Mrs. Kaushik has a square plot with the
measurement as shown in the figure. She wants to
construct a house in the middle of the plot. A garden is developed
around the house. Find the total cost of developing a garden around
the house at the rate of Rs 55 per m2
.Please solve this question step by step
Would be really grateful
Answers
Answer:
Ncert solutions
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Mathematics
Science
Chapters in NCERT Solutions - Mathematics , Class 8
Exercises in Mensuration
Question 3
Q2) Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ` 55 per m2.
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Solution 2:
Area of the garden = Area of square plot - Area of house
Side of square plot=25m
Area of square plot side\times sideside×side
25\times2525×25
=625\ m^2=625 m
2
Area of house =l\times b=l×b
length=20 m
breadth=15 m
Area of House=20\times1520×15
=300\ m^2=300 m
2
Area of garden =625\ m^2-300\ m^2=625 m
2
−300 m
2
=325\ m^2=325 m
2
Cost of developing a garden =Rs\ 55\times325\ m^2Rs 55×325 m
2
=Rs 17,875
Step-by-step explanation:
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Step-by-step explanation:
Required\text{ }number=\left[ \left( Least\text{ }Common\text{ }Multiple \right).n \right]+1\]
Case 1: n = 2
By substituting n value in above equation, we get:
Required number = (60 x 2) + 1
= 121
121 is not divisible by 7.
Case 2: n = 3
By substituting n value in above equation, we get:
Required number = (60 x 3) + 1
= 181
181 is not divisible by 7.
Case 3: n = 4
By substituting the value of n in above equation, we get:
Required number = (60 x 4) + 1
= 241
241 is not divisible by 7.
Case 4: n = 5
By substituting the value of n in above equation, we get:
Required number = (60 x 5) + 1
= 301
301 is divisible by 7.
Therefore 301 is the least number which leaves remainder as 1 when divided by 2, 3, 4, 5, 6 and