Three different containers contain 496 litres,403litres and 713 litres of milk. Find the maximum capacity of a container that can measure the milk of any other tanker on exact of times.
Two bills of rs.6075 and rs.8505 respectively are to be paid by cheques of the same amount. What will be the largest possible amount of each cheque?
Find the largest number that will divide 623,729 and 841 leaving remainder 3,9 and 1 respectively
PLEASE answer these questions it's too urgent for me
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Answered by
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1) Three different containers contain 496 litres, 403 litres and 703 litres of milk
496 = 2×2×2×2×31
403 = 13×31
713 = 23×31
so, HCF of 496, 403, 713 is 31
maximum capacity of a container that can measure the milk of any other tanker on exact of times = 31 litres
2) two bills of 6075 and 8505 respectively are to be paid buy checks of the same amount
6075 = 5×5×243
8505 = 5×7×243
so, HCF of 6075 and 8505 is 5×243 = 1215
largest possible amount of each check is 1215
3) Find the largest number that will divide 623, 729 and 841 leaving remainder of 3, 9, and 1 respectively
largest number that will divide 623, 729 and 841 leaving remainder of 3, 9 and 1 respectively is 10
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496 = 2×2×2×2×31
403 = 13×31
713 = 23×31
so, HCF of 496, 403, 713 is 31
maximum capacity of a container that can measure the milk of any other tanker on exact of times = 31 litres
2) two bills of 6075 and 8505 respectively are to be paid buy checks of the same amount
6075 = 5×5×243
8505 = 5×7×243
so, HCF of 6075 and 8505 is 5×243 = 1215
largest possible amount of each check is 1215
3) Find the largest number that will divide 623, 729 and 841 leaving remainder of 3, 9, and 1 respectively
largest number that will divide 623, 729 and 841 leaving remainder of 3, 9 and 1 respectively is 10
#mark as brainliest
sibhiamar:
just wait for 5 min
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and not pcm LCM is there
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hey
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Answered by
3
Here is the answer for your previous questions
12=2×2×3 = 2^3×3
16=2×2×2×2 = 2^4
24=2×2×2×3 = 2^3×3
36=2×2×3×3 =2^2×3^2
LCM=(12,16,24,36)=product of greatest power of each factor
=>2^4×3^2
=>16×9
=> 144
Greatest 4-digit number is 9,999
If any number is divisible by 12,16,24 and 36 it is also divisible by 144.
On dividing 9,999 by 144 we get 63 as remainder.
Hence, the required number is (9,999-63)=9,936 Ans.
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
iii. For each of the following pairs of numbers,show that product of ther LCM and HCF is equals to their products.
A.14 and 21
14=2×7
21=3×7
HCF(14,21)=poduct of smallest power of each common factor
=>7
Hence ,HCF=7
LCM(21,14)=Product of greatest power of each common factor.
=>2×3×7
=>42
Hence, LCM=42
Now,
LCM×HCF=42×7=>294
Products of the numbers=14×21=>294
Thus,proved that LCM×HCF=product of the numbers
B.27 and 90
27=3×3×3 =3^3
90=2×5×3×3 =2×5×3^2
HCF(27,90)=poduct of smallest power of each common factor
=>3^2
=>9
Hence ,HCF=9
LCM(27,90)=Product of greatest power of each common factor.
=>2×5×3^3
=>270
Hence, LCM=270
Now,
LCM×HCF=9×270=>2,430
Products of the numbers=90×27=>2,430
Thus,proved that LCM×HCF=product of the numbers.
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
3.By using HCF find LCM
i.299 and 391
299=23×13
391=23×17
HCF(299,391)=Product of smallest power of each factore
=>23
Hence, HCF of 299 and 391 is 23
Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=299×391
HCF×LCM=1,16,909
23×LCM=1,16,909
LCM=1,16,909 /23
LCM=5,083 Ans.
ii.1333 and 1767
1,333=31×43
1,767=31×74
Clearly,31 is the HCF
Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=1,333×1,767
HCF×LCM= 23,55,411
31×LCM= 23,55,411
LCM= 23,55,411/31
LCM= 75,981 Ans.
12=2×2×3 = 2^3×3
16=2×2×2×2 = 2^4
24=2×2×2×3 = 2^3×3
36=2×2×3×3 =2^2×3^2
LCM=(12,16,24,36)=product of greatest power of each factor
=>2^4×3^2
=>16×9
=> 144
Greatest 4-digit number is 9,999
If any number is divisible by 12,16,24 and 36 it is also divisible by 144.
On dividing 9,999 by 144 we get 63 as remainder.
Hence, the required number is (9,999-63)=9,936 Ans.
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
iii. For each of the following pairs of numbers,show that product of ther LCM and HCF is equals to their products.
A.14 and 21
14=2×7
21=3×7
HCF(14,21)=poduct of smallest power of each common factor
=>7
Hence ,HCF=7
LCM(21,14)=Product of greatest power of each common factor.
=>2×3×7
=>42
Hence, LCM=42
Now,
LCM×HCF=42×7=>294
Products of the numbers=14×21=>294
Thus,proved that LCM×HCF=product of the numbers
B.27 and 90
27=3×3×3 =3^3
90=2×5×3×3 =2×5×3^2
HCF(27,90)=poduct of smallest power of each common factor
=>3^2
=>9
Hence ,HCF=9
LCM(27,90)=Product of greatest power of each common factor.
=>2×5×3^3
=>270
Hence, LCM=270
Now,
LCM×HCF=9×270=>2,430
Products of the numbers=90×27=>2,430
Thus,proved that LCM×HCF=product of the numbers.
☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆☆
3.By using HCF find LCM
i.299 and 391
299=23×13
391=23×17
HCF(299,391)=Product of smallest power of each factore
=>23
Hence, HCF of 299 and 391 is 23
Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=299×391
HCF×LCM=1,16,909
23×LCM=1,16,909
LCM=1,16,909 /23
LCM=5,083 Ans.
ii.1333 and 1767
1,333=31×43
1,767=31×74
Clearly,31 is the HCF
Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=1,333×1,767
HCF×LCM= 23,55,411
31×LCM= 23,55,411
LCM= 23,55,411/31
LCM= 75,981 Ans.
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