Exercise 5.5
Question 1
(i) Find two consecutive natural numbers such that the sum of their squares
is 61.
(ii) Find two consecutive integers such that the sum of their squares is
61.
Sol :
(i)
Let the first natural number = x
then second natural number = x + 1
According to the condition, (x)² + (x + 1)² = 61
$\Rightarrow x^{2}+x^{2}+2 x+1-61=0$
$\Rightarrow 2 x^{2}+2 x-60=0$
$\Rightarrow x^{2}+x-30=0 $
$\Rightarrow x^{2}+6 x-5 x-30=0$
⇒x(x+6)-5(x+6)=0
⇒(x+6)(x-5)=0
Either x+6=0, then x=-6 or x-5=0, then x=5
∵The numbers are positive
∴x=-6 is not possible
Hence the first natural numbers=5
and second natural number=5+1=6
(ii)
Let first integer=x
Then second integer =x+1
According to the condition, $(x)^{2}+(x+1)^{2}=61$
$\Rightarrow x^{2}+x^{2}+2 x+1=61 $
$\Rightarrow 2 x^{2}+2 x+1-61=0$
$\Rightarrow 2 x^{2}+2 x-60=0 $
$\Rightarrow x^{2}+x-30=0$ (dividing by 2)
$\Rightarrow x^{2}+6 x-5 x-30=0$
⇒x(x+6)-5(x+6)=0
⇒(x+6)(x-5)=0
Either x+6=0 then x=-6 or x-5=0 then x=5
(i) If x=-6 then
First integer =-6 and
second =-6+1=-5
(ii) If x=5, then First integer =5
and second =5+1=6
∴Required integers are(-6,-5) ,(5,6)
Question 2
(i) If the product of two positive consecutive even integers is 288, find
the integers.
(ii) If the product of two consecutive even integers is 224, find the
integers.
(iii) Find two consecutive even natural numbers such that the sum of their
squares is 340.
(iv) Find two consecutive odd integers such that the sum of their squares
is 394.
Sol :
(i) Let first positive even integer = 2x
then second even integer = 2x + 2
According to the condition,
2x × (2x + 2) = 288
⇒ 4x² + 4x – 288 = 0
⇒ x² + x – 72 = 0 (Dividing by 4)
$\Rightarrow x^{2}+9x-8x-72=0$
⇒x(x+9)-8(x+9)=0
Either x+9=0, then x=-9
But it is not possible as it is negative or x-8=0, then x=8
∴ First even integer =2x=2×8=16
and second even integer=16+2=18
(ii) Let first even integer =2 x then
second even integer =2 x+2
According to the condition.
⇒2x×(2 x+2)=224
$\Rightarrow 4 x^{2}+4 x-224=0$
$\Rightarrow x^{2}+x-56=0$
$ \Rightarrow x^{2}+8 x-7 x-56=0$
⇒x(x+8)-7(x+8)=0
⇒(x+8)(x-7)=0
Either x+8=0, then x=-8
∴ First even integer =2×(-8)=-16 and
second even integer =-16+2=-14 or x-7=0, then x=7
∴ First even integer =2x=2×7=14
and second even integer =14+2=16
(iii) Let first even natural number=2 x
Then second number =2 x+2
According to the condition, $(2 x)^{2}+(2 x+2)^{2}=340$
$4 x^{2}+4 x^{2}+8 x+4=340$
$\Rightarrow 8 x^{2}+8 x+4-340=0 \Rightarrow 8 x^{2}+8 x-336=0$
$\Rightarrow x^{2}+x-42=0$ (dividing by 8)
$\Rightarrow x^{2}+7 x-6 x-42=0$
⇒x(x+7)-6(x+7)=0
⇒(x+7)(x-6)=0
Either x+7=0, then x=-7 But it is not a even natural number
or x-6=0, then x=6
∴ First even natural number =2x=2×6=12
and second =12+2=14
∴ Numbers are 12 ,14
(iv) Let first odd integer =2 x+1
Then second odd integer =2 x+3
According to the condition,
$(2 x+1)^{2}+(2 x+3)^{2}=394$
$\Rightarrow 4 x^{2}+4 x+1+4 x^{2}+12 x+9=394$
$\Rightarrow 8 x^{2}+16 x-394+10=0$
$\Rightarrow 8 x^{2}+16 x-384=0$
$\Rightarrow x^{2}+2 x-48=0$ [dividing by 8]
$\Rightarrow x^{2}+8 x-6 x-48=0$
⇒x(x+8)-6(x+8)=0
⇒(x+8)(x-6)=0
Either x+8=0, then x=-8
or x-6=0, then x=6
(i) If x=-8, then first odd integer =2 x+1
=2×(-8)+1=-16+1=-15
and second integer =-15+2=-13
(ii) If x=6, then first odd integer =2x+1
=2×6+1=13
and second integer =13+2=15
Required integers are -15,-13, or 13,15
Question 3
The sum of two numbers is 9 and the sum of their squares is 41. Taking one
number as x, form ail equation in x and solve it to find the numbers.
Sol :
Sum of two numbers = 9
Let first number = x
then second number = 9 – x
Now according to the condition,
$(x)^{2}+(9-x)^{2}=41$
$\Rightarrow x^{2}+81-18 x+x^{2}-41=0$
$ \Rightarrow 2 x^{2}-18 x+40=0$
$\Rightarrow x^{2}-9 x+20=0$ (dividing by 2)
$\Rightarrow x^{2}-4 x-5 x+20=0$
⇒x(x-4)-5(x-4)=0
⇒(x-4)(x-5)=0
Either x-4=0 then x=4 or x-5=0 then x=5
(i) If x=4, then first number =4
and second number =9-4=5
(ii) If x=5, then first number =5
and second number =9-5=4
Hence numbers are 4 and 5
Question 4
Five times a certain whole number is equal to three less than twice the
square of the number. Find the number.
Sol :
Let number = x
Now according to the condition,
5x = 2x² – 3
$\Rightarrow 2 x^{2}-5 x-3=0$
$ \Rightarrow 2 x^{2}-6 x+x-3=0$
⇒2x(x-3)+1(x-3)=0
⇒(x-3)(2 x+1)=0
Either x-3=0, then x=3
or 2x+1=0, then 2 x=-1
$\Rightarrow x=-\frac{1}{2}$
But it is not possible as the number is whole number.
∴ Number =3
Question 5
Sum of two natural numbers is 8 and the difference of their reciprocal is
2/15. Find the numbers.
Sol :
Let x and y be two numbers
Given that, x + y = 8 ……(i)
and $\frac{1}{x}-\frac{1}{y}=\frac{2}{15}$ ...(ii)
From equation (i), we have, y=8-x
Substituting the value of y in equation (ii), we have,
$\frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$
$\frac{8-x-x}{x(8-x)}=\frac{2}{15}$
$\frac{8-2 x}{x(8-x)}=\frac{2}{15}$
$\frac{4-x}{x(8-x)}=\frac{1}{15}$
15(4-x)=x(8-x)
$\Rightarrow 60-15 x=8 x-x^{2}$
$\Rightarrow x^{2}-15 x-8 x+60=0$
$ \Rightarrow x^{2}-23 x+60=0$
$\Rightarrow x^{2}-20 x-3 x+60=0$
⇒x(x-20)-3(x-20)=0
⇒(x-3)(x-20)=0
Either (x-3)=0 or (x-20)=0
⇒x=3 or x=20
Since sum of two natural numbers is 8-x
i.e. 8-20 cannot be equal to 20
Thus, x=3
From equation (i) y=8-x=8-3=5
Thus the values of x and y are 3 and 5 respectively
Question 6
The difference between the squares of two numbers is 45. The square of the
smaller number is 4 times the larger number. Determine the numbers.
