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**LONG ANSWER TYPE QUESTIONS**

**Q1. A two-digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.**

**Sol.** Let the tens digit = x

And the ones digit = y

âˆ´ The number = 10x + y

According to the conditions,

10x + y = 5 (x + y) ...(1)

10x + y = 2xy + 5 ...(2)

From (1), we have

10x + y = 5x + 5y

â‡’ 10x + y âˆ’ 5x âˆ’ 5y =0

â‡’ 5x âˆ’ 4y = 0

â‡’

â‡’ 40x + 5x = 10x^{2} + 20 [Multiplying both sides by 4]

â‡’ 45x = 10x^{2} + 20

â‡’ 10x^{2} âˆ’ 45x âˆ’ 20 = 0

â‡’ 2x^{2} âˆ’ 9x âˆ’ 4= 0

â‡’ 2x^{2} âˆ’ 8x âˆ’ x + 4 = 0

â‡’2x (x âˆ’ 4) âˆ’ 1 (x âˆ’ 4) = 0

â‡’ (x âˆ’ 4) (2x âˆ’ 1) = 0

Either x âˆ’ 4 = 0 â‡’ x = 4

or 2x âˆ’ 1 = 0 â‡’ x = 1/2

But a digit cannot be a fraction,

âˆ´ x = 4

â‡’ The tens digit = 4

âˆ´ x = 4 and y = 5

âˆ´ The required number = 10 Ã— 4 + 5

= 40 + 5 = 45**Q2. The denominator is one more than twice the numerator. If the sum of the fraction and its reciprocal is find the fraction.**

**Sol**. Let the numerator be x

âˆ´ Denominator = (2x + 1)

âˆ´ Fraction = x/(2x +1)

Reciprocal of the fraction = 2x +1/x

According to the condition,

â‡’

â‡’

â‡’

â‡’ 21 (5x^{2} + 4x + 1) = 58 (2x^{2} + x)

â‡’ 105x^{2} + 84x + 21 = 116x^{2}+ 58x

â‡’ 105x^{2} âˆ’ 116x^{2} + 84x âˆ’ 58x + 21 = 0

â‡’ âˆ’ 11x^{2} + 26x + 21 = 0

â‡’ 11x^{2} âˆ’ 26x âˆ’ 21 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0, we have:

a =11

b = âˆ’ 26

c = âˆ’ 21

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 26)^{2} âˆ’ 4 Ã— 11 (âˆ’ 21) = 676 + 924 = 1600

Since,

âˆ´

â‡’

Taking the +ve sign,

Taking the âˆ’ve sign,

But the numerator cannot be -7/11

âˆ´ x = 3 â‡’ The numerator = 3

âˆ´ Denominator = 2 (3) + 1 = 7

Thus, the required fraction = 3/7**Q3. The sum of a number and its reciprocal is 10/3. Find the number.**

**Sol.** Let the required number = x

âˆ´ Its reciprocal = 1/x

According to the condition, we have:

â‡’

â‡’

â‡’ 3 (x^{2} + 1) = 10x

â‡’ 3x^{2} + 3 âˆ’ 10x = 0

â‡’ 3x^{2} âˆ’ 10x + 3 = 0

â‡’ 3x^{2} âˆ’ 9x âˆ’ x + 3 = 0

â‡’ 3x (x âˆ’ 3) âˆ’ 1 (x âˆ’ 3) = 0

â‡’ (x âˆ’ 3) (3x âˆ’ 1) = 0

Either x âˆ’3 = 0 â‡’ x = 3

or 3x âˆ’ 1 = 0 â‡’ x = 1/3

Thus, the required number is 3 or 1/3.**Q4. The hypotenuse of a right triangle is cm. If the smaller side is tripled and the longer side doubled, new hypotenuse will be cm. How long are the sides of the triangle?**

**Sol.** Letthe smaller side = x

âˆ´ Longer side =

According to the condition, we have [3 (Smaller side)]^{2} + [2 (Longer side)]^{2} = [New Hypotenuse]

â‡’ 5x^{2} = 405 âˆ’ 360

â‡’ 5x^{2} = 45

â‡’

â‡’

But x = âˆ’ 3 is not required, because the side of a triangle cannot be negative.

