Q1.If the length and breadth of a rectangle is increased by 10% and 20% respectively. Find the overall change in its area.
(a) 28%
(b) 32%
(c) 36%
(d) 40%
Q2.If the radius of a semicircle is increased by 20%,what will be increment in its area ?
(a) 20%
(b) 40%
(c) 44%
(d) 50%
Q3.If the price of a commodity is increased by 18% then by how much percent one should decrease his consumption in order to keep expenditure same as previous ?
(a) 900/59%
(b) 800/59%
(c) 700/59%
(d) 600/49%
Q4.Find the ratio of ages of A and B after 15 yrs. If the ratio between the present ages of A to B is 2 : 3 and after 5 years is 3 : 4.
(a) 5 : 6
(b) 6 : 5
(c) 3 : 4
(d) 4 : 3
Q5.In what ratio, the shopkeeper should mix water in pure milk so that he can get a profit of 16.66% after selling this mixture of water and milk?
(a) 1 : 7
(b) 1 : 8
(c) 1 : 9
(d) 1 : 6
Q6. If the ratio of area of rectangle to it’s perimeter is 60 : 11. And length and breadth are in the ratio 6 : 5. Find length of rectangle.
(a) 20 units
(b) 22 units
(c) 23 units
(d) 24 units
Q7.If the CP of 20 article is equal to the SP of 16 article, find the profit percentage.
(a) 16%
(b) 20%
(c) 24%
(d) 25%
Q8.If the ratio of MP : SP : CP is 10 : 8 : 5. Then find the ratio of discount percentage to profit/loss percentage.
(a) 1 : 3
(b) 3 : 1
(c) 4 : 1
(d) 1 : 4
Q9.A shopkeeper increases the MP of an article by 20% and again by 30% above the cost price (CP) and further he allows a discount of 25%. Find the overall profit percentage.
(a) 15%
(b) 16%
(c) 17%
(d) 18%
Q10. The difference in the probability of selecting 1 blue Ball and 2 Blue balls is 8/49. If total balls are 50, find the number of blue balls.
(a) 10
(b) 15
(c) 20
(d) 8
Solutions:
S1. Ans.(b)
Sol.
Initial length be ℓ
Final length = 1.1 ℓ
Initial breadth = b
Final breadth = 1.2b
∴ Initial area = ℓ × b
Final area = 1.32 ℓb
∴ Percentage change in area
=((1.32 lb –lb)/lb)×100
= 32%
S2. Ans.(c)
Sol.
Initial radius = r
Final radius = 1.2r
Initial area = πr²
Final area = π(1.2)²
= π 1.44r²
Percentage Increase in area
=((π 1.44r^2 –πr^2)/(πr^2 ))
= 44%
S3. Ans.(a)
Sol.
Let the initial price be Rs.1
New Price=118/100=59/50
Now
59/50×x=1
x=50/59
∴ Decrease in consumption
=1 –50/59
=9/59×100
=900/59%
S4. Ans.(a)
Sol.
Present ages of A and B be 2x and 3x respectively.
∴ age of A after 5 years = 2x + 5
Age of B after 5 years = 3x + 5
Now (2x+5)/(3x+5)=3/4
∴ x = 5
Present age of A = 2x = 10
Present age of B = 3x = 15
Age of A after 15 yrs. = 25
Age of B after 15 yrs. = 30
∴ Required ratio = 25 : 30 = 5 : 6
S5. Ans.(d)
Sol.
Given percentage profit = 16.66%
=100/6%
∴ Required Ratio
=100/6 ∶100
= 1 : 6
Or, let cost price of pure milk be Rs.1 per litre.
∴ C.P. of mixture = 100/(100+50/3)
=Rs. 6/7 per litre
According to law of mixture,
∴ water : milk = 1 : 6
S6. Ans.(d)
Sol.
Ratio of length and breadth is 6 : 5
∴ length = 6x
Breadth = 5x
∴ Area = ℓ × b = 30x²
∴ Perimeter = 2 (ℓ + b)
=2(6x + 5x)
= 22x
∴(30x^2)/22x=60/11
∴ x = 4
∴ length = 6x
= 6 × 4 = 24 units
S7. Ans.(d)
Sol.
Profit percentage
=((20 –16)/16)×100=25%
Or, let cost price of each article be Rs.1
∴ S.P. of each article= 20/16=Rs.1.25
So, required profit =0.25/1×100=25%
S8. Ans.(a)
Sol.
Let MP = 10x
SP = 8x
CP = 5x
∴ Discount = 10x – 8x = 2x
∴ Discount percentage = 2x/10x× 100 = 20%
Profit = 8x – 5x = 3x
Profit percentage = 3x/5x×100
= 60%
∴ Required ratio = 20 : 60 = 1 : 3
S9. Ans.(c)
Sol.
Let CP = x
∴ MP=x×120/100×130/100=156x/100
SP=156x/100×75/100
SP=117x/100=1.17x
∴ P%=((1.17x-x)/x)×100
= 17%
S10. Ans.(a)
Sol.
Total balls = 50
Blue balls = x
Probability of selection of 1 blue ball = x/50
Probability of selection of 2 blue balls =(x×(x –1))/(50×49)
Now x/50-x(x –1)/(50×49)=8/49
∴ x = 10 or 40