Sol :
Let the larger number = x
then smaller number = y
Now according to the condition,
$x^{2}-y^{2}=45$..(i)
$y^{2}=4 x$..(ii)
Substituting the value of $y^{2}$ from (ii) in (i)
$x^{2}-4 x=45$
$ \Rightarrow x^{2}-4 x-45=0$
$\Rightarrow x^{2}-9 x+5 x-45=0$
⇒x(x-9)+5(x-9)=0
⇒(x-9)(x+5)=0
Either x-9=0, their x=9 or x+5 , then x=-5
(i) When x=9 the largest number=9
and smallest number
$y=\sqrt{4 x}=\sqrt{4 \times 9}=\sqrt{36}$
∴y=6
(ii) When x=-5 , then larger number=-5
$y=\sqrt{4 x}=\sqrt{4 \times 5}=\sqrt{-20}$
which is not possible
Hence, numbers are 6,9
Question 7
There are three consecutive positive integers such that the sum of the
square of the first and the product of other two is 154. What are the
integers?
Sol :
Let the first integer = x
then second integer = x + 1
and third integer = x + 2
Now according to the condition,
$x^{2}+(x+1)(x+2)=154$
$\Rightarrow x^{2}+x^{2}+3 x+2-154=0$
$ \Rightarrow 2 x^{2}+3 x-152=0$
$\Rightarrow 2 x^{2}+19 x-16 x-152=0$
⇒x(2x+19)-8(2 x+19)=0
⇒(2x+19)(x-8)=0
Either 2x+19=0, then 2x=-19
$\Rightarrow x=-\frac{19}{2}$
But it is not possible as it is not an positive integer.
x-8=0, then x=8
∴ Numbers are 8,(8+1)=9 and (8+2)=10
Question 8
(i) Find three successive even natural numbers, the sum of whose squares is
308.
(ii) Find three consecutive odd integers, the sum of whose squares is
83.
Sol :
(i) Let first even number = 2x
second even number = 2x + 2
third even number = 2x + 4
Now according to the condition,
$(2 x)^{2}+(2 x+2)^{2}+(2 x+4)^{2}=308$
$\Rightarrow 4 x^{2}+4 x^{2}+8 x+4+4 x^{2}+16 x+16=308$
$\Rightarrow 12 x^{2}+24 x+20-308=0 $
$\Rightarrow 12 x^{2}+24 x-288=0$
$\Rightarrow x^{2}+2 x-24=0$
$\Rightarrow x^{2}+6 x-4 x-24=0$
⇒x(x+6)-4(x+6)=0
⇒(x+6)(x-4)=0
Either x+6=0, then x=-6
But it is not a natural number, hence not possible.
or x-4=0, then x=4
∴First even natural numbers=2x=2×4=8
second number =8+2=10
and the third number =10+2=12
(ii) Let the three numbers be x, x+2, x+4
According to statement $(x)^{2}+(x+2)^{2}+(x+4)^{2}=83$
$\Rightarrow x^{2}+x^{2}+4 x+4+x^{2}+8 x+16=83$
$\Rightarrow 3 x^{2}+12 x+20=83 \Rightarrow 3 x^{2}+12 x+20-83=0$
$\Rightarrow 3 x^{2}+12 x-63=0 \Rightarrow x^{2}+4 x-21=0$
$\Rightarrow x^{2}+7 x-3 x-21=0 \Rightarrow x(x+7)-3(x+7)=0$
⇒(x-3)(x+7)=0
Either x-3=0 then x=3 or x+7=0 then x=-7
∴ Numbers will be 3,3+2,3+4=3,5,7 Ans.
or Numbers will be -7,-7+2,-7+4=-7,-5,-3 Ans.
Question 9
In a certain positive fraction, the denominator is greater than the
numerator by 3. If 1 is subtracted from both the numerator and denominator,
the fraction is decreased by $ \frac{1}{14}$. Find the fraction.
Sol :
Let the numerator of a fraction = x
then denominator = x + 3
then fraction $=\frac{\not 2}{x+3}$
Now according to the condition,
New fraction $\frac{x-1}{x+3-1}=\frac{x}{x+3}-\frac{1}{14}$
$\Rightarrow \frac{x-1}{x+2}=\frac{14 x-x-3}{14(x+3)}$
$ \Rightarrow \frac{x-1}{x+2}=\frac{13 x-3}{14 x+42}$
$\Rightarrow(x-1)(14 x+42)=(13 x-3)(x+2)$
$\Rightarrow 14 x^{2}+42 x-14 x-42=13 x^{2}+26 x-3 x-6$
$\Rightarrow 14 x^{2}+28 x-42-13 x^{2}-23 x+6=0$
$\Rightarrow x^{2}+5 x-36=0 \Rightarrow x^{2}+9 x-4 x-36=0$
⇒x(x+9)-4(x+9)=0
⇒(x+9)(x-4)=0
Either x+9=0, then x=-9, but it is not possible as the fraction is positive.
or x-4=0, then x=4
∴ Fraction $=\frac{x}{x+3}=\frac{4}{4+3}=\frac{4}{7}$
Question 10
The sum of the numerator and denominator of a certain positive fraction is
8. If 2 is added to both the numerator and denominator, the fraction is
increased by $\frac{4}{35}$. Find the fraction.
Sol :
Let the denominator of a positive fraction = x
then numerator = 8 – x
∴ Fraction $=\frac{8-x}{x}$
According to the condition.
$\frac{8-x+2}{x+2}=\frac{8-x}{x}+\frac{4}{35}$
$\Rightarrow \frac{10-x}{x+2}=\frac{8-x}{x}+\frac{4}{35}$
$\Rightarrow \frac{10-x}{x+2}=\frac{8-x}{x}+\frac{4}{35}$
$\Rightarrow \frac{10-x}{x+2}-\frac{8-x}{x}=\frac{4}{35}$
$\Rightarrow \frac{10 x-x^{2}-8 x+x^{2}-16+2 x}{x(x+2)}=\frac{4}{35}$
$\Rightarrow \frac{4 x-16}{x^{2}+2 x}=\frac{4}{35}$
$\Rightarrow 4 x^{2}+8 x=140 x-560$
$\Rightarrow 4 x^{2}+8 x-140 x+560=0$
$\Rightarrow 4 x^{2}-132 x+560=0$
$\Rightarrow x^{2}-33 x+140=0$
$\Rightarrow x^{2}-28 x-5 x+140=0$
⇒x(x-28)-5(x-28)=0
⇒(x-28)(x-5)=0
Either x-28=0, then x=28, but it is not possible as sum of numerator and
denominator is 8
x-5=0 then x=5
∴ Fraction $=\frac{8-x}{x}=\frac{8-5}{5}=\frac{3}{5}$
Question 11
A two digit number contains the bigger at ten’s place. The product of the
digits is 27 and the difference between two digits is 6. Find the number.
