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**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

∴ 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

⇒ 10x + y − 5x − 5y =0

⇒ 5x − 4y = 0

⇒

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

⇒ 45x = 10x^{2} + 20

⇒ 10x^{2} − 45x − 20 = 0

⇒ 2x^{2} − 9x − 4= 0

⇒ 2x^{2} − 8x − x + 4 = 0

⇒2x (x − 4) − 1 (x − 4) = 0

⇒ (x − 4) (2x − 1) = 0

Either x − 4 = 0 ⇒ x = 4

or 2x − 1 = 0 ⇒ x = 1/2

But a digit cannot be a fraction,

∴ x = 4

⇒ The tens digit = 4

∴ x = 4 and y = 5

∴ The required number = 10 × 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

∴ Denominator = (2x + 1)

∴ Fraction = x/(2x +1)

Reciprocal of the fraction = 2x +1/x

According to the condition,

⇒

⇒

⇒

⇒ 21 (5x^{2} + 4x + 1) = 58 (2x^{2} + x)

⇒ 105x^{2} + 84x + 21 = 116x^{2}+ 58x

⇒ 105x^{2} − 116x^{2} + 84x − 58x + 21 = 0

⇒ − 11x^{2} + 26x + 21 = 0

⇒ 11x^{2} − 26x − 21 = 0 ...(1)

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

a =11

b = − 26

c = − 21

∴ b^{2} − 4ac =(− 26)^{2} − 4 × 11 (− 21) = 676 + 924 = 1600

Since,

∴

⇒

Taking the +ve sign,

Taking the −ve sign,

But the numerator cannot be -7/11

∴ x = 3 ⇒ The numerator = 3

∴ 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

∴ Its reciprocal = 1/x

According to the condition, we have:

⇒

⇒

⇒ 3 (x^{2} + 1) = 10x

⇒ 3x^{2} + 3 − 10x = 0

⇒ 3x^{2} − 10x + 3 = 0

⇒ 3x^{2} − 9x − x + 3 = 0

⇒ 3x (x − 3) − 1 (x − 3) = 0

⇒ (x − 3) (3x − 1) = 0

Either x −3 = 0 ⇒ x = 3

or 3x − 1 = 0 ⇒ 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, the new hypotenuse will be cm. How long are the sides of the triangle?**

**Sol.** Let the smaller side = x

∴ Longer side =

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

⇒ 5x^{2} = 405 − 360

⇒ 5x^{2} = 45

⇒

⇒

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

∴ x = 3 ⇒ Smaller side = 3 cm

∴ Longer side

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

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

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

The speed upstream= (22 − x) km/hr

Since, Time = Distance/Speed

∴ Time for going 10 km downstream

Time for returning back 10 km upstream

According to the condition,

⇒

[∵ 55 minutes = 55/60 hours]

⇒

⇒ 11 (484 − x^{2}) = 120 × 44

⇒ 5324 − 11x^{2} = 5280

⇒ 11x^{2} = 5324 − 5280 = 44

⇒

⇒ x = ± 2

But speed of the current cannot be negative,

∴ x = 2

⇒ Speed of current = 2 km/hr**Q6. A motorboat 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 ∴ Downstream speed of the motor boat = (x + 5) km/hr

⇒ Time taken to go 12 km upstream

= 12/5 − x hours

Time taken to return 12 km downstream

= 12/5 + x hours

According to the condition

∴12 (5 + x) − 12 (5 − x) = 1 (5 − x) (5 + x)

⇒ 60 + 12x − 60 + 12x = 25 − x^{2 }

⇒ 24x = 25 − x^{2}

⇒ x^{2} + 24x − 25 = 0

⇒ x^{2 }+ 25x − x − 25 = 0

⇒ x (x + 25) − 1 (x + 25) = 0

⇒ (x − 1) (x − 25) = 0

Either x − 1 = 0 ⇒ x = 1

or x + 25 = 0 ⇒ x = − 25

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

∴ x = 1

⇒ speed of the stream = 1 km/hr.**Q7. The 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 ‘x’ metres

∴ Perimeter of square-I = 4 × x metres = 4x metres

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

∴ Side of the square-II metres

= (6 + x) metres

Now, according to the condition, we have:

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

⇒x^{2} + 36 + x^{2} + 12x − 260 = 0

⇒ 2x^{2} + 12x − 224 = 0

⇒ x^{2} + 6x − 112 = 0 ...(1)

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

a = 1

b = 6

c = − 112

∴^{ b}2 − 4ac = (6)2 − 4 (1) (− 112)

= 36 + 448 = 484

∴

⇒

Taking +ve sign, we have

Taking −ve sign, we have

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

∴ x = 8

⇒ Side of square-I = 8 m

⇒ Side of square-II = 6 + 8 m = 14 m.**Q8. The age of a father is twice the square of the age of 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 ‘x’ years.

∴ Father’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

⇒ 2x^{2} + 8 − 3x − 24 − 4= 0

⇒ 2x^{2} − 3x + 8 − 28 = 0

⇒ 2x^{2} − 3x − 20 = 0 ...(1)

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

a = 2

b = −3

c = −20

∴ b^{2 }− 4ac =(− 3)^{2 }− 4 (2) (− 20)

= 9 + 160 = 169

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

∴ x = 4

⇒ Present age of son = 4 years

Present age of father = 2 × 4^{2} = 32 years.**Q9. A motorboat 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:

∴Downstream speed = (16 + x) km/hr

Upstream speed = (16 − x) km/hr

For going 60 km:

According to the condition:

⇒ 60 (16 + x) − 60 (16 − x) = 2 (16 − x) (16 + x)

⇒ 960 + 60x − 960 + 60x = 2 (256 − x^{2})

