Ppt On Lcm & Hcf Questions For Cat Preparation
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The document explains applications of the Least Common Multiple (LCM) and Highest Common Factor (HCF) in various mathematical problems. It provides examples for finding maximum weights, divisibility conditions, and specific remainders when dividing numbers. The overall focus is on using LCM and HCF to solve problems involving common divisors and remainders.
Ppt On Lcm & Hcf Questions For Cat Preparation
- 1. Applications of LCM & HCF • Greatest Number that will leave no remainder when divides a, b and c Required number = HCF of a, b and c Example A shopkeeper has three cakes of weight 18 kg, 45 kg and 36 kg. If he wants to make these cakes into pieces of equal weight without wastage, what is the maximum possible weight of each piece? (1) 1 (2) 9 (3) 12 (4) 13
- 2. Applications of LCM & HCF • The greatest number that will divide a, b & c leaving remainder x, y & z respectively Required number = HCF of {(a – x), (b – y), (c – z)} Example What is the greatest number that divide 20, 50, and 40 leaving 2, 5 and 4 as remainder? Here a–x= b–y= c–z= Required Number =
- 3. Applications of LCM & HCF • To find the greatest number that will divide x, y and z leaving the same remainder “r” in each case. Required number = HCF of (x – r), (y – r) and (z – r) Example Find the greatest number which will divide 369, 449, 689, 5009 and 729 so as to leave the remainder 9 in each case (A) 2 (B) 49 (C) 35 (D) 40 Required Number is HCF of {(369-9), (449-9), (689-9), (5009 – 9), (729 – 9)} = HCF of {360, 440, 680, 5000, 720} = 40
- 4. Applications of LCM & HCF • To find the greatest number that will divide x, y and z leaving the same remainder in each case. Required number = HCF of |x – y|, |y – z| and |z – x| Example What is the greatest number that will divide 1305, 4665 and 6905 leaving in each case the same remainder? Also calculate the remainder. (1) 1210, 158 (2) 1120, 158 (3) 1120, 185 (4) 1210, 185
- 5. Solution Here we have |4665 – 1305|, |6905 – 4665|, |6905 – 1305| = 3360, 2240, 5600 = 1120 x 3, 1120 x 2, 1120 x 5 Required number = HCF of (3360, 2240, 5600) = 1120 To calculate Remainder 1305 = 1120 x 1 + 185 So the Remainder = 185
- 6. Applications of LCM & HCF • Least number which is exactly divisible by a, b, c Required number = LCM of x, y and z Sequence = n x LCM (Where n is a natural Number) Example Find the greatest number of five digits which is divisible by 32, 36, 40, 42 and 48 (a) 999720 (b) 90702 (C) 90720 (d) 90730
- 7. Solution Step I LCM of (32, 36, 40, 42, 48) = 10080 Step II Find greater number of 5 digit which multiple of 10080 10080 99999 9 90720 9279 Greatest number = 90720
- 8. Applications of LCM & HCF • To find least number which when divided by a, b, c leaves “r” as a remainder Required number = LCM of (a, b, c) + r Example What is the smallest sum which a person can have such that when he distributed @ Rs. 2.5 or Rs 20 or Rs 12 or Rs. 7.5 per person in a group, he is always left with Rs. 2.00? (A) Rs. 62 (B) Rs. 80 (C) Rs. 90 (D) Rs. 100
- 9. Solution According to the problem the person must have the money equal to the LCM of 2.5, 20, 12 and 7.5 and the remainder money always left.
- 10. Applications of LCM & HCF • Least number which when divided by a, b, c leaves x, y, z as remainder, such that a – x = b – y = c – z = k (say) Required number = LCM of (a, b, c) – k Example What is the greatest number of 4 digits that when divided by the numbers 6, 9, 12, 17 leaves 5, 8, 11, 16 as remainders respectively? (a) 9791 (b) 9793 (c) 9895 (d) 9497
- 11. Solution Step 1 Here, 6 – 5 = 9 – 8 = 12 – 11= 17 – 16 = 1 = K (let) Step 2 LCM of 6, 9, 12, 17 = 17 x 9 x 4 = 17 x 36 = 612 Step 3 Find Greatest 4 digit number which is multiple of 612 612 9999 612 16 3879 Greatest Number 3672 = 9999 – 207 = 9792 207 Step 4 Required number = (LCM) n - K = 9792 - 1 = 9791
- 12. Thank you !!