Given numbers are 1.08 , 0.36 and 0.90
H.C.F of 108, 36 and 90 is 18 ( G.C.D is nothing but H.C.F)
So H.C.F of given numbers = 0.18
Workspace
NA
SHSTTON
134
Solv. Corr.
185
Solv. In. Corr.
319
Attempted
1 M:56 S
Avg. Time
2 / 28
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There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
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
123
Solv. Corr.
72
Solv. In. Corr.
195
Attempted
0 M:58 S
Avg. Time
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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
Workspace
NA
SHSTTON
70
Solv. Corr.
58
Solv. In. Corr.
128
Attempted
0 M:27 S
Avg. Time
8 / 28
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Find the largest 4-digit number, which gives the remainder 7 and 13 when divided by 11 and 17?
LCM of 11 and 17 = 187.
When divided by 11 remainder 7,so difference 4.
When devided by 17 remainder 13,so difference 4.
Largest no exactly devide by 11 & 17=9911
The no's = 9911-4 = 9907.
Workspace
NA
SHSTTON
36
Solv. Corr.
48
Solv. In. Corr.
84
Attempted
0 M:57 S
Avg. Time
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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
61
Solv. Corr.
93
Solv. In. Corr.
154
Attempted
4 M:4 S
Avg. Time
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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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