30. The ratio of two numbers is 4 : 5. If the HCF of these numbers is 6, what is their LCM?

A. 30 B. 60 C. 90 D. 120

**Answer : Option D**

Explanation :

Let the numbers be 4k and 5k

HCF of 4 and 5 = 1

Hence HCF of 4k and 5k = k

Given that HCF of 4k and 5k = 6

=> k = 6

Hence the numbers are (4 × 6) and (5 × 6) = 24 and 30

LCM of 24 and 30 = 120

31. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A. 36 minutes 22 seconds B. 46 minutes 22 seconds

C. 36 minutes 12 seconds D. 46 minutes 12 seconds

**Answer : Option D**

Explanation :

LCM of 252, 308 and 198 = 2772

Hence they all will be again at the starting point after 2772 seconds or 46 minutes 12 seconds

32. What is the HCF of 2.04, 0.24 and 0.8 ?

A. 1 B. 2 C. 0.02 D. 0.04

**Answer : Option D**

Explanation :

Step 1 : Make the same number of decimal places in all the given numbers

by suffixing zero(s) in required numbers as needed.

=> 2.04, 0.24 and 0.80

Step 2 : Now find the HCF of these numbers without decimal.

=>HCF of 204, 24 and 80 = 4

Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right.

=> HCF of 2.04, 0.24 and 0.8 = 0.04

33. If HCF of two numbers is 11 and the product of these numbers is 363, what is the the greater number?

A. 9 B. 22 C. 33 D. 11

**Answer : Option C**

Explanation :

Let the numbers be 11a and 11b

11a × 11b = 363

=> ab = 3

co-primes with product 3 are (1, 3)

Hence the numbers with HCF 11 and product 363

= (11 × 1, 11 × 3)

= (11, 33)

Hence numbers are 11 and 33

The greater number = 33

34. What is the greatest number which on dividing 1223 and 2351 leaves remainders 90 and 85 respectively?

A. 1133 B. 127 C. 42 D. 1100

**Answer : Option A**

Explanation :

Required number

= HCF of (1223 - 90) and (2351 - 85)

= HCF of 1133 and 2266

= 1133

35. What is the least multiple of 7 which leaves a remainder of 4 when divided by

6, 9, 15 and 18 ?

A. 364 B. 350 C. 343 D. 371

**Answer : Option A**

Explanation :

LCM of 6, 9, 15 and 18 = 90

Required Number = (90k + 4) which is a multiple of 7

Put k = 1. We get number as (90 × 1) +

4 = 94. But this is not a multiple of 7

Put k = 2. We get number as (90 × 2) +

4 = 184. But this is not a multiple of 7

Put k = 3. We get number as (90 × 3) +

4 = 274. But this is not a multiple of 7

Put k = 4. We get number as (90 × 4) +

4 = 364. This is a multiple of 7

Hence 364 is the answer.

*** MATHS (SOLVED IN STEPS ==> Ratio and Proportion**

*** MATHS (SOLVED IN STEPS ==> Progressions**

*** MATHS (SOLVED IN STEPS ==> Percentage**

*** MATHS (SOLVED IN STEPS ==> Surds and Indices**

*** MATHS (SOLVED IN STEPS ==> Calendar**

*** MATHS (SOLVED IN STEPS ==> Square and cube**

*** MATHS (SOLVED IN STEPS ==> Simplification**

*** MATHS (SOLVED IN STEPS ==> Problems on Average**

*** MATHS (SOLVED IN STEPS ==> Profit and Loss**

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