Three men start together to travel the same way around a circular track of 11 kilometers in circumference. Their speeds are 4, 5 and 8 kilometers per hour respectively. When will they meet at a starting point?
A merchant has three different types of milk: 435 liters, 493 liters and 551 liters. Find the least number of casks of equal size required to store all the milk without mixing.
Let HCF of the numbers be h. The numbers can be taken as ha + hb, where a, b are coprime.
h + ha + hb = 91
h(1 + a + b) = 91
h ≠ 1
h = 7
=> 1 + a + b = 13 a + b = 12
h = 13
=> 1 + a + b = 7
=> a + b = 6
Case 1: h = 7, a + b = 12
(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.
Case 2: h = 13, a + b = 6
(1, 5) only one pair is possible as a, b have to be coprime.
Overall, 3 pairs of numbers are possible – (7, 77) (35, 49) and (13, 65)
The question is “How many such pairs are possible?”
Hence the answer is “3 Pairs”
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Sum of two numbers x, y = 1050. What is the maximum value of the HCF between x and y?
If the question states x, y have to be distinct, then the best solution would be x = 350, y = 700, HCF = 350.
So the HCF is 525.
The question is “What is the maximum value of the HCF between x and y?”
Hence the answer is “525”
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Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?
Note: When a number is divided by 8, a remainder of 7 can be thought of as a remainder of -1. This idea is very useful in a bunch of questions. So, N = 5a – 1 or N + 1 = 5a
N = 6b – 1 or N + 1 = 6b
N = 7c – 1 or N + 1 = 7c
N = 8d – 1 or N + 1 = 8d
N = 9e – 1 or N + 1 = 9e
N + 1 can be expressed as a multiple of (5, 6, 7, 8, 9)
N + 1 = 5a*6b*7c*8d*9e
Or N = (5a*6b*7c*8d*9e) – 1
Smallest value of N will be when we find the smallest common multiple of (5, 6, 7, 8, 9)
or LCM of (5, 6, 7, 8, 9)
N = LCM (5, 6, 7, 8, 9) – 1 = 2520 – 1 = 2519.
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What is the greatest number which divides 639, 1065 and 1491 exactly?
Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b = 174/29 = 6. Values of co-primes (with sum 6) is(1, 5).
So, the possible pairs of numbers is (29 x 1, 29 x 5) i.e. 29 and 145, which will become the answer.
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3 different pieces of iron are of varying length are given to a student which are 44cm, 22 cm,55 cm respectively.He has to form rods of maximum length such that no iron waste is left.Find the maximum length of such rod.
Three numbers which 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. The sum of the three numbers is:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
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A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
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Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together
30
+ 1 = 16 times.
2
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A rectangular courtyard 4.55 meters long and 5.25 meters wide is paved exactly with square tiles of same size. Find the largest size of the tile used for this purpose?
Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F. Step 1: Covert numbers without decimal places i.e 455 cm and 525 cm Step 2: Find the H.C.F. of 455 and 525
H.C.F. of 455 and 525 = 35 cm
Hence, the largest size of the tile is 35 cm
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H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.
H.C.F. of two numbers = 13
2) The numbers are in the ratio of 15 : 11
Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y. So y = 13
The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143 We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
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Find the largest number of 4-digits divisible by 12, 15 and 18.
Largest 4-digit number is 9999. Remember: The question may be asked in a tricky way. Here, largest number does not mean H.C.F.. We have to find a number which is divisible by 12, 15 and 18
Required largest number must be divisible by the L.C.M. of 12, 15 and 18
L.C.M. of 12, 15 and 18
12 = 2 × 2 × 3
15 =5 × 3
18 = 2 × 3 × 3
L.C.M. = 180
Now divide 9999 by 180, we get remainder as 99
The required largest number = (9999 – 99) =9900 Number 9900 is exactly divisible by 180.
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The least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case, but when divided by 9 leaves no remainder, is:
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3) = 1683
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If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to?
Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120
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The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
36 = 22 x 32
84 = 22 x 3 x 7
∴ H.C.F. = 22 x 3 = 12.
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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 :
The numbers are 16 and 24.
