Important Questions Class 10 Mathematics Chapter 1 – Real numbers

 1.The decimal type of 129/225775 is

(a) ending

(b) non-ending

(c) non-ending non-rehashing

(d) nothing from what was just mentioned

Answer: (c) non-ending non-rehashing

2.For some number m, each odd number is of the structure

(a) m

(b) m + 1

(c) 2m

(d) 2m + 1

Answer: (D) 2m + 1

Clarification: A number is called odd in the event that it isn't partitioned by 2. Let me be a number i.e., m = …, −2, −1, 0, 1, 2, …

Duplication of the two sides by 2 gives, 2m = …, - 6, −4, −2, 0, 2, 4, 6 …

Including 1 the two sides, 2m + 1 = …, - 5, −3, −1, 1, 3, 5, 7 …

Thus, for any number m, (2m +1) is generally odd.

Subsequently, the right response is choice D.

3.n 2 - 1 is distinct by 8, assuming n is

(a) a number

(b) a characteristic number

(c) an odd whole number

(d) an even number

Answer: (C) an odd number

Clarification: Let a = n 2 − 1. Here, n can be odd or even.

Case 1: When n = even. Let n = 4k, where k is a whole number.

⇒a=(4k) 2-1

⇒a=16k 2-1

which isn't distinct by 8.

Case 2: When n = odd. Let n = 4k + 1, where k is a whole number.

⇒a= (4k + 1) 2-1

⇒a=16k 2+1+8k-1⇒a=16k 2+8k

⇒a=8(2k 2+k)

which is dependably distinguishable by 8. Hence, on the off chance that n is odd, n 2-1 is distinct by 8.

Subsequently, the right response is choice C.

4.On the off chance that HCF of 65 and 117 is expressible in structure 65m-117, the worth of m is

(a) 4

(b) 2

(c) 1

(d) 3

Answer: (b) 2

Clarification: By Euclid's division lemma, b = aq + r, here 0 ≤r<a

117=65×1+52

65=52×1+13

52=13×4+0

∴ HCF (65,117) = 13 … ... (1)

We see that HCF (65,117) = 65m − 117 … ... (2)

From (1) and (2), we get:

65m−117=13

65m=130

m=2

In this way, the right response is choice (b).

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

(a) 13

(b) 65

(c) 875

(d) 1750

Answer: (a) 13

Explanation: Firstly, we subtract the remainder 5 and 8, respectively, from corresponding numbers and then get the HCF of the resulting numbers using Euclid’s division lemma to get the associated number. After removing these remainders from the numbers, we have:

(70 − 5) = 65

(125 − 8) = 117

Now, the required number is HCF (65,117).

Using Euclid’s division algorithm,

117=65×1+52

65=52×1+13

52=13×4+0

∴ HCF = 13. Thus, 13 is the largest number which is divisible by 70 and 125 leaving remainder 5 and 8. 

Hence, the correct answer is option (a).

6.On the off chance that two positive numbers 'a' as well as 'b' are composed as a = x3y2 and b = xy3; x, y are indivisible numbers, then, at that point, HCF (a, b) is

(a) x y

(b) xy2

(c) x3y3

(d) x2y2

Answer: (b) xy2

Clarification: Given: a= x3y2 = x× x× x× y× y b= xy3= x× y× y× y

HCF has full structure as the most noteworthy normal element of at least two related numbers.

Here, HCF (a, b) = HCF (x3y3, xy3) = x ×y ×y = xy2

Consequently, the right response is choice B.

7.In the event that two positive numbers 'p' too as 'q' can be communicated as p = ab2 and q = a3b; a, b being indivisible numbers, then, at that point, LCM (p, q) is

(a) stomach muscle

(b) a2b3

(c) a3b2

(d) a3b3

Answer: (c) a3b2

Clarification: Considering that,

p=ab2=a ×b ×b

q=a3b=a ×a ×a ×b

The LCM has the full structure as the littlest normal variety that can be isolated by both the numbers.

