Explanation

Explanation

Speed = Distance ÷ Time

= 30 m ÷ 5 s

= 6 m/s

Explanation

Work done by 12 men in 4 days = 1 (full work)

Work done by 1 man in 4 days = 1/12

Work done by 6 men in 4 days = 6/12 = 1/2 (i.e., half of the work)

Work done by 6 men in 2 days = (1/2) × (2/4) = 1/4

Remaining work = 1 - 1/4 = 3/4

15 women complete the full work in 4 days → 1 woman’s 1-day work = 1/(15 × 4) = 1/60

Work needed in 3 days → x women × 3 × (1/60) = 3/4

x = (3/4) × (60/3) = 15 women

Explanation

Work done by 10 women in 7 days = 1

Work done by 1 woman in 7 days = 1/10

Work done by 1 woman in 1 day = 1/(10 × 7) = 1/70

Work done by 10 children in 14 days = 1

Work done by 1 child in 14 days = 1/10

Work done by 1 child in 1 day = 1/(10 × 14) = 1/140

Work done by 5 women in 1 day = 5 × 1/70 = 5/70 = 1/14

Work done by 10 children in 1 day = 10 × 1/140 = 10/140 = 1/14

Total work done per day = 1/14 + 1/14 = 2/14 = 1/7

Total days required = 7 days

Explanation

Step 1: Work Done by A and B Together

• A’s 1-day work = 1/10
• B’s 1-day work = 1/15
• A and B’s 1-day work together:
  1/10​+1/15​=3/30​+2​/30=5/30​=1/6​
• Work done by A and B in 5 days:
  5×1​/6=5/6​

Step 2: Work Left for C

• Total work = 1
• Work remaining after A and B worked for 5 days: $$1 - frac{5}{6} = frac{1}{6}$$
• C completes the remaining 1/6 of the work in 2 days, so C’s 1-day work:
  1/6​÷2=1/12​

Step 3: Wages Calculation

• Total amount = Rs. 1500

• Work done ratio:

  • A's total work = $frac{5}{10} = frac{1}{2}$
  • B's total work = $frac{5}{15} = frac{1}{3}$
  • C's total work = $frac{2}{12} = frac{1}{6}$

• Share of wages:

  • A's share = $frac{1}{2} times 1500 = 750$
    
  • B's share = $frac{1}{3} times 1500 = 500$
    
  • C's share = $frac{1}{6} times 1500 = 250$
    

• Daily wages:

  • B worked 5 days, so B’s daily wage = 500/5 = Rs. 100
  • C worked 2 days, so C’s daily wage = 250/2 = Rs. 125
  • 100+125 = 225

Explanation

We are given:

• Speed while going to school = 3 km/hr

• Speed while returning home = 2 km/hr

• Total time for the round trip = 5 hours

Step 1: Let the Distance Be "d" km

• Time taken to go =d/3​ hours

• Time taken to return = d​/2 hours

Since total time is 5 hours,
d/3​+d/2​=5
Step 2: Solve for d
Taking LCM of 3 and 2 = 6,
2d/6​+3d/6​=5
5d/6​=5
5d=30
$$d=6\text{ km}$$

Explanation

Step 1: Work Done by One Man and One Child in One Day

16 men complete the work in 12 days.
Work done by 16 men in 1 day = 1/12

Work done by 1 man in 1 day = 1/12×16=1/192

24 children complete the work in 18 days.

Work done by 24 children in 1 day = 1/18

Work done by 1 child in 1 day = 1/18×24=1432

Step 2: Work Done by 12 Men and 8 Children in One Day

Work done by 12 men in 1 day = 12×1/192=12/192=1/16

Work done by 8 children in 1 day = 8×1/432=8/432=1/54

Total work done per day by 12 men and 8 children = 1/16+1/54

​ Taking LCM (16, 54) = 432, 

27/432+8/432=35/432

 So, work done per day = 35/432

Step 3: Work Done in 8 Days

Work done in 8 days = 8×35/432=280/432

Remaining work = 1−280/432=152/432

Step 4: Work Done After Adding 3 More Children

Work done by 3 children in 1 day = 3×1/432=3/432

New work rate per day = 35/432+3/432=38/432

Remaining days required = 152/432/38/432=4 days

Explanation

We are given:

24 men complete the work in 16 days → 1 man's 1-day work = 1/24×16=1/38432 women complete the work in 24 days → 1 woman's 1-day work = 1/32×24=1/768 

16 men and 16 women work for 12 days

Step 1: Work done by 16 men and 16 women in 12 days

Work done by 16 men in 1 day = 16×1/384=16/384=1/24

Work done by 16 women in 1 day = 16×1/768=16/768=1/48

Total work done in 1 day = 1/24+1/48=2/48+1/48=3/48=1/16

Total work done in 12 days = 12×1/16=12/16=¾

Step 2: Work remaining

Total work = 1

Work done = ¾

Work remaining = 1−3/4=1/4

Step 3: Additional men required

The remaining work must be completed in 2 days.

Let x be the number of additional men needed.

Work done by (16 + x) men in 1 day = (16+x)×1384(16 + x) 

Work done in 2 days = 2×(16+x)×1384=142

Solving for x:

(16+x)×2/384=¼

16+x=1/4×384/2=384/8=48

x=48−16=32

The closest option is 36 men

Explanation

Let the distance between the police officer's house and office be D km.
Let the actual time required to reach the office on time be T hours.

Step 1: Forming Equations

Using the speed-time formula:

$$text{Time} = frac{text{Distance}}{text{Speed}}$$Speed​

Case 1: Walking at 4 km/h

D/4​​=T+5/60​

(since he is 5 minutes late, which is 5/60 hours)

Case 2: Walking at 5 km/h

D5​=T−10/60​

(since he is 10 minutes early, which is 10/60 hours)

Step 2: Solving for Distance

From the first equation:

T=D/4​−5/60​

From the second equation:

T=D/5​+10/60​

Equating both expressions for T:

D/4​−5/60​=D/5​+10/60​

Multiply everything by 60 to clear the fractions:

$$60 times left( frac{D}{4} - frac{5}{60} right) = 60 times left( frac{D}{5} + frac{10}{60} right)$$15D−5=12D+10
$$15D-12D=10+5$$$$3D = 15$$ $$D = 5 text{ km}$$

Explanation

1. A runs less fast than B → B is faster than A.
2. B runs fast but not faster than C → This means B is either equal to or slower than C.

From this, we can conclude:

• If B is already faster than A and C is at least as fast as B, then C is definitely faster than A.

Final Answer:

C runs faster than A