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Monday, 29 April 2019

MATHS SOLVED IN STEPS: HCM & LCM - 04

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
* MATHS (SOLVED IN STEPS ==> Problems on Trains 
* MATHS (SOLVED IN STEPS ==> Problems on Age 
* MATHS (SOLVED IN STEPS ==> Pipe and Cisterns
* MATHS (SOLVED IN STEPS ==> ODD MAN
* MATHS (SOLVED IN STEPS ==> TIME & WORK



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