Question 1: Find two natural numbers which differ by 3 and the sum of whose squares is 117.

Let the natural numbers be $x$ and $x + 3.$

$x^2 + (x + 3)^2 = 117$

$\Rightarrow x^2+x^2+6x+9 =117$

$\Rightarrow 2x^2+6x-108 -0$

$\Rightarrow x^2+3x-54 = 0$

$\Rightarrow (x+9)(x-6) - 0$

$\Rightarrow x+9= 0, \hspace{0.5cm} \text{ or } \hspace{0.5cm} x-6 = 0$

$\Rightarrow x=-9, \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 6$

Since $-9$ is not a natural number

One number $= 6$ and other number $= 6 + 3 = 9$

Therefore Numbers are $6$ and $9$

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Question 2: Five times a certain whole number is equal to three less than twice the square of the number. Find the number.

Let the number be $x.$

Given, five times the number $= 3$ less than twice the square of the number

$5x = 2x^2 - 3$

$\Rightarrow 2x^2 - 5x - 3 = 0$

$\Rightarrow (x-3)(2x+1) = 0$

$\Rightarrow x-3 = 0 \hspace{0.5cm} \text{ or } \hspace{0.5cm} 2x+1 = 0$

$\displaystyle \Rightarrow x = 3 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = \frac{-1}{2}$

Therefore required whole number is $3.$

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Question 3: Divide 8 into two parts such that the sum of their reciprocals is $\displaystyle \frac{8}{15}.$

Let the two parts be $x$ and $8 - x.$

$\displaystyle \therefore \frac{1}{x} + \frac{1}{8-x} = \frac{8}{15}$

$\displaystyle \Rightarrow \frac{8-x+x}{x(8-x)} = \frac{8}{15}$

$\displaystyle \Rightarrow 120 = 8(8x-x^2)$

$\displaystyle \Rightarrow x^2 - 8x + 15 = 0$

$\displaystyle \Rightarrow x = 5, \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 3$

$\displaystyle x = 5 \Rightarrow \text{ One part } = 5 \text{ and other part } = 8 - 5 = 3$

$\displaystyle x = 3 \Rightarrow \text{ One part } = 3 \text{ and other part } = 8 - 3 = 5$

Therefore Required parts are $3$ and $5.$

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Question 4: For the same amount of work, A takes 6 hours less than B. If together they complete the work in 13 hours 20 minutes; find how much time will B alone take to, I complete the work.

If B alone takes $x$ hours then A alone takes $(x - 6)$ hours for the same work.

$\displaystyle \therefore \frac{1}{x-6} + \frac{1}{x} = \frac{3}{40}$

$\displaystyle \Rightarrow \frac{x+x-6}{x(x-6)} = \frac{3}{40}$

$\Rightarrow 3x^2 - 18x = 80 x - 240$

$3x^2 - 98x +240 = 0$

$3x^2 - 90x - 8 x +240 = 0$

$(x-30)(3x-8) = 0$

$\displaystyle x = 30 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = \frac{8}{3}$

$x = 30$

B along will take 30 hours to complete the work.

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Question 5: The hypotenuse of a right triangle is 13 cm and the difference between the other two sides is 7 cm. Taking x  as the length of the shorter of the two sides, write an equation in x that represents the above statement and also solve the equation to find the two unknown sides of the triangle

Since, the shorter side $= x$ cm.

Therefore Longer side $= (x + 7)$ cm.

Using Pythagoras Theorem, we get :

$x^2+(x+7)^2=13^2$

$\Rightarrow x^2+x^2+14x+49=169$

$\Rightarrow 2x^2+14x-120=0$

$\Rightarrow x^2+7x -60 = 0$

On solving, it gives:

$x=-12, \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 5$

Since the side of triangle cannot be negative, therefore, $x = 5$

Therefore one side of the triangle $= 5$ cm and the other side is $= ( 5+7) = 12$ cm

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Question 6: The length of a verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.
(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement.
(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah.

Since breadth $= x$ m , then  Length $= (x + 3)$m

(i) Given : Area of verandah = its perimeter

$\text{length} \times \text{breadth} = 2 (\text{length + breadth})$

$\Rightarrow (x+3) \times x=2(x +3+x)$

$\Rightarrow x^2 + 3x=4x+6$

$\Rightarrow x^2-x -6 = 0$

(ii) $x^2-x-6 = 0$

$\Rightarrow (x-3)(x+2) = 6$

$\Rightarrow x=3, \hspace{0.5cm} \text{ or } \hspace{0.5cm} x=-2$

Since the breadth cannot be negative, therefore $x = 3$

Hence, the length of verandah $= (x+3) = (3+3) = 6 m$  and its breadth $=x =3 m$

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Question 7: By increasing the speed of a car by 10 km/hr, the time of journey for a distance of 72 km is reduced by 36 minutes. Find the original speed of the car. [ ICSE Board 2005]

