type 1 angle between the hands of clock Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

∴ Angle traced by hour hand per minute =$(1/2)^o$

∴Angle traced by hour hand in 8h 20 min = ( 8 x 60 + 20) ×$ 1/2$ = 205°

Again, angle traced by minute hand per minute = 6°

∴ again traced by minute hand in 20 min = 20 × 6°

= 120° 

Therefore required angle = (250° - 120°) = 130°

Answer: (d)

∴ angle traced by hour hand per minute = $(1/2)^o$

∴ angled traced by hour hand in 8 h 30 min =$[{(8 × 60) + 30}x1/2]^o$

= $[{480 + 30} ×1/2]^o$ = 510 × $(1/2)^o$ = 255°

∴ Angle traced by minute hand per minute = 6°

∴ Angle traced by minute hand in 30 min = 30 × 6° = 180°

∴ required angle = (255° - 180°) = 75°

Answer: (a)

∴ Angle traced by hour hand per minute = $(1/2)^o$

∴ Angle traced by hour hand in 1 h = 1°/2 × 60 = 30°

Time period between 8 O' clock to 2 O' clock = 6h

∴ angle traced by hour hand in 6h = 30° × 6 = 180°

Answer: (b)

Angle traced by hour hand per minute = $(1/2)^o$

∴ Angle traced by hour hand in 7 h 35 min = [(7 × 60) + 35] × $1°/2$

= (420 + 35) × 1°/2 = 455 × $1°/2$ = 227 $1°/2$

∴ angle traced by minute hand per minute = 6°

Angle traced by minute hand in 35 min = 35 × 6° = 210°

∴ required angle = 227 1°/2 - 201° = 17$1°/2$

Answer: (d)

∴ Angle traced by hour hand per minute = $(1/2)^o$

∴ Angled traced by hour hand in 8 h 30 min =$[{(8 × 60) + 30}x1/2]^o$

= $[{480 + 30} ×1/2]° = 510 × (1/2)^o = 255°$

Then,  Angle traced by minute hand per minute = 6°

∴ Angle traced by minute hand in 30 min = 30 × 6° = 180°

Hence, Required angle = (255° - 180°) = 75°