REAL NUMBERS

1. For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy:   1
(a) 0 ≤ r < 3 (b) 1 < r < 3
(c) 0 < r < 3 (d) 0 < r ≤ 3

Solution: (a)

 

2. Find the greatest number of 5 digits, that will give us the remainder of 5, when divided by 8 and 9 respectively.
(a) 99921 (b) 99931
(c) 99941 (d) 99951

Solution: (c) The greatest number will be multiple of LCM (8, 9)
LCM of 8 and 9 = 72
On verification, we find that 99941, when divided by 72, leaves remainder 5.

 

3. A number 10x + y is multiplied by another number 10a + b and the result comes as 100p + 10q + r, where r = 2y, q = 2(x + y) and p = 2x; y < 5, q ≠ 0.
The value of 10a + b may be ________.     2

Solution:  

 (10x + y)(10a + b) = 100p + 10q + r
⇒ (10x + y)(10a + b) = 100 × 2x + 10 × 2(x + y) + 2y
⇒ (10x + y)(10a + b) = 200x + 20(x + y) + 2y
⇒ (10x + y)(10a + b) = 220x + 22y
⇒ (10x + y)(10a + b) = 22(10x + y)
⇒ 10a + b = 22

   

4. Euclid’s division lemma states for any two positive integers a and b, there exist integers q and r such that a = bq + r. If a = 5, b = 8, then write the value of q and r.    1

Solution:  

Using Euclid’s division lemma, we get
a = bq + r
5 = 8 × 0 + 5
∴ q = 0 and r = 5

 

5. If a and b are two positive integers such that a = 14b. Find the HCF of a and b.   1

Solution:  

We can write
a = 14b + 0
∵ remainder is 0
∴ HCF is b.

 

6. For some integers, p and 5, there exist unique integers q and r such that p = 5q + r. Possible values of r are    1
(a) 0 or 1
(b) 0, 1 or 2
(c) 0, 1, 2 or 3
(d) 0, 1, 2, 3 or 4

Solution:  

(d) According to Euclids division lemma,
p = 5q + r, where 0 ≤ r < 5
⇒ r = 0, 1, 2, 3, 4
So, possible values of r are 0, 1, 2, 3 or 4.

 

7. If the HCF of 55 and 99 is expressible in the form 55m – 99, then the value of m is ________.    1

Solution:  

55 = 5 × 11, 99 = 9 × 11
∴ HCF(55, 99) = 11
ATQ, 55m – 99 = 11
55 × 2 – 99 = 11
∵ m = 2

 

8. If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then HCF(a, b) is
(a) xy
(b) xy²
(c) x³y³
(d) x²y²
Also, find LCM of (a, b).

Solution:  

 (b) Here, a = x³y² and b = xy³
⇒ a = x × x × x × y × y and b = xy × y × y
∴ HCF(a, b) = x × y × y = x × y² = xy²
LCM = x³y³

 

9. If two positive integers p and q can be expressed as p = ab2 and q = a³b; where a, b being prime numbers, then LCM (p, q) is equal to     1
(a) ab (b) a²b²
(c) a³b² (d) a²b³

Solution:    

(c) LCM (p, q) = a³b²

 

10. The ratio between the LCM and HCF of 5, 15, 20 is:     1
(a) 9: 1 (b) 4 : 3
(c) 11 : 1 (d) 12 : 1

Solution:  

(d) 5, 15 = 5 × 3, 20 = 2 × 2 × 5
LCM(5, 15, 20) = 5 × 3 × 2 × 2 = 60
HCF(5, 15, 20) = 5

11. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?   1
(a) 12.20 pm
(b) 12.12 pm
(c) 12.11 pm
(d) none of these

Solution:  

(d) LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 p.m., they will beep again for the first time.

