Prove that
Question: Prove that LCM $\{6 !, 7 !, 8 !\}=8 !$ Solution: To Prove : LCM $\{6 !, 7 !, 8 !\}=8 !$ Formula: $n !=n \times(n-1) !$ LCM is the smallest possible number that is a multiple of two or more numbers. Here, we observe that (8!) is the first number which is a multiple of all three given numbers i.e. $6 !, 7 !$ and $8 !$. $1 \times(8 !)=8 !$ $8 \times(7 !)=8 !$ $8 \times 7 \times(6 !)=8 !$ Therefore, $8 !$ is the LCM of $\{6 !, 7 !, 8 !\}$ Conclusion : Hence proved...
Read More →Compute:
Question: Compute: (i) $\frac{9 !}{(5 !) \times(3 !) !}$ (ii) $\frac{32 !}{29 !}$ (iii) $\frac{(12 !)-(10 !)}{9 !}$ Solution: (i) To Find : Value of $\frac{9 !}{(5 !) \times(3 !)}$ Formulae : $n !=n \times(n-1) !$ $n !=n \times(n-1) \times(n-2) \ldots \ldots \ldots \ldots 3 \times 2 \times 1$ Let, $x=\frac{9 !}{(5 !) \times(3 !)}$ By using above formula, we can write, $\therefore x=\frac{9 \times 8 \times 7 \times 6 \times(5 !)}{(5 !) \times(3 \times 2 \times 1)}$ Cancelling ( $5 !$ ) from numer...
Read More →Hamid has three boxes of different fruits.
Question: Hamid has three boxes of different fruits. Box A weighs $2 \frac{1}{2} \mathrm{~kg}$ more than box B and Box C weighs $10 \frac{1}{4} \mathrm{~kg}$ more than box $\mathrm{B}$. The total weight of the three boxes is $48 \frac{3}{4} \mathrm{~kg}$. How many kilograms does box A weigh? Solution: Let the weight of box A be $x \mathrm{~kg}$. According to the question, Weight of Box $A=$ Weight of Box $B+\frac{5}{2} \mathrm{~kg}$ [given] $\Rightarrow \quad$ Weight of box $B=\left(x-\frac{5}{2...
Read More →The sum of four consecutive integers
Question: The sum of four consecutive integers is 266. What are the integers? Solution: Let the four consecutive integers be $x, x+1, x+2$ and $x+3$. According to the question, $x+x+1+x+2+x+3=266$ [given] $\Rightarrow \quad 4 x+6=266$ $\Rightarrow$ $4 x=266-6$ $\Rightarrow$ $4 x=260$ $\Rightarrow$ $x=\frac{260}{4}$ $\therefore$ $x=65$ Hence, the four consecutive integers are $65,65+1,65+2$ and $65+3, i, e, 65,66,67$ and 68 ....
Read More →If 1/2 is subtracted from a number
Question: If 1/2 is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number? Solution: Let the number be $x$. According to the question, $4\left(x-\frac{1}{2}\right)=5 \quad \Rightarrow \quad 4 x-2=5$ $\Rightarrow$ $4 x=5+2 \Rightarrow 4 x=7$ $\therefore$ $x=\frac{7}{4}$ Hence, the required number is $\frac{7}{4}$....
Read More →Anima left one-half of her property
Question: Anima left one-half of her property to her daughter, one-third to her son and donated the rest to an educational institute. If the donation was worth Rs 100000, how much money did Anima have? Solution: Let Anima's property be $₹ x$. Property left for her daughter $=₹ \frac{x}{2}$ Remaining property $=x-\frac{x}{2}=\frac{2 x-x}{2}=₹ \frac{x}{2}$ Property left for her son $=\frac{1}{3}$ of remaining property $=\frac{1}{3} \times \frac{x}{2}=₹ \frac{x}{6}$ Remaining property $=\left[x-\le...
