Real Numbers 2-Mark Questions Class 10: CBSE PYQs with Answers

92 = 2² × 23.

510 = 2 × 3 × 5 × 17.

HCF = 2.

LCM = 23460.

HCF = 2; LCM = 23460

540 = 2² × 3³ × 5.

630 = 2 × 3² × 5 × 7.

HCF = 90.

HCF = 90

15ⁿ = 3ⁿ × 5ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 15ⁿ contains no factor 2.

Hence, 15ⁿ cannot end with the digit 0.

15ⁿ cannot end in 0

576 = 2⁶ × 3².

512 = 2⁹.

LCM = 4608.

LCM = 4608

660 = 2² × 3 × 5 × 11.

704 = 2⁶ × 11.

HCF = 44.

HCF = 44

480 = 2⁵ × 3 × 5.

256 = 2⁸.

LCM = 3840.

LCM = 3840

96 = 2⁵ × 3.

120 = 2³ × 3 × 5.

HCF = 24.

LCM = 480.

HCF = 24; LCM = 480

36 = 2² × 3².

60 = 2² × 3 × 5.

HCF = 12.

LCM = 180.

HCF = 12; LCM = 180

84 = 2² × 3 × 7.

90 = 2 × 3² × 5.

120 = 2³ × 3 × 5.

HCF = 6.

LCM = 2520.

HCF = 6; LCM = 2520

84 = 2² × 3 × 7.

144 = 2⁴ × 3².

HCF = 12.

HCF = 12

231 = 3 × 7 × 11.

396 = 2² × 3² × 11.

LCM = 2772.

LCM = 2772

For two positive integers, HCF × LCM = product of the numbers.

LCM = $\frac{\text{(}\text{306 × 1314}\text{)/(}}{\text{18}\text{)}}$ = 22338.

LCM = 22338

45 = 3² × 5.

54 = 2 × 3³.

270 = 2 × 3³ × 5.

HCF = 9.

HCF = 9

180 = 2² × 3² × 5.

140 = 2² × 5 × 7.

210 = 2 × 3 × 5 × 7.

HCF = 10.

HCF = 10

144 = 2⁴ × 3².

180 = 2² × 3² × 5.

192 = 2⁶ × 3.

HCF = 12.

HCF = 12

35 = 5 × 7.

55 = 5 × 11.

HCF = 5.

LCM = 385.

Check: 5 × 385 = 35 × 55 = 1925.

HCF = 5; LCM = 385

51 = 3 × 17; 85 = 5 × 17.

LCM = 3 × 5 × 17 = 255.

6z − 9 = 255 ⇒ 6z = 264 ⇒ z = 44.

z = 44

480 = 2⁵ × 3 × 5.

720 = 2⁴ × 3² × 5.

HCF = 240.

LCM = 1440.

HCF = 240; LCM = 1440

85 = 5 × 17; 238 = 2 × 7 × 17 ⇒ HCF = 17.

17 = 85m − 238 ⇒ 85m = 255.

m = 3.

m = 3

Subtract the remainders: 85 − 1 = 84; 72 − 2 = 70.

84 = 2² × 3 × 7; 70 = 2 × 5 × 7.

Required number = HCF(84, 70) = 14.

14

Let the numbers be 2x and 3x, where x is a positive integer.

Since 2 and 3 are coprime, HCF = x and LCM = 6x.

6x = 180 ⇒ x = 30.

HCF = 30

18 = 2 × 3²; 24 = 2³ × 3; 36 = 2² × 3².

LCM = 2³ × 3² = 72.

999 = 72 × 13 + 63.

Required greatest three-digit multiple = 72 × 13 = 936.

936

6 = 2 × 3.

72 = 2³ × 3².

120 = 2³ × 3 × 5.

HCF = 6.

LCM = 360.

HCF = 6; LCM = 360

They meet at the starting point after a common multiple of their lap times.

30 = 2 × 3 × 5; 40 = 2³ × 5; 48 = 2⁴ × 3.