⇒ 2x^{2} + 120x = 2 × 256 −2x^{2}

⇒ x^{2} + 60x = 256

⇒ x^{2} + 60x − 250 = 0 ...(1)

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

a = 1

b = 60

c = − 256

∴ b^{2} − 4ac = (60)^{2} − 4 (1) (− 256)

1 = 3600 + 1024 = 4624

∴

⇒

⇒

Taking +ve sign,

Taking −ve sign,

Since, the speed of a stream cannot be negative,

∴x = − 64 is not admissible

∴ x = 4

⇒ 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

∵ Total distance travelled = 288 km

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

According to the condition,

⇒ 288x + 1152 − 288x = 1 (x) (x + 4)

⇒ 288x + 1152 − 288x = x^{2} + 4x

⇒ 1152 = x^{2} + 4x

⇒ x^{2} + 4x − 1152 = 0 ...(1)

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

a = 1

b = 4

c = − 1152

∴ b^{2} − 4ac = (4)^{2} − 4 (1) (− 1152) = 16 + 4608

⇒

⇒

Taking +ve sign,

Taking −ve sign,

∵ speed cannot be negative,

∴ x ≠− 36

∴ x = 32

⇒ 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. If takes 3 hours to complete the total journey, what is its original average speed?**

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

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

Time taken to cover 63 km = 63/x hours

Since, total time = 3 hours

∴

⇒

⇒

⇒

⇒

⇒ 3[15x + 42] = x^{2} + 6x

⇒ 45x + 126 – x^{2} – 6x = 0

⇒ x^{2} – 39x – 126 = 0

⇒ x^{2 }– 42x + 3x – 126 = 0

⇒ x(x – 42) + 3(x – 42) = 0

⇒ (x + 3) (x – 42) = 0

Either x + 3 = 0 ⇒ x = – 3

or

x – 42 = 0 ⇒ x = 42

Since, speed cannot be negative,

∴ x = –3 is not desired.

Thus, the original speed of the train is 42 km/hr.**Q12. If − 5 is a root of the quadratic equation 2x ^{2} + px − 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 − 5 is a root of 2x^{2} + px − 15 = 0,

∴ Substituting x = − 5 in the given equation, we get

2 (− 5)^{2} + p (− 5) − 15 = 0

⇒ 2 (25) + (− 5p) − 15 = 0

⇒ 50 − 5p − 15 = 0

⇒ − 5p + 35 = 0

⇒ − 5p = − 35

⇒

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} − 4ac =^{ }p^{2} − 4 (p) (k)

= p^{2 }− 4pk

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

∴ p^{2} − 4pk =0

⇒ (7)^{2 }− 4 (7) k = 0 |∵ p = 7

⇒ 49 − 28k = 0

⇒

Thus, the required values of p = 7 and k = 7/4.**Q13. In a class test, the sum of Gagan’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 Gagan’s marks in maths = x and Marks in English = (45 − x)

∴ According to the condition, (x + 1) × (45 − x + 1) = 500

⇒ (x + 1) × (44 − x) = 500

⇒ 44x − x^{2} + 44 − x = 500

⇒− x^{2} + 44x − 456 − x =0

⇒ x^{2} − 43x + 456 = 0

⇒ x^{2} − 19x − 24x + 456 = 0

⇒ x (x − 19) − 24 (x − 19) = 0

⇒ (x − 19) (x − 24) = 0

Either x − 19 = 0 ⇒ x = 19

or x − 24 = 0 ⇒ x = 24

When x = 19, then 45 − 19 = 26

When x = 24, then 45 − 24 = 21

∴ **Gagan’s marks in Maths = 19 and in English = 26**

Or

**Gagan’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.

∴ The perimeter of square I = 4x metres

⇒ Perimeter of square II = (64 + 4x) m

= (16 + x) m

Now Area of square I = x × x = x^{2}

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

= 256 + x^{2} + 32x

According to the condition, (x + 1) × (45 − x + 1) = 500

⇒ x^{2} + [256 + x^{2} + 32x] = 640

⇒ x^{2} + x^{2} + 32x + 256 − 640 = 0

⇒ 2x^{2} + 32x − 384 = 0

⇒ x^{2} + 16x − 192 = 0

⇒ x^{2} + 24x − 8x − 192 = 0

⇒ x (x + 24) − 8 (x + 24) = 0

⇒(x + 24) (x − 8) = 0

Either x + 24 =0

⇒ x = − 24 or x − 8 = 0 ⇒ x = + 8

∵ side of a square cannot be negative,

∴ Rejecting x = − 24, we have x = 8

⇒ Side of smaller square = 8 m

Side of larger square = 8 + 16 m = 24 m.

**Q15. In a class test, the sum of Kamal’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 Kamal’s marks in Maths = x

∴ His marks in English = (40 − x) According to the condition,

(x + 3) [40 − x − 4] = 360

⇒ (x + 3) (36 − x) = 360

⇒ 36x − x^{2} + 108 − 3x − 360 = 0

⇒ − x^{2} + 33x − 252 = 0

⇒ x^{2 }− 33x + 252 = 0

⇒ x^{2} − 21x − 12x + 252 = 0

⇒ x (x − 21) − 12 (x − 21) = 0

⇒(x − 21) (x − 12) = 0

Either (x − 21) = 0

⇒ x = 21 = 0

or (x − 12) = 0

⇒ x = 12

For x = 21, Marks of Kamal

in Maths = 21

in English = 40 − 21 = 19

For x = 12, Marks of Kamal

in Maths = 12

in English = 40 − 12

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

**Sol ^{. }**2x

⇒ 2x^{2} + 2ax – ax – a^{2} =0

⇒ 2x[x + a] – a[x + a]=0

⇒ (x + a) (2x – a) = 0

x = –a or x = a/2

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