The ratio of boys and girls in a school is 18:12. To make the number of girls is equal to boys, how much percentage should be increased
Answers
Answer:
Here, 2m = 6 \Rightarrow\ m=\frac{6}{2}=3⇒ m=
2
6
=3
Therefore, Second number \left(m^2-1\right)=\left(3\right)^2-1=9-1=8(m
2
−1)=(3)
2
−1=9−1=8
Third number m^2+1=\left(3\right)^2+1=9+1=10m
2
+1=(3)
2
+1=9+1=10
Hence, Pythagorean triplet is (6, 8, 10).
(ii) There are three numbers
2m,\ m^2-1\ and\ m^2+12m, m
2
−1 and m
2
+1 in a Pythagorean Triplet.
Here, 2m = 14 \Rightarrow m=\frac{14}{2}=7⇒m=
2
14
=7
Therefore, Second number m^2-1=\left(7\right)^2-1=49-1=48m
2
−1=(7)
2
−1=49−1=48
Third number m^2+1=\left(7\right)^2+1=49+1=50m
2
+1=(7)
2
+1=49+1=50
Hence, Pythagorean triplet is (14, 48, 50).
(iii) There are three numbers 2m,m^2-1\ and\ m^2+12m,m
2
−1 and m
2
+1 in a Pythagorean Triplet.
Here, 2m = 16 \Rightarrow m=\frac{16}{2}=8⇒m=
2
16
=8
Therefore, Second number \left(m^2-1\right)=\left(8\right)^2+1=64+1=65(m
2
−1)=(8)
2
+1=64+1=65
Hence, Pythagorean triplet is (16, 63, 65).
(iv) There are three numbers 2m,\ m^2-1\ and\ m^2+12m, m
2
−1 and m
2
+1 in a Pythagorean Triplet.
Here, 2m = 18 \Rightarrow m=\frac{18}{2}=9⇒m=
2
18
=9
Therefore, Second number \left(m^2-1\right)=\left(9\right)^2-1=81-1=80(m
2
−1)=(9)
2
−1=81−1=80
Third number m^2+1=\left(9\right)^2+1=81+1=81m
2
+1=(9)
2
+1=81+1=81
Hence, Pythagorean triplet is (18, 80, 82)
Answer
Correct option is
C
21 : 22
let the number of boys and girls in the college be 7x and 8x
their increased number is (120% of 7x) =
⇒
100
120
×7x
⇒
5
42x
And (10% of 8x).=⇒
100
110
×8x
⇒
5
44x
∴Required Ratio=
5
42x
:
5
44x
21:22