Required time = LCM(30, 40, 48) = 240 minutes.

240 minutes

The required smallest positive number is LCM(8, 9, 10).

8 = 2³; 9 = 3²; 10 = 2 × 5.

LCM = 2³ × 3² × 5 = 360.

360

Required greatest divisor = HCF(65, 117).

65 = 5 × 13; 117 = 3² × 13.

HCF = 13.

13

285 − 9 = 276; 1249 − 7 = 1242.

276 = 2² × 3 × 23; 1242 = 2 × 3³ × 23.

Required number = HCF(276, 1242) = 138.

138

644 = 2² × 7 × 23.

462 = 2 × 3 × 7 × 11.

LCM = 21252.

LCM = 21252

Since 4 and 5 are coprime, the numbers are 4 × 11 = 44 and 5 × 11 = 55.

HCF × LCM = product of the two numbers.

LCM = $\frac{\text{(}\text{44 × 55}\text{)/(}}{\text{11}\text{)}}$ = 220.

LCM = 220

Let the numbers be 3x and 5x, where x is a positive integer.

Since 3 and 5 are coprime, HCF = x and LCM = 15x.

15x = 180 ⇒ x = 12.

HCF = 12

24 = 2³ × 3; 36 = 2² × 3², so LCM = 72.

The smallest five-digit number is 10000.

10000 = 72 × 138 + 64.

Required number = 10000 + (72 − 64) = 10008.

10008

37 − 2 = 35; 66 − 3 = 63; 89 − 5 = 84.

35 = 5 × 7; 63 = 3² × 7; 84 = 2² × 3 × 7.

Required number = HCF(35, 63, 84) = 7.

7

Let the numbers be x and y, with x > y.

x + y = 256; x − y = 224 ⇒ x = 240 and y = 16.

240 = 2⁴ × 3 × 5; 16 = 2⁴.

HCF = 16; LCM = 240.

HCF = 16; LCM = 240

Required time = LCM(9, 12, 15).

9 = 3²; 12 = 2² × 3; 15 = 3 × 5.

LCM = 180 minutes = 3 hours.

180 minutes, or 3 hours

The number of books must be divisible by both 40 and 48.

40 = 2³ × 5; 48 = 2⁴ × 3.

Minimum number = LCM(40, 48) = 240.

240 books

Required time = LCM(48, 72, 108).

48 = 2⁴ × 3; 72 = 2³ × 3²; 108 = 2² × 3³.

LCM = 2⁴ × 3³ = 432 seconds.

432 seconds

The required number plus 17 is LCM(520, 468).

520 = 2³ × 5 × 13; 468 = 2² × 3² × 13.

LCM = 4680.

Required number = 4680 − 17 = 4663.

4663

25ⁿ = 5²ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 25ⁿ contains no factor 2.

Hence, 25ⁿ cannot end with the digit 0.

25ⁿ cannot end in 0

Convert to centimetres: 975 cm, 675 cm and 525 cm.

975 = 3 × 5² × 13; 675 = 3³ × 5²; 525 = 3 × 5² × 7.

Longest ruler = HCF(975, 675, 525) = 75 cm.

75 cm, or 0.75 m

LCM(5, 6, 8) = 120 minutes = 2 hours.

After 10 a.m., they toll together at 12 noon, 2 p.m., 4 p.m. and 6 p.m.

Exclude the opening toll because the question asks how many more times.

4 more times

6ⁿ = 2ⁿ × 3ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 6ⁿ contains no factor 5.

Hence, 6ⁿ cannot end with the digit 0.

6ⁿ cannot end in 0

72 = 2³ × 3².

120 = 2³ × 3 × 5.

HCF = 24.

LCM = 360.

HCF = 24; LCM = 360

8ⁿ = 2³ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 8ⁿ contains no factor 5.

Hence, 8ⁿ cannot end with the digit 0.

8ⁿ cannot end in 0

96 = 2⁵ × 3.

160 = 2⁵ × 5.

