time and work Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (b)

Let speed of jogging be x km/h.

Total time taken = $(9/6 hrs + 1.5 hrs)$ = 3 hrs.

Total distance covered = (9 + 1.5x) km.

∴ ${9 + 1.5x}/3$ = 9⇒9 + 1.5x = 27

⇒$3/2 x = 18 ⇒ x = (18 × 2/3)$ = 12 kmph.

Answer: (a)

Let A and B together work for x minutes than amount of water filled in the period

= $(1/{30} + 1/{40}) = {7x}/{120}$

Remaining part = 1 - ${7x}/{120} = ({120 - 7x}/{120})$

Work done by A in (10 – x) minutes = ${120 - 7x}/{120}$

${7x}/{120} + {10 - x}/{30}$ = 1 or 7x + 40 – 4x = 120

3x = 120 – 40 = 80

x = 26 $2/3$ min

Answer: (a)

Let man's rate upstream = x km/hr.

Then, man's rate downstream = 2x km/hr.

∴ Man's rate in still water = $1/2$(x + 2x) km/hr.

∴ ${3x}/2$ = 6 or x = 4 km/hr.

Thus, man's rate upstream = 4 km/hr.

Man's rate downstream = 8 km/hr.

∴ Rate of stream = $1/2$(8 - 4) km/hr = 2 km/hr.

Answer: (d)

After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80 soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts = 10 – 4 = 6 days.

Answer: (a)

Let the distance be x km. Let speed of train be y km/h.

Then by question, we have

$x/{y + 4} = x/y - {30}/{60}$ ...(i)

and $x/{y - 2} = x/y + {20}/{60}$ ...(ii)

On solving (i) and (ii), we get x = 3y

Put x = 3y in (i) we get

${3y}/{y + 4} = 3 - 1/2$⇒ y = 20

Hence, distance = 20 × 3 = 60 km.