Sol :
Let unit’s digit = x
then tens digit = x + 6
Number = x + 10(x + 6)
= x + 10x + 60
= 11x + 60
According to the condition,
x(x+6)=27
$\Rightarrow x^{2}+6 x-27=0$
$\Rightarrow x^{2}+9 x-3 x-27=0$
⇒x(x+9)-3(x+9)=0
⇒(x+9)(x-3)=0
Either x+9=0, then x=-9, but it is not possible as it is negative.
or x-3=0, then x=3
∴Number=11x+60
=11×3+60
=33+60=93
Question 12
A two digit positive number is such that the product of its digits is 6. If
9 is added to the number, the digits interchange their places. Find the
number. (2014)
Sol :
Let 2-digit number = xy = 10x + y
Reversed digits = yx = 10y + x
According to question,
x y=6
$y=\frac{6}{x}$...(i)
and 10x+y+9=10y+x
$\Rightarrow 10 x+\frac{6}{x}+9=10 \times \frac{6}{x}+x\left(\right.$ From
$\left.(i) y=\frac{6}{x}\right)$
$\Rightarrow 10 x^{2}+6+9 x=60+x^{2}$
$\Rightarrow 10 x^{2}-x^{2}+9 x+6-60=0$
$\Rightarrow 9 x^{2}+9 x-54=0$
$\Rightarrow x^{2}+x-6=0$
$\Rightarrow x^{2}+3 x-2 x-6=0$
⇒ x(x+3)-2(x+3)=0
⇒(x-2)(x+3)=0
⇒x=2 or -3 (rejecting -3)
putting the value of x in (i)
$y=\frac{6}{2}=3$
∴2-digit=10x+y=10×2+3=23
Question 13
A rectangle of area 105 cm² has its length equal to x cm. Write down its
breadth in terms of x. Given that the perimeter is 44 cm, write down an
equation in x and solve it to determine the dimensions of the rectangle.
Sol :
Perimeter of rectangle = 44 cm
length + breadth $=\frac{44}{2}=22 \mathrm{~cm}$
Let length = x
then breadth = 22 – x
According to the condition,
x(22-x)=105
$\Rightarrow 22 x-x^{2}=105$
$\Rightarrow \quad x^{2}-22 x+105=0$
$\Rightarrow \quad x^{2}-15 x-7 x+105=0$
⇒x(x-15)-7(x-15)=0
⇒(x-15)(x-7)=0
Either x-15=0, then x=15
or x-7=0, then x=7
As length > breadth, x=7 is not admissible.
∴Length=15 cm
and breadth=22-15=7 cm
Question 14
A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of
uniform width. Given that the area of the walk is 120 square metres,
assuming the width of the walk to be x, form an equation in x and solve it
to find the value of x. (1992)
Sol :
Length of garden = 16 m
and width = 10 m
Let the width of walk = x m
Outer length = 16 + 2x
and outer width = 10 + 2x
Now according to the condition,
(16+2 x)(10+2 x)-16 ×10=120
$\Rightarrow \quad 160+32 x+20 x+4 x^{2}-160=120$
$\Rightarrow \quad 4 x^{2}+52 x-120=0$
$\Rightarrow \quad x^{2}+13 x-30=0 \quad$ (Dividing by $\left.4\right)$
$\Rightarrow \quad x^{2}+15 x-2 x-30=0$
⇒x(x+15)-2(x+15)=0
⇒(x+15)(x-2)=0
Either x+15=0, then x=-15
But it is not possible.
or x-2=0 then x=2
Question 15
(i) Harish made a rectangular garden, with its length 5 metres more than
its width. The next year, he increased the length by 3 metres and decreased
the width by 2 metres. If the area of the second garden was 119 sq m, was
the second garden larger or smaller ?
(ii) The length of a rectangle exceeds its breadth by 5 m. If the breadth
were doubled and the length reduced by 9 m, the area of the rectangle would
have increased by 140 m². Find its dimensions.
Sol :
In first case,
Let length of the garden = x m
then width = (x – 5) m
Area = l × b = x(x – 5) sq. m
In second case,
Length =(x+3) m
and width =x-5-2=(x-7) m
According to the condition,
(x+3)(x-7)=119
$\Rightarrow \quad x^{2}-7 x+3 x-21=119$
$\Rightarrow \quad x^{2}-4 x-21-119=0$
$\Rightarrow \quad x^{2}-4 x-140=0$
$\Rightarrow \quad x^{2}-14 x+10 x-140=0$
⇒x(x-14)+10(x-14)=0
⇒(x-14)(x+10)=0
Either x-14=0, then x=14
or x+10=0, then x=-10, but it is not possible as it is negative.
∴ Length of first garden =14 m
and width =14-5=9 m
Area $=l \times b=14 \times 9=126 \mathrm{~m}^{2}$
Difference of areas of two rectangles
=126-119=7 sq. m
∴ Area of second garden is smaller than the area of the first
garden by 7 sq. m
Let length of the rectangle =x m
then width =(x-5) m
Area =x(x-5)sq. m
In second case,
Length of the second rectangle =x-9
and width =2(x-5) m
Area =(x-9) 2(x-5)=2(x-9)(x-5) sq. m
According to the condition,
2(x-9)(x-5)=x(x-5)+140
$\Rightarrow \quad 2\left(x^{2}-14 x+45\right)=x^{2}-5 x+140$
$\Rightarrow \quad 2 x^{2}-28 x+90=x^{2}-5 x+140$
$\Rightarrow \quad 2 x^{2}-28 x+90-x^{2}+5 x-140=0$
$\Rightarrow \quad x^{2}-23 x-50=0$
$\Rightarrow \quad x^{2}-25 x+2 x-50=0$
⇒x(x-25)+2(x-25)=0
⇒(x-25)(x+2)=0
Either x-25=0, then x=25
or x+2=0, then x=-2, but it is not possible as it is negative.
∴ Length of the rectangle =25 m
and width =25-5=20 m
Question 16
The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take
the length of the plot as x m. Use the perimeter 180 m to write the value of
the breadth in terms of x. Use the values of length, breadth and the area to
write an equation in x. Solve the equation to calculate the length and
breadth of the plot. (1993)
Sol :
The perimeter of a rectangular field = 180 m
and area = 1800 m²
Let length = x m
But length+breadth $=\frac{180}{2}=90 \mathrm{~m}$
breadth =(90-x) m
According to the condition,
x(90-x)=1800
$\Rightarrow \quad 90 x-x^{2}-1800=0$
$\Rightarrow \quad x^{2}-90 x+1800=0$
$\Rightarrow \quad x^{2}-60 x-30 x+1800=0$
⇒x(x-60)-30(x-60)=0
⇒(x-60)(x-30)=0
Either x-60=0, then x=60
or x-30=0 , then x=30
∵Length is greater than its breadth
∴Length= 60 m
and breadth=90-60=30 m
Question 17
The lengths of the parallel sides of a trapezium are (x + 9) cm and (2x –
3) cm and the distance between them is (x + 4) cm. If its area is 540 cm²,
find x.
Sol :
Area of trapezium$=\frac{1}{2}$
(sum of parallel sides) × height
Lengths of parallel sides are (x + 9) and (2x – 3)
and height = (x + 4)
According to the condition,
$\frac{1}{2}(x+9+2 x-3) \times(x+4)=540$
$\Rightarrow(3 x+6)(x+4)=540 \times 2$
$\Rightarrow 3 x^{2}+12 x+6 x+24-1080=0$
$\Rightarrow 3 x^{2}+18 x-1056=0$
$\Rightarrow x^{2}+6 x-352=0 \quad$ (Dividing by
$\Rightarrow x^{2}+22 x-16 x-352=0$
⇒x(x+22)-16(x+22)=0
⇒(x+22)(x-16)=0
Either x+22=0, then x=-22
But its is not possible as it is negative
or x-16=0, then x=16
Question 18
If the perimeter of a rectangular plot is 68 m and the length of its
diagonal is 26 m, find its area.
