# GSEB Solutions for ધોરણ ૧૦ English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

જવાબ : Let us say, the speed of the train be x km/hr. Time taken to cover 360 km = 360/x hr. As per the question given, (x + 5)(360-1/x) = 360 360 – x + 1800-5/x = 360  x+ 5x + 10x – 1800 = 0  x(x + 45) -40(x + 45) = 0 (x + 45)(x – 40) = 0  x = 40, -45 As we know, the value of speed cannot be negative. Therefore, the speed of train is 40 km/h.

Find the roots of the equation x2 – 3x – m (m + 3) = 0, where m is a constant. (2011OD)

જવાબ : x2 – 3x – m(m + 3) = 0
D = b2 – 4ac
D = (- 3)2 – 4(1) [-m(m + 3)]
= 9 + 4m (m + 3)
= 4m2 + 12m + 9 = (2m + 3)2

If 1 is a root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0, then find the value of ab. (2012D)

જવાબ : ay2 + ay + 3 = 0
a(1)2 + a(1) + 3 = 0
2a = -3
a = −32
y2 + y + b = 0
12 + 1 + b = 0
b = -2
ab =(−32)(−2) = 3

If x = – 12 , is a solution of the quadratic equation 3x2 + 2kx – 3 = 0, find the value of k. (2015D)

જવાબ : The given quadratic equation can be written as, 3x2 + 2kx – 3 = 0

If the quadratic equation px\frac{1}{2} – 25–√ px + 15 = 0 has two equal roots, then find the value of p. (2015OD)

જવાબ : The given quadratic equation can be written as px\frac{1}{2} – 25–√ px + 15 = 0
Here a = p, b = – 25–√ p, c = 15
For equal roots, D = 0
D = b2 – 4ac – 0 …[
Equal roots
0 = (-25–√p)2 – 4 × p × 15
0 = 4 × 5p2 – 60p
0 = 20p2 – 60p => 20p2 = 60p
p = 60p20p = 3
p = 3

Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots. (2011OD)

જવાબ : We have, mx (x – 7) + 49 = 0
mx2 – 7mx + 49 = 0
Here, a = m, b = – 7m, c = 49
D = b2 – 4ac = 0 …[For equal roots
(-7m)2 – 4(m) (49) = 0
49m2 – 4m (49) = 0
49m (m – 4) = 0
49m = 0 or m – 4 = 0
m = 0 (rejected) or m = 4
m = 4

Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the squares. (2011D)

જવાબ : Here 4x2 + 3x + 5 = 0
But (2x+34)2 cannot be negative for any real value of x.

Find the value of p so that the quadratic equation px(x – 3) + 9 = 0 has two equal roots. (2011D, 2014OD)

જવાબ : We have, px (x – 3) + 9 = 0
px2 – 3px + 9 = 0 Here a = p, b = -3p,
D = 0
b2 – 4ac = 0
(-3p)2 – 4(p)(9) = 0
9p2 – 36p = 0
9p (p – 4) = 0
9p = 0 or p – 4= 0
p = 0 (rejected) or p = 4

p = 4 ……..( Coeff. of x2 cannot be zero

Solve for x:
36x2 – 12ax + (a2 – b2) = 0 (2011OD)

જવાબ : We have, 36x2 – 12ax + (a2 – b2) = 0
(36x2 – 12ax + a2) – b2 = 0
[(6x)2 – 2(6x)(a) + (a)2] – b2 = 0
(6x – a)2 – (b)2 = 0 …[ x2 – 2xy + y2 = (x – y)2
(6x – a + b) (6x – a – b) = 0 „[ x2 – y2 = (x + y)(x – y)
6x – a + b = 0 or 6x – a – b = 0
6x = a – b or 6x = a + b
x = a−b6 or a+b6

Find the value of p for which the roots of the equation px(x – 2) + 6 = 0, are equal. (2012OD)