HCF = 32.

LCM = 480.

HCF = 32; LCM = 480

12ⁿ = 2²ⁿ × 3ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 12ⁿ contains no factor 5.

Hence, 12ⁿ cannot end with the digit 0.

12ⁿ cannot end in 0

(i) √2 and −√2 are irrational, and their sum is 0, a rational number.

(ii) √2 and √3 are irrational, and their product √6 is irrational.

(i) √2 and −√2; (ii) √2 and √3

7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009.

Each is a product of two integers greater than 1.

Both numbers are composite

4ⁿ = 2²ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 4ⁿ contains no factor 5.

Hence, 4ⁿ cannot end with the digit 0.

4ⁿ cannot end in 0

15ⁿ = 3ⁿ × 5ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 15ⁿ contains no factor 2.

Hence, 15ⁿ cannot end with the digit 0.

No, 15ⁿ cannot end in 0

8ⁿ = 2³ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 8ⁿ contains no factor 5.

Hence, 8ⁿ cannot end with the digit 0.

No, 8ⁿ cannot end in 0

14ⁿ = 2ⁿ × 7ⁿ.

A number ending in 0 or 5 is divisible by 5.

The unique prime factorisation of 14ⁿ contains no factor 5.

Hence, 14ⁿ cannot end with 0 or 5.

No; 14ⁿ ends in neither 0 nor 5

45ⁿ = 3²ⁿ × 5ⁿ, which contains no factor 2.

Ending in 0 requires both 2 and 5, so 45ⁿ cannot end in 0.

Multiplying by the prime 2 supplies the missing factor.

45ⁿ cannot end in 0; a = 2

15ⁿ × 2ⁿ = (15 × 2)ⁿ = 30ⁿ.

For n ≥ 1, 30ⁿ is divisible by 10.

Hence, the number ends with 0.

Yes; 15ⁿ × 2ⁿ ends in 0

6ⁿ = 2ⁿ × 3ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 6ⁿ contains no factor 5.

Hence, 6ⁿ cannot end with the digit 0.

Multiplying by the prime 5 supplies the missing factor, so 5 × 6ⁿ is divisible by 10.

6ⁿ cannot end in 0; multiply by 5

2 × 5 × 7 × 11 + 11 × 13 = 11(2 × 5 × 7 + 13).

= 11 × 83.

Both factors exceed 1, so the number is composite.

Composite; 11 × 83

306 = 2 × 3² × 17.

657 = 3² × 73.

LCM = 22338.

LCM = 22338

Assume that 2 + √3 is rational.

Then 2 + √3 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √3 = $\frac{\text{(}\text{p − 2q}\text{)/(}}{\text{q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √3.

Hence, 2 + √3 is irrational.

2 + √3 is irrational

Assume that 6 − √7 is rational.

Then 6 − √7 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √7 = $\frac{\text{(}\text{6q − p}\text{)/(}}{\text{q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √7.

Hence, 6 − √7 is irrational.

6 − √7 is irrational

Assume that 5 − 2√2 is rational.

Then 5 − 2√2 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √2 = $\frac{\text{(}\text{5q − p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √2.

Hence, 5 − 2√2 is irrational.

5 − 2√2 is irrational

6ⁿ = 2ⁿ × 3ⁿ.

A natural number ending in 0 is divisible by 10, so its prime factorisation contains both 2 and 5.

The unique prime factorisation of 6ⁿ contains no factor 5.

Hence, 6ⁿ cannot end with the digit 0.

No, 6ⁿ cannot end in 0

Assume that 6 − 4√5 is rational.

Then 6 − 4√5 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √5 = $\frac{\text{(}\text{6q − p}\text{)/(}}{\text{4q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √5.

Hence, 6 − 4√5 is irrational.

6 − 4√5 is irrational

11 × 19 × 23 + 3 × 11 = 11(19 × 23 + 3).

= 11 × 440.

Both factors exceed 1, so the number is composite.

Composite; 11 × 440

Assume that −7 − 2√3 is rational.

