A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is [#1711]

5² + X² = 13² => 25 + X² = 169 => X² = 196 - 25 => X² = 144 => X = 12

---

5² + X² = 13² => 25 + X² = 169 => X² = 196 - 25 => X² = 144 => X = 12

11

---

11

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

60

---

60

25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years

---

25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years

6*4 = 24 8*3 = 24 12*2 = 24

---

6*4 = 24 8*3 = 24 12*2 = 24

33

---

33

$\frac{X}{3}=\frac{Y}{4}=\frac{Z}{7}$ Hence, X:Y:Z = 3:4:7 $\frac{X+Y+Z}{Z}$ = $\frac{3+4+7}{7}$ = $\frac{14}{7}$ = 2

---

$\frac{X}{3}=\frac{Y}{4}=\frac{Z}{7}$ Hence, X:Y:Z = 3:4:7 $\frac{X+Y+Z}{Z}$ = $\frac{3+4+7}{7}$ = $\frac{14}{7}$ = 2

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

---

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.