Maths Quiz for 2016-17 Exams


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%