5 : 27 :: 9 : ________ [#1134]

5 : (5² + 2) = 5 : 27 Hence 9 : (9² + 2) = 9 : 83

---

5 : (5² + 2) = 5 : 27 Hence 9 : (9² + 2) = 9 : 83

A : B = 3 => $\frac{A}{B}$ = 3 $\frac{144A}{36B}$ = $\frac{144}{36}\times \frac{A}{B}$ = $\frac{144}{36}$ x 3 = 4 x 3 = 12

---

A : B = 3 => $\frac{A}{B}$ = 3 $\frac{144A}{36B}$ = $\frac{144}{36}\times \frac{A}{B}$ = $\frac{144}{36}$ x 3 = 4 x 3 = 12

41

---

41

2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20

---

2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20

12 ---- x 60 = 48 KMPH 15

---

12 ---- x 60 = 48 KMPH 15

The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec

---

The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec

7 and 8

---

7 and 8

LCM will be r. px= q and qy = r, hence r = pxy.

---

LCM will be r. px= q and qy = r, hence r = pxy.

(360°/15) * (7-1) = 144°

---

(360°/15) * (7-1) = 144°

$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

---

$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

39

---

39