Question: 36 -

605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20 % of the total number of children. How many sweets did each child receive?

11

24

15

45

Answer:

11

Solution:

Let the total number of children be x. 
Then x × 20/100 × x = 605 
=> x2 = 3025 
=> x = √3025 = 55 
Number of sweets received by each child 
= (20/100 ×55) = 11 

Question: 37 -

A and B are two fixed points 5 cm apart and c is a point on AB such that AC = 3 cm. If the length of AC is increased by 6 % the length of CB is decreased by: 

7%

6%

8%

9%

Answer:

9%

Solution:

AC = 3cm, CB = (5 - 3)cm = 2 cm 
New length AC = 106 % of 3 cm 
= (106/100 ×3) cm = 3.18 cm. 
New length CB = (5 – 3.18) cm = 1.82 cm 
CB Decreased on 2cm = (2 – 1.82)cm = 0.18cm
CB Decrease % = (0.18/2 ×100) % = 9 % 

Question: 38 -

A dishonest dealer claims to sell his goods at the cost price but uses a false weight of 900 gm for 1 kg what is his gain percent? 

 13 %

11.25 % 

12 1/9 %

11 1/9 % 

Answer:

11 1/9 % 

Solution:

Let the C.P of the article be Rs 1 per gram 
C.P of 900gm = Rs 900, 
S.P of 1000gm = Rs 1000 
Gain % = (100/900 × 100)% = 100/9 % = 11 1/9 %

Question: 39 -

A period of 4 hrs 30 min is what percent of a day?

19 %

16 3/4 % 

20 %

 18 3/4 %

Answer:

 18 3/4 %

Solution:

Required % = (9/48 × 100)% = 75/4 % = 18 ¾ %

Question: 40 -

The length of a rectangle is increased by 60 %. By what percent would the width have to be decreased to maintain the same area? 

37 ½ %

None

75%

60%

Answer:

37 ½ %

Solution:

Let length = 100 m. 
Breadth = 100 m. 
New length = 160 m.
New breath = x meters 
Then, = 160 × x = 100 × 100 
(or) X = (100 ×100)/160 × 125/2 
Decrease in breadth = (100- 125/2) % = 37 ½ %