Sol :
Perimeter = 68 m and diagonal = 26 m
Length + breadth $=\frac{68}{2}=34 \mathrm{~m}$
then breadth = (34 – x) m
According to the condition,
$l^{2}+b^{2}=h^{2}$
$(x)^{2}+(34-x)^{2}=(26)^{2}$
$\Rightarrow x^{2}+1156+x^{2}-68 x=676$
$\Rightarrow 2 x^{2}-68 x+1156-676=0$
$\Rightarrow 2 x^{2}-68 x+480=0$
$\Rightarrow x^{2}-34 x+240=0 \quad$ (Dividing by 2)
$\Rightarrow x^{2}-24 x-10 x+240=0$
⇒x(x-24)-10(x-24)=0
⇒(x-24)(x-10)=0
Either x-24=0, then x=24
or x-10=0, then x=10
∵Length is greater than breadth
∴Length =24 m
and breadth =(34-24)=10m
and Area =l×b=24×10=240m
Question 19
If the sum of two smaller sides of a right – angled triangle is 17cm and
the perimeter is 30cm, then find the area of the triangle.
Sol :
The perimeter of the triangle = 30 cm.
Let one of the two small sides = x
then, other side = 17 – x
∴Length of hypotenuse
=perimeter-sum of other two sides
=30 cm-17 cm=13 cm
According to the problem
$x^{2}+(17-x)^{2}=(13)^{2}$ (Pythagoras theorem)
$\Rightarrow \quad x^{2}+289+x^{2}-34 x=169$
$\Rightarrow 2 x^{2}-34 x+289-169=0$
$\Rightarrow 2 x^{2}-34 x+120=0$
$\Rightarrow \quad x^{2}-17 x+60=0 \quad$ (Dividing by $\left.2\right)$
$\Rightarrow \quad x^{2}-12 x-5 x+60=0$
⇒x(x-12)-5(x-12)=0
⇒(x-12)(x-5)=0
Either x-12=0, then x=12
Or x-5=0, then x=5
(i) When x=12, then first side=12cm
and second side=17-12=5 cm
(ii) When x=5, then first side=5
and second side=17-5=12
∴sides are 5cm, 12cm
Now , area of the triangle
$=\frac{5 \times 12}{2}=\frac{60}{2}=30 \mathrm{~cm}^{2}$
Question 20
The hypotenuse of grassy land in the shape of a right triangle is 1 metre
more than twice the shortest side. If the third side is 7 metres more than
the shortest side, find the sides of the grassy land.
Sol :
Let the shortest side = x
Hypotenuse = 2x + 1
and third side = x + 7
According to the condition,
$(2 x+1)^{2}=x^{2}+(x+7)^{2}$
$\Rightarrow \quad 4 x^{2}+4 x+1=x^{2}+x^{2}+14 x+49$
$\Rightarrow \quad 4 x^{2}+4 x+1-2 x^{2}-14 x-49=0$
$\Rightarrow \quad 2 x^{2}-10 x-48=0$
$\Rightarrow \quad x^{2}-5 x-24=0 \quad$ (Dividing by 2)
$\Rightarrow \quad x^{2}-8 x+3 x-24=0$
⇒x(x-8)+3(x-8)=0
⇒(x-8)(x+3)=0
Either x-8=0, then x=8
or x+3=0, then x=-3,
but it is not possible as it is negative.
∴Shortest side=8 m
Third side=x+7=8+7=15m
and hypotenuse=2x+1
=8×2+1=16+1=17 m
Question 21
Mohini wishes to fit three rods together in the shape of a right triangle.
If the hypotenuse is 2 cm longer than the base and 4 cm longer than the
shortest side, find the lengths of the rods.
Sol :
Let the length of hypotenuse = x cm
then base = (x – 2) cm
and shortest side = x – 4
According to the condition,
$(x)^{2}=(x-2)^{2}+(x-4)^{2}$
$\Rightarrow x^{2}=x^{2}-4 x+4+x^{2}-8 x+16$
$\Rightarrow x^{2}=2 x^{2}-12 x+20$
$\Rightarrow 2 x^{2}-12 x+20-x^{2}=0 $
$\Rightarrow x^{2}-12 x+20=0$
$\Rightarrow x^{2}-10 x-2 x+20=0$
⇒x(x-10)-2(x-10)=0
⇒(x-10)(x-2)=0
Either x-10=0, then x=10
or x-2=0, then x=2, but it is not possible as the hypotenuse is the longest
side.
∴Hypotenuse=10 cm
Base=10-2=8
and Shortes side=10-4=6 cm
Question 22
In a P.T. display, 480 students are arranged in rows and columns. If there
are 4 more students in each row than the number of rows, find the number of
students in each row.
Sol :
Total number of students = 480
Let the number of students in each row = x
then the number of rows $=\frac{480}{x}$
According to the condition, $x=\frac{480}{x}+4$
$\Rightarrow \quad x^{2}=480+4 x$
$\Rightarrow \quad x^{2}-4 x-480=0$
$\Rightarrow \quad x^{2}-24 x+20 x-480=0$
⇒x(x-24)+20(x-24)=0
⇒(x-24)(x+20)=0
Either x-24=0 or x+20=0
⇒x=24
which is not possible as it negative
∴Number of students in each row=24
Question 23
In an auditorium, the number of rows are equal to the number of seats in
each row.If the number of rows is doubled and number of seats in each row is
reduced by 5, then the total number of seats is increased by 375. How many
rows were there?
Sol :
Let the number of rows = x
then no. of seats in each row = x
and total number of seats = x × x = x²
According to the condition,
$\quad 2 x \times(x-5)=x^{2}+375$
$\Rightarrow \quad 2 x^{2}-10 x=x^{2}+375$
$\Rightarrow \quad 2 x^{2}-10 x-x^{2}-375=0$
$\Rightarrow \quad x^{2}-10 x-375=0$
$\Rightarrow \quad x^{2}-25 x+15 x-375=0$
⇒x(x-25)+15(x-25)=0
⇒(x-25)(x+15)=0
Either x-25=0, then x=25
or x+15=0, then x=-15,
but it is not possible as it is negative.
∴ Number of rows =25
Question 24
At an annual function of a school, each student gives the gift to every
other student. If the number of gifts is 1980, find the number of
students.
Sol :
Let the number of students = x
then the number of gifts given = x – 1
Total number of gifts = x (x – 1)
According to the condition,
x (x – 1) = 1980
$\Rightarrow \quad x^{2}-x-1980=0$
$\Rightarrow \quad x^{2}-45 x+44 x-1980=0$
⇒x(x-45)+44(x-45)=0
⇒(x-45)(x+44)=0
Either x-45=0, then x=45
or x+44=0, then x=-44, but it is not
possible as it is negative.
Hence number of students =45
Question 25
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain,
its speed gets reduced by 10 km/h and as such it takes two hours longer to
cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an
equation and solve it to evaluate x. (2016)
Sol :
Distance = 240 km
Let speed of a bus = x km/hr
∴ Time taken $=\frac{\mathrm{D}}{\mathrm{S}}=\frac{240}{x}$ hours
Due to heavy rains
Speed of the bus $=(x-10) \mathrm{km} / \mathrm{hr}$
∴ Time taken $=\frac{240}{x-10}$
According to the condition,
$\frac{240}{x}=\frac{240}{x-10}-2$
$\frac{240}{x-10}-\frac{240}{x}=2$
$\frac{240 x-240 x+2400}{x(x-10)}=2$
$=\frac{2400}{x^{2}-10 x}=2$
$\Rightarrow 2400=2 x^{2}-20 x$
$\Rightarrow 2 x^{2}-20 x-2400=0$
$\Rightarrow x^{2}-10 x-1200=0$
$\Rightarrow x^{2}-40 x+30 x-1200=0$
⇒x(x-40)+30(x-40)=0
⇒(x-40)(x+30)=0
Either x-40=0, then x=40
or x+30=0, then x=-30 which is not
possible, speed being negative.
