Solve the following pair of equations 2 /x - 3 /y = 15 8 /x + 5 /y = 77

Solve the following pair of equations 7x - 2y = 5 xy 8 x + 7y = 15xy

7x - 2y = 5xy ................... (1) 8 x + 7y = 15xy ................. ( 2) Dividing eq1 and 2 with xy, 7/y - 2/x = 5 ............. (3) 8/y + 7/x = 15 ............ (4) Taking m =1/y and n = 1/x 7m - 2n = 5 ...................(5) 8m + 7n = 15 ....................(6) Multiplying eq. 5 with 7 and eq 6 with 2, 49m - 14n = 35 ........... (7) 16m + 14n = 30 ..............(8) Adding Eq. 7 and 8, 65 m = 65, m =1 Substituting value of m in Eq.5, 7 - 2n = 5 , - 2 n = -2, n =1 Resubstuting,1/x = 1, x = 1 1/y = 1, y =1 So correct option is Option 1.

13/(2x +1) + 7 / (y + 2) = 33 ................... (1) 7 / (2x+1) + 13/ (y + 2) = 27 ................. ( 2) Taking m =1/(2x + 1) and n = 1/(y + 2) 13m + 7n = 33 ...................(3) 7m + 13n = 27 ....................(4) Adding eq. 3 and eq 4, 20m + 20n = 60 m + n = 3 ..............(5) Subtracting Eq. 3 and 4, 6m - 6n = 6 m - n =1 ................. ( 6) Adding Eq. 5 and 6, 2m =4, m = 2 Substituting value of m in Eq.5, 2 + n = 3 , n =1 Resubstituting,1/2x+ 1 = 2, 4x + 2 =1 4x = -1, x = - 1/4 1/y+2 = 1, y+2 =1, y = -1 So correct option is Option 2

During lockdown, Amar and Birju decided to walk to their village 30 kms away. Amar took 3 hours more than Birju to reach the village. However,if Amar had walked twice as fast he would have reached 3/2 hours before Birju. Find their speed of walking.

Let Amar's speed = x Kmph Let Birju's speed = y Kmph As Time = Distance / Speed, we get, 30/x - 30/y = 3 ................(1) 30/y - 30/2x = 3/2 ...........(2) Taking 1/x = m , 1/y = n, 30m - 30n =3 10m - 10 n = 1 .............(3) 30n - 15m = 3/2 2n - m = 1/10 ...........(4) From eq, 4, m = 2n - 1/10 Substituting value of m in eq. 1, 10 ( 2n - 1/10) - 10 n = 1 20n - 1 -10n =1, 10n = 2 n = 1/5 and m = 2/5 - 1/10 = 3/10 Resubstituting, 1/x = 3/10, x = 10/3 1/y = 1/5, y = 5 So the correct answer is Option 2