1. Find the LCM of

(a) 280

(b) 360

(c) 420

(d) None of these

**Answer: (d)**

**Explanation:**

Step 1: The fraction must be reduced to their lowest terms.

For Example which is its lowest term i.e. , the factors common to the numerator and the denominator have to be cancelled out.

Step 2: Next, we convert mixed number into fraction.

For example,

LCM of

LCM of Numerators 14, 48, 13 and 3;

LCM=2×3×7×8×13=4368

HCF of denominators 3, 5, 4, 4:

We can clearly see that there is no factor which is common to all. Hence, HCF will be 1.

Required LCM=4368.

2. The greatest number that divides 690 and 875 leaving remainders 10 and 25 respectively is:

(a) 170

(b) 130

(c) 150

(d) 120

**Answer: (a)**

**Explanation:**

690 -10= 680, 875 – 25 = 850

Highest number that can divide 680 and 850 exactly is HCM of the numbers.

HCF of 680 and 850 = 170

3. An officer was appointed on maximum daily wages on contract money of Rs. 4956. But on being absent for some days, he was paid Rs. 3894. For how many days was he absent? [SSC 2008]

(a) 1

(b) 2

(c) 3

(d) 4

**Answer: (c)**

**Explanation:**

HCF (4956, 3894) = 354

Here 354 are the maximum daily wages.

The officer was appointed on contract money of Rs. 4956 = 354 × 14.

i.e., he was appointed for 14 days.

But he was paid Rs. 3894 = 354 × 11,

i.e., he was present for 11 days.

Hence he was absent for 3 days.

4. What is the LCM of [CDS 2006]

(a) 15

(b)

(c)

(d)

**Answer: (a)**

**Explanation:**

5. HCF of two numbers is 15 and their LCM is 180. If their sum is 105, then the numbers are: [CDS 1998]

(a) 30 and 75

(b) 35 and 70

(c) 40 and 65

(d) 45 and 60

**Answer: (d)**

**Explanation:**

Let the numbers be 15a and 15b. Then,

15a + 15b = 105 or a+ b = 7 .. (i)

LCM = 15ab = 180

ab = 12..(ii)

Solving equations (i) and (ii), we get a=4, b=3

So, the numbers are 15 ×4 and 15 ×3, i.e., 60 and 45

6. If the HCF of two numbers is 2 and their LCM is 70, then the numbers are: [CDS 1999]

(a) 2, 235

(b) 6, 70

(c) 4, 70

(d) 14, 10

**Answer: (d)**

**Explanation:**

HCF of the two numbers is 2

Let us take two numbers are x and y

Their LCM= 70

LCM is the product of two numbers.

xy=70×2 = 140

xy = 140

x = 14, y = 10 or x = 70, y = 2.

But according to the options, the numbers are x = 14 and y = 10.

7. The HCF and LCM of two numbers are 12 and 72 respectively. If the sum of the two numbers is 60, then one of the two numbers will be: [CDS 2000]

(a) 12

(b) 24

(c) 60

(d) 72

**Answer: (b)**

**Explanation:**

HCF= 12, LCM= 72. The sum of the two numbers = 60

Let one of the number be = x.

Then the other number = 60+x.

x (60 + x) = 12 × 72

(x-36) (x-24)=0

x=36, 24

8. The HCF and LCM of two numbers are 18 and 3780 respectively. If one of them is 540, then the second one is:

(a) 142

(b) 126

(c) 118

(d) 112

**Answer: (b)**

**Explanation:**

Product of HCF1 and LCM1 = product of HCF2 and LCM2

18× 3780 = 540× HCF2

9. Find the side of the largest possible square slabs which can be paved on the floor of a room 2 m 50 cm long and 1 m 50 cm broad. Also find the number of such slabs to pave the floor. [LIC 2007]

(a) 25, 20

(b) 30, 15

(c) 50, 15

(d) None of these

**Answer: (c)**

**Explanation:**

HCF (250, 150) = 50 cm

The number slabs =

10. There are 408 boys and 312 girls in a school which are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed. [LIC 2007]

(a) 31

(b) 32

(c) 35

(d) 30

**Answer: (d)**

**Explanation:**

HCF (408, 312) = 24

The number of boys or girls that can be placed in a section = 24.

Thus the total number of sections is given by