eSaral - JEE | NEET | Class 8-10 Exam Prep App

If x and y vary inversely as each other and

Question: Ifxandyvary inversely as each other and (i)x= 3 wheny= 8, findywhenx= 4 (ii)x= 5 wheny= 15, findxwheny= 12 (iii)x= 30, findywhen constant of variation = 900. (iv)y= 35, findxwhen constant of variation = 7. Solution: (i) Since $x$ and $y$ vary inversely, we have: $x y=k$ For $x=3$ and $y=8$, we have : $3 \times 8=k$ $\Rightarrow k=24$ For $x=4$, we have $:$ $4 y=24$ $\Rightarrow y=\frac{24}{4}$ $=6$ $\therefore y=6$ (ii) Since $x$ and $y$ vary inversely, we have: $x y=k$ For $x=5$ and $...

Raghu has enough money to buy 75 machines worth Rs 200 each.

Question: Raghu has enough money to buy 75 machines worth Rs 200 each. How many machines can he buy if he gets a discount of Rs 50 on each machine? Solution: Letxbe the number of machines he can buy if a discount of Rs. 50 is offered on each machine. Since Raghu is getting a discount of Rs 50 on each machine, the cost of each machine will get decreased by Rs 50 . If the price of a machine is less, he can buy more number of machines. It is a case of inverse variation. Therefore, we have: $75 \tim...

Prove that angle bisector of any angle of

Question: Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. Solution: Given $\triangle A B C$ is inscribed in a circle. Bisecter of $\angle A$ and perpendicular bisector of $B C$ intersect at point $Q$. To prove $A, B, Q$ and $C$ are con-cyclic. Construction Join $B Q$ and $Q C$. Proof We have assumed that, $Q$ lies outside the circle. In $\triangle B M Q$ and $\triangle C M ...

A square tank has an area of 1600 cm2.

Question: A square tank has an area of 1600cm2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs 12.50 per m2. Solution: Letam be the side of the square. Area of the square $=a^{2}$ Thus, we have: $a^{2}=1600$ $\Rightarrow a=40$ Area of the plots = 4(Area of the semicircle of radius 20 m) $=\left|4\left(\frac{1}{2} \pi r^{2}\right)\right| \mathrm{m}^{2}$ $=\left|4\left(\frac{1}{2} \times 3.14 \times 20 \times 20\right)\right| \mathrm{m}^{2}$ $=2512 \mathrm{~m...

A square tank has an area of 1600 cm2.

Question: A square tank has an area of 1600cm2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs 12.50 per m2. Solution: Letam be the side of the square. Area of the square $=a^{2}$ Thus, we have: $a^{2}=1600$ $\Rightarrow a=40$ Area of the plots = 4(Area of the semicircle of radius 20 m) $=\left|4\left(\frac{1}{2} \pi r^{2}\right)\right| \mathrm{m}^{2}$ $=\left|4\left(\frac{1}{2} \times 3.14 \times 20 \times 20\right)\right| \mathrm{m}^{2}$ $=2512 \mathrm{~m...

Solve this

Question: If $A=\left[\begin{array}{lll}3 -3 4 \\ 2 -3 4 \\ 0 -1 1\end{array}\right]$, show that $A^{-1}=A^{3}$ Solution: We have, $A=\left[\begin{array}{lll}3 -3 4\end{array}\right.$ $A^{2}=\left[\begin{array}{lll}3 -3 4\end{array}\right.$ $\begin{array}{lll}2 -3 4\end{array}$ $0-1 \quad 1 \quad 1]\left[\begin{array}{lll}3 -3 4\end{array}\right.$ $2 \quad-3 \quad 4$ $0-1 \quad 1]=\left[\begin{array}{lll}9-6+0 -9+9-4 12-12+4\end{array}\right.$ $6-6+0 \quad-6+9-4 \quad 8-12+4$ $0-2+0 \quad 0+3-1 ...

A person has money to buy 25 cycles worth Rs 500 each.

Question: A person has money to buy 25 cycles worth Rs 500 each. How many cycles he will be able to buy if each cycle is costing Rs 125 more? Solution: Letxbe the number of cycles bought if each cycle costs Rs 125 more. It is in inverse variation. Therefore, we get: $500 \times 25=625 \times x$ $\Rightarrow x=\frac{500 \times 25}{625}$ $500 \times 25=625 \times x$ $\Rightarrow x=\frac{500 \times 25}{625}$ $=20$ $\therefore$ The required number of cycles is 20 ....

Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area.

