**Q1. The side of a triangle are in the ratio of 1/2:1/3:1/4 if the perimeter is 104 cm. What is the sum of the largest side & smallest side?**

(a) 80

(b) 84

**(c) 72**

(d) 56

**Sol.**

**Ratio of side = 1/2:1/3:1/4**

**= 6 : 4 : 3**

**Let the side of triangle is 6x , 4x and 3x respectively.**

**Given that perimeter = 104 cm**

**6x + 4x + 3x = 104**

**x = 104/13**

**x = 8**

**sides are = 48 cm, 32 cm and 24 cm**

**Required answer = 48 cm + 24 cm**

**= 72 cm**

**Q2. The exterior angle of a polygon is 1/5 of the its interior angle. How many sides does the polygon have**

(a) 8

(b) 10

**(c) 12**

(d) 14

**Sol.**

**Interior angle of polygon having n sides**

**=((n-2))/n×π**

**Exterior angle = 2π/n**

**According to question,**

**(interior angle)/(exterior angle)=((n-2)π)/2π=5**

**n – 2 = 10**

**n = 12**

**Q3. What is (in unit square) the area at a triangle with side length 4 unit, 6 unit and 10 unit?**

(a) 24

(b) 30

(c) 48

**(d) None of these**

**Sol.**

**The triangle is not possible (since sum of length of two sides > length of third side**

**Q4. If the perimeter of right angle triangle is 30 cm and its hypotenuse is 13 cm. What is the radius of the circle inscribed in a triangle?**

**(a) 2 cm**

(b) 5 cm

(c) 1 cm

(d) None of these

**Sol.**

**In a right angle triangle,**

**In radius = semi perimeter-hypotenuse**

**= 15 – 13 = 2 cm**

**Q5. What is the relation between the circum radius and in radius in any triangle?**

**(a) R ≥ 2r**

(b) R = 2r

(c) R ≤ 2r

(d) R < 2r

**Sol.**

**Euler’s Formula**

**d^2=R(R-2r) …(1)**

**When R = circum radius**

**r = in radius**

**d = distance between circum centre and in centre**

**From equation (1) be obtain the inequality**

**R ≥ 2r**

**Q6. A seller fixes the marked price 50% more than cost price and gives a discount of 15%, and so he gets a profit of Rs. 165. Find the marked price of that article.**

(a) Rs. 900

(b) Rs. 800

**(c) Rs. 700**

(d) Rs. 1000

**Sol.**

**Let the cost price = 100**

**Mark price = 150 Selling price = 150 × 85/100 = 127.5**

**Profit = 27.5**

**In a mark price of 150 profit = 27.5**

**Mark price for profit of 165 = (150 × 165)/27.5 = 900**

**Q7. If the area of a square is increased by 100%, then the percentage increase in the length of its diagonal is**

(a) 10%

**(b) 41.4%**

(c) 50%

(d) 55.5%

**Sol.**

**Assume some values and solve the question.**

**Assume original area = 100 ⇒ Diagonal = 2√10**

**New area = 200 ⇒ Diagonal = 20**

**Percentage increase = 41.4%**

**Q8. On Independence Day, if 30 children were made to stand in a column, 16 columns could be formed. If 24 children were made to stand in a column, how many columns could be formed?**

**(a) 20**

(b) 21

(c) 22

(d) 18

**Sol.**

**Let the number of column =x, when 24 student stands in one column.**

**Total students = 30 × 16 = 480 children**

**24 × x = 480**

**x = 20 column**

**Q9. At the start of seminar, the ratio of the number of male participants to the number of female participants was 3:1. During the tea break, 16 participants left and 6 more female participants registered. The ratio of the male to the female participants became 2 : 1. What was the total number of participants at the start of the seminar?**

(a) 164

(b) 148

(c) 154

**(d) 112**

**Sol.**

**Let the number of male participants and the number of female participants is 3n and n respectively**

**During the tea break number of participants left,**

**Male participants=3n-16**

**And female participants=n+6**

**A.T.Q.**

**3n-16/n+6=2/1**

**n=28**

**so total number of participants=4*28=112**

**Q10. Each edge of a cube is increased by 40%. What is the percentage increase in its volume?**

(a) 120%

(b) 40%

**(c) 174.4%**

(d) 146%

**Sol.**

**Volume of cube = a3 (where a = length of a edge)**

**When each edge is increased by 40%**

**⇒ Length of the new edge = 1.4a**

**⇒ Volume of new cube = (1.4a)3 = 2.744a3**

**⇒ Required % increase = [(2.744a3 – a3)/a3] × 100%**

**= 174.4%**