# Type 1 Angle Between The Hands Of Clock Practice Questions Answers Test With Solutions & More Shortcuts

#### CLOCKS PRACTICE TEST [5 - EXERCISES]

Question : 1 [Hotel Mgmt 012]

Find the angle between the hands of clock at 8 : 20.

a) 130°

b) 115°

c) 125°

d) 140°

∴ 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°

Question : 2

What will be the angle between the hands of clock at 7 : 10?

a) 185°

b) 155°

c) 170°

d) 165°

∴ 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°

Question : 3

Find the angle traced by hour hand of a correct clock between 8 O' clock and 2O' clock.

a) 180°

b) 168°

c) 150°

d) 144°

∴ 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°

Question : 4 [M.A.T. 2010]

What angle will be traced by the hands of a clock at 7 : 35?

a) \$7 1°/2\$

b) \$17 1°/2\$

c) \$27 1°/2\$

d) \$37 1°/2\$

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\$

Question : 5 [S.S.C. (CPO) 2015]

What will be the angle between the hands of clock at 8 : 30?

a) 15°

b) 30°

c) 45°

d) 75°

∴ 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°

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