RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 10. Quadratic Equations

Subject: Maths - Class 10th

Q. No. 48 of Exercise 10E

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48

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Let the original speed of the train be x km/h

Time taken to cover 90 km =


New speed of train = x + 15 km/h


Time taken to cover 90 km =


According to the question






2700 = x2 + 15x


x2 + 15x – 2700 = 0


Using the splitting middle term – the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 1 b = 15 c = – 2700


= 1. – 2700 = – 2700


And either of their sum or difference = b


= 15


Thus the two terms are – 45 and 60


Difference = 60 – 45 = 15


Product = 60. – 45 = – 2700


x2 + 60x – 45x – 2700 = 0


x(x + 60) – 45(x + 60) = 0


(x + 60) (x – 45) = 0


x = – 60 or x = 45 (but x cannot be negative)


x = 45


Hence the original speed of the train is 45 km/h


Chapter Exercises

Exercise 10A
Exercise 10B
Exercise 10C
Exercise 10D
Exercise 10E
Exercise 10F

More Exercise Questions

1

The sum of a natural number and its square is 156. Find the number.
2

The sum of a natural number and its positive square root is 132. Find the number.
3

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4

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5

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6

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7

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8

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9

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10

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11

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12

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13

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14

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15

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16

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17

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18

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19

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20

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21

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22

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23

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24

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25

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26

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27

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28

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29

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30

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31

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32

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33

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34

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35

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36

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37

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38

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39

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40

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41

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42

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43

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44

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45

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46

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47

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48

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49

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50

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51

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52

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53

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54

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55

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56

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57

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58

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59

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60

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61

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