quick answer please
Q. 1 What percent is 13 weeks of one year ?
Q. 2 Mr. Gupta purchases a house for ₹ 2,50,000 and spends ₹ 50,000 on repairs. If he sells it for ₹ 4,00,000, find his gain percent.
Q3 Find the simple interest on ₹ 2000 for six months at the rate of 8% per annum.
Q.4 Find the amount Meena gets on depositing ₹ 1500 at 8% per annum for 73 days.
Q5 What percent is 1/2 dozen of one score?
Q6 Write the definitions of : line segment , Collinear points, intersecting lines, parallel lines and concurrent lines.
with regards
minerva78
Answers
Step-by-step explanation:
1. 1 year=52 weeks
13 weeks /1 year×100
=13/52 ×100
=1/4×100
=25%
2. amount he used on house=2,50,000+50,000
=3,00,000
sells at=4,00,000
gain %=(4,00,000-3,00,000)÷3,00,000×100
=1/3×100
=33.33%
3. P=2000
T=6months
R=8% per Annum
simple interest =PRT÷100
=2000×1/2×8÷100
=1000×8÷100
=8000÷100
=Rs.80
4. P=1500
T=73 days=1 term
R=8%÷5=1.6%for 73 days
A=P+I
I=PRT÷100
=1500×1× 1.6÷100
=15×1.6
=24
A=1500+24
=1524
5.1/2 dozen=6
1 score=20
6/20×100
=6×5
=30%
6. line segment is a part of a line with finite no. of points
collinear points are the points that fall on a straight line
intersecting lines are the lines that have a common point
parallel lines are the lines that do not have ant points in common.they are lines that never intersect
regards
sumanthbhat99
Answer:
The percentage of 13 weeks of one year(leap year) = 14.2855 %
q2 Mr Gupta purchased a house for Rs.250000
So,
Cost Price (C.P) of the house be Rs 250000
Mr Gupta spends Rs 50000 for repairing the house
So, total C.P will be
Rs 250000 + Rs 50000 = Rs 300000
Now,
Mr Gupta sells the house at Rs 400000
So,
Selling Prices (S.P) of the house = Rs 400000
Now,
C.P < S.P
Than its profit
So,
Profit = S.P - C.P
Profit = Rs 400000 - Rs 300000
Profit = Rs 100000
To be found :-
The profit percent
Profit% = \frac{Profit}{C.P} \times 100C.PProfit×100
Profit% = \frac{100000}{300000} \times 100300000100000×100
Profit% = 33.33%
Hence
Mr Gupta gained 33.33% profit
q3
Principal (P) = Rs 2000
Rate of Interest (R) = 4 whole \frac{1}{2}21 % per annum = 4.5 % p.a
Time (T) = 6 months = \frac{6}{12}126 years = \frac{1}{2}21 years = 0.5 years
To be found :-
The Simple Interest (S.I)
S.I = \frac{P \times R \times T }{100}100P×R×T
S.I = \frac{2000 \times 4.5 \times 0.5}{100}1002000×4.5×0.5
S.I = Rs 45
Hence
The Simple Interest (S.I) = Rs 45
q4 incipal (P) = Rs 1600
Rate of Interest (R) = 15% per annum
Time (T) = 73 days = \frac{73}{365}36573 years = 0.2 years
[ as there are 365 days in a normal year, and to convert days into year we have to divide the given days by 365 day (you can also divide by 366 is in the question leap years would be given) ]
To be found :-
The Simple Interest (S.I)
and Amount (A)
q5 1/2 dozen of one score is 30%.
Step-by-step explanation:
We have to find the percent of 1/2 dozen in one score.
As we know that 1 dozen = 12 units.
\frac{1}{2}\text{ Dozen}=\frac{1}{2}\times 12=6units21 Dozen=21×12=6units
1 score = 20 units.
By using this conversion, we get
\frac{6}{20}\times 100=30206×100=30
Therefore the 1/2 dozen of one score is 30%.
q6 (i) Line-segment- Give two points A and B on a line l, the connected part (segment) of the line with end points at A and B is called the line segment AB. ... The common point is called point of intersection. (v) Concurrent lines – Three or more lines are said to be concurrent if there is a point which lies on all of them.