Staff Selection Commission Mathematics - LCM and HCF

Q.1

The LCM and the HCF of the numbers 28 and 42 are in the ratio :

6 : 1

2 : 3

3 : 2

7 : 2

Ans .

Explanation :

2   28, 42
2   14, 21
7   7, 21
     1, 3
= 2 × 2 × 7 × 3 = 84 H.C. F. of 28 and 42
             28 ) 42 ( 1
                    28
                 14 ) 28 ( 2
                      28
                 0
H.C. F = 14
Required ratio = 84 = 6:1
    14

Q.2

If the ratio of two numbers is 2 : 3 and their L.C.M. is 54, then the sum of the two num- bers is

270

Ans .

Explanation :
Let the two numbers are 2x and 3x respectively. According to question, LCM = 54 x ( ) 3 2 54 ´ = Þ x = 9 Numbers = 2x = 2 × 9 = 18 and, 3x = 3 × 9 = 27 Sum of the two numbers = 18 + 27 = 45

Q.3

The ratio of two numbers is 4 : 5 and their L.C.M. is 120. The numbers are

30, 40

40, 32

24, 30

36, 20

Ans .

Explanation :
Suppose the numbers are 4x and 5x respectively According to question x × 4 × 5 = 120 Þ x = 6 Required numbers = 4 × 6 = 24 and = 5 × 6 = 30

Q.4

Three numbers are in the ratio 2 : 3 : 4 and their H.C.F. is 12. The L.C.M. of the numbers is

144

192

Ans .

Explanation :
Let the numbers be 2x, 3x and 4x respectively. HCF = x = 12 Numbers are : 2 ×12 = 24 3 ×12 = 36, 4 ×12 = 48 LCM of 24, 36, 48 = 2 × 2 × 2 × 3 × 3 × 2 = 144

Q.5

Two numbers are in the ratio 3 : 4. If their LCM is 240, the smaller of the two number is

100

Ans .

Explanation :
Let the number be 3x and 4x. Their LCM = 12x According to the question, 12x = 240
x = 240 = 20
12
Smaller number = 3x = 3 × 20 = 60

Q.6

Two numbers are in the ratio 3 : 4. Their L.C.M. is 84. The greater number is

Ans .

Explanation :
Let the numbers be 3x and 4x. Their LCM = 12x 12x = 84
x = 84 = 7
12
Larger number = 4x = 4 × 7 = 28

Q.7

Two numbers are in the ratio 3 : 4. If their HCF is 4, then their LCM is

Ans .

Explanation :
Numbers = 3x and 4x HCF = x = 4
LCM = 12x = 12 × 4 = 48

Q.8

The ratio of the sum to the LCM of two natural numbers is 7 : 12. If their HCF is 4, then the smaller number is :

Ans .

Explanation :
Let the numbers be 4x and 4y where x and y are prime to each other. LCM = 4xy
(4x + 4y) = 7
4xy 12
12 (x + y) = 7 xy
x = 3, y = 4
Smaller number = 4 × 3 = 12

Q.9

Two numbers are in the ratio 3 : 4. The product of their H.C.F. and L.C.M. is 2028. The sum of the numbers is

Ans .

Explanation :
Using Rule 1, Let the numbers be 3x and 4x respectively First number × second number
= HCF × LCM
3x × 4x = 2028
x² = 2028 = 169
3 x 4
x = √169 = 13 Sum of the numbers = 3x + 4x = 7x = 7 × 13 = 91

Q.10

The LCM of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is

Ans .

Explanation :
If the numbers be 2x and 3x, then LCM = 6x 6x = 48
x = 8 Required sum = 2x + 3x = 5x = 5 × 8 = 40

Q.11

The ratio of two numbers is 4 : 5 and their H.C.F. is 8. Then their L.C.M. is

130

140

150

160

Ans .

Explanation :
Let the numbers be 4x and 5x. H.C.F. = x = 8 Numbers = 32 and 40 Their LCM = 160

Q.12

The ratio of two numbers is 3 : 4 and their HCF is 5. Their LCM is:

Ans .

Explanation :
If the numbers be 3x and 4x, then HCF = x = 5 Numbers = 15 and 20 LCM = 60

Q.13

Three numbers are in the ratio 1 : 2 : 3 and their HCF is 12. The numbers are

12, 24, 36

5, 10, 15

4, 8, 12

10, 20, 30

Ans .

