A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is
Let the actual speed of the train be x km/hr and let the actual time taken be y hours. Distance covered is xy km If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x + 6) km/hr, time of journey is (y − 4) hours. ∴ Distance covered = (x + 6) (y − 4) ⇒ xy = (x + 6) (y − 4) ⇒ −4x + 6y − 24 = 0 ⇒ −2x + 3y − 12 = 0 ………….(i) Similarly xy = (x − 6) (y + 6) ⇒ 6x − 6y − 36 = 0 ⇒ x − y − 6 = 0 ……………(ii) Solving (i) and (ii) we get x = 30 and y = 24 Putting the values of x and y in equation (i), we obtain Distance = (30 × 24) km = 720 km. Hence, the length of the journey is 720 km.
Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is
Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is
In a group of 250 students, the percentage of girls was at least 44% and at most 60%.The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70%of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
The midpoints of sides AB, BC, and AC in ΔABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ΔABC is 1440 sq cm, then the area, in sq cm, of ΔXYZ is
In any triangle, the medians divide the triangle into six smaller triangles of equal area. The area of the medial triangle formed by the midpoints of the sides of the original triangle (in this case, △XYZ) is exactly one-fourth of the area of the original triangle. Since the area of △ABC is given as 1440 sq cm, the area of △XYZ, which is the medial triangle, is: Area of △XYZ = 14 × 1440 = 360 sq cm However, we must consider the area of the triangle formed by the medians, which is half the area of the medial triangle. Therefore, the area of △XYZ is: {90} sq cm
Here we need to check in three ranges Case 1⇒ x ≥ 2, we get x – 2 = x + 2 – (x – 2) x = 6 (Satisfied) Case 2 ⇒ – 2 ≤ x < 2, we get 2 – x = x + 2 – [–(x – 2)] x = 23 (Satisfied) Case 3 ⇒ x < –2, we get 2 – x = –x – 2 + (x – 2) x = 6 (Not Satisfied) Number of distinct real values of x, satisfying the equation will be 2.
A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was
Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
Ravi gets 10% discount on this marked price, and thus saves Rs 15 Means 10% of MP = 15Rs MP = 150 Rs. Discount is 15 Rs, SP = 150-15 =135 Gopi marks a price on a product in order to make 20% profit Means 1.2 CP = MP 1.2 CP = 150 CP = 125 Rs Profit = SP – CP Profit = 135 – 125 Profit = 10 Rs
The number of all positive integers up to 500 with non-repeating digits is
We will consider the cases of 1 digit, 2 digit and 3 digit numbers satisfying the above condition separately. 1 digit numbers with non-repeating digits = 9 2 digit numbers with non-repeating digits = 9 × 9 = 81 3 digit numbers with non-repeating digits = 4 × 9 × 8 = 288 Total 378 such numbers exist
After two successive increments, Gopal’s salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna is the same job. The undertake a job with as arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is
If Mohit is 2 times as efficient as Sam then Efficiency of Mohit : Efficiency of Sam = 2 : 1 i.e. (M : S = 2 : 1) If Mohit is 3 times as efficient as Ayna (M : A = 3 : 1) therefore S : M : A = 3 : 6 : 2 Total work = 3 × 20 = 60 According to the question Sam & Mohit work on the 1st day (So, S + M will complete 3 + 6 = 9 unit work in one day.) Sam & Ayna work on the 2nd day (So, S + A will complete 3 + 2 = 5 unit work in one day.) Mohit & Ayna work on the 2nd day (So, M + A will complete 6 + 2 = 8 unit work in one day.) So, in 3 days 9 + 5 + 8 = 22 unit work gets finished & in 6 days 22 × 2 = 44 unit gets completed On 7th day S + M will complete 9 unit work On 8th day S + A will do 5 unit work On 9th day M + A will do 2 unit work. So, in total the work gets completed on the 9th day but S works on 6 days in total i.e. 1st, 2nd, 4th, 5th, 7th and 8th day Therefore the fraction of the work done by Sam in those 6 days will be 6x3/60 = 3/10
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