Subsequently,

LCM (p, q)

= LCM (ab2, a3b)

Thus,

=a ×b ×b ×a ×a =a3b2

Thusly, the right response is choice ©

8.The result of an unreasonable number and a non-zero objective is

(a) it's dependably nonsensical

(b) it's dependably objective

(c) judicious or nonsensical

(d) one

Answer: (a) consistently silly

Clarification: Think about the judicious number as 7/2 and the nonsensical number as √5/2

Their item is given as:

7/2×√5/2=7√5/4, which is silly.

Thus, the right response is choice A.

9.The most un-number which is detachable by every one of the numbers from 1 to 10 (both 1 and 10 comprehensive) is

(a) 10

(b) 100

(c) 504

(d) 2520

Answer: (d) 2520

Clarification: The LCM is the littlest positive number that is distinguishable by the given numbers.

Here, we really want to find the LCM that is the least normal different of the numbers beginning from 1 to 10 (both comprehensive). Variables of 1 to 10 numbers:

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

Hence, LCM of numbers from 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Subsequently, LCM = 1×2×2×2×3×3×5×7=2520

Hence, the right response is choice (d)

10.The decimal development of normal number 14587/1250 will end later:

(a) one decimal spot

(b) two decimal spots

(c) three decimal spots

(d) four decimal spots

Answer: (d) four decimal spots

Clarification: For ending decimal extension of a levelheaded number, the denominator is of the structure 2m×5n.

Here, 14587/1250=14587/21×54

⇒ [14587/10×53] × [23/23\ = [14587×8]/ [10×1000] = 116696/10000=11.6696

Subsequently, the decimal development of the given objective number ends after four decimal spots.

In this way, the right response is choice (d).

11.Express each number as a result of its great elements:

(I) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(I) 140

Clarification:

With the assistance of division of the number by the indivisible number's technique, we notice the different of prime elements of 140.

subsequently,

140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7

(ii) 156

With the assistance of division of the number by the indivisible number's technique, we notice the different of prime elements of 156.

here, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

(iii) 3825

With the assistance of division of the number by the indivisible number's strategy, we notice the various of prime variables of 3825.

here, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) 5005

With the assistance of division of the number by the indivisible number's technique, we notice the different of prime elements of 5005.

12.. Considering that HCF (306, 657) = 9, track down LCM (306, 657).

Answer:

You know about,

HCF × LCM = Increase of the two given numbers

Along these lines,

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Thus, LCM (306,657) = 22338

here, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429

With the assistance of division of the number by the indivisible number's technique, we notice the different of prime elements of 7429.

here, 7429 = 17 × 19 × 23 = 17 × 19 × 23

13.Show that any sure odd number is in the structure 6q + 1, or 6q + 3, or 6q + 5, where q is some number.

Answer:

Think about any sure number and b = 6. Thus, by Euclid's calculation, a = 6q + r, if some number q ≥ 0, and r = 0, 1, 2, 3, 4, 5, due to 0≤r<6.

Presently subbing the worth of r, we notice,

In the event that r = 0, a = 6q

On the off chance that, for r= 1, 2, 3, 4 and 5, we consider the worth of an is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, separately.

for a = 6q, 6q+2, 6q+4, then, at that point, and is much number and partitioned by 2. A positive whole number can be even or not odd, subsequently, any sure odd whole number is of the type of 6q+1, 6q+3 and 6q+5, where q is some whole number.

14.Check whether 6n can end with the digit 0 for any regular number n.

Answer:

In the event that any number 6n closes with the digit zero (0), it will be separable by 5, as we probably are aware any number with a unit place as 0 or 5 can be isolated by 5.

Prime factorization of 6n = (2 × 3) n

Thus, the superb factorization of 6n does exclude the indivisible number 5.

Consequently, obviously for any regular number n, 6n isn't detachable by 5 and in this way, it demonstrates that 6n can't end with the digit 0 for any normal number n.

15.Compose whether each sure whole number can be of the structure 4q + 2, where q is a number. Legitimize your response.

Answer:

No, every positive number can't take the structure 4q + 2, where q is the whole number.

explanation:

All the connected quantities of the structure 4q + 2, where 'q' is the whole number, are even numbers which we cannot separate by '4'.

