MCQ | CHAPTER 1 | REAL NUMBERS
d) 1
Ans : d
b) irrational
(c) both of above
d) none of these
Ans : b
(a) 4
(b) 28
(c) 38
(d) 48
Ans: c
Solution Hint
LCM of two prime numbers = product of the numbers
221 = 13 x 17.
So p = 17 & q = 13
⸫ 3p – q = 51 – 13 = 38
(a) rational
(b) irrational
(c) both of above
(d) none of above
(b) 3
(c) 4
(d) 5
Ans: c
Solution Hint
Since HCF = 81, two numbers can be taken as 81x and 81y,
ATQ 81x + 81y = 1215 Or x + y = 15
(b) 3
(c) 4
(d) 5
(a) whole numbers
(b) integers
(c) natural numbers
(d) even numbers
(a) 1516
(b) 1452
(c) 1011
(d) 1121
(a) x
(b) y
(c) xy
(d) x / y
(a) 17
(b) 11
(c) 34
Ans a
(a) 48
(b) 56
(c) 52
Ans c
(a) 36 minutes
(b) 18 minutes
(c) 6 minutes
(d) They will not meet
(a) 2² × 7
(b) 2² × 7²
(c) 2 × 7²
(d) 23 × 7
(a) 350 kg
(b) 290 kg
(c) 200 kg
Ans d
Q16) The ratio of LCM and HCF of the least composite and the least prime numbers is
(b) 2:1
(c) 1:1
(d) 1:3
Ans: b
Solution Hint
Least composite number is 4 and the least prime number is 2. LCM(4, 2) : HCF(4, 2) = 4 : 2 = 2 : 1
(b) irrational
(c) whole number
(d) integer
(a) 2, 3
(b) 2, 3, 5
(c) 2, 5
(d) 3, 5
Ans: c
Solution Hint
Since it has a terminating decimal expansion, so prime factors of the denominator will be 2,5
and B = x4y2z ; x, y being prime numbers, the LCM (A, B)
is
(a) xy²
(b) x4y²z
(c) x4y3
(d) x4y3z
Ans d
Q20) If two positive integers A and B can be
expressed as A = xy3 and B = x4y2z ; x, y
being prime numbers then HCF (A, B) is
(a) xy²
(b) x4y²z
(c) x4y3
(d) x4y3z
Ans a
Q21) The largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is
(a) 260
(b) 75
(c) 65
(d) 13
(a) 5
(b) 20
(c) 60
(d) 100
Ans c
Q 23) The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is
(a) 840
(b) 2520
(c) 8
(d) 420
Ans a
b) 13
c) 7
d) 11
b) 3a3b3c2
c) 5a3b2c3
d) 15a3b3c3
(a) prime number
(b) even number
(c) odd number
(d) even or odd
(a) greater than 7
(b) at least 7
(c) less than 7
(d) at most 7
(a) 24
(b) 16
(c) 8
(d) 48
(a) 22
(b) 7
(c) 9
(d) 11
(a) 99921
(b) 99931
(c) 99941
(d) 99951
Ans c
Hint: The greatest number will be multiple of LCM (8, 9)
LCM of 8 and 9 = 72
Now divide the greatest 5 digit number by 72.
Q31) The ratio between the LCM and HCF of 5, 15, 20 is
(a) 9 : 1
(b) 4:3
(c) 11:1
(d) 12:1
Ans: d
Hint : LCM(5, 15, 20) = 5 × 3 × 2 × 2 = 60
HCF(5, 15, 20) = 5
b) 84
c) 72
d) 96
a) x + y
b) 1
c) xy
d) x – y
Q34) Two numbers are in the ratio 3 : 5 and their HCF = 8. Find their LCM
a) 960
b) 120
c) 480
d) 240
Ans b
Hint: Let numbers are 3x and 5x. Common factor is x.