∴Speed of bus=40 km/hr
Question 26
The speed of an express train is x km/hr and the speed of an ordinary train
is 12 km/hr less than that of the express train. If the ordinary train takes
one hour longer than the express train to cover a distance of 240 km, find
the speed of the express train.
Sol :
Let the speed of express train = x km
Then speed of the ordinary train = (x – 12) km
Time is taken to cover 240 km by the express
train $=\frac{240}{x}$ hours
Time taken to cover 240 km by the ordinary
$\operatorname{train}=\frac{240}{x-12}$ hours
According to the condition,
$\frac{240}{x-12}-\frac{240}{x}=1$
$\Rightarrow 240\left[\frac{1}{(x-12)}-\frac{1}{x}\right]=1$
$\Rightarrow 240\left[\frac{x-x+12}{x(x-12)}\right]=1$
$\Rightarrow 240\left[\frac{12}{x^{2}-12 x}\right]=1$
$\Rightarrow 2880=x^{2}-12 x$
$\Rightarrow x^{2}-12 x-2880=0$
$\Rightarrow x^{2}-60 x+48 x-2880=0$
$\Rightarrow x(x-60)+48(x-60)=0$
$\Rightarrow(x-60)(x+48)=0$
$\Rightarrow x=60$ or $x=-48$
$\Rightarrow x=60$
Rejecting x=-48, as speed can't be negative
Hence, speed of the express train=60km/hr
Question 27
A car covers a distance of 400 km at a certain speed. Had the speed been 12
km/h more, the time taken for the journey would have been 1 hour 40 minutes
less. Find the original speed of the car. (1996)
Sol :
Let the original speed of the car = x km/h.
Distance covered = 400 km
Time taken to cover $400 \mathrm{~km}=\frac{400}{x} \mathrm{~h}$
In second case, speed of car =(x+12) km/hr
New time taken to cover $400 \mathrm{~km}=\frac{400}{x+12} \mathrm{~h}$
According to the condition
$\frac{400}{x}-\frac{400}{x+12}=1 \frac{40}{60}=1 \frac{2}{3}=\frac{5}{3}$
$\Rightarrow 400\left(\frac{x+12-x}{x(x+12)}\right)=\frac{5}{3} \Rightarrow
\frac{400 \times 12}{x^{2}+12 x}=\frac{5}{3}$
$400 \times 12 \times 3=5 x^{2}+60 x $
$\Rightarrow 14400=5 x^{2}+60 x$
$\Rightarrow 5 x^{2}+60 x-14400=0$
$\Rightarrow x^{2}+12 x-2880=0$ (dividing both side by $\left.5\right)$
$\Rightarrow x^{2}+60 x-48 x-2880=0 $
$\Rightarrow x(x+60)-48$
(x+60)=0
⇒(x+60)(x-48)=0
⇒ x=48 or x=-60
⇒ x=48 (Rejecting x=-60, being speed)
Hence, original speed of the car =48km/hr
Question 28
An aeroplane travelled a distance of 400 km at an average speed of x km/hr.
On the return journey, the speed was increased by 40 km/hr. Write down an
expression for the time taken for
(i)the onward journey,
(ii) the return journey.
If the return journey took 30 minutes less than the onward journey, write
down an equation in x and find its value. (2002)
Sol :
Distance = 400 km
Speed of aeroplane = x km/hr
(i) ∴ Time taken $=\frac{400}{x}$ hours
On increasing the speed by 40 km/hr on the return journey, the speed =(x+40)
km/hr
(ii) Time taken $=\frac{400}{x+40}$ hours
Now according to the condition,
$\frac{400}{x}-\frac{400}{x+40}=30$ minutes $=\frac{1}{2}$
$400\left[\frac{1}{x}-\frac{1}{x+40}\right]=\frac{1}{2}$
$\Rightarrow 400\left[\frac{x+40-x}{x(x+40)}\right]$
$ \Rightarrow \frac{400 \times 40}{x^{2}+40 x}=\frac{1}{2}$
$\Rightarrow x^{2}+40 x=400 \times 40 \times 2$
$\Rightarrow x^{2}+40 x-32000=0$
$\Rightarrow x^{2}+200 x-160 x-32000=0$
⇒x(x+200)-160(x+200)=0
⇒(x+200)(x-160)=0
Either x+200=0, then x=-200, which is not possible as it is negative. or
x-160=0, then x=160
Question 29
The distance by road between two towns A and B, is 216 km, and by rail it
is 208 km. A car travels at a speed of x km/hr, and the train travels at a
speed which is 16 km/hr faster than the car. Calculate :
(i) The time taken by the car, to reach town B from A, in terms of x ;
(ii) The time taken by the train, to reach town B from A, in terms of x
;
(iii) If the train takes 2 hours less than the car, to reach town B, obtain
an equation in x and solve it.
(iv) Hence find the speed of the train. (1998)
Sol :
The distance by road between A and B = 216 km
and the distance by rail = 208 km
speed of car = x km/hr
and speed of train =(x+16) km/hr
(i) Time taken by car $=\frac{216}{x}$ hours.
(ii) Time taken by train $=\frac{208}{x+16}$ hours
(iii) According to the condition,
$\frac{216}{x}-\frac{208}{x+16}=2$
$\Rightarrow \frac{216 x+216 \times 16-208 x}{x(x+16)}=\frac{2}{1}$
$\Rightarrow \frac{8 x+3456}{x^{2}+16 x}=\frac{2}{1} $
$\Rightarrow 8 x+3456=2 x^{2}+32 x$
$\Rightarrow 2 x^{2}+32 x-8 x-3456=0 \Rightarrow 2 x^{2}+24 x-3456=0$
$\Rightarrow x^{2}+12 x-1728=0 \quad$ (Dividing by 2)
$\Rightarrow x^{2}+48 x-36 x-1728=0$
⇒x(x+48)-36(x+48)=0
⇒(x+48)(x-36)=0
Either x+48=0, then x=-48 which is not possible as it is negative. or x-36=0,
then x=36
(iv) Speed of the train =(x+16)km/hr
=(36+16)km/hr =52km/hr
Question 30
An aeroplane flying with a wind of 30 km/hr takes 40 minutes less to fly
3600 km, than what it would have taken to fly against the same wind. Find
the planes speed of flying in still air.
Sol :
Let the speed of the plane in still air = x km/hr
Speed of wind = 30 km/hr
Distance = 3600 km
∴ Time taken with the wind $=\frac{3600}{x+30}$ and time taken against the
wind $=\frac{3600}{x-30}$ According to the condition,
$\frac{3600}{x-30}-\frac{3600}{x+30}=40$ minutes $=\frac{2}{3}$ hour
$\Rightarrow 3600\left(\frac{1}{x-30}-\frac{1}{x+30}\right)=\frac{2}{3}$
$\Rightarrow 3600\left(\frac{x+30-x+30}{(x-30)(x+30)}\right)=\frac{2}{3}$
$\Rightarrow \frac{3600 \times 60}{x^{2}-900}=\frac{2}{3}$
$ \Rightarrow 2 x^{2}-1800=3 \times 3600 \times 60$
$\Rightarrow 2 x^{2}-1800=648000$
$ \Rightarrow 2 x^{2}-1800-648000=0$
$\Rightarrow 2 x^{2}-649800=0$
$\Rightarrow x^{2}+324900=0$ (dividing by 2)
$\Rightarrow x^{2}-(570)^{2}=0$
⇒(x+570)(x-570)=0
Either $x+570 \div 0,$ then x=-570
which is not possible as it is negative or x-570=0, then x=570
Hence speed of plane in still air =570 km/hr
Question 31
A school bus transported an excursion party to a picnic spot 150 km away.
