Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
let the present ages of father and his son be \[~x\text{ }years\text{ }and\text{ }\left( 45\text{ }\text{ }x \right)\text{ }years\] hence five years ago, Father’s age \[=\text{ }\left( x\text{...
According to ques, \[S\text{ }=\text{ }n\left( n\text{ }+\text{ }1 \right)\] Also, \[S\text{ }=\text{ }420\] So, \[n\left( n\text{ }+\text{ }1 \right)\text{ }=\text{ }420\] Or, \[{{n}^{2}}~+\text{...
Let the speed of the second train be \[~x\text{ }km/hr.\] Then, the speed of the first train is \[\left( x\text{ }+\text{ }5 \right)\text{ }km/hr\] Let O be the position of the railway station,...
Let the usual speed of the plane to be \[x\text{ }km/hr\] The distance to travel \[=\text{ }1500km\] since, Time = Distance/ Speed As the ques suggests, \[{{x}^{2}}~+\text{ }250x\text{ }\text{...
let the original number of persons to be x. According to ques, Total money which was divided is \[=\text{ }Rs\text{ }6500\] Each person’s share is \[=\text{ }Rs\text{ }6500/x\] Then, as the question...
According to ques, Distance \[=\text{ }400\text{ }km\] Average speed of the airplane \[=\text{ }x\text{ }km/hr\] Also, speed while returning \[=\text{ }\left( x\text{ }+\text{ }40 \right)\text{...
Let the number of people staying overnight as x. According to ques, total hotel bill \[~=\text{ }Rs\text{ }4800\] Now,hotel bill for each person \[=\text{ }Rs\text{ }4800/x\] therefore,...
According to ques, Number of articles \[=\text{ }x\] And, the total cost of articles \[=\text{ }Rs\text{ }600\] Again, (i) Cost of one article \[=\text{ }Rs\text{ }600/x\] (ii) also,...
(iii) According to the question, \[4x\text{ }+\text{ }1728\text{ }=\text{ }{{x}^{2}}~+\text{ }16x\] Or, \[{{x}^{2}}~+\text{ }12x\text{ }\text{ }1728\text{ }=\text{ }0\] Or, \[{{x}^{2}}~+\text{...
According to ques, Speed of car = \[x\text{ }km/hr\] Speed of train = \[\left( x\text{ }+\text{ }16 \right)\text{ }km/hr\] Time = \[Distance/\text{ }Speed\] (i)Time taken by the car to reach town B...
We should expect the current age of the child to be x years. Thus, the current age of the dad = \[2x2\text{ }years\] Eight years henceforth, Child's age = \[\left( x\text{ }+\text{ }8 \right)\] a...
ACCORDING TO QUES, How about we think about the current age of the child to be x years. Along these lines, the current age of the man \[=\text{ }x2\text{ }years\] One year prior, Child's age...
Given, the ages of two sisters are 11 years and 14 years. Leave x alone the quantity of years some other time when their result of their ages become 304. In this way, \[\left( 11\text{ }+\text{ }x...
How about we accept the ten's and unit's digit of the necessary number to be x and y individually. Then, at that point, from the inquiry we have \[x\text{ }\times \text{ }y\text{ }=\text{ }24\]...
As indicated by the inquiry, \[16t2\text{ }+\text{ }4t\text{ }=\text{ }420\] \[4t2\text{ }+\text{ }t\text{ }\text{ }105\text{ }=\text{ }0\] \[4t2\text{ }\text{ }20t\text{ }+\text{ }21t\text{ }\text{...
From the inquiry, we have \[n\left( n\text{ }+\text{ }2 \right)\text{ }=\text{ }168\] \[n2\text{ }+\text{ }2n\text{ }\text{ }168\text{ }=\text{ }0\] \[n2\text{ }+\text{ }14n\text{ }\text{ }12n\text{...
