Question
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.

Answer

Let the length and breadth of the rectangle be x and y units respectively.
Then area of rectangle = xy square units
It is given that if length is increared and breadth reduced each by 2 units. then the area is reduced by 28 square units.
⇒ (x + 2) (y - 2) = xy - 28
⇒ xy - 2x + 2y - 4 = xy - 28
⇒ -2x + 2y - 4 + 28 = 0
⇒ -2x + 2y + 24 = 0
⇒ 2x - 2y - 24 = 0 ......(i)
It is also given that the length is reduced by 1 unit and breath is increcred by 2 units then the area is increared by 33 square units.
⇒ (x - 1) (y + 2) = xy + 33
⇒ xy + 2x - y - 2 = xy + 33
⇒ xy + 2x - y - 2 - xy - 33 = 0
⇒ 2x - y - 35 = 0 ......(ii)
Now, subtracting eq. (ii) from eq. (i) and we get
⇒ 2x - 2y - 24 - (2x - y - 35) = 0
⇒ 2x - 2y - 24 - 2x + y + 35 = 0
⇒ -y + 11 = 0
⇒ y = 11
Putting the value of y in eq. (i)
⇒ 2x - 2y - 24 = 0
⇒ 2x - 2 × 11 - 24 = 0
⇒ 2x - 22 - 24 = 0
⇒ 2x - 46 = 0
⇒ x = 23
Hence, the length of the rectangle is 23 and breadth of the rectangle is 11
Area os rectangle = length × breadth
= 23 × 11
= 253 square units.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Solve the following systems of equations graphically:
2x + 3y + 5 = 0
3x - 2y - 12 = 0
Solve the following equations by using the method of completing the square:
$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
In triangles BMP and CNR it is given that PB = 5cm, MP = 6cm, BM = 9cm and NR = 9cm. If $\triangle\text{BMP}\sim\triangle\text{CNR}$ then find the perimeter of $\triangle\text{CNR}.$
Draw a circle of radius 6cm. From a point 10cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
The following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours)
Number of lamps
1500-2000
2000-2500
2500-3000
3000-3500
3500-4000
4000-4500
4500-5000
14
56
60
86
74
62
48
Find the median life.
In $\triangle ABC , \angle BAC =90^{\circ}$, seg $B L$ and seg $C M$ are medians of $\triangle A B C$. Then prove that: $4\left( BL ^2+C M^2\right)=5 B C^2$
Solve the following quadratic equation by factorization:
$3\Big(\frac{3\text{x}-1}{2\text{x}+3}\Big)-2\Big(\frac{2\text{x}+3}{3\text{x}-1}\Big)=5,$ $\text{x}\neq\frac{1}{3},-\frac{3}{2}$
Mr. Deepak Pal invested ₹ 1,00,354 in shares of FV ₹ 100 , when the market value is ₹ 50 . Rate of brokerage is $0.3 \%$ and rate of GST on brokerage is $18 \%$, then how many shares were purchased?
Show that $(x + 2)$ is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
Write the expression $a_n - a_k$​​​​​​​ for the A.P. $a, a + d, a + 2d, ...$
Hence, find the common difference of the A.P. for which,
$20^{th}​​​​​​​$​​​​​​​ term is $10$ more than the $18^{th}​​​​​​​$​​​​​​​ term.