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Applied Maths Chapter 3 Solutions
Quantitative Aptitude
EXERCISE- 3.5
Q.1 A person crosses a bridge 900 m long in 3 minutes. Find his speed in km/h and m/s.
Ans. D = 900m, S = ?, T = 3min = 3 x 60 = 180sec
Speed = Distance/Time
Therefore,
Speed of person = 900/180
= 5 m/s
now, m/s in km/h
5m/s = 5 x 18/5 km/h
= 18 km/h.
Q.2 Rishabh walks at 6 km/h and reaches his school in 1 hour 30 minutes. How far is his school?
Ans. Speed of Rishabh = 6 km*h
Time = 1.5 hr
D =?
Distance = Speed x Time
= 6 x 1.5
= 9 km
Q.3 Archi rides at 20 km/h to her office, which is 15 km away. How long will it take to reach the office?
Ans. Speed = 20 km/h
Distance = 15 km
Time =?
Time = Distance/Speed
= 15/20
= 3/4 hr
= 3/4 x 60 = 45 min.
Q.4 A man covers 1/3rd of his journey at 15 km/h and the remaining at 20 km/h. If he takes 10 hours to complete his journey, find the length of his journey.
Ans. Let, the total length be x km
ATQ,
Speed of 1/3rd distance = 15 km/h
D1 = x/3
S1 = 15 km/h
T1 = D/S
= (x/3)/15
= x/(3 x 15)
= x/45
Now, ATQ
Speed for 2/3rd distance = 20 km/h
D2 = 2X/3
S2 = 20 km/h
T2 = D/S
= (2x/3)/20
= 2x/60
= x/30
Total Time = 10 hrs
T1 + T2 = 10
x/45 + x/30 = 10
(2x + 3x)/90 = 10
5x/90 = 10
x/90 = 2
Therefore, x = 180 km
Q.5 Harpreet covered a certain distance at 5 km/h in 4 hours and 12 minutes. Find the time required to cover the same distance at the speed of 7 km/h.
Ans. According to the question,
Speed of Harpreet = 5 km/h
Time of Harpreet = 4hrs 12min
= 4hrs 12/60 min
= 4.2 hrs
Now, Distance = Speed X Time
= 5 x 4.2
= 21 km
Now, Distance = 21 km
New speed of Harpreet = 7 km/h
A new time of Harpreet =?
Now, Time = Distance/Speed
= 21/7 = 3 hrs
Q.6 A boy leaves for his school at 7:00 am. He spends 6 hours 30 minutes in the school and comes back at 3:00 p.m. on the same day. If he walks at a constant speed of 4 km/h, find how far is his school.
Ans. Speed of Boy = 4 km/h, Distance = 2x
Time of Boy = 1.5 hrs [8 – 6.5]
Let, the distance of the school from the boy’s house be x
Noe, Distance = Speed x Time
2x = 4 x 1.5
x = 2 x 1.5
x = 3 km.
Q.7 Walking at 3/4th of his usual speed, a man is 1 hour late. Find his usual travel time.
Ans. Let, the original speed be x
New speed = 3x/4
Let, the original time be y hrs
New time = y + 1
Now, we know that
Original speed x Original time = New speed x New time
x X y = 3x/4 (y + 1)
4y = 3y + 3
y = 3 hrs.
Q.8 Pulkit walks at 5/4th of his usual speed and reaches 20 minutes early. Find his usual time of journey.
Ans. Let, the original speed be x km/h
New speed = 5x/4
Let the original time be y hrs
New time = y – 20/60
= y – 1/3
Now, we know that
Original speed x Original time = New speed x New time
x X y = 5x/4 (y – 1/3)
4y = 5y – 5/3
5/3 = 5y – 4y
5/3 = y
5/3 x 60 = y
y = 100 min
Therefore, y = 1hr 40min
Q.9 A car is running at 5/7th of its actual speed and covers a distance of 45 km in 2 hours 20 minutes. Find the actual speed of the car.
Ans. According to the question,
Distance = 45 km
Time = 2hr 20min
= 2hr 20/60min
= 2 1/3 hrs
We know that,
Speed = Distance/Time
Here, New speed = 45/(7/3)
= (45 x 3)/7
= 135/7 km/h
ATQ, New speed = 5/7 Actual speed
135/7 = 5/7 Actual speed
135/5 = Actual speed
Therefore, Actual speed = 27 km/h
Q.10 Rohit covers a distance of 30 km in 6 hours. He travelled partly on foot at 3 km/h and partly on bicycle at 9 km/h. What is the distance travelled on a bicycle?
Ans. Total distance = 30 km
Total time = 6 hrs
Speed by foot = 3 km/h
Speed by bicycle = 9 km/h
Let, the distance travelled by bicycle be x km
So, Distance travelled by foot = (30 – x) km
We know that,
Time = Distance/Speed
Therefore, Distance travelled by foot = (30 – x)/3
Time taken by bicycle = x/9
Now, Time taken by foot + Time taken by bicycle = Total time
[(30 – x)/3] + x/9 = 6
(90 – 3x + x)/9 = 6
90 – 2x = 54
90 – 54 = 2x
36 = 2x
x = 18 km
Q.11 A man covered a certain distance at some speed. If he had moved 5 km/h faster, he would have taken 1 hour less. If he had moved 3 km/h slower, he would have taken 1 hour more. What is the distance in km?
