
3

Solution

A number is divisible by 4 if the number formed by the last two digits of the number is divisible by 4.
Number formed by the last two digits of 457N8 = N8
For N8 to be a multiple of 4, the least possible number that can be formed is 08, which is a multiple of 4.
Question 18
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The least number, which when increased by 7 is exactly divisible by 24, 28, 32 and 48, is ____.

A
661

B
665

C
667

D
679

Solution

The least number which is divisible by 24, 28, 32 and 48 is the LCM of 24, 28, 32 and 48.
24 = 2 × 2 × 2 × 3
28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
LCM = 2 × 2 × 2 × 2 × 3 × 2 × 7 = 672
Now, the required number is 7 less than the LCM.
I.e. 672 - 7 = 665
Question 19
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Find the largest two-digit number that will divide the smallest five-digit and six-digit numbers.

A
20

B
40

C
50

D
80

Solution

Smallest five-digit number = 10,000; which is exactly divisible by 50 because any number that ends with two or more zeros is divisible by 50.
Smallest six-digit number = 1,00,000; which is exactly divisible by 50 because any number that ends with two or more zeros is divisible by 50.
Question 20
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The HCF and the LCM of two numbers are 32 and 1344, respectively. If one of the numbers is 96, what is the other number?

A
396

B
420

C
436

D
448

Solution

Product of the numbers = 32 x 1344 = 43008
One of the numbers is 96.
Let the other number be X.
So, 96 x X = 43008
X = 43008 ÷ 96
X = 448
Question 21
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A person has 114 sweets and 171 chocolates. He wants to pack them into baskets such that no sweet or chocolate is left, making sure that each basket has same number of sweets and chocolates. How many maximum baskets can he make?

A
56

B
51

C
53

D
57

Solution

To make sure that each sweet and chocolate is used such that each basket has same number of sweets and chocolates, we will find the HCF of 114 and 171.
HCF of 114 and 171 =
HCF of 114 and 171 = 57
Therefore, 57 such baskets can be made with each basket having 2 sweets (57 × 2 = 114) and 3 chocolates (57 × 3 = 171).
Question 22
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What is the least number of square sheets of cloth required to cover a land having dimensions 64 m x 32 m?

A
4

B
6

C
8

D
12

Solution

Area of the land = 64 x 32 = 2048 m²

Out of the given numbers, when we divide 2048 by 8, we get a perfect square number.
2048 ÷ 8 = 256
Therefore, the area of each sheet will be 256 m².

Square root of 256 = 16
Therefore, the sheets will have dimensions 16 m x 16 m, and the number of sheets required is 8.
Question 23
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There are three types of sweets in a shop. Type A sweets come in packets of 36, type B sweets come in packets of 48 and type C sweets come in packets of 57. What is the minimum number of sweets a retailer must buy of each type so that he has equal number of sweets of each type?

A
1368

B
2736

C
4104

D
5472

Solution

To find the minimum number of sweets the person must buy of each type so that he has equal number of sweets of each type, we will find the LCM of 36, 48 and 57 because we know that the sweets come in packets of 36, 48 and 57.

LCM of 36, 48 and 57 =
= 2 × 2 × 2 × 2 × 3 × 3 × 19
= 2736
Question 24
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There are two containers in a factory, one container can hold 1875 kg of sugar and another container can hold 2625 kg of sugar. What can be the capacity of the biggest tool that can be used to fill the containers fully?

A
315 kg

B
355 kg

C
375 kg

D
405 kg

Solution

Capacity of the two containers = 1875 kg and 2625 kg
Maximum capacity of the tool used to fill the containers fully can be found out by determining the HCF of 1875 and 2625.
HCF of 1875 and 2625 =
= 375
Therefore, capacity of the biggest tool that can be used to fill the containers fully = 375 kg
Question 25
1 / -0

There are 475, 513 and 551 balls in three sacks. The balls are to be packed into different boxes of same size. How many maximum balls can each box hold such that after packing no ball is left?

