Co-prime numbers:Two numbers having only 1 as a common factor are called co-prime numbers. Thus, 4 and 15 are co-prime numbers.
If a number is divisible by two co-prime numbers then it is divisible by their product also.
If two given numbers are divisible by a number, then their sum is also divisible by that number.
If two given numbers are divisible by a number, then their difference is also divisible by that number.
HCF of two consecutive numbers is 1. HCF of two consecutive even numbers is 2. HCF of two consecutive odd numbers is 1. HCF of co-prime numbers is 1.
To get the lowest form of a fraction, find the HCF of the numerator and the denominator and divide both by the HCF. So to reduce the fraction 36 : 24 we find the HCF which is 12 and then divide both by 12 to get 3:2.
Least Common Multiple or L.C.M. of 12 and 18 is calculated as:
12 = 2 * 2 * 3 and 18 = 2 * 3 * 3.
The prime factors are 2 and 3 and both occur twice in the factorization. Hence the LCM is 2*2 * 3*3 = 36.
When the two numbers are co-prime their LCM is the product of those numbers. Also when one number is the factor of the other number, their LCM is the larger of the numbers.
Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of a square = 4 × length of a side
Perimeter of an equilateral triangle = 3 × length of a side
Area of a rectangle = (length × breadth)
Area of the square = side × side
When two proper fractions i.e. numerator smaller than denominator are multiplied the result is less than both the fractions.
When two improper fractions i.e. numerator greater than denominator are multiplied the result is greater than both the fractions.
When one proper fraction and one improper fraction is multiplied The product obtained is less than the improper fraction and greater than the proper fraction involved in the multiplication.
Q.Example 12 : Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.
Ans:The required container has to measure both the tankers in a way that the count is an exact number of times. So its capacity must be an exact divisor of the capacities of both the tankers. Moreover, this capacity should be maximum. Thus, the maximum capacity of such a container will be the HCF of 850 and 680.
The common factors of 850 and 680 are 2, 5 and 17. Thus, the HCF of 850 and 680 is 2 × 5 × 17 = 170. Therefore, maximum capacity of the required container is 170 litres. It will fill the first container in 5 and the second in 4 refills.
Q.In a morning walk, three persons step off together. Their steps measure 80 cm, 85 cm and 90 cm respectively . What is the minimum distance each should walk so that all can cover the same distance in complete steps?
Ans : The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps. Can you describe why? Thus, we find the LCM of 80, 85 and 90. The LCM of 80, 85 and 90 is 12240. The required minimum distance is 12240 cm.
Q.Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case.
Ans : first find the LCM of 12, 16, 24 and 36. We get LCM = 2 × 2 × 2 × 2 × 3 × 3 = 144. 144 is the least number which when divided by the given numbers will leave remainder 0 in each case. But we need the least number that leaves remainder 7 in each case. Therefore, the required number is 7 more than 144. The required least number = 144 + 7 = 151.
Q.Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
Ans : Find the HCF of the weights. the maximum value of weight which can measure the weight of the fertiliser exact number of times is the HCF (75, 69) = 3.
Q.Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Ans : the minimum distance each should cover so that all can cover the distance in complete steps is LCM (63,70,77) = 6930.
Q.The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.?
Ans : longest tape which can measure the three dimensions of the room exactly is HCF (825,675,450) = 75 cm.
Q.Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12. ?
Ans : smallest number which is exactly divisible by 6, 8 and 12 is LCM (6,8,12) = 24. So 24*3=120 is the smallest 3 digit number.
Q.Determine the greatest 3-digit number exactly divisible by 8, 10 and 12. ?
Ans : greatest 3-digit number exactly divisible by 8, 10 and 12 is multiple of the LCM (10,8,12) = 120. So 120*8=960 is the greatest 3 digit number.
Q.The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again? ?
Ans : Time they will change again is LCM of 48,72,108 i.e 432 = 7 min and 12 secs. So they will change again at 7:07:12 am.
Q.Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
Ans : HCF of 403,434,465 = 31.
Q.Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Ans : LCM of 6,15,18 = 90. So answer is 90+5=95
Q.Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Ans : LCM of 18,24,32 = 288. So answer is 288*4=1152.
Q.A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
length = (45 + 2.5 + 2.5) m = 50 m
breadth = (30 + 2.5 + 2.5) m = 35 m
Area of the rectangle inner rectangle = l * b = 45 × 30 m 2
Area of the rectangle outer rectangle = l * b = 50 × 35 m 2
Area of the path = Area of the outer rectangle − Area of the inner rectangle = (1750 − 1350) m 2 = 400 m 2
Q.Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the roads. Also find the cost of constructing the roads at the rate of Rs 105 per m2
Area of horizontal rectangle = 70 * 5 ;
Area of vertical rectangle = 45 * 5;
Area of overlapped portion = 5 * 5;
Total area = 225 + 350 - 25 = 550.
Cost of constructing the path = Rs 105 × 550 = Rs 57,750