Ans. Let, speed by x km/h
Time be y km/h
Distance = x y km/h
Case 1- When he moved 5 km/h faster, he would have taken 1 hour less
x y = (x + 5) (y – 1)
x y = x y – x + 5y – 5
x – 5y = -5……..(i)
Case 2- When he moved 3 km/h slower, he would have taken 1 hour more
x y = (x – 3) (y + 1)
x y = x y + x – 3y -3
x – 3y = 3 ………(ii)
From eq. (i) & (ii)
x – 5y = -5
x – 3y = 3
(-) (+) (-)
-2y = -8
y = 4
x = 15
Therefore, distance = 15 x 4 = 60 km.
Q.12 In covering a distance of 40 km Sachin takes 3 hours more than Utkarsh. If Sachin doubles his speed then he would take 1 hour less than Utkarsh. Find their speeds.
Ans. Distance = 40 km
Time taken by Utkarsh + 3 = Time taken by Sachin
Let, ‘s original speed of Sachin be x km/h
The new speed of Sachin = 2x
Time taken by Utkarsh – 1 = New time taken by Sachin
For Sachin:
Distance = 40 km
Speed = x km/h
Time = 40/x
Now, New speed = 2x
Distance = 40 km
Time = 40/2x = 20/x
For Utkarsh:
Time = (40/x) – 3
ATQ, after Sachin doubles his speed,
Time of Utkarsh – 1 = New time of Sachin
(40/x) – 3 – 1 = 20/x
(40/x) – (20/x) = 4
(40 – 20)/x = 4
20/x = 4
20/4 = x
x = 5 km/h
Therefore,
Speed of Utkarsh = 40/5 = 8 km/h
Time of Utkarsh = 40/5 – 3 = 5 hours
Q.13 A train travels 25% faster than a car. Both travel between A and B which are 120 km apart. Both start from A at the same time and reach point B at the same time. However, the train lost about 18 minutes while stopping at the stations. Find the speeds of the train and the car.
Ans. Let, the speed of the car be x km/h
Therefore, speed train = x + 25/100 * x
= x + 1/4 x
= 5/4x km/h
Distance = 120 km
Time of car = distance/speed
= 120/x
Time of train = 120/(5x/4)
= 120 x 4/5x = 96/x
ATQ,
T1 = T2
96/x + 18/60 = 120/x
3/10 = 120/x – 96/x
3/10 = 24/x
3x = 240
x = 80 km/h
Therefore,
Speed of car = 80 km/h
Speed of Train = 80 + 80 x 1/4
= 80 + 20 = 100 km/h
Q.14 A person has to cover a distance of 8 km in 45 minutes. He covers half of the distance in two-thirds of the total time. What should be his speed to cover the remaining distance in time?
Ans. Distance = 8 km
Time = 45 min = 3/4 hr
Now, ATQ
Distance Covered = 4 km
Time Taken = 3/4 x 2/3 = 1/2 hr
Remaining distance = 4 km
Remaining Time = 3/4 – 13 = 1/4 hr
Therefore,
Required Speed = Distance/Time
= 4/(1/4)
= 16 km/h
Q.15 A car travels the first 180 km at 80 km/h and the next 180 km at 70 km/h. Find its average speed.
Ans. Speed of car at first 180 km = 80 km/h
Speed of car at next 180 km = 70 km/h
Average speed = 2xy/(x + y)
= (2 x 80 x 70)/(80 + 70)
= 11200/150
= 75.67 km/h
Q.16 Manish goes to the office from his house at a speed of 3 km/h and returns to his house at 2 km/h. If he takes 50 minutes to go to the office and returns to his house, find the distance between his house and office.
Ans. Speed of Manish from house to office = 3 km/h
Speed of Manish from office to house = 2 km/h
Total Time taken = 50 min = 50/60
= 5/6 h
Average speed = 2xy/(x + y)
= (2 x 3 x 2)/(3 + 2)
= 12/5 km/h
Distance = Speed x Time
= (12/5) x (5/6)
= 2 km
Therefore,
Distance between his house and office is 1 km.
Q.17 A person travels from A to B at 56 km/h and returns from B to A by decreasing his speed by 25%. Find his average speed.
Ans. Speed from A to B = 56 km/h
Speed from B to A = 56 x 25/100 = 14
= 56 – 14 = 42 km/h
Average speed = 2xy/(x + y)
= (2 x 56 x 48)/(56 + 42)
= (2 x 56 x 42)/98
= 48 km/h
Q.18 A man rides at 40 km/h for the first 3 hours and at 60 km/h for the next 3 hours. Find his average speed.
