406. In a regiment the ratio between the number of officers to soldiers was 3 : 31 before battle. In a battle 6 officers and 22 soldiers were killed and the ratio became 1 : 13, the number of officers in the regiment before battle was

1. 1. 31

   
   
   

   2. 38

   
   
   

   3. 21

   
   
   

   4. 28

   
   
   

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1. View Hint View Answer Discuss in Forum

   Before battle,
   Officers ⇒ 3x
   Soldiers ⇒ 31x
   According to the question,
   After battle,
   

   3x − 6	=	1
   31x − 22		13

   
   ⇒  39x – 78 = 31x – 22
   ⇒  39x – 31x = 78 – 22
   ⇒  8x = 56
   

   ⇒  x =	56	= 7
   	8

   
   ∴  Required number of officers
   = 3 × 7 = 21

   Correct Option: C

   Before battle,
   Officers ⇒ 3x
   Soldiers ⇒ 31x
   According to the question,
   After battle,
   

   3x − 6	=	1
   31x − 22		13

   
   ⇒  39x – 78 = 31x – 22
   ⇒  39x – 31x = 78 – 22
   ⇒  8x = 56
   

   ⇒  x =	56	= 7
   	8

   
   ∴  Required number of officers
   = 3 × 7 = 21

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407. If (a + b) : √ab = 4 : 1, where a > b > 0, then a : b is

1. 1. (2 + √3) : (2 − √3)

   
   
   

   2. (2 − √3) : (2 + √3)

   
   
   

   3. (3 + √2) : (3 − √2)

   
   
   

   4. (3 − √2) : (3 + √2)

   
   
   

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1. View Hint View Answer Discuss in Forum

   a + b	=	4	⇒	a + b	=	2
   √ab		1		2√ab		1

   
   By componendo and dividendo,
   

   a + b + 2 √ab	=	3
   a + b − 2 √ab		1

   
   

   ⇒	(√a + √b)2	=	(√3)2
   	(√a − √b)2		(1)2

   
   Again using componendo and dividendo,
   

   2√a	=	√3 + 1
   2√b		√ 3 − 1

   
   

   ⇒	√a	=	√3 + 1
   	√b		√ 3 − 1

   
   On squaring both sides
   

   a	=		√3 + 1		2	=	3 + 1 + 2 √3
   b			√ 3 − 1				3 + 1 − 2 √3

   
   

   =	4 + 2√3	=	2 + √3
   	4 − 2√3		2 − √3

   
   2 + √3 : 2 − √3

   Correct Option: A

   a + b	=	4	⇒	a + b	=	2
   √ab		1		2√ab		1

   
   By componendo and dividendo,
   

   a + b + 2 √ab	=	3
   a + b − 2 √ab		1

   
   

   ⇒	(√a + √b)2	=	(√3)2
   	(√a − √b)2		(1)2

   
   Again using componendo and dividendo,
   

   2√a	=	√3 + 1
   2√b		√ 3 − 1

   
   

   ⇒	√a	=	√3 + 1
   	√b		√ 3 − 1

   
   On squaring both sides
   

   a	=		√3 + 1		2	=	3 + 1 + 2 √3
   b			√ 3 − 1				3 + 1 − 2 √3

   
   

   =	4 + 2√3	=	2 + √3
   	4 − 2√3		2 − √3

   
   2 + √3 : 2 − √3

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408. The number of pupils of a class is 55. The ratio of the number of male pupils to the number of female pupils is 5: 6. The number of female pupils is

1. 1. 11

   
   
   

   2. 25

   
   
   

   3. 30

   
   
   

   4. 35

   
   
   

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1. View Hint View Answer Discuss in Forum

   Boys : Girls = 5 : 6
   Sum of the terms of ratio = 5 + 6 = 11
   

   ∴  Number of girls =	6	× 55 = 30
   	11

   Correct Option: C

   Boys : Girls = 5 : 6
   Sum of the terms of ratio = 5 + 6 = 11
   

   ∴  Number of girls =	6	× 55 = 30
   	11

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409. In a parade of school students, the number of boys and girls are in the ratio of 9 : 7 respectively and the number of students is 256. Find the number of girls.

1. 1. 102

   
   
   

   2. 112

   
   
   

   3. 118

   
   
   

   4. 128

   
   
   

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1. View Hint View Answer Discuss in Forum

   Boys : Girls = 9 : 7,
   Sum of the terms of the ratio = 9 + 7 = 16
   Number of students = 256
   

   ∴  Number of girls =	25 × 7	= 112
   	16

   Correct Option: B

   Boys : Girls = 9 : 7,
   Sum of the terms of the ratio = 9 + 7 = 16
   Number of students = 256
   

   ∴  Number of girls =	25 × 7	= 112
   	16

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410. Sum of two numbers is thrice their difference. Their ratio is

1. 1. 1:2

   
   
   

   2. 2:1

   
   
   

   3. 3:1

   
   
   

   4. 1:3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the numbers be x and y.
   According to the question,
   x + y = 3(x – y)
   ⇒  x + y = 3x – 3y
   ⇒  3x – x = y + 3y
   ⇒  2x = 4y
   ⇒  x = 2y
   

   ⇒	x	=	2
   	y		1

   Correct Option: B

   Let the numbers be x and y.
   According to the question,
   x + y = 3(x – y)
   ⇒  x + y = 3x – 3y
   ⇒  3x – x = y + 3y
   ⇒  2x = 4y
   ⇒  x = 2y
   

   ⇒	x	=	2
   	y		1

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