જવાબ : We have , px(x – 2) + 6 = 0
px2 – 2px + 6 = 0, p ≠ 0
Two equal roots …[Given
b2 – 4ac = 0 ….[a = p, b = -2p, c = 6
(-2p)2 – 4(p)(6) = 0
4p2 – 24p = 0 4p(p – 6) = 0
4p = 0 or p – 6 = 0
p = 0 (rejected) or p = 6
Since p cannot be equal to 0.
…[Standard form of a quad. eq. ax2 + bx + c = 0, a ≠ 0
P = 6

Find the value(s) of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots. (2012D)

જવાબ : We have, x2 – 4kx + k = 0
Here a = 1, b = -4k:, c = k D = 0 …[Since, Equal roots
As b2 – 4ac = 0
(-4k)2 – 4(1) (k) = 0
16k2 – 4k = 0 4k(4k – 1) = 0
4k = 0 or 4k – 1 = 0
k = 0 (rejected) or 4k = 1
k = ¼

Solve the following quadratic equation for x: 43–√x2 + 5x – 23–√ = 0 (2013D)

જવાબ : 43–√x2 + 5x – 23–√ = 0
43–√x2 + 8x – 3x – 23–√ =0
4x(3–√x + 2) – 3–√(3–√x + 2) = 0
(3–√x + 2)(4x – 3–√) = 0
3–√x + 2 = 0 or 4x – 3–√ = 0

Solve the following for x: 2–√x2 + 7x + 52–√ = 0 (2017D )

જવાબ : 2–√x2 + 7x + 52–√ = 0
2–√x2 + 5x + 2x + 52–√ = 0
x(2–√x + 5) + 2–√(2–√x + 5) = 0
(2–√x + 5)(x + 2–√) = 0

Solve the quadratic equation 2x2 + ax – a2 = 0 for x. (2014D)

જવાબ : We have, 2x2 + ax – a2 = 0
2x2 + 2ax – ax – a2 = 0
2x(x + a) – a(x + a) = 0
(x + a) (2x – a) = 0
x + a = 0 or 2x – a = 0

x = -a or x = a2
Alternatively:
First calculate D = b2 – 4ac
Then apply x = −b±D√2a
We get x = -a, x = a/2

Find the value of k for which the equation x2 + k(2x + k – 1) + 2 = 0 has real and equal roots. (2017D)

જવાબ : We have, x2 + k(2x + k – 1) + 2 = 0
x2 + 2kx + k2 – k + 2 = 0
Here a = 1, b = 2k, c = k2 – k + 2
D = 0 …[real and equal roots

b2 – 4ac = 0
(2k)2 – 4 × 1(k2 – k + 2) = 0
4k2 – 4 (k2 – k + 2) = 0
4(k2 – k2 + k – 2) = 0 4(k – 2) = 0
k – 2 = 0 k = 2

Solve the following quadratic equation for x: 4x2– 4a2x + (a4 – b4) = 0. (2015D)

જવાબ : The given quadratic equation can be written as,
4x2 – 4a2x + (a44 – b4) = 0
(4x2 – 4a2x + a4) – b4 = 0
or (2x – a2)2 – (b2)2 = 0
(2x – a2 + b2) (2x – a2 – b2) = 0
(2x – a2 + b2) = 0 or (2x – a2 – b2) = 0
x = a2−b22 or x = a2+b22

Find the values of p for which the quadratic equation 4x2 + px + 3 = 0 has equal roots. (2014OD)

જવાબ : Given: 4x2 + px + 3 = 0
Here a = 4, b = p. (= 3 … [Equal roots
D = 0 (Equal roots)
As b2 – 4ac = 0
(p)2 – 4(4)(3) = 0
= p2 – 48 = 0 p2 = 48
p = ±16×3−−−−−√=±43–√

Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0) (2015OD)