âˆ´ x = 3 â‡’ Smaller side = 3 cm

âˆ´ Longer side

Thus, the required sides of the triangle are 3 cm and 9 cm.**Q5. A motor-boat goes 10 km upstream and returns back to the starting point in 55 minutes. If the speed of the motor boat in still water is 22 km/hr, find the speed of the current.**

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

âˆ´ The speed downstream = (22 + x) km/hr

The speed upstream= (22 âˆ’ x) km/hr

Since, Time = Distance/Speed

âˆ´ Time for going 10 km downstream

Time for returning back 10 km upstream

According to the condition,

â‡’

[âˆµ 55 minutes = 55/60 hours]

â‡’

â‡’ 11 (484 âˆ’ x^{2}) = 120 Ã— 44

â‡’ 5324 âˆ’ 11x^{2} = 5280

â‡’ 11x^{2} = 5324 âˆ’ 5280 = 44

â‡’

â‡’ x = Â± 2

But speed of the current cannot be negative,

âˆ´ x = 2

â‡’ Speed of current = 2 km/hr**Q6. A motor boat whose speed in still water is 5 km/hr, takes 1 hour more to go 12 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 âˆ´ Downstream speed of the motor boat = (x + 5) km/hr

â‡’ Time taken to go 12 km upstream

= 12/5 âˆ’ x hours

Time taken to return 12 km downstream

= 12/5 + x hours

According to the condition

âˆ´12 (5 + x) âˆ’ 12 (5 âˆ’ x) = 1 (5 âˆ’ x) (5 + x)

â‡’ 60 + 12x âˆ’ 60 + 12x = 25 âˆ’ x^{2 }

â‡’ 24x = 25 âˆ’ x^{2}

â‡’ x^{2} + 24x âˆ’ 25 = 0

â‡’ x^{2 }+ 25x âˆ’ x âˆ’ 25 = 0

â‡’ x (x + 25) âˆ’ 1 (x + 25) = 0

â‡’ (x âˆ’ 1) (x âˆ’ 25) = 0

Either x âˆ’ 1 = 0 â‡’ x = 1

or x + 25 = 0 â‡’ x = âˆ’ 25

But x = âˆ’ 25 is not admissible, because the speed of the stream cannot be negative

âˆ´ x = 1

â‡’ speed of the stream = 1 km/hr.**Q7. Sum of the areas of two squares is 260 m ^{2.} If the difference of their perimeters is 24 m, then find the sides of the two squares.**

**Sol.** Let the side of one of the squares be â€˜xâ€™ metres

âˆ´ Perimeter of square-I = 4 Ã— x metres = 4x metres

âˆ´ Perimeter of square-II = (24 + 4x) metres

âˆ´ Side of the square-II metres

= (6 + x) metres

Now, according to the condition, we have:

x^{2} + (6 + x)^{2} = 260

â‡’x^{2} + 36 + x^{2} + 12x âˆ’ 260 = 0

â‡’ 2x^{2} + 12x âˆ’ 224 = 0

â‡’ x^{2} + 6x âˆ’ 112 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0, we get,

a = 1

b = 6

c = âˆ’ 112

âˆ´^{ b}2 âˆ’ 4ac = (6)2 âˆ’ 4 (1) (âˆ’ 112)

= 36 + 448 = 484

âˆ´

â‡’

Taking +ve sign, we have

Taking âˆ’ve sign, we have

But x = âˆ’ 14 is not required, as the length of a side cannot be negative.

âˆ´ x = 8

â‡’ Side of square-I = 8 m

â‡’ Side of square-II = 6 + 8 m = 14 m.**Q8. The age of a father is twice the square of the age his son. Eight years hence, the age of the father will be 4 years more than three times the age of his son. Find their present ages.****Sol.** Let the present of son be â€˜xâ€™ years.

âˆ´ Fatherâ€™s present age = 2x^{2} years 8 years hence:

Age of son = (x + 8) years

Age of father = (2x^{2} + 8) years

**According to the condition:**

(2x^{2} + 8) = 3 (x + 8) + 4

â‡’ 2x^{2} + 8 âˆ’ 3x âˆ’ 24 âˆ’ 4= 0

â‡’ 2x^{2} âˆ’ 3x + 8 âˆ’ 28 = 0

â‡’ 2x^{2} âˆ’ 3x âˆ’ 20 = 0 ...(1)

Comparing (1) with a^{x2} + bx + c = 0, we get

a = 2

b = âˆ’3

c = âˆ’20

âˆ´ b^{2 }âˆ’ 4ac =(âˆ’ 3)^{2 }âˆ’ 4 (2) (âˆ’ 20)

= 9 + 160 = 169

But x = -15/2 is not required, as the agecannot be negative.

