Question

a number of boys approached a gentleman for donation to a famine fund the gentleman found that he would have to pay rupees 80 more if he pays to each boy as many 50 paise as there were boys than if he pays to each thrice as many 10 paise as the number of boys find the number of boys​

Answer 1

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)