Given, The young lady covers a distance of 6 km at a speed x km/hr. Along these lines, the time taken to cover initial 6 km \[=\text{ }6/x\text{ }hr\] [Since, Time = Distance/Speed] Likewise given,...
We should accept x km/h to be the first speed of the vehicle. We realize that, Time = Distance/Speed From the inquiry, The time taken by the vehicle to finish 400 km = \[400/x\text{ }hrs\]...
How about we think about the first speed of the plane to be x km/hr. Presently, the time taken to cover a distance of \[1200\text{ }km\text{ }=\text{ }1200/x\text{ }hrs\] [Since, Time =...
(i) Given, Speed of the conventional train \[=\text{ }x\text{ }km/hr\] Speed of the express train \[=\text{ }\left( x\text{ }+\text{ }25 \right)\text{ }km/hr\] Distance \[=\text{ }300\text{ }km\] We...
We should take the length and the expansiveness of the square shape be x m and y m. Thus, the edge \[=\text{ }2\left( x\text{ }+\text{ }y \right)\text{ }m\] \[104\text{ }=\text{ }2\left( x\text{...
How about we think about the more limited side of the square shape to be x m. Then, at that point, the length of the opposite side \[=\text{ }\left( x\text{ }+\text{ }30 \right)\text{ }m\] Length of...
Leave the hypotenuse of the right triangle alone x cm. From the inquiry, we have Length of one side \[=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }cm\] Length of opposite side \[=\text{ }\left(...
Given, The more drawn out side = \[Hypotenuse\text{ }=\text{ }\left( 3x\text{ }+\text{ }1 \right)\text{ }cm\] Furthermore, the lengths of other different sides are\[\left( x\text{ }\text{ }1...
Given, a right triangle \[Hypotenuse\text{ }=\text{ }26\text{ }cm\] and the amount of other different sides is \[34\text{ }cm.\] Presently, let believe the other different sides to be \[x\text{...
Given, the space of triangle \[=\text{ }30\text{ }cm2\] As, x can't be negative, just \[x\text{ }=\text{ }3\] is substantial. Thus, we have \[AB\text{ }=\text{ }4\text{ }\times \text{ }3\text{...
Leave the two back to back certain even numbers alone taken as \[x\text{ }and\text{ }x\text{ }+\text{ }2.\] Now, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }2 \right)2\text{ }=\text{ }52\]...
How about we expect the two numbers to be x and y, y being the bigger of the two numbers. Then, at that point, from the inquiry \[x2\text{ }+\text{ }y2\text{ }=\text{ }208\text{ }\ldots \text{...
We should expect the two sections to be \[x\text{ }and\text{ }15\text{ }\text{ }x.\] now, \[150\text{ }=\text{ }45x\text{ }\text{ }3x2\] \[3x2\text{ }\text{ }45x\text{ }+\text{ }150\text{ }=\text{...
How about we believe the two normal numbers to be\[x\text{ }and\text{ }x\text{ }+\text{ }3\] . (As they vary by 3) now, \[20x\text{ }+\text{ }30\text{ }=\text{ }7x2\text{ }+\text{ }21x\] \[7x2\text{...
Leave the number alone x. In this way, its complementary is 1/x now, \[4x2\text{ }\text{ }17x\text{ }+\text{ }4\text{ }=\text{ }0\] \[4x2\text{ }\text{ }16x\text{ }\text{ }x\text{ }+\text{...
We should expect the two normal numbers to be\[x\text{ }and\text{ }x\text{ }+\text{ }5\] . (As given they contrast by 5) So from the inquiry, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }5...
Allow us to take the two successive regular numbers as x and x + 1. So from the inquiry, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }1 \right)2\text{ }=\text{ }41\] \[2x2\text{ }+\text{ }2x\text{...
let two continuous numbers to be $$ \[x\text{ }and\text{ }x\text{ }+\text{ }1.\] So from the inquiry, \[x\left( x\text{ }+\text{ }1 \right)\text{ }=\text{ }56\] \[x2\text{ }+\text{ }x\text{...