Read More →If a young man rides his motorcycle at 40 km per hour,
Question: If a young man rides his motorcycle at $40 \mathrm{~km}$ per hour, he has to spend $6 \mathrm{per} \mathrm{km}$ on petrol and if he rides it at $50 \mathrm{~km}$ hour, the petrol cost rises to 10 per $\mathrm{km}$. He has 500 to spend on petrol and wishes to find the maximum distance he can travel within one hour. Formulate the data in the form of inequation and draw a graph representing the solution of these inequations. Solution: Let the distance covered with speed $40 \mathrm{~km} /...
Read More →A furniture dealer deals in only two items : tables and chairs.
Question: A furniture dealer deals in only two items : tables and chairs. He has 30000 to invest and a space to store at most 60 pieces. A table costs him 1500 and a chair 300 . Formulate the data in the form of inequations and draw a graph representing the solution of these inequation. Solution: Let the number of tables and chairs be $x$ and $y$ respectively. Therefore $x \geq 0, y \geq 0$ Now the maximum number of pieces he can store $=60$. Therefore, $x+y \leq 60 \ldots \ldots$ (1) Also it is...
Read More →Find the linear inequalities for which the shaded area is the solution set in the figure given below.
Question: Find the linear inequalities for which the shaded area is the solution set in the figure given below. Solution: We have seen that the shaded region and origin are on the opposite side of the line $6 x+2 y=8$ For $(0,0)$ we have $0+0-80$. So the shaded region satisfies the inequality $6 x+2 y \geq 8$. We have seen that the shaded region and origin are on the opposite side of the line $x+5 y=4$ For $(0,0)$ we have $0+0-40$. So the shaded region satisfies the inequality $x+5 y \geq 4$. We...
Read More →Find the linear inequalities for which the shaded area is the solution set in the figure given below.
Question: Find the linear inequalities for which the shaded area is the solution set in the figure given below. Solution: We have seen that the shaded region and origin are on the same side of the line $3 x+4 y=12$ For $(0,0)$ we have $0+0-120$. So the shaded region satisfies the inequality $3 x+4 y \leq 12$. We have seen that the shaded region and origin are on the same side of the line $4 x+3 y=12$ For $(0,0)$ we have $0+0-120$. So the shaded region satisfies the inequality $4 x+3 y \leq 12$. ...
Read More →After 12 years, Kanwar shall be 3 times
Question: After 12 years, Kanwar shall be 3 times as old as he was 4 years ago. Find his present age. Solution: Let Kanwar's present age be $x$ yr. After 12.yr, Kanwar's age $=(x+12)$ y and 4 yr ago, Kanwar's age $=(x-4)$ yr According to the question, $x+12=3(x-4) \Rightarrow x+12=3 x-12$ $\Rightarrow \quad 3 x-x=12+12 \Rightarrow 2 x=24$ $\Rightarrow$ $x=\frac{24}{2}$ $\therefore \quad x=12$ Hence, Kanwar's present age is $12 \mathrm{yr}$....
Read More →Two equal sides of a triangle are each 4 m
Question: Two equal sides of a triangle are each 4 m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55 m. Solution: Let the third side of triangle be $x \mathrm{~m}$. Then, two equal sides of triangle $=(3 x-4) \mathrm{m}$ Given, perimeter of triangle $=55 \mathrm{~m}$ $\because$ Perimeter of a triangle $\simeq$ Sum of the sides of the triangle $\therefore \quad x+3 x-4+3 x-4=55$ $\Rightarrow \quad 7 x-8=55 \Rightarrow 7 x=55+8$ $\Rightarrow$ $7 x=...
Read More →Solve the given inequalities
Question: Solve the given inequalities $3 x+y \geq 12, x+y \geq 9, x \geq 0, y \geq 0$.graphically in two - dimensional plane. Solution: The graphical representation of $3 x+y \geq 12, x+y \geq 9$ $x \geq 0, y \geq 0$ is given by common region in the figure below. $3 x+y \geq 12 \ldots \ldots$ (1) $x+y \geq 9 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0 \ldots \ldots$ (4) Inequality (1) represents the region above line $3 x+y=12$ (including the line $3 x+y=12$ ). Inequality (2) repr...