A
19

B
38

C
57

D
76

Solution

Capacity of three sacks = 475, 513 and 551
Maximum size of each box can be found out by determining the HCF of 475, 513 and 551.
HCF of 475, 513 and 551 =

= 19
Question 26
1 / -0

Which of the following statements are true?

A. The divisor can't be greater than the remainder.
B. All prime numbers are odd.
C. The LCM of two even numbers is also even.
D. 1 is a multiple of every number.

A
A and B

B
A and C

C
A, B and C

D
B and C

Solution

A - True - Divisor is the number we divide a number by. Remainder is the amount left over after dividing the number. The divisor can't be greater than the remainder because if the remainder is bigger than the divisor, the divisor can divide the remainder, which is not possible.
B - False - 2 is a prime number and is even.
C - True - LCM is the lowest possible common number in the times table of two numbers, and the numbers in the times table of even numbers are also even.
D - False - 1 is in the times table of 1 only. It doesn't come in the times table of any other number.
Question 27
1 / -0

Directions: Read the following statements and choose the correct option.

Statement 1: If two numbers are divisible by a number, then their product will also be divisible by that number.
Statement 2: The sum of two perfect square numbers can be equal to another perfect square number.

Which of the above is true?

A
Both statements 1 and 2 are true.

B
Only statement 1 is true.

C
Only statement 2 is true.

D
Both statements 1 and 2 are false.

Solution

Statement 1: If two numbers are divisible by a number, then their product will also be divisible by that number. (TRUE)
Let the two numbers be 8 and 12, which are in the table of 4, hence divisible by 4.
Product of 8 and 12 = 96
96 is also in the table of 4, hence divisible by 4.

Statement 2: The sum of two perfect square numbers can be equal to another perfect square number. (FALSE)
Square of 3 = 9
Square of 5 = 25

Square of 3 + Square of 5 = 9 + 25 = 34
34 is not a square number.
Question 28
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Fill in the blanks.

(A) The HCF of two consecutive even numbers below 10 is ________.
(B) The natural numbers which have more than two factors are called ________.
(C) If a number is divisible by 6, it must be divisible by ______.

A
A = 2, B = composite, C = 2

B
A = 2 or 4, B = composite, C = 2

C
A = 2, B = composite, C = 2 and 3

D
A=2 or 4, B = composite, C = 2 and 3

Solution

Even numbers below 10 are 2, 4, 6 and 8.
The HCF of two numbers could be either 2 or 4.
The numbers which have more than two factors are called composite numbers.
2 and 3 are prime factors of 6. If a number is divisible by 2 and 3, it is also divisible by 6.
Question 29
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Find the values of X, Y and Z, if 24X8 is divisible by 4, 447Y is divisible by 3 and 724Z is divisible by 6.

A
X = 0, Y = 2, Z = 2

B
X = 1, Y = 3, Z = 2

C
X = 0, Y = 3, Z = 2

D
X = 0, Y = 3, Z = 4

Solution

A number is divisible by 4 if the number formed by the last two digits of the number is divisible by 4.
Number formed by the last two digits of 24X8 = X8
Here, the value of X can be 0, 2, 4, 6, or 8 because 08, 28,48,68 and 88 are all in the times table of 4.

A number is divisible by 3 if the sum of its digits is divisible by 3.
Sum of the digits of 447Y = 4 + 4 + 7 + Y
= 15 + Y
Here, the value of Y can be 0, 3, 6, or 9 because 15, 18, 21 and 24 are all in the times table of 3.

A number is divisible by 6 if it is divisible by both 2 and 3.
To be divisible by 2, 724Z has to be an even number.
Therefore, the value of Z can be 0, 2, 4, 6, or 8 because 7240, 7242, 7244, 7246 and 7248 are all even numbers.

A number is divisible by 3 if the sum of its digits is divisible by 3.
Out of these 5 numbers, only the sum of digits of 7242 and 7248 is divisible by 3.
Therefore, the value of Z can be 2 or 8.