96. 8 children and 12 men complete a certain piece of work in 9 days. Each child takes twice the time taken by a man to finish the work. In how many days will 12 men finish the same work ?

1. 1. 9 days

   
   
   

   2. 13 days

   
   
   

   3. 12 days

   
   
   

   4. 15 days

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   2 children ≡ 1 man
   ∴ 8 children + 12 men ≡ 16 men
   ∴ M₁D₁ = M₂D₂
   ⇒ 16 × 9 = 12 × D₂
   

   ⇒ D2 =	16 × 9	= 12 days.
   	12

   Correct Option: C

   Using Rule 1,
   2 children ≡ 1 man
   ∴ 8 children + 12 men ≡ 16 men
   ∴ M₁D₁ = M₂D₂
   ⇒ 16 × 9 = 12 × D₂
   

   ⇒ D2 =	16 × 9	= 12 days.
   	12

---

97. If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in

1. 1. 21 days

   
   
   

   2. 24 days

   
   
   

   3. 26 days

   
   
   

   4. 33 days

   
   
   

---

1. View Hint View Answer Discuss in Forum

   1 man ≡ 2 women ≡ 3 boys
   ∴ 1 man + 1 woman + 1 boy
   

   ≡ 3 boys +	3	boys + 1 boy
   	2

   
   

   =		3 +	3	+ 1		boys
   			2

   
   

   =	11	boys
   	2

   
   ∴ By M₁ D₁ = M₂ D₂
   

   3 × 44 =	11	× D2
   	2

   
   

   ⇒ D2 =	2 × 3 × 44	= 24 days
   	11

   
   Aliter : Using Rule 13,
   Here, A = 1, B= 2, C = 3, a = 44
   A₁ = 1, B₁ = 1, C₁ = 1
   Required time
   

   =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	44
   	1	+	1	+	1
   	1		2		3

   
   

   =	44 × 6	= 24 days
   	11

   Correct Option: B

   1 man ≡ 2 women ≡ 3 boys
   ∴ 1 man + 1 woman + 1 boy
   

   ≡ 3 boys +	3	boys + 1 boy
   	2

   
   

   =		3 +	3	+ 1		boys
   			2

   
   

   =	11	boys
   	2

   
   ∴ By M₁ D₁ = M₂ D₂
   

   3 × 44 =	11	× D2
   	2

   
   

   ⇒ D2 =	2 × 3 × 44	= 24 days
   	11

   
   Aliter : Using Rule 13,
   Here, A = 1, B= 2, C = 3, a = 44
   A₁ = 1, B₁ = 1, C₁ = 1
   Required time
   

   =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	44
   	1	+	1	+	1
   	1		2		3

   
   

   =	44 × 6	= 24 days
   	11

---

---

98. 3 men or 7 women can do a piece of work in 32 days. The number of days required by 7 men and 5 women to do a piece of work twice as large is

1. 1. 19

   
   
   

   2. 21

   
   
   

   3. 27

   
   
   

   4. 36

   
   
   

---

1. View Hint View Answer Discuss in Forum

   ∵ 3 men ≡ 7 women
   

   ∴ 7 men ≡	7 × 7
   	3

   
   

   =	49	women
   	3

   
   ∴ 7 men + 5 women
   

   =			49	+ 5		women
   			3

   
   

   =		49 + 15		women
   		3

   
   

   =	64	women
   	3

   
   

   ∴	M1D1	=	M2D2
   	W1		W2

   
   

   =	7 × 32	=	64 × D2
   	1		3 × 2

   
   

   ⇒ D2 =	7 × 32 × 3 × 2	= 21 days
   	64

   
   Aliter : Using Rule 12,
   Here, A = 3, B = 7, a = 32
   A₁ = 7, B₁ = 5
   Required time
   

   =	a
   	A1	+	B1
   	A		B

   
   

   =	32
   	7	+	5
   	3		7

   
   

   =	32	× 21 =	21
   	64		2

   
   They do the twice work in
   

   	21	× 2 = 21 days
   	2

   Correct Option: B

   ∵ 3 men ≡ 7 women
   

   ∴ 7 men ≡	7 × 7
   	3

   
   

