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
ATQ, 9 × [ Actual marks + 10] = [Square of actual marks]
∴ 9 × (x + 10) = x2
⇒ 9x + 90 = x2
⇒ x2 – 9x – 90 = 0
⇒ x2 – 15x + 6x – 90 = 0
⇒ x(x – 15) + 6(x – 15) = 0
⇒(x + 6) (x – 15) = 0
Either x + 6 = 0
⇒ x = – 6 or x – 15 = 0
⇒ x = 15
But marks cannot be less than 0.
∴ x = –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 be x km/hr
Given that, the speed boat in still water is 18 km/hr.
Sspeed of the boat in upstream = (18 - x) km/hr
Speed of the boat in downstream = (18 + x) km/hr
It is mentioned that the boat takes 1 hour more to go 24 km upstream than to return downstream to the same spot
Therefore, One-way Distance traveled by boat (d) = 24 km
Hence, Time in hour
Tupstream = Tdownstream + 1
[distance / upstream speed ] = [distance / downstream speed] + 1
[ 24/ (18 - x) ] = [ 24/ (18 + x) ] + 1
[ 24/ (18 - x) - 24/ (18 + x) ] = 1
24 [1/ (18 - x) - 1/(18 + x) ] = 1
24 [ {18 + x - (18 - x) } / {324 - x2} ] = 1
24 [ {18 + x - 18 + x) } / {324 - x2} ] = 1
⇒ 24 [ {2}x / {324 - x2} ] = 1
⇒ 48x = 324 - x2
⇒ x2 + 48x - 324 = 0
⇒ x2 + 54x - 6x - 324 = 0 ----------> (by splitting the middle-term)
⇒ x(x + 54) - 6(x + 54) = 0
⇒ (x + 54)(x - 6) = 0
⇒ x = -54 or 6
As speed to stream can never be negative, we consider the speed of the stream (x) as 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 ‘x’.
∴ His marks in Science = (28 − x)
According to the condition,
(x + 3) (28 − x − 4) = 180
⇒ (x + 3) (− x + 24) = 180
⇒ 24x − x2 + 72 − 3x = 180
⇒ − x2 + 21x + 72 − 180 = 0
⇒ − x2 + 21x − 108 = 0
⇒ x2 − 21x + 108 = 0
⇒ x2 − 12x − 9x + 108 = 0
⇒ x (x − 12x) − 9(x − 12) = 0
⇒ (x − 9) (x − 12) = 0
⇒ (x − 9) (x − 12) = 0
Either x − 9 = 0
⇒ x = 9 or x − 12 = 0
⇒ x = 12 When x = 9 then 28 − x = 28 − 9 = 19
When x = 12 then 28 − x = 28 − 12 = 16
Thus P’s marks in Maths = 9 and Science = 19
Or
P’s marks in Maths = 12 and Science = 16.
Q4: At ‘t’ 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 ‘t’.
Sol: For a minute-hand time needed to show 2 pm to 3 pm is ‘60’ minutes. It has already covered ‘t’ minutes. ∴ Time required by the minute-hand to reach to 12 (at 3 pm) = (60 – t) minutes.
60 - t = (t²/4 - 3)
60 - t = (t² - 12)/4
240 - 4t = t² - 12
t² + 4t - 12 - 240 = 0
t² + 4t - 252 = 0
On factorizing,
t² - 14t + 18t - 252 = 0
t(t - 14) + 18(t - 14) = 0
(t + 18)(t - 14) = 0
Now, t - 14 = 0
t = 14
Also, t + 18 = 0
t = -18
Solving, we get, t = 14 or – 18
But t = – 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
∴ Original time taken = 360/x.
New speed = (x + 5) km/hr
New time = 360/ (x + 5) km / hr
According to the condition,
360x = 360x + 5 + 4860
= 360[ x + 5 -xx(x + 5)] = 45
= 360[ 5x2 + 5x] = 45
= x2 + 5x - 2250 = 0
Solving for x, we get x = – 50 or 45
Speed cannot be negative
∴ Rejecting x = – 50, we have x = 45
Thus, the original speed of the train = 45 km/hr.
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