Explanation :
Numbers = x , 2 x and 3 x (let) Their H.C.F. = x = 12 Numbers = 12, 24 and 36

Q.14

If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two num- bers is

xz/y

xy/z

xyz

Ans .

Explanation :
Using Rule 1, Product of two numbers = HCF × LCM
Numbers = zx and zy
zx × zy = z × LCM
LCM = xyz

Q.15

The H.C.F. and L.C.M. of two numbers are 21 and 84 respec- tively. If the ratio the two num- bers is 1 : 4, then the larger of the two numbers is

108

Ans .

Explanation :
HCF of numbers = 21 Numbers = 21x and 21y Where x and y are prime to each other. Ratio of numbers = 1 : 4
Larger number = 21 × 4 = 84

Q.16

The product of the LCM and HCF of two numbers is 24. The difference of the two numbers is 2. Find the numbers ?

8 and 6

8 and 10

2 and 4

6 and 4

Ans .

Explanation :
Using Rule 1, Let the numbers be x and (x + 2). Product of numbers = HCF × LCM x (x + 2) = 24
x² + 2x – 24 = 0
x² + 6x – 4x – 24 = 0
x (x + 6) – 4 (x + 6) = 0
(x – 4) (x + 6) = 0
x = 4, as x ¹ – 6 = 0
Numbers are 4 and 6.

Q.17

The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is

Ans .

Explanation :
Using Rule 1, Suppose 1st number is x then, 2nd number = 100 – x
LCM × HCF = 1st number × 2nd number
495 × 5 = x × (100 – x)
495 × 5 = 100x – x²
x² – 55x – 45x – 2475 = 0
(x – 45) (x – 55) = 0
x = 45 or x = 55

Then, difference = 55 – 45 = 10

Q.18

Two numbers, both greater than 29, have HCF 29 and LCM 4147. The sum of the numbers is

966

696

669

666

Ans .

Explanation :
Let the number be 29x and 29y respectively where x and y are prime to each other.
LCM of 29x and 29y = 29xy
Now, 29xy = 4147
xy = 4147 = 143
29
Thus xy = 11 × 13
Numbers are 29 × 11
= 319 and 29 × 13 = 377
Required sum
= 377 + 319 = 696

Q.19

The sum of the H.C.F. and L.C.M of two numbers is 680 and the L.C.M. is 84 times the H.C.F. If one of the number is 56, the other is :

Ans .

Explanation :
Let HCF be h and LCM be l. Then, l = 84h and l + h = 680 84h + h = 680 h = 680 = 8
85 l = 680 – 8 = 672
Other number
= 672 x 8
56
= 96

Q.20

The sum of two numbers is 84 and their HCF is 12. Total num- ber of such pairs of number is

Ans .

Explanation :
HCF = 12 Numbers = 12x and 12y where x and y are prime to each other.
12x + 12y = 84
12 (x + y) = 84
x + y = 84 = 7
12
Possible pairs of numbers satisfying this condition = (1,6), (2,5) and (3,4). Hence three pairs are of required numbers

Q.21

The sum of a pair of positive in- teger is 336 and their H.C.F. is 21. The number of such possi- ble pairs is

Ans .

Explanation :
Let the numbers be 21x and 21y where x and y are prime to each other. 21x + 21y = 336
21 (x + y) = 336
x + y = 336 =16 21 Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)

Q.22

The sum of two numbers is 45.Their difference is 1/9 of their sum. Their L.C.M. is

200

250

100

150

Ans .

Explanation :
Let the number be x and y. According to the question, x + y = 45 ......... (i)
Again, x – y = 1( x + y)
9
or x – y = 145
9 or x – y = 5 ..... (ii)
By (i) + (ii) we have,
x + y = 45
x – y = 5
= 50 2x
or, x = 25
y = 45 – 25 = 20.
Now, LCM of 25 and 20 = 100

Q.23

The H.C.F. of two numbers, each having three digits , is 17 and their L.C.M. is 714. The sum of the numbers will be

289

391

221

731

Ans .

