Free Playing with Numbers Subjective Test 01 Practice Test - 6th grade

What is a factor of a number? [1 MARK]

SOLUTION

Solution :

A factor of a number is an exact divisor of that number.

Oshin has 36 chocolates. Among how many possible sets of children can she distribute the chocolate so that each child gets an equal number of chocolates without breaking the chocolates? [2 MARKS]

SOLUTION

Solution :

Steps: 1 Mark

Oshin can distribute the chocolates by finding the factors of 36: Each of the factors will divide the chocolates in an equal number of parts, So each child will get an equal number of chocolates.
We can find the factors of 36 in the following way:

36=1×36

36=2×18

36=3×12

36=4×9

36=6×6

Stop here, because both the factors (6) are same.

Thus, the factors are 1, 2, 3, 4, 6, 9, 12, 18 and 36. So she can distribute in 9 possible sets, depending on how many children form one set.

The area of a rectangular garden is 256 square meters. List all possible whole number dimensions that the garden can have. [2 MARKS]

SOLUTION

Solution : Steps: 1 Mark
Solution: 1 Mark

The dimensions of the garden can be found by finding the factors of the number 256.
The various factors of 256 are as follows:
256=1×256
256=2×128
256=4×64
256=8×32
256=16×16
We will stop at (16,16), because they are equal.
The various dimensions of the garden can be, (1,256),(2,128),(4,64),(8,32),(16,16).

What are the common multiples? If a number is divisible by 16, which numbers will definitely be a factor of that number? [3 MARKS]

SOLUTION

Solution :

Definition: 1 Mark
Steps: 1 Mark
Solution: 1 Mark

A Common Multiple is a number that is a multiple of two or more numbers.
We can find all the factors of 16, as shown below.
16=1×16
16=2×8
16=4×

Since the number is divisible by 16, it will also be divisible by its factor i.e. 1, 2, 4, 8, 16.
So, clearly, the number which is divisible by 16, will also be divisible by 1, 2, 4, 8.

What are prime numbers? Find the prime factorization of 2048. [3 MARKS]

SOLUTION

Solution :

Definition: 1 Mark
Prime Factorisation: 2 Marks

All the numbers other than 1, whose only factors are 1 and the number itself are called prime numbers.
The prime factorization of 2048 is shown below.

2048=2×2×2×2×2×2×2×2×2×2×2

Find the smallest number which is divisible by the first four prime numbers. What is the highest common factor of the first four prime numbers? [3 MARKS]

SOLUTION

Solution :

Smallest Number: 1 Mark
Highest Common factor: 2 Marks

The first four prime numbers are 2, 3, 5, 7.
1 is neither a prime nor a composite number so we will not include that.
Now, all these numbers are prime numbers, so the smallest number possible which is a factor of these numbers is the lowest common multiple of these numbers.
Since these are all prime numbers, their lowest common multiple is the product of these numbers, which is:
2 × 3 × 5 × 7 = 210
Hence, 210 is the required number.

The given numbers are prime numbers, so all of them will have the highest common factor equal to 1.

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? [4 MARKS]

SOLUTION

Solution :

Concept: 1 Mark
Steps: 2 Marks
Solution: 1 Mark

The given numbers are 2,4,6,8,10 and 12.
The factors of the given numbers are:
2=1×2
2=2×2
6=2×3
8=2×2×2
10=2×5
12=2×2×3
Now ,the maximum number of times prime factor 2 occurs is 3,
the maximum number of times prime factor 3 occurs is 1,
the maximum number of times prime factor 5 occurs is 1.
So, LCM=2×2×2×3×5 = 120
L.C.M. of 2, 4, 6, 8, 10 and 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 30 minutes, they will toll together 302 + 1 = 16 times.

What is the greatest four digit number which is divisible by 15, 25, 40 and 75? [4 MARKS]

SOLUTION

Solution :

Steps: 3 Marks

The greatest number of 4-digits is 9999.
The prime factors of the given numbers are:
15=3×5
25=5×5
40=2×2×2×5
75=3×5×5

The maximum number of times the prime factor 2 occurs is 3.
The maximum number of times the prime factor 3 occurs is 1.
The maximum number of times the prime factor 5 occurs is 2.
The LCM of the given numbers=2×2×2×3×5×5

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

Three brands A, B and C of biscuits are available in packets of 12, 15, and 21 biscuits respectively.
If a shopkeeper wants to buy an equal number of biscuits of each brand, then what is the minimum number of packets of each brand he should buy? [4 MARKS]

SOLUTION

Solution : Formula: 1 Mark
Steps: 2 Marks

For a minimum number of biscuits, first, we find the LCM of 12, 15 and 21.
The prime factors of the following numbers are:
12=2×2×3
15=3×5
21=3×7
The maximum number of times the prime factor 2 occurs is 2.
The maximum number of times the prime factor 3 occurs is 1.
The maximum number of times the prime factor 5 occurs is 1.
The maximum number of times the prime factor 7 occurs is 1.
LCM= 3 × 4 × 5 × 7 = 420
Minimum number of packets of brand A =42012=35
Minimum number of packets of brand B =42015=28
And minimum number of packets of brand C =42021=20

A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of the third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin? [4 MARKS]

SOLUTION

Solution : Concept: 1 Mark
Steps: 2 Marks