Question: Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area. What area will be left ungrazed? Solution: Letrbe the radius of the circle.Thus, we have: $r=\frac{50}{2} \mathrm{~m}$ $=25 \mathrm{~m}$ Area left ungrazed $=$ (Area of the square) $-4$ (Area of the sector where $r=25 \mathrm{~m}$ and $\theta=90^{\circ}$ ) $=\left|(50 \times 50)-4\left(3.14 \times 25 \times 25 \times \frac{90}{360}\right)\right| \mathrm{m}^{...

Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area.

Question: Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area. What area will be left ungrazed? Solution: Letrbe the radius of the circle.Thus, we have: $r=\frac{50}{2} \mathrm{~m}$ $=25 \mathrm{~m}$ Area left ungrazed $=$ (Area of the square) $-4$ (Area of the sector where $r=25 \mathrm{~m}$ and $\theta=90^{\circ}$ ) $=\left|(50 \times 50)-4\left(3.14 \times 25 \times 25 \times \frac{90}{360}\right)\right| \mathrm{m}^{...

ABCD is a parallelogram. A circle through A,

Question: ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic. Solution: Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ Proof 1 = A [exterior angle property of cyclic quadrilateral] But A = C [opposite angles of a parallelogram] 1 = C ,..(i) But C+ D = 180 [sum of coin...

18 men can reap a field in 35 days.

Question: 18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required? Solution: Let the number of men required to reap the field in 15 days bex. Since the number of days and the number of men required ro reap the field are in inverse variation, we have: $35 \times 18=15 \times x$ $\Rightarrow x=\frac{35 \times 18}{15}$ $=42$ Thus, the required number of men is 42 ....

If P, Q and R are the mid-points of the sides,

Question: If $P, Q$ and $R$ are the mid-points of the sides, $B C, C A$ and $A B$ of a triangle and $A D$ is the perpendicular from $A$ on $B C$, then prove that $P, Q, R$ and $D$ are concyclic. Solution: Given in $\triangle A B C, P, Q$ and $R$ are the mid-points of the sides $B C, C A$ and $A B$ respectively Also, $A D \perp B C$. . To prove $P, Q, R$ and $D$ are concyclic. Construction Join $D R, R Q$ and $Q P$ Proof in right angled $\triangle A D P, R$ is the mid-point of $A B$. $\therefore ...

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle.

Question: A chord of a circle of radius 30 cm makes an angle of 60 at the centre of the circle. Find the area of the minor and major segments. Solution: Let $A B$ be the chord of a circle with centre $O$ and radius $30 \mathrm{~cm}$ such that $\angle A O B=60^{\circ}$. Area of the sector $O A C B O=\frac{\pi r^{2} \theta}{360}$ $=\left(3.14 \times 30 \times 30 \times \frac{60}{360}\right) \mathrm{cm}^{2}$ $=471 \mathrm{~cm}^{2}$ Area of $\Delta O A B=\frac{1}{2} r^{2} \sin \theta$ $=\left(\frac{...

55 cows can graze a field in 16 days.

Question: 55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days? Solution: Letxbe the number of cows that can graze the field in 10 days . Since the number of cows and the number of days taken by them to graze the field are in inverse variation, we have: $16 \times 55=10 \times x$ $\Rightarrow x=\frac{16 \times 55}{10}$ $=88$ $\therefore$ The required number of cows is 88...

A group of 3 friends staying together, consume 54 kg of wheat every month.

Question: A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How many new members are there in this group now? Solution: Letxbe the number of new members in the group. Since more members can finish the wheat in less number of days, it is a case of inverse variation. Therefore, we get: $3 \times 30=x \times 18$ $\Rightarrow 90=18 x$ $\Rightarrow x=\frac{90}{18}$ Thus, the numbe...

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m.

Question: A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track. Solution: Letrm andRm be the inner and outer boundaries, respectively.Thus, we have: $2 \pi r=352$ $\Rightarrow r=\frac{352}{2 \pi}$ Also, $2 \pi \mathrm{R}=396$ $\Rightarrow R=\frac{396}{2 \pi}$ Width of the track $=(R-r)$ $=\left(\frac{396}{2 \pi}-\frac{352}{2 \pi}\right) \mathrm{m}$ $=\frac{1}{2 \pi}(396-352) \mathrm{m}$ $=\left(\frac{1}...

Three spraying machines working together can finish painting a house in 60 minutes.

Question: Three spraying machines working together can finish painting a house in 60 minutes. How long will it take for 5 machines of the same capacity to do the same job? Solution: Let the time taken by 5 spraying machines to finish a painting job bexminutes. Since the number of spraying machines and the time taken by them to finish a painting job are in inverse variation, we have: $3 \times 60=5 \times x$ $\Rightarrow 180=5 x$ $\Rightarrow x=\frac{180}{5}$ $=36$ Thus, the required time will be...