Explanation :
Let the numbers be 17x and 17y where x and y are co-prime.
LCM of 17x and 17y = 17 xy
According to the question,
17xy = 714
xy = 714 = 42 = 6 x 7
17
x = 6 and y = 7
or, x = 7 and y =6.
First number = 17x
= 17 × 6 = 102
Second number = 17y
=17 × 7 = 119
Sum of the numbers
= 102 + 119 = 221

Q.24

The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is

Ans .

Explanation :
Using Rule 1, Let the larger number be x. Smaller number = x – 2
First number × Second number = HCF × LCM x (x – 2) = 24
x² – 2x – 24 = 0
x² – 6x + 4x – 24 = 0
x (x – 6) + 4 (x – 6) = 0
(x – 6) (x + 4) = 0
x = 6 because x not equal to 4

Q.25

The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

Ans .

Explanation :
HCF of two numbers = 27 Let the numbers be 27x and 27y
where x and y are prime to each other.
According to the question,
27x + 27y = 216
27 (x + y) = 216
x + y = 216 = 8
27
Possible pairs of x and y = (1, 7) and (3, 5)
Numbers =(27, 189) and (81, 135)

Q.26

The LCM of two numbers is 12 times their HCF. The sum of the HCF and the LCM is 403. If one of the number is 93, then the other number is

124

128

134

138

Ans .

Explanation :
Using Rule 1, Let the HCF of numbers = H
Their LCM = 12H
According to the question,
12H +H = 403
13H = 403
H =403 = 31
13
LCM = 12 × 31
Now,
First number × second number
= HCF × LCM
= 93 × Second Number
= 31 × 31 × 12
Second number =
31 x 31 x 12
93
= 124

Q.27

Sum of two numbers is 384. H.C.F. of the numbers is 48. The difference of the numbers is

100

192

288

336

Ans .

Explanation :
Let the numbers be 48x and 48y where x and y are co-primes.
48x + 48y = 384
48 ( x + y) = 384
x + y =384 =8
48
Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).
Numbers are : 48 × 1 = 48
and 48 × 7 = 336
and 48 × 3 = 144 and 48 × 5 = 240
Required difference
= 336 – 48 = 288

Q.28

The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is

2/35

3/25

4/35

2/25

Ans .

Explanation :
Let the numbers be 3x and 3y.
3x + 3y = 36
x + y = 12 ... (i)
and 3xy = 105 ... (ii)
Dividing equation (i) by (ii), we have
x + y =12
3xy 3xy 105
1 + 1 =4
3xy 3xy 35

Q.29

L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers ?

140

Ans .

Explanation :
Let the numbers be 10x and 10y where x and y are prime to each other.
LCM = 10 xy
10xy = 120
xy = 12
Possible pairs = (3, 4) or (1, 12)
Sum of the numbers
= 30 + 40 = 70

Q.30

Three numbers which are co- prime to one another are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is

Ans .

Explanation :
Let the numbers be x, y and z which are prime to one another. Now, xy = 551
yz = 1073
y = HCF of 551 and 1073
y = 29
x = 551 =19
29
and
z = 1073 =37
29
Sum = 19 + 29 + 37 = 85

Q.31

The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are pos- sible ?

Ans .

Explanation :
HCF of two numbers = 4.
Hence, the numbers can be given by 4x and 4y where x and y are co-prime. Then,
4x + 4y = 36
4 (x + y) = 36
x + y = 9
Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)

Q.32

If the HCF and LCM of two con- secutive (positive) even numbers be 2 and 84 respectively, then the sum of the numbers is

Ans .

Explanation :
Let the numbers be 2x and 2y where x and y are prime to each other.
LCM = 2xy
2xy = 84
xy = 42 = 6 × 7
Numbers are 12 and 14.
Hence Sum = 12 + 14 = 26

Q.33

The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :

Ans .

Explanation :
Let the numbers be x H and yH where H is the HCF and yH x H.
LCM = xy H
xyH = 2yH
x = 2 Again, x H – H = 4
2H – H = 4
H = 4
Smaller number = x H = 8

Q.34

The L.C.M. of two numbers is 20 times their H.C.F. The sum of H.C.F. and L.C.M. is 2520. If one of the number is 480, the other number is :

400

480

520

600

Ans .