Let the original speed of the car $= x$ km/hr

$\displaystyle \text{Time taken by it to cover 72 km } = \frac{72}{x}$ hrs

New speed of the car $= (x+10)$ km/hr

$\displaystyle \text{Therefore new time taken by the car to cover 72 km } = \frac{72}{x+ 10}$ hrs

Given : Time is reduced by 36 minutes :

$\displaystyle \Rightarrow \frac{72}{x} - \frac{72}{x+ 10} = \frac{36}{60}$

$\Rightarrow 60[ 72(x+10) -72x] = 36 x ( x+10)$

$\Rightarrow 60[ 72x + 720 - 72x] = 36 x(x+10)$

$\Rightarrow x^2 + 10 x - 1200 = 0$

$\Rightarrow (x-30)(x+40) = 0$

$\Rightarrow x= 30 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = -40$

Since speed cannot be negative hence the value of $x = 30$

The original speed of the car $= 30$ km/hr

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Question 8: Car A travels x km for every liter of petrol, while car B travels (x + 5) km for every liter of petrol.
(i) Write down the number of liters of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 liters of petrol more than car B in covering the 400 km, write down an equation in x and solve it to determine the number of liters of petrol used by car B for the journey. [ ICSE Board 1997]

$\displaystyle \text{(i) No. of liters of petrol used by car A } = \frac{400}{x} \text{liters }$

$\displaystyle \text{No. of liters of petrol used by car A } = \frac{400}{x+5} \text{liters }$

$\displaystyle \text{(ii) Given: } \frac{400}{x} - \frac{400}{x+5} = 4$

$\displaystyle \Rightarrow \frac{400x + 2000 - 400x}{x(x+5)} = 4$

$\displaystyle \Rightarrow 4 ( x^2 + 5x) = 2000$

$\displaystyle \Rightarrow x^2 + 5x - 500 = 0$

$\displaystyle \Rightarrow x = - 25 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 20$

$\displaystyle \text{Therefore No. of litres of petrol used by car B } = \frac{400}{x+5} = \frac{400}{20 + 5} = 16 \text{ liters }$

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Question 9: By selling an article for Rs. 24, a trader loses as much per cent as the cost price of the article. Calculate the cost price.

Let Cost Price of the article be Rs. $x$

$\displaystyle \text{Therefore Loss } = \frac{x}{100} \times x = \frac{x^2}{100}$

$\displaystyle \Rightarrow x - \frac{x^2}{100} = 24$

$\displaystyle \Rightarrow 100x - x^2 = 24000$

$\displaystyle \Rightarrow x^2 - 100 x + 2400 = 0$

$\displaystyle \Rightarrow (x-60)(x-40) = 0$

$\displaystyle \Rightarrow x = 40 \text{ and } x= 60$

Therefore the cost price of the article is $60$ or $40$

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Question 10: The sum S of first n natural numbers is given by the relation: $S = \frac{1}{2} n ( n+1).$ Find $n$ if the sum is $276.$

Given:  $S = 276$

$\displaystyle \Rightarrow \frac{1}{2} n (n+1) = 276$

$\displaystyle \Rightarrow n^2 + n - 552 = 0$

$\displaystyle \Rightarrow (n+24)(n-23) = 0$

$\displaystyle \Rightarrow n = -24, \hspace{0.5cm} \text{ or } \hspace{0.5cm} n = 23$

Since $n$ is a natural number, $n$ cannot be $-24$

Therefore $n = 23$

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Question 11: A two digit number is such that the product of its digits is 6. When 9 is added to this number; the digits interchange their places. Find the number. [ICSE Board 2014]

Let the required two digit number be $10x + y$

$\text{Given: } xy = 6 \text{ and } 10x + y + 9 = 10y + x$

$\Rightarrow 10x+ y + 9 = 10y +x$

$\Rightarrow 9y =9x + 9$

$\Rightarrow y = x + 1$

Now, $xy = 6$

$\Rightarrow x(x+1)=6$

$\Rightarrow x^2 + x - 6 = 0$

$\Rightarrow (x+3)(x-2) = 0$

$x = -3 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 2$

Since, $-3$ is not a digit, $x = 2$

$y = x+1 = 2+1 = 3$

Therefore The required two digit number $= 10x + y = 10(2) + 3 = 23$

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Question 12: Five years ago, a woman’s age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find :
(i) the age of the son five years ago.
(ii) the present age of the woman [ICSE Board 20007]

Let the age of the son 5 years ago $= x$ years

The woman’s age 5 years ago $= x^2$ years

The present age of the woman $= (x^2+5)$ years

The present age of her son $= (x+5)$ years

The woman’s age will be $= ( x^2 + 5) + 10 = ( x^2 + 15)$ years

Her son’s age will be $= (x+5) + 10 = (x+15)$ years

According to the given statement :

10 years hence, woman’s age = twice her son’s age

$\Rightarrow x^2 + 15 = 2 ( x + 15)$

$\Rightarrow x^2 - 2x - 15 = 0$

$\Rightarrow (x+3)(x-5) = 0$

$\Rightarrow x = -3 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 5$

Age cannot be negative so $x = 5$

(i) The age of the son 5 years ago $= x = 5$ years

(i0 The present age of the woman $= (x^2 + 5) = 25 + 5 = 30$ years

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Question 13: A motor-boat, whose speed is 9 km/h in still water, goes 12 km downstream and comes back in a total time of 3 hours. Find the speed of the stream.