જવાબ : Given quadratic equation can be written as
x2 – 2ax – 4b2 + a2 = 0.
(x2 – 2ax + a2) – 4b2 = 0 or (x – a)2 – (2b)2 = 0
As we know,
[a2 – b2 = (a + b)(a – b)]

(x – a + 2b) (x – a – 2b) = 0
x – a + 2b = 0 or x – a – 2b = 0
x = a – 2b or x = a + 2b
x = a – 2b and x = a + 2b

Solve the following quadratic equation for x: 4x2 + 4bx – (a2 – b2) = 0 (20150D)

જવાબ : The given quadratic equation can be written as
4x2 + 4bx + b2 – a22 = 0

(2x + b)2 – (a)2 = 0
(2x + b + a) (2x + b – a) = 0 …[x2 – y2 = (x + y)(x – y)
(2x + b + a) = 0 or (2x + b – a) = 0
2x = -(a + b) or 2x = (a – b)

Solve the following quadratic equation for x: 9x2 – 6b2x – (a4 – b4) = 0 (2015D)

જવાબ : The given quadratic equation can be written as
(9x2 – 6b2x + b4) – a4 = 0

(3x – b2)2 – (a2)2 = 0
(3x – b2 + a2) (3x – b2 – a2) = 0 …[:: x2 – y2 = (x + y) (x – y)
3x – b2 + a2 = 0 or 3x – b2 – a2 = 0
3x = b2 – a2 or 3x = b2 + a2

If x = 23 and x = -3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b. (2016D)

જવાબ : We have, ax2 + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
Now, α = 23 and β = -3 … [Given

Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other. (20170D)

જવાબ : Given equation is px2 – 14x + 8 = 0.
Here a = p b = -14 c = 8
Let roots be a and 6α.

If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k. (2016OD)

જવાબ : We have, 2x2 + px – 15 =0
Since (-5) is a root of the given equation

2(-5)2 + p(-5) – 15 = 0
2(25) – 5p – 15 = 0
50 – 15 = 5p
35 = 5p p = 7 …(i)
Now, p(x
2 + x) + k px2 + px + k = 0
7x2 + 7x + k = 0 …[From (i)
Here, a = 7, b = 7, c = k
D = 0 …[Roots are equal
b2 – 4ac = 0

(7)2 – 4(7)k = 0 49 – 28k = 0
49 = 28k k = 4928=74

Solve for x:

જવાબ :

Solve for x:√3x– 2√2x-2√3=0

જવાબ :

A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.

જવાબ :

Solve for x:  + x= 13

જવાબ :

If- 5 is a root of the quadratic equation 2x2 + px -15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.

જવાબ :

If x= 2/3 and x = – 3 are roots of the quadratic equations ax2 + lx + b = 0, find the values of a and b.

જવાબ :

If- 5 is a root of the quadratic equation 2x2 + px -15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.

જવાબ :

A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.

જવાબ :

Solve for x:√3x– 2√2x-2√3=0

જવાબ :

Solve for x:

જવાબ :

જવાબ :

જવાબ :

If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

જવાબ :

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.

જવાબ :

If the quadratic equation px2 – 2√5px + 15 = 0 has two equal roots, then find the value of p.

જવાબ :

If x =-1/2 , is a solution of the quadratic equation 3x2 + 2kx -3 = 0, find the value of K .

જવાબ :

Solve the following quadratic equation for x: 4x2 – 4a2 x + (a4 – b4 ) = 0

જવાબ :

Solve the following quadratic equation for x: 9x2 – 6b2 x – (a4 – b4 ) = 0

જવાબ :

Solve the following quadratic equation for x: 4x2 + 4bx – (a2 – b2) = 0

જવાબ :

Solve the following quadratic equation for x: x2 – 2ax – (4b2 -a2) = 0

જવાબ :

Solve for x: √3x– 2√2x-2√3=0

જવાબ :

Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2 (k – 1)x + x2= 0 has equal roots. Hence, find the roots of the equation.