Explanation :
Using Rule 1, Let the H.C.F. be H.
L.C.M. = 20H
Then, H + 20H = 2520
21 H = 2520
H = 2520 = 120
21
L.C.M. = 20H = 20×120= 2400
As,
First number × Second number
= L.C.M. × H.C.F.
480 × Second number = 2400 × 120
Second number = 2400 x 120 = 600
480

Q.35

The LCM of two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the oth- er number is

1100

975

900

800

Ans .

Explanation :
Using Rule 1, If the HCF = H, then
LCM = 44 H
44 H + H = 1125
45 H = 1125
H = 1125 = 25
45
LCM = 44 × 25 = 1100
Now
First number × Second number
= LCM × HCF
25 × Second number
= 1100 × 25
Second number
= 1100 x 25 = 1100
25

Q.36

If A and B are the H.C.F. and L.C.M. respectively of two al- gebraic expressions x and y, and A + B = x + y, then the value of A 3 + B 3 is

x qube- y qube

x qube

y qube

x qube + y qube

Ans .

Explanation :
Let no. are x and y and HCF = A, LCM = B
Using Rule, we have
xy = AB
x + y = A + B (given) ...(i)
(x–y)² = (x + y)² – 4xy
or, (x–y)² = (A + B)² – 4 AB
or, (x–y)² = (A – B)2
or, (x–y) = A – B ...(ii)
Using (i) and (ii), we get
x = A and y = B
A³ + B³ = x³ + y³

Q.37

HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is :

Ans .

Explanation :
Let the numbers be 7x and 7y where x and y are co-prime.
Now, LCM of 7x and 7y = 7xy
7xy = 140
xy = 140 = 20
7
Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.
Numbers are 28 and 35 which lie between 20 and 45.
Required sum = 28 + 35 = 63

Q.38

The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is

3625

3250

3500

3600

Ans .

Explanation :

Firstly, we find the LCM of 30, 36 and 80.
2   30, 36, 80
2   15, 18, 40
3   15, 9, 20
5    5, 3, 20
     1, 3, 4
LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
Required number = Multiple of 720 = 720 × 5 = 3600; because 3000 < 3600 < 4000

Q.39

Let x be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is

Ans .

Explanation :
LCM of 5, 6, 7 and 8 = 840
2 5, 6, 7, 8
5, 3, 7, 4
LCM = 2 × 5 × 3 × 7 × 4 = 840
Required number = 840x + 3
which is divisible by 9 for a certain least value of x.
Now,
840x + 3 = 93x × 9 + 3x + 3
3x + 3, is divisible by 9 for x = 2
Required number = 840 × 2 + 3
= 1680 + 3 = 1683
Sum of digits = 1 + 6 + 8 + 3
= 18

Q.40

The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is

9828

9288

9882

9928

Ans .

Explanation :

2   12, 18, 21, 28
2   6, 9, 21, 14
3   3, 9, 21, 7
7   1, 3, 7, 7
     1, 3, 1, 1
LCM = 2 × 2 × 3 × 3 × 7= 252 The largest 4-digit number = 9999
                 252 ) 9999 ( 39
                  756
                 2439
                 2268
                 171
Required number = 9999 – 171 = 9828

Q.41

A number x is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remain- der 3 in each case. The least value of x is:

148

149

150

147

Ans .

Explanation :
LCM of 8, 12 and 16 = 48 Required number = 48a + 3 which is divisible by 7.
x = 48a + 3
= (7 × 6a) + (6a + 3) which is divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3
= 21 which is divisible by 7.
x = 48 × 3 + 3 = 144 + 3 = 147

Q.42

Let x be the smallest number, which when added to 2000 makes the resulting number di- visible by 12, 16, 18 and 21. The sum of the digits of x is

Ans .

Explanation :

2   12, 16, 18, 21
2   6, 8, 9, 21
3   3, 4, 9, 21
     1, 4, 3, 7
LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008
Multiple of 1008 = 2016
Required number = 2016 – 2000 = 16 = x
Sum of digits of x = 1 + 6 = 7

Q.43

The smallest five digit number which is divisible by 12,18 and 21 is :

10224

30256

10080

50321

Ans .

Explanation :

2   12, 18, 21
3   6, 9, 21
     2, 3, 7
LCM of 12, 18 and 21
= 2 × 3 × 2 × 3 × 7 = 252
Of the options, 10080 ÷ 252 = 40

Q.44

A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder 11 in each case is

1451

1641

1712

1523

Ans .

