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The G.C.D. of 1.08, 0.36 and 0.9 is:
A0.03
B0.9
C0.18
D0.108
Answer: Option C
Explanation: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
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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 :
A80
B82
C85
D87
Answer: Option C
Explanation:Given that numbers are co primes,
and two products have the middle number in common.
=> Middle number = H.C.F. of 551 and 1073 = 29
so first number is = 551/29 = 19
=> Third number = 1073/29 = 37
Therefore, sum of these numbers is = (19 + 29 + 37) = 85
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3 / 28
The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275, then the other is:
A308
B208
C318
D283
Answer: Option A
Explanation:lcm*hcf = product of 2 nos
11*7700=275*?
ans =308
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4 / 28
Let p and q be two prime numbers such that p is greater than q. If 319 is their LCM then the difference of thrice of q and p is:
A1
B0
C4
D9
Answer: Option C
Explanation:Solution: We know that HCF of two prime numbers is 1.
Product of HCF and LCM = 1 x 319 = 319.
Remember that, Product of two number = Product of their HCF and LCM
pq = 319
Now, co-primes with product 319 are (1,319) and (29,11)
Since p > q, p = 29 and q = 11
Then 3q - p = 3 - 29 = 4.
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5 / 28
Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?
A121
B91
C75
D131
Answer: Option B
Explanation: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.
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6 / 28
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?
A567
B576
C765
DNone of these
Answer: Option B
Explanation:Find the LCM of 64,72
i.e 2*2*2*8*9=576
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Sum of squares of two numbers is 2754, their HCF is 9, LCM is 135, Find the numbers
A45 and 36
B45 and 27
C54 and 27
DNone of these
Answer: Option B
Explanation: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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8 / 28
Find the largest 4-digit number, which gives the remainder 7 and 13 when divided by 11 and 17?
A9900
B9907
C9807
DNone of these
Answer: Option B
Explanation: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.
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9 / 28
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.
A8:49:14 AM
B7:49:14 AM
C9:49:14 AM
D7:49:14 PM
Answer: Option B
Explanation: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
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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?
A0
B2
C3
D5
Answer: Option C
Explanation: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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