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m.

Question: A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track. Solution: Letrm andRm be the inner and outer boundaries, respectively.Thus, we have: $2 \pi r=352$ $\Rightarrow r=\frac{352}{2 \pi}$ Also, $2 \pi \mathrm{R}=396$ $\Rightarrow R=\frac{396}{2 \pi}$ Width of the track $=(R-r)$ $=\left(\frac{396}{2 \pi}-\frac{352}{2 \pi}\right) \mathrm{m}$ $=\frac{1}{2 \pi}(396-352) \mathrm{m}$ $=\left(\frac{1}...

Solve this problem

Question: If $A=\frac{1}{9}\left[\begin{array}{ccc}-8 1 4 \\ 4 4 7 \\ 1 -8 4\end{array}\right]$, prove that $A^{-1}=A^{3}$ Solution: $A=\frac{1}{9}\left[\begin{array}{lll}-8 1 4\end{array}\right.$ $\begin{array}{lll}4 4 7\end{array}$ $\left.\begin{array}{lll}1 -8 4\end{array}\right]=\left[\begin{array}{lll}-8 / 9 1 / 9 4 / 9\end{array}\right.$ $\begin{array}{ccc}4 / 9 4 / 9 7 / 9 \\ 1 / 9 -8 / 9 4 / 9]\end{array}$ $\Rightarrow A^{T}=\left[\begin{array}{lll}-8 / 9 4 / 9 1 / 9\end{array}\right.$ $...

1200 soldiers in a fort had enough food for 28 days.

Question: 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort? Solution: It is given that after 4 days, out of 28 days, the fort had enough food for 1200 soldiers for $(28-4=24)$ days. Let $x$ be the number of soldiers who left the fort. Since the number of soldiers and the number of days for which the food lasts are in inverse variation, we have : $1200 \t...

The minute hand of a clock is 7.5 cm long.

Question: The minute hand of a clock is 7.5 cm long. Find the area of the face of the clock described by the minute hand in 56 minutes. Solution: The length of minute hand of clock $=7.5 \mathrm{~cm}$. The angle made by minute hand in 60 minutes $=360^{\circ}$. The angle made by minute hand in 1 minute $=6^{\circ}$. The angle made by minute hand in 56 minutes $=56 \times 6^{\circ}=336^{\circ}$. So, the area of clock described by minute hand in 56 minutes = area of sector with angle $336^{\circ}=...

A car can finish a certain journey in 10 hours at the speed of 48 km/hr.

Question: A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance? Solution: Letthe increased speed bex km/h. Since speed and time taken are in inverse variation, we get: $10 \times 48=8(x+48)$ $\Rightarrow 480=8 x+384$ $\Rightarrow 8 x=480-384$ $\Rightarrow 8 x=96$ $=12$ Thus, the speed should be increased by $12 \mathrm{~km} / \mathrm{h}$....

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm.

Question: In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region. Solution: Draw $O D \perp B C$. Because $\Delta A B C$ is equilateral, $\angle A=\angle B=\angle C=60^{\circ}$. Thus, we have: $\angle O B D=30^{\circ}$ $\Rightarrow \frac{O D}{O B}=\sin 30^{\circ}$ $\Rightarrow \frac{O D}{O B}=\frac{1}{2}$ $\Rightarrow O D=\left(\frac{1}{2} \times 4\right) \mathrm{cm} \quad[\because O B=$ radius $]$ $\Rightarrow O D=2 \mathrm...

In a hostel of 50 girls, there are food provisions for 40 days.

Question: In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel, how long will these provisions last? Solution: Let $x$ be the number of days with food provisions for 80 (i.e., $50+30$ ) girls. Since the number of girls and the number of days with food provisions are in inverse variation, we have: $50 \times 40=80 x$ $\Rightarrow x=\frac{50 \times 40}{80}$ $=\frac{2000}{80}$ $=25$ Thus, the required number of days is 25 ....

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm.

Question: In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region. Solution: Draw $O D \perp B C$. Because $\Delta A B C$ is equilateral, $\angle A=\angle B=\angle C=60^{\circ}$. Thus, we have: $\angle O B D=30^{\circ}$ $\Rightarrow \frac{O D}{O B}=\sin 30^{\circ}$ $\Rightarrow \frac{O D}{O B}=\frac{1}{2}$ $\Rightarrow O D=\left(\frac{1}{2} \times 4\right) \mathrm{cm} \quad[\because O B=$ radius $]$ $\Rightarrow O D=2 \mathrm...

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