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**Q1. Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?**

**Sol.** Let actual marks be x

âˆ´ 9 Ã— [ Actual marks + 10] = [Square of actual marks]

or 9 Ã— (x + 10) = x^{2}

â‡’ 9x + 90 = x^{2}

â‡’ x^{2} â€“ 9x â€“ 90 = 0

â‡’ x^{2} â€“ 15x + 6x â€“ 90 = 0

â‡’ x(x â€“ 15) + 6(x â€“ 15) = 0

â‡’(x + 6) (x â€“ 15) = 0

Either x + 6 = 0

â‡’ x = â€“ 6 or x â€“ 15 = 0

â‡’ x = 15

But marks cannot be less than 0.

âˆ´ x = â€“6 is not desired.

Thus, Ravita got 15 marks in her Mathematics test.**Q2. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.**

**Sol. **Let the speed of the stream = x km/hr

âˆ´ speed of the motor boat:

upstream = (18 âˆ’ x) km/hr

downstream = (18 + x) km/hr

â‡’ Time taken by the motor boat in going:

24 km upstream = 20/18-x hours

According to the condition:

â‡’ 24 Ã— (18 + x) âˆ’ 2(18 âˆ’ x) = 1 (18 âˆ’ x) (18 + x)

â‡’ 24 [18 + x âˆ’ 18 + x] = 182 âˆ’ x^{2}

â‡’ 24 [2x] = 324 âˆ’ x^{2}

â‡’ 48x = 324 âˆ’ x^{2}

â‡’ x^{2} + 48x âˆ’ 324 = 0

â‡’ x^{2} + 54x âˆ’ 6x âˆ’ 324 = 0

â‡’ x (x + 54) âˆ’ 6 (x + 54) = 0

â‡’ (x âˆ’ 6) (x + 54) = 0

Either x âˆ’ 6 = 0 â‡’ x = 6

or x + 54 = 0 â‡’ x = âˆ’ 54

But speed cannot be negative

âˆ´ Rejecting x =âˆ’ 54, we have

x = 6 â‡’ Speed of the boat = 6 km/hr.**Q3. In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately.**

**Sol.** Let marks obtained by P in Maths be â€˜xâ€™.

âˆ´ His marks in Science = (28 âˆ’ x)

According to the condition,

(x + 3) (28 âˆ’ x âˆ’ 4) = 180

â‡’ (x + 3) (âˆ’ x + 24) = 180

â‡’ 24x âˆ’ x^{2} + 72 âˆ’ 3x = 180

â‡’ âˆ’ x^{2} + 21x + 72 âˆ’ 180 = 0

â‡’ âˆ’ x^{2} + 21x âˆ’ 108 = 0

â‡’ x^{2} âˆ’ 21x + 108 = 0

â‡’ x^{2} âˆ’ 12x âˆ’ 9x + 108 = 0

â‡’ x (x âˆ’ 12x) âˆ’ 9(x âˆ’ 12) = 0

â‡’ (x âˆ’ 9) (x âˆ’ 12) = 0

â‡’ (x âˆ’ 9) (x âˆ’ 12) = 0

Either x âˆ’ 9 = 0

â‡’ x = 9 or x âˆ’ 12 = 0

â‡’ x = 12 When x = 9 then 28 âˆ’ x = 28 âˆ’ 9 = 19

When x = 12 then 28 âˆ’ x = 28 âˆ’ 12 = 16

Thus Pâ€™s marks in Maths = 9 and Science = 19

Or

Pâ€™s marks in Maths = 12 and Science = 16.**Q4. At â€˜tâ€™ minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than minutes. Find â€˜tâ€™.**

**Sol. **For a minute-hand time needed to show 2 pm to 3 pm is â€˜60â€™ minutes. It has already covered â€˜tâ€™ minutes. âˆ´ Time required by the minute-hand to reach to 12 (at 3 pm) = (60 â€“ t) minutes.

âˆ´

â‡’

â‡’ t^{2}+ 4t â€“ 252 = 0

Solving, we get, t = 14 or â€“ 18

But t = â€“ 18 is not desirable (being negative)

Thus, t = 14 minutes.**Q5. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/ hr more. Find the original speed of the train.**

**Sol.** Let the original speed be x km/hr

âˆ´ Original time taken = 360/x.

New speed = (x + 5) km/hr

According to the condition,

Solving for x, we get x = â€“ 50 or 45

Speed cannot be negative

âˆ´ Rejecting x = â€“ 50, we have x = 45

Thus, the original speed of the train = 45 km/hr.

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