Then −7 − 2√3 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √3 = $\frac{\text{(}\text{−7q − p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √3.

Hence, −7 − 2√3 is irrational.

−7 − 2√3 is irrational

7 × 11 × 13 + 2 × 11 = 11(7 × 13 + 2).

= 11 × 93.

Both factors exceed 1, so the number is composite.

Composite; 11 × 93

Assume that 5 − 2√3 is rational.

Then 5 − 2√3 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √3 = $\frac{\text{(}\text{5q − p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √3.

Hence, 5 − 2√3 is irrational.

5 − 2√3 is irrational

5 × 11 × 17 + 3 × 11 = 11(5 × 17 + 3).

= 11 × 88.

Both factors exceed 1, so the number is composite.

Composite; 11 × 88

Assume that 7 − 3√5 is rational.

Then 7 − 3√5 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √5 = $\frac{\text{(}\text{7q − p}\text{)/(}}{\text{3q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √5.

Hence, 7 − 3√5 is irrational.

7 − 3√5 is irrational

Assume √3 = $\frac{\text{(}\text{a}\text{)/(}}{\text{b}\text{)}}$, where a and b are coprime integers and b ≠ 0.

Then a² = 3b². Since 3 is prime, 3 divides a.

Write a = 3k. Substituting gives b² = 3k².

Thus 3 also divides b, contradicting that a and b are coprime.

Hence, √3 is irrational.

√3 is irrational

(i) (√5 − √3)² = 5 + 3 − 2√15 = 8 − 2√15.

Since √15 is irrational, 8 − 2√15 is irrational.

(ii) (5 + √3)(5 − √3) = 25 − 3 = 22.

22 is rational.

(i) Irrational; (ii) Rational, 22

15 = 3 × 5; 24 = 2³ × 3; 36 = 2² × 3².

LCM = 2³ × 3² × 5 = 360.

1000 = 360 × 2 + 280, so the next multiple is 360 × 3 = 1080.

1080

Assume that 2√7 is rational.

Then 2√7 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √7 = $\frac{\text{(}\text{p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √7.

Hence, 2√7 is irrational.

2√7 is irrational

Assume that 2 + 3√5 is rational.

Then 2 + 3√5 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √5 = $\frac{\text{(}\text{p − 2q}\text{)/(}}{\text{3q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √5.

Hence, 2 + 3√5 is irrational.

2 + 3√5 is irrational

210 = 2 × 3 × 5 × 7; 55 = 5 × 11, so HCF = 5.

5 = 210 × 5 + 55m.

55m = −1045 ⇒ m = −19.

m = −19

Assume that 2 − 5√3 is rational.

Then 2 − 5√3 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √3 = $\frac{\text{(}\text{2q − p}\text{)/(}}{\text{5q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √3.

Hence, 2 − 5√3 is irrational.

2 − 5√3 is irrational

Assume that 4 − 2√5 is rational.

Then 4 − 2√5 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √5 = $\frac{\text{(}\text{4q − p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √5.

Hence, 4 − 2√5 is irrational.

4 − 2√5 is irrational

Assume that 14 − 2√3 is rational.

Then 14 − 2√3 = $\frac{\text{(}\text{p}\text{)/(}}{\text{q}\text{)}}$, where p and q are integers and q ≠ 0.

On rearranging, √3 = $\frac{\text{(}\text{14q − p}\text{)/(}}{\text{2q}\text{)}}$.

The right-hand side is rational, contradicting the given irrationality of √3.

Hence, 14 − 2√3 is irrational.

14 − 2√3 is irrational

Assume √2 = $\frac{\text{(}\text{a}\text{)/(}}{\text{b}\text{)}}$, where a and b are coprime integers and b ≠ 0.

Then a² = 2b². Since 2 is prime, 2 divides a.

Write a = 2k. Substituting gives b² = 2k².

Thus 2 also divides b, contradicting that a and b are coprime.

Hence, √2 is irrational.

√2 is irrational