âˆ´ x = 4

â‡’ Present age of son = 4 years

Present age of father = 2 Ã— 4^{2} = 32 years.**Q9. A motor boat whose speed in still water is 16 km/h, takes 2 hours more to go 60 km upstream than to return to the same spot. Find the speed of the stream.**

**Sol.** Let the speed of the stream = x km/hr For the motor boat, we have:

âˆ´Downstream speed = (16 + x) km/hr

Upstream speed = (16 âˆ’ x) km/hr

For going 60 km:

According to the condition:

â‡’ 60 (16 + x) âˆ’ 60 (16 âˆ’ x) = 2 (16 âˆ’ x) (16 + x)

â‡’ 960 + 60x âˆ’ 960 + 60x = 2 (256 âˆ’ x^{2})

â‡’ 2x^{2} + 120x = 2 Ã— 256 âˆ’2x^{2}

â‡’ x^{2} + 60x = 256

â‡’ x^{2} + 60x âˆ’ 250 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0,

a = 1

b = 60

c = âˆ’ 256

âˆ´ b^{2} âˆ’ 4ac = (60)^{2} âˆ’ 4 (1) (âˆ’ 256)

1 = 3600 + 1024 = 4624

âˆ´

â‡’

â‡’

Taking +ve sign,

Taking âˆ’ve sign,

Since, the speed of a stream cannot be negative,

âˆ´x = âˆ’ 64 is not admissible

âˆ´ x = 4

â‡’ speed of the stream = 4 km/hr.**Q10. A train travels 288 km at a uniform speed. If the speed had been 4 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.**

**Sol.** Let the speed of the train be x km/hr

âˆµ Total distance travelled = 288 km

In the other case, Speed of the train = (x + 4) km/hr

According to the condition,

â‡’ 288x + 1152 âˆ’ 288x = 1 (x) (x + 4)

â‡’ 288x + 1152 âˆ’ 288x = x^{2} + 4x

â‡’ 1152 = x^{2} + 4x

â‡’ x^{2} + 4x âˆ’ 1152 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0,

a = 1

b = 4

c = âˆ’ 1152

âˆ´ b^{2} âˆ’ 4ac = (4)^{2} âˆ’ 4 (1) (âˆ’ 1152) = 16 + 4608

â‡’

â‡’

Taking +ve sign,

Taking âˆ’ve sign,

âˆµ speed cannot be negative,

âˆ´ x â‰ âˆ’ 36

âˆ´ x = 32

â‡’ speed of the train = 32 km/hr.**Q11. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed of. If takes 3 hours to complete the total journey, what is its original average speed?**

**Sol.** Let the original average speed = x km/hr

âˆ´ Time taken to cover 72 km = 72/ x + 6 hours

Time taken to cover 63 km = 63/x hours

Since, total time = 3 hours

âˆ´

â‡’

â‡’

â‡’

â‡’

â‡’ 3[15x + 42] = x^{2} + 6x

â‡’ 45x + 126 â€“ x^{2} â€“ 6x = 0

â‡’ x^{2} â€“ 39x â€“ 126 = 0

â‡’ x^{2 }â€“ 42x + 3x â€“ 126 = 0

â‡’ x(x â€“ 42) + 3(x â€“ 42) = 0

â‡’ (x + 3) (x â€“ 42) = 0

Either x + 3 = 0 â‡’ x = â€“ 3

or

x â€“ 42 = 0 â‡’ x = 42

Since, speed cannot be negative,

âˆ´ x = â€“3 is not desired.

Thus, the original speed of the train is 42 km/hr.**Q12. If âˆ’ 5 is a root of the quadratic equation 2x ^{2} + px âˆ’ 15 = 0 and the quadratic equation p (x^{2} + x) + k = 0 has equal roots, then find the values of p and k.**

**Sol. **Since âˆ’ 5 is a root of 2x^{2} + px âˆ’ 15 = 0,

âˆ´ Substituting x = âˆ’ 5 in the given equation, we get

2 (âˆ’ 5)^{2} + p (âˆ’ 5) âˆ’ 15 = 0

â‡’ 2 (25) + (âˆ’ 5p) âˆ’ 15 = 0

â‡’ 50 âˆ’ 5p âˆ’ 15 = 0

â‡’ âˆ’ 5p + 35 = 0

â‡’ âˆ’ 5p = âˆ’ 35

â‡’

Now, comparing the another quadratic equation p (x^{2} + x) + k = 0, i.e., px^{2} + px + k = 0 with ax^{2} + bx + c = 0, we have:

a = p

b = p

c = k

b^{2} âˆ’ 4ac =^{ }p^{2} âˆ’ 4 (p) (k)

= p^{2 }âˆ’ 4pk

Since p (x^{2} âˆ’ x) + k = 0 has equal roots,

âˆ´ p^{2} âˆ’ 4pk =0

â‡’ (7)^{2 }âˆ’ 4 (7) k = 0 |âˆµ p = 7

â‡’ 49 âˆ’ 28k = 0

â‡’