While returning, it was raining and the bus had to reduce its speed by 5
km/hr, and it took one hour longer to make the return trip. Find the time
taken to return.
Sol :
Distance = 150 km
Let the speed of bus = x km/hr
∴ Time taken $=\frac{150}{x}$ hour
On returning speed of the bus =(x-5) km/hr
∴ Time taken $=\frac{150}{x-5}$
According to the condition
$\frac{150}{x-5}-\frac{150}{x}=1$
$\Rightarrow 150\left(\frac{1}{x-5}-\frac{1}{x}\right)=1 $
$\Rightarrow 150\left(\frac{x-x+5}{x(x-5)}\right)=1$
$\Rightarrow \frac{150 \times 5}{x^{2}-5 x}=1$
$ \Rightarrow x^{2}-5 x=750$
$\Rightarrow x^{2}-5 x-750=0 $
$\Rightarrow x^{2}-30 x+25 x-750-0$
⇒x(x-30)+25(x-30)=0
⇒(x-30)(x+25)=0
Either x-30=0, then x=30 or x+25=0, then x=-25,
but it is not possible as it is negative.
∴ Speed of bus =30 km and time taken while returning
$=\frac{150}{x-5}=\frac{150}{30-5}-\frac{150}{25}=6$ hours
Question 32
A boat can cover 10 km up the stream and 5 km down the stream in 6 hours.
If the speed of the stream is 1.5 km/hr. find the speed of the boat in still
water.
Sol :
Distance up stream = 10 km
and down stream = 5 km
Total time is taken = 6 hours
Speed of stream = 1.5 km/hr
Let the speed of a boat in still water = x km/hr
According to the condition,
$\frac{10}{x-1.5}+\frac{5}{x+1.5}=6$
$\Rightarrow 10 x+15+5 x+5 x-7.5=6(x-15)(x+1.5)$
$\Rightarrow 15 x+7.5=6\left(x^{2}-2.25\right)$
$ \Rightarrow 15 x+7.5=\left(x^{2}-13.5\right.$
$\Rightarrow 6 x^{2}-15 x-13.5-7.5$
$ \Rightarrow 6 x^{2}-15 x-21=0$
$\Rightarrow 2 x^{2}-5 x-7=0$
$\Rightarrow 2 x^{2}-7 x+2 x-7=0$ (Dividing by 3)
⇒x(2 x-7)+1(2 x-7)=0 (dividing by 3)
⇒(2 x-7)(x+1)=0 $\left\{\begin{array}{l}2 \times(-7)=\cdot \cdot 14 \\
-14=-7 \times 2 \\ -5=-7+2\end{array}\right.$
⇒x(2 x-7)+1(2 x-7)=0
⇒(2 x-7)(x+1)=0
Either 2x-7=0 then 2x=7
$\Rightarrow x=\frac{7}{2}$
or x+1=0 , then x=-1
But it is not possible being negative
∴$x=\frac{7}{2}=3.5$
∴Speed of boat=.5 km/h
Question 33
Two pipes running together can fill a tank in $11^{\frac{1}{9}}$
minutes. If one pipe takes 5 minutes more than the other to fill the tank,
find the time in which each pipe would/fill the tank.
Sol :
Let the time taken by one pipe = x minutes
Then time taken by second pipe = (x + 5) minutes
Time taken by both pipes = $11^{\frac{1}{9}}$ minutes
Now according to the condition.
$\frac{1}{x}+\frac{1}{x+5}=\frac{9}{100} $
$\Rightarrow \frac{(x+5)+x}{x(x+5)}=\frac{9}{100}$
$\Rightarrow \frac{x+5+x}{x^{2}+5 x}=\frac{9}{100}$
$ \Rightarrow \frac{2 x+5}{x^{2}+5 x}=\frac{9}{100}$
$\Rightarrow 9 x^{2}+45 x=200 x+500$
$\Rightarrow 9 x^{2}+45 x-200 x-500=0 $
$\Rightarrow 9 x^{2}-155 x-500=0$
$\Rightarrow 9 x^{2}-180 x+25 x-500=0$
⇒9x(x-20)+25(x-20)=0
⇒(x-20)(9 x+25)=0
$\Rightarrow x=\frac{-25}{9}$ but is not possible as it is in negative.
∴x=20
Hence the first pipe can fill the tank in 20 minutes and second pipe can do
the same in 20+5=25 mints.
Question 34
(i) Rs. 480 is divided equally among ‘x’ children. If the number of
children was 20 more then each would have got Rs. 12 less. Find ‘x’.
(ii) Rs. 6500 is divided equally among a certain number of persons. Had
there been 15 more persons, each would have got Rs. 30 less. Find the
original number of persons.
Sol :
(i) Share of each child $=\operatorname{Rs} \frac{480}{x}$
According to the question
$\frac{480}{x+20}=\frac{480}{x}-12 $
$\Rightarrow \frac{480}{x+20}=\frac{480-12 x}{x}$
$\Rightarrow \frac{480}{x+20}=\frac{12(40-x)}{x}$
$\Rightarrow(x+20)(40-x)=40 x $
$\Rightarrow 40 x-x^{2}+800-20 x=40 x$
$\Rightarrow x^{2}+20 x-800=0$
$ \Rightarrow x^{2}+40 x-20 x-800=0$
⇒x(x+40)-20(x+40)=0
⇒(x+40)(x-20)=0
⇒ x=-40, x=20
-ve value of x is not possible
∴No. of children=20
(ii) Total amount=Rs 6500
Let number of persons in first case=x
Then share of each person Rs. $=\frac{6500}{x}$
Number of persons increased =15
∴ Total number =x+15
Now share of each person$=\frac{6500}{x+15}$
According to the condition,
$\frac{6500}{x}-\frac{6500}{x+15}=30 $
$\Rightarrow 6500\left[\frac{1}{x}-\frac{1}{x+15}\right]=30$
$\Rightarrow 6500\left[\frac{x+15-x}{x(x+15)}\right]=30$
$ \Rightarrow \frac{6500 \times 15}{x^{2}+15 x}=30$
$\Rightarrow 6500 \times 15=30\left(x^{2}+15 x\right)$
$\Rightarrow 3250=x^{2}+15 x$ (dividing by 30)
$\Rightarrow x^{2}+15 x-3250=0$
$ \Rightarrow x^{2}+65 x-50 x-3250=0$
⇒ x(x+65)-50(x+65)=0
⇒ (x+65)(x-50)=0
Either x+65=0, then x=-65 which is not possible being negative or x-50=0, then
x=50
∴ No. of persons in the beginning =50
Question 35
2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x –
4), find x.
