RD SHARMAS SOLUTION
REAL NUMBERS
(part 2 )
QUESTION
The LCM and HCF of two numbers are 180 and 6 respectively . If one of the number is 30 ,find the other number?
ANSWER
Given ; HCF and LCM of two numbers are 180 and 6 respectively
If one number is 30
To find other number
we know that,
LCM*HCF = first number*second number
180*6 = 30* second number
second number = 180*6/30
=36
QUESTION
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468
ANSWER
To find smallest number which when increased by 17 is exactly divisible by both 520 and 468
LCM of 520 and 468
520 =23*5*13
468 =2*2*3*3*13
LCM = 23*32*5*13 =4680
Hence 4680 is the least number which exactly divides 520 and 468
i.e, we will get a remainder of 0 in this case.
But we need the smallest number which when increased by 17 is exactly divided by 520 and 468
Therefore 4680-17 =46631
Hence 4663 is smallest number which when increased by 17 is exactly divisible by both .
QUESTION
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively?
ANSWER
LCM of 28 and 32
28 =2*2*7
32=2*2*2*2*2
LCM of 28 and 32 =2*2*2*2*2*7 =224
Hence 224 is the least number which exactly divides 28 and 32.
i.e, we will get the remainder 0 in this case.
But we need the smallest number which leaves remainder 8 and 12 when divided by 8 and 12 when divided by 28 and 32 respectively.
Therefore 224-8-12= 204
Hence 204 is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively.
QUESTION
What is the smallest number that ,when divisible by 35 ,56 and 91 leaves remainders of 7 in each case?
ANSWER
To find smallest number that when divided by 35,56 and 91 leaves remainder of 7 in each case
LCM of 35,56 and 91
35 = 5*7
56 =2*2*2*7
91 = 13*7
LCM =2*2*2*7*5*13 =3640
Hence 84 is the least number which exactly divides 28,42 and 84.
i.e, we will get a remainder of 0 in this case
But we need the smallest number, that when divided by 35,56 and 91 leave remainder of 7 in each case.
therefore 3640+7 =3647
QUESTION
A Rectangular courtyard is 18m 72cm long and 13n20cm broad. It is to be paved with square tiles of the same size. Find the least number of such tiles?
ANSWER
Length of the yard = 18m 72cm =1872cm
Breadth of the yard =13m 20cm= 1320cm
The size of the square tile of the same size needed to the pave the rectangular yard is equal to the HCF of the length and breadth of the rectangular yard.
1872 =24*32*13
1320 = 23*5*3*11
HCF of 1872 and 1320 = 23*3 =24
Therefore ,length of side of the square tile =24cm
Number of tiles required = Area of the courtyard, Area of each tile = Length * Breadth
side*side = 1872*1320cm /24 cm =4290
Thus ,the least possible number of tiles required is 4290
QUESTION
Find the greatest no. of 6 digits exactly divisible by 24,15 and 36
ANSWER
To find greatest number of 6 digits exactly divisible by 24,15 and 36
The greatest 6 digit number be 999999
24 = 2*2*2*3
15 = 3*5
36 = 2*2*3*3
LCM of 24, 15 and 36 =360
since 999999= 2777*360+279
Therefore, the remainder is 279. Hence the desired number is 999999 -279 =9997201
Hence 9997201 is the greatest number of 6 digits exactly divisible by 24,15 and 36
QUESTION
Determine the number nearest to 110000 but greater100000 which is exactly divisible by each of 8,15 and 21?
ANSWER
LCM of 8, 15 and 21
8 = 2*2*2
15 = 3*5
21 =3*7
LCM of 8,15 and 21 =2*2*2*3*5*71 =840
When 110000 is divided by 840 , the remainder obtained is 800.
Now, 110000 -800 =109200 is divisible by each of 8,15 and 21
Also ,110000 +40 =110040 is divisible by each of 8, 15 and 21
109200 and 110040 are greater than 100000
Hence 110040 is the number nearest to 100000 which is exactly divisible by each of 8, 15 and 21
QUESTION
Find the least number that is divisible by all the numbers between 1 and 10 ( both inclusive)?
ANSWER
To find least number that is divisible by all the numbers between 1 and 10(both inclusive)
Let us first find the LCM of all the numbers between 1 and 10(both inclusive)
1= 1
2 =2
3 =3
4 =2*2
5 =5
6 = 2*3
7 = 7
8 = 2*2*2
9 = 3*5
10 =2*5
LCM =2520
Hence 2520 is the least number that is divisible by all the numbers between 1 and 10 (both inclusive)
QUESTION
A circular field has a circumference of 360 km. Three cyclist start together and can cycle 48,60 and 72 km a day, round the field. When will they meet again?
ANSWER
Given
circumference of a circular field = 360 km
To calculate the time they meet,
find out the time taken by each cyclist in covering the distance
No. of days the first cyclist to cover 360 km = Total distance covered in 1 day
=360/48 =7.5
=75/10 =152 days
similarly,
No. of days taken by second cyclist to cover this distance = 360/60
= 6 days
Also, no. of days taken by third cyclist to cover this distance =360/72
= 5 days
Now, LCM of 152, 6 and 5 = LCM of numerators
HCF of denominators =30 *1 =30 days
Thus all of them will take 30 days to meet again
QUESTION
In a morning walk three persons step of together, their steps measure 80cm,85cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?
ANSWER
To find the minimum distance each should walk, so that all can cover the same distance in complete steps
LCM of 80,85,90
80 =2*2*2*2*5
85 =17*5
90 =2*2*2*2*3*3*5*17
LCM = 2*2*2*2*3*3*5*17 =12240cm
Hence minimum 12240 cm distance each should walk so that all can cover the same distance in complete steps
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