Ans. Speed 1 (x) = 40 km/h
Speed 2 (y) = 60 km/h
Average speed = (x + y)/2
= (40 + 60)/2
= 100/2
= 50 km/h
Q.19 The distance between two stations X and Y is 520 km. A train starts from X at 9:00 a.m. and travels towards Y at 70 km/h. Another train starts from Y at 10:00 a.m. and travels towards X at 80 km/h. At what time will the trains meet?
Ans. Distance travelled by x in hour = 70 km
At 10:00 a. m. distance between both train = 520 – 70
= 450 km
Average speed = 70 + 80 = 150 km/h
Time Taken to meet = 450/150
= 3 hours
Therefore,
Both train meet at 1:00 p. m.
Q.20 Two persons start from the same place riding in the same direction at 20 km/h and 15 km/h respectively. After how many hours will they be 30 km apart?
Ans. Relative speed = 20 – 15
= 5 km/h
Distance = 30 km
Time = Distance/Speed
= 30/5
= 6 hours
Q.21 A goods train runs at the speed of 72 km/h and crosses a platform 120 m long in 15 seconds. Find the length of the goods train.
Ans. Speed = 72 km/h = 72 x 5/18 = 20 m/s
Length of platform = 120 m
Time = 15 sec
let, the length of goods time be x m
Distance = Time x Speed
= 20 x 15
= 300 m
Therefore,
x + 120 = 300
x = 300 – 120
= 180 m
Q.22 Two trains running at 48 km/h and 54 km/h in the same direction. The length of the second train is 120 m and the time taken by them to cross each other is 150 seconds from the time they meet. Find the length of the first train.
Ans. Relative speed = 54 – 48 = 6 km/h = 6 x 5/18 = 5/3 m/s
Length of second train = 120 m
let, the length of the first train be x m
Time Taken = 150 sec
Distance = Speed x Time
= 5/3 x 150 = 250 m
Now,
x + 120 = 250
x = 250 – 120
x = 130 m
Q.23 Two trains of length 140 m and 110 m are running in opposite directions. They cross each other in 6 seconds. If the speed of the first train is 80 km/h, find the speed of the second train.
Ans. Length of first train (T1) = 140 m
Length of second train (T2) = 110 m
Time Taken = 6 sec
Distance Covered = 140 + 110 = 250 m
Speed = Distance/Time
= 250/6
= 125/3 m/s
= 125/3 x 18/5
= 150 km/h
So, Relative speed = 150 km/h
Speed of T1 + Speed of T2 = 150
80 + Speed of T2 = 150
Speed of T2 = 150 – 80
= 70 km/h
Q.24 A train 150 m long is running at a speed of 120 km/h. A man standing on the roof of the train runs from one end of the train to the other end at 10 km/h in the opposite direction. Find the time taken by the man to reach the other end.
Ans. Train 1:
Length = 150 m
Speed = 120 km/h
= 120 x 5/18
= 100/3 m/s
Man:
Speed = 10 km/h
= 10 x 5/18 = 25/9 m/s
Distance = 150 m
Speed = 25/9 m/s
Time = Distance/Speed
= 150/(25/9)
= 150 x 9/25
= 54 seconds
Q.25 Two trains 130 m and 110 m in length move in the same direction at the same time. The faster train completely overtakes the slower train in 15 seconds. If the slower train moves at half of its speed, the overtaking would take 10 seconds. Find the speed of two trains in m/s.
Ans. Let, the faster train be x m/s
slower train be y m/s
Relative speed = (x – 9) m/s
Total length = 130 + 110 = 240 m
Time = 15 seconds
Distance = Speed x Time
240 = (x – y)15
240/15 = (x – y)
x – y = 16…….. (i)
If the speed of a slower train = y/2 m/s
Relative speed = x – y/2, Time = 10 sec
Distance = Speed x Time
240 = (x – y/2) 10
24 = x – y/2
2x – y = 48………(ii)
From eq. (i) & (ii)
x – y = 16
2x – y = 48
(-) (+) (-)
-x = – 32
x = 32 m/s
y = 16 m/s
FAQ’s related to Applied Maths Chapter 3 on Quantitative Aptitude:
Q.1 Why is Quantitative Aptitude important?
Ans. Quantitative aptitude skills are crucial for various competitive exams, entrance tests, and job interviews. They are also essential for everyday tasks such as budgeting, financial planning, and decision-making.
Q.2 What topics are covered in Applied Maths Chapter 3?
Ans. Applied Maths Chapter 3 on Quantitative Aptitude typically covers topics such as percentages, profit and loss, clock, time and distance, time and work, averages, ratio and proportion, etc.
Q.3 How will mastering Chapter 3 help you in future exams?
Ans. Mastering Chapter 3 will provide you with a strong foundation in quantitative aptitude, which is essential for scoring well in various competitive exams such as GRE, GMAT, SAT, CAT, banking exams, etc. These exams often contain sections dedicated to quantitative aptitude, and a good score in these sections can significantly enhance your overall performance.
These are few Frequently Asked Questions relating to Applied Maths Chapter 3
In Applied Maths chapter 3, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 3 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 3, we delve deep into advanced mathematical concepts that are crucial for understanding.