Sol :
Cost of 2x articles = 5x + 54
Cost of 1 article $=\frac{5x+54}{2x}$..(i)
Again cost of x + 2 articles = 10x – 4
∴ Cost of 1 article $=\frac{10 x-4}{x+2}$..(ii)
From (i) and (ii)
$\frac{5 x+54}{2 x}=\frac{10 x-4}{x+2}$
$\Rightarrow(5 x+54)(x+2)=2 x(10 x-4)$
$\Rightarrow 5 x^{2}+10 x+54 x+108=20 x^{2}-8 x$
$\Rightarrow 5 x^{2}+10 x+54 x+108-20 x^{2}+8 x=0$
$\Rightarrow-15 x^{2}+72 x+108=0$
$\Rightarrow 5 x^{2}-24 x-36=0$ (Dividing by -3)
$\Rightarrow 5 x^{2}-30 x+6 x-36=0$
⇒ 5x(x-6)+6(x-6)=0
⇒ (x-6)(5x+6)=0
Either x-6=0, then x=6
or 5x+6=0 then 5x=-6
$x=\frac{-6}{5}$ b
it is not possible as it is in negative.
∴ x=6
Question 36
A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x. If the cost per
article were Rs. 5 more, the number of articles that can be bought for Rs.
600 would be four less.
(ii) Write down the equation in x for the above situation and solve it to
find x. (1999)
Sol :
Total cost = Rs. 600,
No. of articles = x
(i) ∴Cost of one article $=\operatorname{Rs.} \frac{600}{x}$
In second case price of one article =Rs. $\frac{600}{x}+5=\frac{600}{x-4}$
(ii) $\frac{600+5 x}{x}=\frac{600}{x-4}$
$\Rightarrow \quad(x-4)(600+5 x)=600 x$
$\Rightarrow \quad 600 x-2400+5 x^{2}-20 x=600 x$
$\Rightarrow \quad 5 x^{2}+600 x-20 x-600 x-2400=0$
$\Rightarrow \quad 5 x^{2}-20 x-2400=0$
$\Rightarrow \quad x^{2}-4 x-480=0 \quad$ (Dividing by $\left.5\right)$
$\Rightarrow \quad x^{2}-24 x+20 x-480=0$
⇒x(x-24)+20(x-24)=0
⇒(x-24)(x+20)=0
Either x-24=0, then x=24
or x+20=0, then x=-20 but it is not possible as it is in negative
∴x=24
Question 37
A shopkeeper buys a certain number of books 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 an
equation in x and solve it to find the original cost of each book.
Sol :
Let original cost = Rs x
No. of books bought $=\frac{960}{x}$
New cost of books = Rs (x – 8)
∴ No. of books bought $=\frac{960}{x-8}$
If no. of books bought is 4 more then cost $=\frac{960}{x}+4$
∴ According to condition,
$\frac{960}{x-8}-\frac{960}{x}=4 $
$\Rightarrow 960\left(\frac{1}{x-8}-\frac{1}{x}\right)=4$
$\Rightarrow \frac{x-(x-8)}{x(x-8)}=\frac{4}{960}$
$ \Rightarrow \frac{x-x+8}{x^{2}-8 x}=\frac{4}{960}$
$\Rightarrow \frac{8}{x^{2}-8 x}=\frac{1}{240}$
$ \Rightarrow x^{2}-8 x=8 \times 240$
$\Rightarrow x^{2}-8 x-1920=0$
$x=\frac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(-1920)}}{2}$
$=\frac{8 \pm \sqrt{64+7680}}{2}=\frac{8 \pm \sqrt{7744}}{2}$
$=\frac{8 \pm 88}{2}=\frac{8+88}{2}, \frac{8-88}{2}=\frac{96}{2},
\frac{-80}{2}$
=48, -40 (rejecting)
∴Cost of book=48
Question 38
A piece of cloth costs Rs. 300. If the piece was 5 metres longer and each
metre of cloth costs Rs. 2 less, the cost of the piece would have remained
unchanged. How long is the original piece of cloth and what is the rate per
metre?
Sol :
The total cost of cloth piece = Rs. 300
Let the length of the piece of cloth in the beginning = x m
Then cost of 1 metre = Rs. $\frac{300}{x}$
In second case, length of cloth =(x+5)
Cost of 1 metre = Rs. $\frac{300}{x+5}$
According to the condition. $\frac{300}{x}-\frac{300}{x+5}=2$
$ \Rightarrow 300\left(\frac{1}{x}-\frac{1}{x+5}\right)=2$
$\Rightarrow 300\left(\frac{x+5-x}{x(x+5)}\right)=2 $
$\Rightarrow \frac{300 \times 5}{x(x+5)}=2$
$ \Rightarrow \frac{150 \times 5}{x(x+5)}=1$ (dividing by 2)
$750=x^{2}+5 x$
$ \Rightarrow x^{2}+5 x-750=0$
$\Rightarrow x^{2}+30 x-25 x-750=0$
⇒x(x+30)-25(x+30)=0
⇒(x+30)(x-25)=0
Either x+30=0, then x=-30 which is not possible being negative
or x-25=0, then x=25
∴Length of cloth piece in the beginning =25 metres
and rate per metre $=$ Rs. $\frac{300}{25}=$ Rs. 12
Question 39
The hotel bill for a number of people for an overnight stay is Rs. 4800. If
there were 4 more, the bill each person had to pay would have reduced by Rs.
200. Find the number of people staying overnight. (2000)
Sol :
Let the number of people = x
Amount of bill = Rs. 4800
Then bill for each person = Rs. $\frac{4800}{x}$
In second case, the number of people =x+4
then bill of each person $=\frac{4800}{x+4}$
According to the condition,
$\frac{4800}{x}-\frac{4800}{x+4}=200$
$\Rightarrow 4800\left(\frac{x+4-x}{x(x+4)}\right)=200 $
$\Rightarrow \frac{4800 \times 4}{x(x+4)}=200$
$\Rightarrow 19200=200 x^{2}+800 x$
$\Rightarrow 200 x^{2}+800 x-19200=0$
$\Rightarrow x^{2}+4 x-96=0$ (Dividing by $\left.200\right)$
$\Rightarrow x^{2}+12 x-8 x-96=0$
⇒x(x+12)-8(x+12)=0
⇒(x+12)(x-8)=0
Either x+12=0, then x=-12,
but it is not possible as it is in negative :
or x-8=0, then x=8
∴Number of peoples=8
Question 40
A person was given Rs. 3000 for a tour. If he extends his tour programme by
5 days, he must cut down his daily expenses by Rs. 20. Find the number of
days of his tour programme.
Solution:
Let the number of days of tour programme = x
Amount = Rs. 3000
∴ Expenses for each day $=\frac{3000}{x}$
In second case, no. of days =x+5
Then expenses of each day $=\frac{3000}{x+5}$
Now according to the condition,
$\frac{3000}{x}-\frac{3000}{x+5}=20$
$\Rightarrow 3000\left(\frac{1}{x}-\frac{1}{x+5}\right)=20$
$\Rightarrow 3000 \frac{(x+5-x)}{x^{2}+5 x}=20$
$\Rightarrow 3000 \times 5=20 x^{2}+100 x$
$\Rightarrow 20 x^{2}+100 x-15000=0$
Either x-25=0, then x=25 or x+30=0, then x=-30,
but it is not possible as it is in negative.
∴Number days=25
Question 41
Ritu bought a saree for Rs. 60 x and sold it for Rs. (500 + 4x) at a loss
of x%. Find the cost price.