Read More →Sum of the digits of a two-digit number is 11.
Question: Sum of the digits of a two-digit number is 11. The given number is less than the number obtained by interchanging the digits by 9. Find the number. Solution: Let the unit's digit be $x$. Then, the ten's digit $=11-x$ $\therefore$ Number $=10(11-x)+x=110-10 x+x=110-9 x$ Number obtained by interchanging the digits $=10 x+(11-x)=9 x+11$ According to the question, $9 x+11-(110-9 x)=9$ $\Rightarrow$ $9 x+11-110+9 x=9$ $\Rightarrow \quad 18 x=9-11+110$ $\Rightarrow$ $18 x=108$ $\Rightarrow$ ...
Read More →Solve the given inequalities
Question: Solve the given inequalities $2 x-y \leq-2, x-2 y \geq 0, x \geq 0, y \geq 0$ graphically in two - dimensional plane. Solution: The graphical representation of $2 x-y \leq-2, x-2 y \geq 0$ $x \geq 0, y \geq 0$ is given by common region in the figure below. $2 x-y \leq-2 \ldots \ldots$ (1) $x-2 y \geq 0 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0 \ldots \ldots$ (4) Inequality (1) represents the region above line $2 x-y=-2$ (including the line $2 x-y=-2$ ). Inequality (2) r...
Read More →Solve the given inequalities
Question: Solve the given inequalities $3 x+2 y \geq 24,3 x+y \leq 15, x \geq 4$, graphically in two - dimensional plane. Solution: The graphical representation of $3 x+2 y \geq 24,3 x+y \leq 15$ $x \geq 4$ is given by common region in the figure below. $3 x+2 y \geq 24 \ldots \ldots$ (1) $3 x+y \leq 15 \ldots \ldots$ (2) $x \geq 4 \ldots \ldots$ (3) Inequality (1) represents the region above line $3 x+2 y=24$ (including the line $3 x+2 y=24$ ). Inequality $(2)$ represents the region below line ...
Read More →Solve the given inequalities
Question: Solve the given inequalities $x+2 y \leq 2000, x+y \leq 1500, y \leq 600, x \geq 0, y \geq 0$ graphically in two dimensional plane. Solution: The graphical representation of $x+2 y \leq 2000, x+y \leq 1500$ $y \leq 600, y \geq 0, x \geq 0$ is given by common region in the figure below. $x+2 y \leq 2000 \ldots \ldots$ (1) $x+y \leq 1500 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0 \ldots \ldots$ (4) $y \leq 600 \ldots \ldots$ (5) Inequality (1) represents the region below l...
Read More →Find a number whose fifth part increased
Question: Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30. Solution: Let the number be $x$. According to tne question, $\frac{x}{5}+30=\frac{x}{4}-30$ $\Rightarrow$ $\frac{x}{5}-\frac{x}{4}=-30-30 \quad\left[\right.$ transposing $\frac{x}{4}$ to LHS and 30 to RHS $]$ $\Rightarrow$ $\frac{4 x-5 x}{20}=-60$ $\Rightarrow$ $-x=-60 \times 20$ [by cross-multiplication] $\therefore$ $x=1200$ Hence, the required number is 1200 ....
Read More →Solve the given inequalities
Question: Solve the given inequalities $x+2 y \leq 100,2 x+y \leq 120, x+y \leq 70, x \geq 0, y \geq 0$ graphically in two dimensional plane. Solution: The graphical representation of $x+2 y \leq 100,2 x+y \leq 120$ $x+y \leq 70, y \geq 0, x \geq 0$ is given by common region in the figure below. $x+2 y \leq 100 \ldots \ldots$ (1) $2 x+y \leq 120 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0 \ldots \ldots$ (4) $x+y \leq 70 \ldots \ldots$ (5) Inequality (1) represents the region below ...
