Real Numbers Test

Result

Real Numbers Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

For some integer m, every even integer is of the form

A
m

B
m+1

C
2m

D
2m+1

Solution

We know that, even integers are 2, 4, 6 …

Where, m = integer [Since, here integer is represented by m]

m = ..., −1, 0, 1, 2, 3,...

∴ 2m = ... , −2, 0, 2, 4, 6,...
Question 2
1 / -0

For some integer q, every odd integer is of the form

A
q

B
q+1

C
2q

D
2q+1

Solution

We know that, odd integers are 1, 3, 5…

So, it can be written in the form of 2q + 1.

Where,

q = integer

Or q = ..., −1, 0, 1, 2, 3,...

∴ 2q + 1 = ..., −3, −1, 1, 3, 5,...
Question 3
1 / -0

If the HCF of 65 and 117 is expressible in the form 65m − 117, then the value of m is

A
4

B
2

C
1

D
3

Solution

Using Euclid’s division algorithm,

b = aq + r, 0 ≤ r < a [∵ dividend = divisor × quotient + remainder]

⇒ 117 = 65 × 1 + 52

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

∴ HCF (65, 117) = 13 …. (i)

Also, given that, HCF (65, 117) = 65m – 117 … (ii)

From Equations (i) and (ii), we get

65m − 117 = 13

⇒ 65m = 130

⇒ m = 2
Question 4
1 / -0

The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is

A
13

B
65

C
875

D
1750

Solution

Since, 5 and 8 are the remainders of 70 and 125, respectively.

Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70 – 5),

117 = (125 − 8), which is divisible by the required number.

Now, required number = HCF of 65, 117 [Since we need the largest number]

For this, 117 = 65 × 1 + 52 [∵ dividend = divisor × quotient + remainder]

⇒ 65 = 52 × 1 + 13

⇒ 52 = 13 × 4 + 0

∴ HCF = 13

Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.
Question 5
1 / -0

The least number that is divisible by all the numbers from 1 to 10 (both inclusive)

A
10

B
1000

C
504

D
2520

Solution

Factors of 1 to 10 numbers are as follows:

1 = 1

2 = 1 × 2

3 = 1 × 3

4 = 1 × 2 × 2

5 = 1 × 5

6 = 1 × 2 × 3

7 = 1 × 7

8 = 1 × 2 × 2 × 2

9 = 1 × 3 × 3

10 = 1 × 2 × 5

∴ LCM of number 1 to 10

= LCM (1, 2, 3, 4, 5, 6, 8, 9, 10)

= 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520