HCF is the product of common factors. So HCF of 3x and 5x = x = 8
So Numbers are 3 x 8 = 24 and 5 X 8 = 40
Now LCM X HCF = Product of numbers LCM X 8 = 24 X 40 LCM = (24 X 40) / 8 = 120
a) 40
b) 800
c) 8
d) 80
a) 25
b) 10
c) 4
d) 5
Ans d
Hint: Let two numbers are 2x and 7x. LCM is the product of common and non-common factors
So LCM of 2x and 7x = 14x = 70 x = 5 As x is the common factor of 2x and 7x. Therefore HCF (2x and 7x ) = x = 5
a) 12
b) 18
c) 16
d) 24
Ans c
a) 21 / 12
b) 15/ 200
c) 21/45
d) 6/15
a) 13
b) 11
c) 19
d) 15
a) 4
b) 8
c) 2
d) 12
a) 12
b) 36
c) 24
d) 48
a) 9960
b) 9999
c) 9980
d) 9940
Ans a
Hint: Find LCM of 12,15 and 24 = 120
Divide largest 4 digit number (9999) by 120 and find remainder = 39 Required largest number is 9999 – 39 = 9960
a) 9
b) 27
c) 36
d) 18
Ans d
Hint: Subtract 7 and 4 from 205 and 238 respectively we get 198 and 234 Requires number id the HCF (198 and 234) = 18
a) 10040
b) 10080
c) 10020
d) 10060
Ans b
Hint: Find LCM of 8,10,12 = 120
Divide smallest 5 digit number (10000) by 120 and find the remainder = 40
Divisor – Remainder = 120 – 40 = 80 Required number = 10000 + 80 = 10080
Ans d
Hint:
Ans: c
Hint: Required Number
a) 24
b) 26
c) 22
d) 28
Ans: b
a) 255
b) 240
c) 235
d) 245
a) 15
b) 25
c) 18
d) 20
a) 3
b) 9
c) 6
d) 12
a) 2565
b) 2655
c) 2555
d) 2465
(a)
3 x 52 x 21
(b) 32 x 52 x 35
(c) 32 x 52 x 17
(d) 32 x 25 x 17
(a) 100
(b) 1000
(c) 2520
(d) 5040
Ans: c
Solution Hint
1 = 1, 2 = 2 × 1, 3 = 3 × 1, 4 = 2 × 2
5 = 5 × 1, 6 = 2 × 3, 7 = 7 × 1, 8 = 2 × 2 × 2
9 = 3 × 3, 10 = 2 × 5
So, LCM of these numbers = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 Hence, least number divisible by all the numbers from 1 to 10 is 2520
ring together again?
(a) 6:07 AM
(b) 6:14 AM
(c) 6:28 AM
(d) 6:25 AM
Ans: c
Solution Hint
LCM 0f 4,7,14 = 28 Bells will they ring together again at 6 : 28 AM
x 32 and 22 x 33 is
a) 23
b) 33
c) 23 x 33
d) 22 x 33
Ans: c
(a) 420
(b) 600
(c) 720
(d) 800
a) 177
b) 183
c) 60
d) 63
a) 22 x 3 x 13
b) 22 x 32 x 13
c) 62 x 22
a) 2
b) 3
c) 6
d) 4
a) 6:04 a.m.
b) 6:07 a.m.
c) 6:14 a.m.
d) 6:28 a.m.
Q 61) If HCF (39, 91) = 13, then LCM (39, 91) is:
a) 91
b) 273
c) 39
d) 3549
b) Rational number
c) Natural Number
d) Irrational Number
a) 2
b) 3
c) 4
d) 1
Ans: a
Solution Hint
Let two numbers be 12x and 12y where x and y are coprime numbers.
ATQ 12x X 12y = 6336 ⇒ 144 xy = 6336 ⇒ xy = = 44
But x and y are co-primes so 2, 22 cannot be possible Hence required numbers are 1, 44 or 4, 11 ⇒ Two pair of numbers are possible.
a) 2
b) 4
c) 8
d) 0
a) 5 x 11 x 13
b) 5 x 7 x 11
c) 5 x 7 x 13
d) 5 x 11 x 17