 

12. If A = 2n + 13, B = n + 7, where n is a natural number, then HCF of A and B is:    1
(a) 2 (b) 1
(c) 3 (d) 4

Solution:  

(b) Taking different values of n we find that A and B are coprime.
∴ HCF = 1

 

13. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:    1
(a) 22 (b) 16
(c) 36 (d) 21

Solution:  

(b) HCF of 576 and 448 = 64
  ∴ Number of sections     = 

                                          = 9 + 7

                                          = 16

14. The HCF of 2472, 1284 and a third number N is 12. If their LCM is 2³ × 3² × 5 × 10³ × 107, then the number N is:    1
(a) 22 × 32 × 7 (b) 22 × 33 × 103
(c) 22 × 32 × 5 (d) 24 × 32 × 11

Solution:  

(c) 2472 = 2³ × 3 × 103
1284 = 2² × 3 × 107
∵ LCM = 2³ × 3² × 5 × 103 × 107
∴ N = 2² × 3² × 5 = 180

15. Two natural numbers whose difference is 66 and the least common multiple is 360, are:   1
(a) 120 and 54 (b) 90 and 24
(c) 180 and 114 (d) 130 and 64

Solution:

(b) Difference of 90 and 24 = 66 and LCM of 90 and 24 = 360
∴ Numbers are 90 and 24

16. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is ________________.   1

Solution: 

17. HCF of 5² × 3² and 3⁵ × 5³ is:    1
(a) 5³ × 3⁵ (b) 5 × 3³
(c) 5³ × 3² (d) 5² × 3²

Solution:  

(d) HCF of 5² × 3² and 3⁵ × 5³ = 5² × 3²

 

18. 4 Bells toll together at 9.00 am. They toll after 7, 8, 11 and 12 seconds respectively. How many times will they toll together again in the next 3 hours?   1

(a) 3 (b) 4
(c) 5 (d) 6

Solution:  

(c) LCM of 7, 8, 11, 12 = 1848
∴ Bells will toll together after every 1848 sec.
∴ In the next 3 hrs, number of times the bells will toll together = 

= 5.84
⇒ 5 times.

19. Given that LCM (91, 26) = 182, then HCF (91, 26) is _________.  1

Solution:  

LCM (91, 26) × HCF (91, 26) = 91 × 26
182 × HCF (91, 26) = 91 × 26
⇒ HCF (91, 26) = 

⇒ HCF (91, 26) = 13
 

20. The decimal expansion of the rational number will terminate after    1
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Solution:  

∴ The given rational number will terminate after four decimal places.

21. Which of the following rational numbers have a terminating decimal expansion?   1

Solution:  

(c)  

The denominator 2⁶ × 5² is of the form 2^m × 5ⁿ, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.

22. The decimal expansion of number  has ____________ decimal representation.   1

Solution:  

The denominator 2² × 5³ is of the form 2^m × 5ⁿ, where m and n are non-negative integers. Hence, it has a terminating decimal representation.

 

23. From the following, the rational number whose decimal expansion terminating is:   1

Solution:  

The denominator 2⁵ × 5¹ is of the form 2^m × 5ⁿ, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.

24. For any integer a and 3, there exists unique integers q and r such that a = 3q + r. Find the possible values of r.   1

Solution:  

According to Euclid's division lemma for two positive number a and b; there exist integer q and r such that
a = b × q + r where 0 ≤ r < b
Here b = 3
 0 ≤ r < 3
So values of r are 0, 1 or 2.

25. If a and b are two positive integers such that a = 14b. Find the HCF of a and b.   1

Solution:  

We can write
a = 14b + 0
  remainder is 0
  HCF is b.

26. Decompose 32760 into prime factors.   1

Solution:  

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13

27. Write the sum of exponents of prime factors in the prime factorisation of 250.  1

Solution:  

250 = 2 × 5³
  Sum of exponents = 1 + 3 = 4

28. The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, then find the other number.  1

Solution:  

HCF × LCM = Product of two numbers
  145 × 2175 = 725 × other number 

  Other number =

= 435