Sol :
The cost price of saree = Rs. 60x
and selling price = Rs. (500 + 4x)
Loss = x%
Now according to the condition
S.P.=$C.P \times \frac{100-\text { Loss } \%}{100}$
$500+4 x=\frac{60 x(100-x)}{100}$
$\Rightarrow 50000+400 x=6000 x-60 x^{2}$
$\Rightarrow 60 x^{2}-6000 x+400 x+50000=0$
$\Rightarrow 60 x^{2}-5600 x+50000=0$
$\Rightarrow 3 x^{2}-280 x+2500=0$ (Dividing by 20)
$\Rightarrow 3 x^{2}-30 x-250 x+2500=0$
3x(x-10)-250(x-10)=0
(x-10)(3 x-250)=0
Either x-10=0, then x=10
Or 3x-250=0, then 3x=250
$\Rightarrow x=\frac{250}{3}$
But it is not possible
∴ Loss =10 %
Cost price =60x=60×10= Rs. 600
Question 42
(i) The sum of the ages of Vivek and his younger brother Amit is 47 years.
The product of their ages in years is 550. Find their ages. (2017)
(ii) Paul is x years old and his father’s age is twice the square of Paul’s
age. Ten years hence, the father’s age will be four times Paul’s age. Find
their present ages.
Sol :
(i) Let Vivek’s present age be x years.
His brother’s age = (47 – x) years
According to question,
x(47 – x) = 550
⇒ 47x – x² = 550
⇒ x² – 47x + 550 = 0
⇒ x² – 25x – 22x + 550 = 0
⇒ x(x – 25) – 22(x – 25) – 0
⇒ (x-25)(x-22)=0$
⇒ x-25=0 or x-22=0
⇒ x=25 or x=22
When x=25, then 47-x=47-25=22
When x=22, then 47-x=47-22=25
(does not satisfy the given condition)
∴Vivek's age =x=25 years.
His younger brother's age =22 years.
(ii) Age of Paul x years.
Father's age $=2 x^{2}$ 10 years hence,
Age of Paul =x+10
and father's age $=2 x^{2}+10$
According to the conditions,
$2 x^{2}+10=4(x+10)$
$\Rightarrow 2 x^{2}+10=4 x+40$
$\Rightarrow 2 x^{2}+10-4 x-40=0$
$\Rightarrow 2 x^{2}-4 x-30=0$
$\Rightarrow x^{2}-2 x-15=0$ (Dividing by 2)
$\Rightarrow x^{2}-5 x+3 x-15=0$
⇒x(x-5)+3(x-5)=0
⇒(x-5)(x+3)=0
Either x-5=0, then x=5
or x+3=0, then x=-3, but it is not possible as it is in negative.
∴ Age of Paul =5 years.
and his father's age $=2 x^{2}=2(5)^{2}=2 \times 25$ =50 years
Question 43
The age of a man is twice the square of the age of his son. Eight years
hence, the age of the man will be 4 years more than three times the age of
his son. Find the present age.
Sol :
Let the present age of the son = x years
then, the present age of the man = 2x² years.
8 years hence,
The age of son will be = (x + 8) years and the
age of man = (2x² + 8) years
According to the problem,
$2 x^{2}+8=3(x+8)+4 $
$\Rightarrow 2 x^{2}+8=3 x+24+4$
$\Rightarrow 2 x^{2}-3 x-24-4+8=0$
$\Rightarrow 2 x^{2}-3 x-20=0 $
$\Rightarrow 2 x^{2}-8 x+5 x-20=0$
⇒ 2x(x-4)+5(x-4)=0
⇒(x-4)(2 x+5)=0
Either x-4=0 then x=4
or 2x+5=0 then 2x=-5
$\Rightarrow x=-\frac{5}{2}$
But, it is not possible.
Present age of the son =4 years
and present age of the man $=2 x^{2}$
$=2(4)^{2}$ years
=32 years Ans.
Question 44
Two years ago, a man’s age was three times the square of his daughter’s
age. Three years hence, his age will be four times his daughter’s age. Find
their present ages.
Sol :
2 years ago,
Let the age of daughter = x
age of man = 3x²
then present age of daughter = x + 2
and mean = 3x² + 2
and 3 years hence, the age of the daughter
=x+2+3=x+5 and man$=3 x^{2}+2+3=3 x^{2}+5$
According to the condition. $3 x^{2}+5=4(x+5) $
$\Rightarrow 3 x^{2}+5=4 x+20$
$3 x^{2}-4 x+5-20=0$
$ \Rightarrow 3 x^{2}-4 x-15=0$
$3 x^{2}-9 x+5 x-15=0$
⇒3x(x-3)+5(x-3)=0
⇒(x-3)(3 x+5)=0
Either x-3=0, then x=3
or 3x+5=0, then 3 x=-5
$\Rightarrow x=\frac{-5}{3}$
which is not possible, as age can't be negative
If x=3, then Present age of $\operatorname{man}=3 x^{2}+2$
$=3(3)^{2}+2=27+2=29$ years
and age of daughter =x+2=3+2=5 years.
Question 45
The length (in cm) of the hypotenuse of a right-angled triangle exceeds the
length of one side by 2 cm and exceeds twice the length of another side by 1
cm. Find the length of each side. Also, find the perimeter and the area of
the triangle.
Sol :
Let the length of one side = x cm
and other side = y cm.
then hypotenuse = x + 2, and 2y + 1
∴x+2=2y+1
⇒x-2y=1-2
⇒ x-2 y=-1
⇒x=2y-1..(i)
and by using Pythagoraus
theorem,
$x^{2}+y^{2}=(2 y+1)^{2}$
$\Rightarrow x^{2}+y^{2}=4 y^{2}+4 y+1$
$\Rightarrow(2 y-1)^{2}+y^{2}=4 y^{2}+4 y+1$ [From(i)]
$\Rightarrow 4 y^{2}-4 y+1+y^{2}=4 y^{2}+4 y+1$
$\Rightarrow 4 y^{2}-4 y+1+y^{2}-4 y^{2}-4 y-1=0$
$\Rightarrow y^{2}-8 y=0$
⇒y(y-8)=0
Either y=0 but it is not possible. or y-8=0 then y=8
Substituting the value of y in (i)
x=2(8)-1=16-1=15
∴ Length of one side =15 cm
and length of other side =8 cm and
hypotenuse =x+2=15+2=17
∴ Perimeter =15+8+17=40cm
and Area $=\frac{1}{2} \times$ one side × other side
$=\frac{1}{2} \times 15 \times 8=60 \mathrm{~cm}^{2}$
Question 46
If twice the area of a smaller square is subtracted from the area of a
larger square, the result is 14 cm². However, if twice the area of the
larger square is added to three times the area of the smaller square, the
result is 203 cm². Determine the sides of the two squares.
Sol :
Let the side of smaller square = x cm
and side of bigger square = y cm
According to the condition,
$y^{2}-2 x^{2}=14$..(i)
$2 y^{2}+3 x^{2}=203$...(i)
Multiplying (i) by 2 and (ii) by 1
$\begin{aligned}2 y^{2}-4 x^{2}=&28 \\2 y^{2}+3 x^{2}=&203\\
-\phantom{2 y^{2}}-\phantom{3 x^{2}}=&-\phantom{203}\\
\hline-7x^2=&-175\end{aligned}$
$\Rightarrow x^{2}=\frac{-175}{-7}=25$
$x^{2}-25=0$
⇒(x+5)(x-5)=0
Either x+5=0 then x=-5, but it is not possible, or x-5=0,
then x=5
Substitute the value of x in (i)
$y^{2}-2(5)^{2}=14 $
$\Rightarrow y^{2}=14+2 \times 25$
$y^{2}=14+50=64=(8)^{2}$
∴y=8
Hence side of smaller square 5cm and side of bigger square 8cm