   =	49	women
   	3

   
   ∴ 7 men + 5 women
   

   =			49	+ 5		women
   			3

   
   

   =		49 + 15		women
   		3

   
   

   =	64	women
   	3

   
   

   ∴	M1D1	=	M2D2
   	W1		W2

   
   

   =	7 × 32	=	64 × D2
   	1		3 × 2

   
   

   ⇒ D2 =	7 × 32 × 3 × 2	= 21 days
   	64

   
   Aliter : Using Rule 12,
   Here, A = 3, B = 7, a = 32
   A₁ = 7, B₁ = 5
   Required time
   

   =	a
   	A1	+	B1
   	A		B

   
   

   =	32
   	7	+	5
   	3		7

   
   

   =	32	× 21 =	21
   	64		2

   
   They do the twice work in
   

   	21	× 2 = 21 days
   	2

---

99. One man or two women or three boys can do a piece of work in 88 days. One man, one woman and one boy will do it in

1. 1. 44 days

   
   
   

   2. 24 days

   
   
   

   3. 48 days

   
   
   

   4. 20 days

   
   
   

---

1. View Hint View Answer Discuss in Forum

   1 man ≡ 2 women ≡ 3 boys
   ∴ 1 man + 1 woman + 1 boy
   

   =		3 +	3	+ 1		boys
   			2

   
   

   =		6 + 3 + 2		boys
   		2

   
   

   =	11	boys
   	2

   
   ∴ M₁D₁ = M₂D₂
   

   ⇒ 3 × 8 =	11	× D2
   	2

   
   

   ⇒ D2 =	3 × 2 × 88	= 48 days
   	11

   
   Aliter : Using Rule 13,
   Here, A = 1, B = 2, C = 3, a= 88
   A₁ = 1, B₁ = 1, C₁ = 1
   Required time
   

   =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	88
   	1	+	1	+	1
   	1		2		3

   
   

   =	88	= 48 days
   	6 + 3 + 2
   	6

   Correct Option: C

   1 man ≡ 2 women ≡ 3 boys
   ∴ 1 man + 1 woman + 1 boy
   

   =		3 +	3	+ 1		boys
   			2

   
   

   =		6 + 3 + 2		boys
   		2

   
   

   =	11	boys
   	2

   
   ∴ M₁D₁ = M₂D₂
   

   ⇒ 3 × 8 =	11	× D2
   	2

   
   

   ⇒ D2 =	3 × 2 × 88	= 48 days
   	11

   
   Aliter : Using Rule 13,
   Here, A = 1, B = 2, C = 3, a= 88
   A₁ = 1, B₁ = 1, C₁ = 1
   Required time
   

   =	a
   	A1	+	B1	+	C1
   	A		B		C

   
   

   =	88
   	1	+	1	+	1
   	1		2		3

   
   

   =	88	= 48 days
   	6 + 3 + 2
   	6

---

---

100. A man is twice as fast as a woman and a woman is twice as fast as a boy in doing a work. If all of them, a man, a woman and a boy can finish the work in 7 days, in how many days a boy will do it alone ?

1. 1. 49

   
   
   

   2. 7

   
   
   

   3. 6

   
   
   

   4. 42

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Using Rule 11,
   According to the question,
   1 man ≡ 2 women ≡ 4 boys
   ∴ 1 man + 1 woman + 1 boy
   = (4 + 2 +1) boys = 7 boys
   ∴ M₁ D₁ = M₂ D₂
   ⇒ 7 × 7 = 1 × D₂
   ⇒ D₂ = 49 days

   Correct Option: A

   Using Rule 11,
   According to the question,
   1 man ≡ 2 women ≡ 4 boys
   ∴ 1 man + 1 woman + 1 boy
   = (4 + 2 +1) boys = 7 boys
   ∴ M₁ D₁ = M₂ D₂
   ⇒ 7 × 7 = 1 × D₂
   ⇒ D₂ = 49 days

---