CRE 1 - Angle between hands of a clock

At 6’O clock the hands of the clock are exactly opposite to each other. 

∴ The angle between the hands will be 180°.

Hence, option (b).

At 3’O clock the hands of the clock are exactly perpendicular to each other. 

∴ The angle between the hands will be 90°.

Hence, option (c).

The clock can be divided into 12 sectors of 30° each. (A sector would cover the area between any two consecutive numbers and the center.)

At 4’O clock the area between the two hands would cover 4 such sectors.

∴ The angle between the hands will be 4 × 30° = 120°.

Hence, option (a).

At 4:30 the minute hand points towards 6, while the hour hand will be exactly between 4 and 5.

∴ The angle between the two hands will be equivalent to that of one and a half sector.

∴ Angle between the two hands = 1.5 × 30° = 45°

Alternately,
Angle between the two hands at h : m = $\left|30h-\frac{11}{2}m\right|$

∴ At 4 : 30 the angle will be = $\left|30\times 4-\frac{11}{2}\times 30\right|$ = |120 – 165| = 45°.

Hence, option (d).

Angle between the two hands at h : m = $\left|30h-\frac{11}{2}m\right|$

∴ At 6 : 40 the angle will be = $\left|30\times 6-\frac{11}{2}\times 40\right|$ = |180 – 220| = 40°.

Hence, option (c).

Angle between the two hands at h : m = $\left|30h-\frac{11}{2}m\right|$

∴ At 9 : 20 the angle will be = $\left|30\times 9-\frac{11}{2}\times 20\right|$ = |270 – 110| = 160°.

Hence, option (b).

Hands should be in straight line but not together ⇒ hands should be opposite each other.

Angle between the two hands at h : m = |$\left|30h-\frac{11}{2}m\right|$| = 180°.

Between 7 and 8, h = 7.

∴ |$\left|30\times 7-\frac{11}{2}m\right|$| = 180°

⇒ |$\left|210-\frac{11}{2}m\right|$| = 180°

⇒  210 - $\frac{11}{2}$m = ± 180°

Case 1: 210 - $\frac{11}{2}$m = 180°

⇒ $\frac{11}{2}$m = 30°

⇒ m = 60/11 minutes = 5$\frac{5}{11}$  minutes

∴ Hands will be opposite at 5$\frac{5}{11}$ minutes past 7.

Case 2: 210 - $\frac{11}{2}$m = - 180°

⇒ $\frac{11}{2}$m = 370°

⇒ m = $\frac{740}{11}$ (not possible since m should be ≤ 60).

Hence, option (d).

Hands of a clock overlap once every hour except between 11 and 1 O’clock when they overlap only once in these 2 hours (at 12 O'clock).

∴ The hands of a clock overlap 11 times in every 12 hours.

∴ So, in a day the hands overlap 22 times.

Hence, option (d).

Hands of a clock are opposite to each other once every hour except between 5 and 7 O’clock when they are opposite only once in these 2 hours (at 6 O'clock).

∴ The hands of a clock are opposite 11 times in every 12 hours.

∴ So, in a day the hands are opposite 22 times.

Hence, option (d).

Hands of a clock are perpendicular twice every hour except between 2 - 4 O’clock and 8 – 10 O’clock when they are perpendicular thrice in there 2 hours.

∴ In 12 hours they are at right angles 22 times (because two positions 3 O'clock and 9 O'clock are common)

∴ In a day they are at right angle 44 times.

Hence, option (d).

Hands overlap when the angle between the two hands is 0°.

We know, angle between the hands of a clock (θ°) = |$\left|30\mathrm{h}-\frac{11}{2}\mathrm{m|}\right|$

⇒ 0° = |$\left|30\times 3-\frac{11}{2}\mathrm{m|}\right|$

⇒ 0° = 90 - $\frac{11}{2}$m

⇒ m = $\frac{180}{11}$ = 16$\frac{4}{11}$minutes

∴ The hands will coincide at 3 : 16$\frac{4}{11}$.

Hence, option (d).

We know, angle between the hands of a clock (θ°) = |$\left|30\mathrm{h}-\frac{11}{2}\mathrm{m|}\right|$

⇒ 50° = |$\left|30\times 7-\frac{11}{2}\mathrm{m|}\right|$

Case 1:
⇒ 50° = 210 - $\frac{11}{2}$m

⇒ $\frac{11}{2}$m = 160°

⇒ m = $\frac{320}{11}$ = 29$\frac{1}{11}$minutes

∴ The hands will be at an angle of 50° at 7 : 29$\frac{1}{11}$.

Case 2:
⇒ -50° = 210 - $\frac{11}{2}$m

⇒ $\frac{11}{2}$m = 260°

⇒ m = $\frac{520}{11}$ = 47$\frac{3}{11}$minutes

∴ The hands will be at an angle of 50° at 7 : 47$\frac{3}{11}$.

Hence, option (d).

We know, angle between the hands of a clock (θ°) = |$\left|30\mathrm{h}-\frac{11}{2}\mathrm{m|}\right|$

⇒ 69° = |$\left|30\times 10-\frac{11}{2}\mathrm{m|}\right|$

Case 1:
⇒ 69° = 300 - $\frac{11}{2}$m

⇒ $\frac{11}{2}$m = 231°

⇒ m = 42 minutes

∴ The hands will be at an angle of 69° at 10 : 42.

Case 2:
⇒ -69° = 300 - $\frac{11}{2}$m

⇒ $\frac{11}{2}$m = 369°

⇒ m = $\frac{738}{11}$ = 67$\frac{1}{11}$minutes

∴ The hands will be at an angle of 69° at 10 : 67$\frac{1}{11}$.

Hence, option (b).