Explanation :
2   30, 36, 80
2   15, 18, 40
3   15, 9, 20
5   5, 3, 20
     1, 3, 4

LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720 Required number = 2 × 720 + 11 = 1440 + 11 = 1451

Q.45

The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is

4023

4032

4302

4203

Ans .

Explanation :
2   12, 18, 21, 32
2   6, 9, 21, 16
3   3, 9, 21, 8
     1, 3, 7, 8
LCM = 2 × 2 × 3 × 3 × 7 × 8 = 2016 Required number = 2016 × 2 = 4032

Q.46

If the product of three consecu- tive numbers is 210 then sum of the smaller number is

Ans .

Explanation :
2   210
3   105
5   35
     7
210 = 2 × 3 × 5 × 7 = 5 × 6 × 7 Required answer = 5 + 6 = 11

Q.47

The LCM of two multiples of 12 is 1056. If one of the number is 132, the other number is

132

Ans .

Explanation :
Let the numbers be 12x and 12y.
Their LCM = 12xy when x and y are prime to each other.
y = 1056 = 8
132 Other number = 12y
= 12 × 8 = 96

Q.48

The least number to be sub- tracted from 36798 to get a num- ber which is exactly divisible by 78 is

Ans .

Explanation :
When 36798 is divided by 78, remainder = 60
The least number to be subtracted = 60

Q.49

Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

3013

3024

3002

3036

Ans .

Explanation :
LCM of 18, 21 and 24
2   18, 21, 24
3   9, 21, 12
     3, 7, 4
LCM = 2 × 3 × 3 × 7 × 4 = 504 Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504 K – 11. Where K is a positive integer Since 23 × 21 = 483 We can write 504 K – 11 = (483 + 21) K – 11 = 483 K + (21K – 11) 483 K is multiple of 23, since 483 is divisible by 23.
So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.
Put the value of K = 1, 2, 3, 4, 5, 6, ..... and so on successively.
We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11) = 115 which is divisible by 23.
Therefore, the required least number = 504 × 6 – 11 = 3013

Q.50

The greatest number, that divides 122 and 243 leaving respectively 2 and 3 as remainders, is

120

Ans .

Explanation :
Using Rule 7, Clearly, 122 – 2 = 120 and 243 – 3 = 240 are exactly divisible by the required number.
Required number = HCF of 120 and 240 = 120

Q.51

If P = 2³ . 3¹⁰ . 5 ; Q = 2⁵.3.7, then HCF of P and Q is :

2.3.5.7

3.2³

2² . 3⁷

2⁵ . 3¹⁰. 5 . 7

Ans .

Explanation :
P = 2³ × 3¹⁰ × 5 Q = 2⁵ × 3 × 7 HCF = 2³ × 3

Q.52

A fraction becomes ¹⁄₆ when 4 is subtracted from its numerator and 1 is added to its denominator. If 2 and 1 are respectively added to its numerator and the denominator, it becomes ¹⁄₃ . Then, the LCM of the numerator and denominator of the said fraction, must be

350

Ans .

Explanation :
Let the original fraction be x.
y
x+2 = 1
y+1 6
6x – 24 = y + 1
6x – y = 25 .......(i) Again, x+2 = 1
y+1 3
3x + 6 = y + 1
3x – y = –5 .......(ii) By equation (i) – (ii),
6x – y – 3x + y = 25 + 5
= 3x = 30
x = 10 From equation (i), 60 – y = 25
= y = 35 LCM of 10 and 35 = 70

Q.53

The HCF (GCD) of a, b is 12, a, b are positive integers and a > b > 12. The smallest values of (a, b) are respectively

12, 24

24, 12

24, 36

36, 24

Ans .

Explanation :
HCF of a and b = 12
Numbers = 12x and 12y where x and y are prime to each other. Q a > b > 12
a = 36; b = 24

Q.54

The number of pair of positive integers whose sum is 99 and HCF is 9 is

Ans .

Explanation :
Let the numbers be 9x and 9y where x and y are prime to each other. According to the question, 9x + 9y = 99 Þ 9(x + y) = 99
x + y = 11 Possible pairs = (1, 10) (2, 9), (3, 8), (4, 7), (5, 6

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