Let the speed of the stream $= x$ km/hr

Speed of the boat down stream $= (9+x)$ km/hr

Speed of the boat upstream $= (9-x)$ km/hr

$\displaystyle \text{Also, time taken to go 12 km downstream } = \frac{12}{9+x}$ hr

$\displaystyle \text{Time taken to come back } = \frac{12}{9-x}$ hr

$\displaystyle \text{ Given: } \frac{12}{9+x} + \frac{12}{9-x} = 3$

$\displaystyle \Rightarrow \frac{108-12x+108+12x}{(9+x)(9-x)} = 3$

$\Rightarrow 3 ( 81-x^2) = 216$

$\Rightarrow 81 - x^2 = 72$

$\Rightarrow x^2 = 9$

$\Rightarrow x = \pm 3$

Since, speed can not be negative. Therefore $x = 3$

Hence, the speed of the stream $= 3$ km/hr

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Question 14: A piece of cloth costs Rs. 200. If the piece was 5 m longer and each meter of cloth costs Rs. 2 less; the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter ?

Let the length of the piece be $x$ meters.

Since the cost of x meters cloth $= \text{ Rs. } 200$

$\displaystyle \text{ Therefore the cost of each meters cloth } = \text{ Rs. } \frac{200}{x}$

New length of cloth $= (x+5)$ m

$\displaystyle \text{ Therefore New cost of each meter of cloth } = \text{ Rs. } \frac{200}{x+5}$

$\displaystyle \text{ Given: } \frac{200}{x} - \frac{200}{x+5} = 2$

$\displaystyle \Rightarrow \frac{200x + 1000 - 200x}{x(x+5)} = 2$

$\Rightarrow 2 ( x^2 + 5x) = 1000$

$\Rightarrow x^2 + 5x = 500$

$\Rightarrow x^2 + 5x - 500 = 0$

$\Rightarrow (x+25)(x-20) = 0$

$\Rightarrow x = -25 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 20$

Since length cannot be negative, therefore $x = 20$

The length of the piece $= 20$ m

$\displaystyle \text{And the original rate per meter } = \frac{200}{20} = \text{ Rs. } 10$

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Question 15: A shopkeeper buys a certain number of boots for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be bought for Rs. 960 would be 4 more. Taking the original cost of each book to be Rs. x,  write and equation in x and solve it. [ICSE Board 2013]

Original cost of each book $= \text{ Rs. } x$

$\displaystyle \text{No. of books bought for} \text{ Rs. } 960 = \frac{960}{x}$

In 2nd Case:

The cost of each book $= \text{ Rs. } (x-8)$

$\displaystyle \text{No. of books bought for} \text{ Rs. } 960 =\frac{960}{x-8}$

$\displaystyle \text{Given: } \frac{960}{x-8} - \frac{960}{x} = 4$

$\displaystyle \Rightarrow \frac{960x- 960x+7680}{x(x-8)} = 4$

$\Rightarrow 7680 = 4 ( 4x^2 - 8x)$

$\Rightarrow x^2 - 8x = 1920$

$\Rightarrow x^2 - 8x - 1920 = 0$

$\Rightarrow (x-48)(x+40) = 0$

$\Rightarrow x = 48 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = -49$

Since cost cannot be negative, therefore $x = 48$

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Question 16: Some students planned a picnic. The budget for the food was Rs. 480. As eight of them failed to join the party, the cost of the food for each member increased by Rs. 10. Find how many students went for the picnic. [ICSE Board 2008]

Let the number of students who planned the picnic $= \text{ Rs. } x$

Since, the budget for food for all $= \text{ Rs. } 480$

$\displaystyle \text{Share of each in it }= \text{ Rs. } \frac{480}{x}$

Given eight of them failed to join the party. Therefore $(x-8)$ students went for the picnic

$\displaystyle \text{So, the share of each will be } = \text{ Rs. } \frac{480}{x-8}$

Since, the cost of food for each member (student) is increased by $\text{ Rs. } 10$

$\displaystyle \frac{480}{x-8} - \frac{480}{x} = 10$

$\displaystyle \Rightarrow \frac{480x - 480x + 3340}{x(x-8)}= 10$

$\displaystyle \Rightarrow 3840 = 10 (x^2 - 8x)$

$\displaystyle \Rightarrow x^2 - 8x - 3840 = 0$

On solving, it gives $x = 24 \text{ and } x = - 16$

But the number of students cannot be negative, therefore $x =24$

The number of student who went for picnic $= x-8 = 24-8 = 16$