Can start by understanding that the numbers will be close to the cube root of 190740 which should lie between 50 and 60 (between 125000 and 216000). Can start with the smallest prime numbers and check. It will be satisfied for 53 * 59 * 61. Total of 53 + 59 + 61 = 173.

100 km in 10 hours at 10 km/hr. 50 km in 5 hours covered. 50 km back at 15 km/hr in 10/3 hours. 100 km to be covered in 5/3 hours at 60 km/hr. Increase by 500%

@Naman-Jain-0 Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17 6 dice and each dice can give 6 options. Total cases = 6^6 Now we will find the number of ways to get a sum lesser than or equal to 17a + b + c + d + e + f ≤ 17a , b, c ... f can take values from 1 to 6 Because of ≤, add a dummy variable. (say, g)so we have a + b + c + d + e + f + g = 17 Now we will discuss a very important point. a, b, c .. f are all natural numbers and for g - 0 is possible. Means this is neither in case 1 or case 2. To resolve this let subtract 1 from a, b, c .. f so that they can be given a value of 0 also. Our equation becomes a + b + c + d + e + f +g = 17 - 6 = 11 Concept : a + b + c ... (k terms) = n has (n + k - 1) C (k - 1) non negative integer solutions Number of ways = (11 + 7 - 1) C (7 - 1) = 17C6 now we have to remove cases where a > 6.say a = A + 6So, A + b + c + d + e + f + g = 11 - 6 = 5Number of ways which a can be higher than 6 (invalid cases for a) = 11C6 ways.6 x 11C6 invalid cases if we include for b, c, d, e and f also. So Final - 17C6 - 6 x 11C6 Now coming back to the original question,Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17= 6^6 - (17C6 - 6 x 11C6) If you need a clear understanding of this concept, refer the below article. Happy Learning!(adsbygoogle = window.adsbygoogle || []).push({}); Solving Combinatorics Problems Using Stars & Bars Concept

@Vikrant-Garg Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?) This is possible in 2 ways - if we get DD or if we get GG as the result of first 2 tests. In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases. So total number of ways = 4!/(2! x 2!) = 6 ways. Favourable cases = 2 Probability = 2/6 = 1/3 Is it clear ?

@Naman-Jain-0 ___ | ___ | ___ | ___ | ___ | ___ | ___ | ___There are a total of 8 steps represented by (___) and let the dash ('|') represent if he is stopping at the previous step. There are 7 such dashes. Each dash can take a 0 or 1. 0 indicates, he is stopping at the step immediately before it. As each dash can take a 0 or 1, the number of ways is 128.And as the maximum steps u can take is 6, the cases where he takes 7 steps at a time - given by (1,0,0,0,0,0,0) and (0,0,0,0,0,0,1)- and 8 steps at a time - given by (0,0,0,0,0,0,0)- are eliminated. Therefore it is 125.

@Naman-Jain-0 Partition method is applicable for cases like dividing n identical objects among r distinguishable parts right ? How would you apply it here ?

@zabeer thanks for the explanation.

@Sonam-Priya it is called chord theorem. This theorem states that A × B is always equal to C × D no matter where the chords are.

@shashank_prabhu @zabeer sure sir will post my solution too , thanks for ur approach

@naman-jain-0 yeah 23 s the answer

Q96) From a rectangular sheet of dimensions 30 cm × 20 cm, four squares of equal size are cut from the four corners. Then the resulting four sides are bent upwards to give it the shape of an open box. If the volume of the box is 1056 cm^3, what is the length of the side of the squares cut from the corners? [OA: 4]

@shreyasnegi13 There are 10 primes in the given range. 4 also satisfies 9 and 25 won't work because we would already have two 3s and 5s respectively in the numerator So 10 + 1 = 11 values.

@shashank_prabhu 1

@Rohit-Rathore Q3) Let say we had x coins in the beginning After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3 After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9 After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27 Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3. So we can write, (8x - 65)/27 = 3k x = (81k + 65)/8 x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0 We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7. Least value would be K = 7 For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer. Let me know in case of any Logical/Calculation errors.

OA: a, c, b, d and c

Q91) A four-letter code has to be formed using the alphabets from the set (a, b, c, d) such that the codes formed have odd number of a’s. How many different codes can be formed satisfying the mentioned criteria?

Q97) The percentage profit earned by selling an article at Rs. 1,920 is equal to the percentage loss incurred by selling the same article at Rs. 1,280. At what price (in Rs.) should the article be sold to make a profit of 25%? [OA: 2000]