ANSWERS Quiz Ques 1: Uday travels one third of its journey by train with speed of 60kmph and the rest of journey by car with speed of 80kmph. Find his average speed of his journey? Ans: Know this that in this, the distance travelled is unequal. SO, we have to apply basics. Let total distance be 3D so that, 1/3rd of distance will be D, the rest will be 2D S1 = 60kmph S2 = 80kmph So, we have T1 = D/60 T2 = 2D/80 = D/40 Now, average speed = total distance / total time Total distance = distance for 1st time + Distance for 2nd time = 3D Total time = D/60 + D/40 = [D/24] Average speed will be = [3D] / ( [D/24] ) = 72 kmph Quiz Ques 2: The distance of the college and home of Rajeev is 80km. One day, he was late by 1 hour than normal time to leave for college, so he increased his speed by 4kmph and thus he reached to college at the normal time. What is the changed speed of Rajeev? Ans: Distance is 80km. It is constant. Only speed is changed. Now, let’s say his normal speed is X kmph, Then he will reach the college in 80/X hour time. (Equation 1) www.bankersadda.com | www.careerpower.in | www.careeradda.co.in 1 With [X+4] speed, he will reach the college in 80/[X+4] hour time. (Equation 2) Now the question says, he is late 1 hour but with X+4 speed, he reaches the college on time. That means time in (Equation 1) must be 1 hour more than the time in (Equation 2) I.e. 80/X = 1 + 80/[X+4] This becomes a quadratic equation, X^2 + 4X - 320 = 0  X^2 + 20X - 16X - 320 = 0  (X + 20) * (X - 16) = 0  X = -20 and +16 But Speed can't be negative  X = 16 kmph And his new speed = [X+4] = 20 kmph Quiz Ques 3: A lives at P and B lives at Q. A usually goes to meet B at Q. He covers the distance in 3 hour at 150kmph. On a particular day, B started moving away from A While A was moving towards Q, thus A took 5 hours to meet B. What is the speed of B? Ans: Distance = 3*150 = 450 km. Let Sb = Speed of B They’re moving in the same direction. Relative speed = 150 –Sb Time = Distance/Speed Hence, 5 = 450 / [150 – Sb] 150 – Sb = 90 Sb = 60 kmph Quiz Ques 4: A man can row at 10 kmph in still water. If the river flows at 3 kmph and, it takes 12 hours more in upstream than to go downstream for the same distance. How far is the place? www.bankersadda.com | www.careerpower.in | www.careeradda.co.in 2 Ans: Let distance be = x Time taken for upstream = x/(10 - 3) Time taken for downstream = x/(10 + 3) According to question, x/(10 + 3) + 12 = x/(10 - 3) x/7 - x/13 = 12  (13x - 7x)/ 91 = 12  6x/91 = 12 x = 182 km Quiz Ques 5: In a 1600m rade, A beats B by 80m and C by 60m. If they run at the same time, then by what distance will C beat B in a 400 m race? Ans: Now, the race is given of 1600m and question is asked of 400m. First we will find their speeds. A has covered 1600m, B has covered how much? Ans is 1600-80 = 1520m A has covered 1600m, C has covered how much? Ans is 1600-60 = 1540m Now, distances covered in the same time is known. Speed will be directly proportional to this distance. So, their speed ratio will be? Speed of A : B : C = 1600 : 1520 : 1540 = 80 : 76 : 77 Now, in the second race, only B and C have run. Their speed ratio we know, i.e. B : C = 76 : 77 C's speed is more. Means to say, when C covers 77m, B covers 76m. The distance B covers when C has covered 400m = [76/77]*400m And C has beaten B by 400 - [76/77]*400 = 400/77 meters = answer Quiz Ques 6: A beats B by 100m in a race of 1200m and B beats C by 200m in a race of 1600m. Approximately by how many meters can A beat C in a race of 9600m? Ans: Ratio of their speed would be equal to the ratio of their distances covered. Ratio of speed of A:B = ratio of their distances covered = 1200:1100 = 12:11 www.bankersadda.com | www.careerpower.in | www.careeradda.co.in 3 Ratio of speed of B:C = ratio of their distances covered = 1600:1400 = 8:7 A:B = 12:11 = 96:88 B:C = 8:7 = 88:77 A:B:C = 96:88:77 When A covers 96m, C covers 77m When A will cover 9600m, C will cover 7700m A will beat C by 9600-7700 = 1900 meters = Answer www.bankersadda.com | www.careerpower.in | www.careeradda.co.in 4