જવાબ :

Find the value of p for which the quadratic equation (p + l)x2 – 6(p + 1)x + 3(p + 9) = 0,p ≠ 1 has equal roots. Hence, find the roots of the equation.

જવાબ :

Solve for x  .x2 -(√3+1)x+√3=0

જવાબ :

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

જવાબ : Let us say, the marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

As per the given question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

-x2 + 25x + 54 = 210

x2 – 25x + 156 = 0

x2 – 12x – 13x + 156 = 0

x(x – 12) -13(x – 12) = 0

(x – 12)(x – 13) = 0

x = 12, 13

Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18 and the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

જવાબ : Let us say, present age of Rahman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

3(2x + 2) = (x-3)(x+5)

6x + 6 = x2 + 2x – 15

x2 – 4x – 21 = 0

x2 – 7x + 3x – 21 = 0

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7, -3

As we know, age cannot be negative.

Therefore, Rahman’s present age is 7 years.

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

જવાબ : Let us say, the number of articles produced be x.

Therefore, cost of production of each article = Rs (2x + 3)

Given, total cost of production is Rs.90

x(2x + 3) = 90

2x+ 3x – 90 = 0

2x+ 15x -12x – 90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x – 6 = 0

x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15.

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

જવાબ : Let us say, the base of the right triangle be x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base2 + Altitude2 = Hypotenuse2

x+ (x – 7)2 = 132

x+ x+ 49 – 14x = 169

2x– 14x – 120 = 0

x– 7x – 60 = 0

x– 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

Find two consecutive positive integers, the sum of whose squares is 365.

જવાબ : Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given questions,

x2 + (x + 1)2 = 365

xx+ 1 + 2x = 365

2x2 + 2x – 364 = 0

x– 182 = 0

x+ 14x – 13x – 182 = 0

x(x + 14) -13(x + 14) = 0

(x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Find two numbers whose sum is 27 and product is 182.

જવાબ : Let us say, first number be x and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

x2 – 27x – 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

A motor boat whose speed is 20 km/h in still water, takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.

જવાબ : Let the speed of the stream = km/h
Speed of the boat in still water = 20 km/h
Now, speed of boat during downstream = (20 + x) km/h

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.

જવાબ :

Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.

જવાબ :

Solve for x:

જવાબ :

Solve for x:

જવાબ :

Solve for x :

જવાબ :

A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

જવાબ :

A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the toal distance in 5 hours, find the first speed of the truck.

જવાબ :

The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original  fraction is 29/20. Find the original fraction.

જવાબ :

Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

જવાબ : Let the length and breadth of the park be and b.

Perimeter of the rectangular park = 2 (l + b) = 80

So, l + b = 40

Or, b = 40 – l

Area of the rectangular park = l×b = l(40 – l) = 40– l= 400

l2   40l + 400 = 0, which is a quadratic equation.

Comparing the equation with ax2 + bx c = 0, we get

a = 1, b = -40, c = 400

Since, Discriminant = b2 – 4ac

=(-40)2 – 4 × 400

= 1600 – 1600 = 0

Thus, b2 – 4ac = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

Root of the equation,

l = –b/2a

l = (40)/2(1) = 40/2 = 20

Therefore, length of rectangular park, = 20 m

And breadth of the park, = 40 – = 40 – 20 = 20 m.

Solve for x:

જવાબ :

Two pipes running together can fill a tank in 11. 1/9 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

જવાબ :

The time taken by a person to cover 150 km was 2. 1/2 hours more than the time taken in the return journey. If he returned at a speed of 10 km/hour more than the speed while going, find the speed per hour in each direction.

જવાબ :

Find x in terms of a, b and c:

જવાબ :

A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for treatment. Due to this, the plane got delayed by half an hour. To reach the destination 1500 km away in time, so that the passengers could catch the connecting flight, the speed of the plane was increased by 250 km/hour than the usual speed. Find the usual speed of the plane. What value is depicted in this question?

