Solution: Product of two number = Product of their HCF and LCM
pq = 13*273
pq = 3549.
Now, co-primes with product 3549 are (1,3549), (21,169), (39,91) and (13,273)
Acc. to the condition given in the question only one satisfy them that is (39,91)
So the answer is 91.
Workspace
NA
SHSTTON
78
Solv. Corr.
38
Solv. In. Corr.
116
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3 / 20
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Sudhir goes to the market once every 64 days and Sushil goes to the same market once every 72 days. They met each other one day. How many days later will they meet each other again?
Product of two no. = H.C.F*L.C.M
So,x*y=135*9=1215 -----(1)
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184
So,x+y=72 ----------- (2)
By solving eq. (1) & (2)
nos. are 45 and 27
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NA
SHSTTON
23
Solv. Corr.
35
Solv. In. Corr.
58
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Printer A prints 8192 character per min and printer B prints 13862 character per min four character are equal to one word. Printer A starts at 7:15 am while Printer B starts at 7:29 am then at what time both will have same no. of words printed.
Four character are equal to one word
8192/4*(t+14)=13862/4*t
8192*(t+14)=13862*t yields the same result
t=20.2272 minutes = about 20 min 14 sec
7:49:14 am
Workspace
NA
SHSTTON
35
Solv. Corr.
46
Solv. In. Corr.
81
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If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?
Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a
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SHSTTON
63
Solv. Corr.
47
Solv. In. Corr.
110
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find the number between 100 to 400 which is divisible by either 2,3,5,7..
2472 = 2^3×3×103
1284 = 2^2×3×107
HCF = 2^2×3
LCM = 2^3×3^2×5×103×107
HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 2^2×3
LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.
So N = 2^2×3^2×5^1
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SHSTTON
26
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28
Solv. In. Corr.
54
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The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
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