Thus, the required values of p = 7 and k = 7/4.**Q13. In a class test, the sum of Gaganâ€™s marks in Mathematics and English is 45. If he had 1 more mark in Mathematics and 1 less in English, the product of marks would have been 500. Find the original marks obtained by Gagan in Mathematics and English separately.****Sol.** Let Gaganâ€™s marks in maths = x and Marks in English = (45 âˆ’ x)

âˆ´ According to the condition, (x + 1) Ã— (45 âˆ’ x + 1) = 500

â‡’ (x + 1) Ã— (44 âˆ’ x) = 500

â‡’ 44x âˆ’ x^{2} + 44 âˆ’ x = 500

â‡’âˆ’ x^{2} + 44x âˆ’ 456 âˆ’ x =0

â‡’ x^{2} âˆ’ 43x + 456 = 0

â‡’ x^{2} âˆ’ 19x âˆ’ 24x + 456 = 0

â‡’ x (x âˆ’ 19) âˆ’ 24 (x âˆ’ 19) = 0

â‡’ (x âˆ’ 19) (x âˆ’ 24) = 0

Either x âˆ’ 19 = 0 â‡’ x = 19

or x âˆ’ 24 = 0 â‡’ x = 24

When x = 19, then 45 âˆ’ 19 = 26

When x = 24, then 45 âˆ’ 24 = 21

âˆ´ **Gaganâ€™s marks in Maths = 19 and in English = 26**

Or

**Gaganâ€™s marks in Maths = 24 and in English = 21.****Q14. The sum of areas of two squares is 640 m ^{2}. If the difference of their perimeters is 64 m, find the sides of two squares.**

**Sol. **Let the side of square I be x metres.

âˆ´ Perimeter of square I = 4x metres

â‡’ Perimeter of square II = (64 + 4x) m

= (16 + x) m

Now Area of square I = x Ã— x = x^{2}

Area of square II = (16 + x) Ã— (16 + x) = (16 + x)^{2}

= 256 + x^{2} + 32x

According to the condition, (x + 1) Ã— (45 âˆ’ x + 1) = 500

â‡’ x^{2} + [256 + x^{2} + 32x] = 640

â‡’ x^{2} + x^{2} + 32x + 256 âˆ’ 640 = 0

â‡’ 2x^{2} + 32x âˆ’ 384 = 0

â‡’ x^{2} + 16x âˆ’ 192 = 0

â‡’ x^{2} + 24x âˆ’ 8x âˆ’ 192 = 0

â‡’ x (x + 24) âˆ’ 8 (x + 24) = 0

â‡’(x + 24) (x âˆ’ 8) = 0

Either x + 24 =0

â‡’ x = âˆ’ 24 or x âˆ’ 8 = 0 â‡’ x = + 8

âˆµ side of a square cannot be negative,

âˆ´ Rejecting x = âˆ’ 24, we have x = 8

â‡’ Side of smaller square = 8 m

Side of larger square = 8 + 16 m = 24 m.**Q15. In a class test, the sum of Kamalâ€™s marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of his marks would have been 360. Find his marks in two subjects separately.**

**Sol.** Let Kamalâ€™s marks in Maths = x

âˆ´ His marks in English = (40 âˆ’ x) According to the condition,

(x + 3) [40 âˆ’ x âˆ’ 4] = 360

â‡’ (x + 3) (36 âˆ’ x) = 360

â‡’ 36x âˆ’ x^{2} + 108 âˆ’ 3x âˆ’ 360 = 0

â‡’ âˆ’ x^{2} + 33x âˆ’ 252 = 0

â‡’ x^{2 }âˆ’ 33x + 252 = 0

â‡’ x^{2} âˆ’ 21x âˆ’ 12x + 252 = 0

â‡’ x (x âˆ’ 21) âˆ’ 12 (x âˆ’ 21) = 0

â‡’(x âˆ’ 21) (x âˆ’ 12) = 0

Either (x âˆ’ 21) = 0

â‡’ x = 21 = 0

or (x âˆ’ 12) = 0

â‡’ x = 12

For x = 21, Marks of Kamal

in Maths = 21

in English = 40 âˆ’ 21 = 19

For x = 12, Marks of Kamal

in Maths = 12

in English = 40 âˆ’ 12

= 28.**Q16. Solve the quadratic equation 2x ^{2} + ax â€“ a^{2} = 0**

**Sol ^{. }**2x

â‡’ 2x^{2} + 2ax â€“ ax â€“ a^{2} =0

â‡’ 2x[x + a] â€“ a[x + a]=0

â‡’ (x + a) (2x â€“ a) = 0

x = â€“a or x = a/2

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