Read More →The sum of three consecutive numbers is 156.
Question: The sum of three consecutive numbers is 156. Find the number which is a multiple of 13 out of these numbers. Solution: Let three consecutive numbers be $x_{1}(x+1)$ and $(x+2)$. According to the question, $x+(x+1)+(x+2)=156$ [given] $\Rightarrow$ $3 x+3=156 \Rightarrow 3 x=156-3=153$ $\therefore$ $x=153 \times \frac{1}{3}=51$ Thus, we get the numbers $51,51+1$ and $51+2, i . e, 51,52$ and 53 . Out of these, only 52 is a multiple of $13 .$...
Read More →Solve the given inequalities
Question: Solve the given inequalities $x-2 y \leq 2, x+y \geq 3,-2 x+y \leq 4, x \geq 0, y \geq 0 g r a p h i c a l l y$ in two - dimensional plane. Solution: The graphical representation of $x-2 y \leq 2, x+y \geq 3, y \geq 0$ $-2 x+y \leq 4, x \geq 0$ is given by common region in the figure below. $x-2 y \leq 2 \ldots \ldots$ (1) $x+y \geq 3 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0 \ldots \ldots$ (4) $-2 x+y \leq 4 \ldots \ldots \ldots$ (5) Inequality (1) represents the regio...
Read More →The sum of three consecutive
Question: The sum of three consecutive odd natural numbers is 69. Find the prime number out of these numbers. Solution: Let the three consecutive odd natural numbers be $x,(x+2)$ and $(x+4)$. According to the question, $x+(x+2)+(x+4)=69$ [given] $\Rightarrow$ $3 x+6=69$ $\Rightarrow$ $3 x=69-6 \Rightarrow 3 x=63$ $\therefore$ $x=63 \times \frac{1}{3}=21$ Thus, the three consecutive odd natural numbers are $21,(21+2)$ and $(21+4), i, e .21,23$ and $25 .$ Out of these, only 23 is the prime number....
Read More →Solve the given inequalities
Question: Solve the given inequalities $4 x+3 y \leq 60, y \geq 2 x, x \geq 3, x \geq 0, y \geq 0$ graphically in two - dimensional plane. Solution: The graphical representation of $4 x+3 y \leq 60, y \geq 2 x, y \geq 0$ $x \geq 3, x \geq 0$ is given by common region in the figure below. $4 x+3 y \leq 60$ ........(1) $y \geq 2 x \ldots \ldots(2)$ $x \geq 0$.............(3) $y \geq 0$ .........(4) $x \geq 3$ ...........(5) Inequality (1) represents the region below line $4 x+3 y=60$ (including th...
Read More →Solve the given inequalities
Question: Solve the given inequalities $x+2 y \leq 10, x+y \geq 1, x-y \leq 0, x \geq 0, y \geq 0$ graphically in two - dimensional plane. Solution: The graphical representation of $x+2 y \leq 10, x+y \geq 1, y \geq 0$ $x-y \leq 0, x \geq 0$ is given by common region in the figure below. $x+2 y \leq 10 \ldots \ldots$ (1) $x+y \geq 1 \ldots \ldots$ (2) $x \geq 0 \ldots \ldots$ (3) $y \geq 0$ . (4) $x-y \leq 0 \ldots \ldots(5)$ Inequality (1) represents the region below line $x+2 y=10$ (including ...
Read More →Anushka and Aarushi are friends.
Question: Anushka and Aarushi are friends. They have equal amount of money in their pockets. Anushka gave 1/3 of her money to Aarushi as her birthday gift. Then, Aarushi gave a party at a restaurant and cleared the bill by paying half of the total money with her. If the remaining money in Aarushis pocket is Rs 1600, then find the sum gifted by Anushka. Solution: Let Anushka and Aarushi have equal amount of money in their packet, which is $₹ x$. After giving $\frac{1}{3}$ of the money of Anushka ...
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