જવાબ :

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

જવાબ : Let’s say, the age of one friend be x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of First friend = (x – 4) years

Age of Second friend = (20 – x – 4) = (16 – x) years

As per the given question, we can write,

(x – 4) (16 – x) = 48

16x – x2 – 64 + 4x = 48

– x2 + 20x – 112 = 0

x2 – 20x + 112 = 0

Comparing the equation with ax2 + bx c = 0, we get

a = 1b = -20 and c = 112

Discriminant = b2 – 4ac

= (-20)2 – 4 × 112

= 400 – 448 = -48

b2 – 4ac < 0

Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

જવાબ : Let the breadth of mango grove be l.

Length of mango grove will be 2l.

Area of mango grove = (2l) (l)= 2l2

2l= 800

l= 800/2 = 400

l– 400 =0

Comparing the given equation with ax2 + bx c = 0, we get

a = 1, b = 0, c = 400

As we know, Discriminant = b2 – 4ac

=> (0)2 – 4 × (1) × ( – 400) = 1600

Here, b2 – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

= ±20

As we know, the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

જવાબ : (i) 2x2 + kx + 3 = 0

Comparing the given equation with ax2 + bx c = 0, we get,

a = 2, b = k and c = 3

As we know, Discriminant = b2 – 4ac

= (k)2 – 4(2) (3)

k2 – 24

For equal roots, we know,

Discriminant = 0

k2 – 24 = 0

k2 = 24

k = ±√24 = ±2√6

(ii) kx(x – 2) + 6 = 0

or kx2 – 2kx + 6 = 0

Comparing the given equation with ax2 + bx c = 0, we get

a = kb = – 2k and c = 6

We know, Discriminant = b2 – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For equal roots, we know,

b2 – 4ac = 0

4k2 – 24k = 0

4k (k – 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘.

Therefore, if this equation has two equal roots, k should be 6 only.

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

જવાબ : Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x2 and y2 respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6

Also, x+ y2 = 468

(6 + y2) + y2 = 468

36 + y2 + 12y + y2 = 468

2y2 + 12y + 432 = 0

y2 + 6y – 216 = 0

y2 + 18y – 12y – 216 = 0

y(+18) -12(y + 18) = 0

(y + 18)(y – 12) = 0

y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

જવાબ : Let us say, the average speed of passenger train =  x km/h.

Average speed of express train = (x + 11) km/h

Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

132 × 11 = x(x + 11)

x2 + 11x – 1452 = 0

x2 +  44x -33x -1452 = 0

x(x + 44) -33(x + 44) = 0

(x + 44)(x – 33) = 0

x = – 44, 33

As we know, Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Two water taps together can fill a tank in
hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

જવાબ : Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/(– 10)

As given, the tank can be filled in
= 75/8 hours by both the pipes together.

Therefore,

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

2x-10/x(x-10) = 8/75

75(2x – 10) = 8x2 – 80x

150x – 750 = 8x2 – 80x

8x2 – 230x +750 = 0

8x2 – 200x – 30x + 750 = 0

8x(x – 25) -30(x – 25) = 0

(x – 25)(8x -30) = 0

x = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

જવાબ : Let us say, the larger and smaller number be x and y respectively.

As per the question given,

x– y2 = 180 and y2 = 8x

x– 8x = 180

x– 8x – 180 = 0

x– 18x + 10x – 180 = 0

x(x – 18) +10(x – 18) = 0

(x – 18)(x + 10) = 0

x = 18, -10

However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be 18 only.

x = 18

y2 = 8x = 8 × 18 = 144

y = ±√144 = ±12

Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

જવાબ : Let us say, the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

x2 + (x + 30)2 = (x + 60)2

x2 + x2 + 900 + 60x = x2 + 3600 + 120x

x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0

x(– 90) + 30(x -90) = 0

(– 90)(x + 30) = 0

= 90, -30

However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.

and the length of the larger side will be (90 + 30) m = 120 m.