As given model,

in the event that q=1,

4q+2 = 4(1) + 2= 6.

in the event that q=2,

4q+2 = 4(2) + 2= 10

if q=0,

4q+2 = 4(0) + 2= 2 and so on.

Accordingly, any number which has the structure 4q+2 will give just even numbers which cannot be increased by 4.

In this manner, each certain whole number can't be written in the structure 4q+2

16.Demonstrate that √5 is unreasonable.

Answer:

Consider that √5 is a levelheaded number.

for example, √5 = x/y (as well as x and y are co-primes)

y√5= x

Figuring out the both the sides, we notice,

(y√5)2 = x2

⇒5y2 = x2… … … … … … … … … … … … ... (1)

Along these lines, x2 is separated by 5, so x is additionally partitioned by 5.

Consider, x = 5k, for a few worth of k and placing the worth of x in condition (1), we notice,

5y2 = (5k)2

⇒y2 = 5k2

is detachable by 5 it implies y is distinct by 5.

Obviously, x and y are not co-primes. Subsequently, our supposition about √5 is reasonable is inaccurate.

Subsequently, √5 is an unreasonable number.

17.Demonstrate that 3 + 2√5 is silly.

Answer:

Allow us to expect to be 3 + 2√5 to be a judicious number.in the event that the co-primes x and y of the necessary objective number where (y ≠ 0) is to such an extent that:

3 + 2√5 = x/y

By revamping, we get,

2√5 = (x/y) - 3

√5 = 1/2[(x/y) - 3]

Presently x and y are whole numbers, in this way, 1/2[(x/y) - 3] is a judicious number.

Consequently, √5 is likewise a reasonable number. In any case, this defies the way that √5 is silly.

Consequently, our supposition that 3 + 2√5 is a judicious number is off base.

Thus, 3 + 2√5 is unreasonable.

18.On the off chance that n is an odd whole number, show that n2 - 1 is detachable by 8.

Answer:

We know that any odd positive number n can be written in the structure 4q + 1 or 4q + 3.

consequently, as per the inquiry, on the off chance that n = 4q + 1, then, at that point, we get n2 - 1 = (4q + 1)2 - 1 = 16q2 + 8q + 1 - 1 = 8q (2q + 1), separated by 8.

in the event that n = 4q + 3,

Then, at that point, we get n2 - 1 = (4q + 3)2 - 1 = 16q2 + 24q + 9 - 1 = 8(2q2 + 3q + 1), isolated by 8.

consequently, from the above conditions, obviously, assuming n is an odd positive number.

n2 - 1 is detachable by 8.

Subsequently Demonstrated.

19. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

Answer:

Assume the two odd positive numbers x and y be 2r + 1 and 2s + 1, that is

 = x2 + y2 = (2r + 1)2 +(2s + 1)2

= 4r2 + 4r + 1 + 4s2 + 4s + 1

= 4r2 + 4s2 + 4r + 4s + 2

= 4 (r2 + s2 + r + s) + 2

we conclude, the sum of square is even the number is not divided by 4

Thus, if x and y are odd positive integers, then x2 + y2 is even but not divisible by four.

Hence proved.

20.Show that the square of any odd number is of the structure 4q + 1, for some number q.

Answer:

Expect a be any odd number and b = 4.

According to Euclid's calculation,

a = 4m + r, here some number m ≥ 0

likewise, r = 0,1,2,3 if 0 ≤ r < 4.

consequently, we view as here,

a = 4m or 4m + 1 or 4m + 2 or 4m + 3 Thus, a = 4m + 1 or 4m + 3

We definitely know that a can't be 4m or 4m + 2, as they are partitioned by 2.

(4m + 1)2 = 16m2 + 8m + 1

= 4(4m2 + 2m) + 1

= 4q + 1, here q is some whole number and q = 4m2 + 2m.

(4m + 3)2 = 16m2 + 24m + 9

= 4(4m2 + 6m + 2) + 1

= 4q + 1, here q is some number and q = 4m2 + 6m + 2

Subsequently, Square of any odd whole number is of the structure 4q + 1, for some whole number q.

Subsequently demonstrated.