There are 3 variants: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. From this we get,
Popcorn 7x + 17x + 16x = 40 x
Chips 6y + 15y + 14y = 35y
Now let us bring them together into the same metric by taking the LCM of 40 and 35 which is equal to 280.
So the ratio of the number of jumbo popcorn packets to the chips packets is
= 16 Γ 7 : 14 Γ 8
= 1 : 1
Jumbo popcorn packets and jumbo chips packets are in the ratio of 1 : 1.
In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is:
The price of medium quality mangoes is half that of good mangoes. He buys 80 kg good mangoes and 40 kg medium quality mangoes from the market.
Then sells all these at a common price which is 10% less than the price at which he bought the good ones.
i.e. 40 + 80 = 120 kgs is sold at 1.8x.
His overall profit can be found as follows,
Cost price βΉ 40x + 160x = 200x
Selling price βΉ 120 Γ 1.8x = 216x
Hence there is 8% increase from cost price to selling price.
His overall profit was 8%.
If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?
Given that if Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire.
If she needs to make the same amount of profit by selling the remaining toys, she needs to sell them at,
βΉ 60 Γ 0.6x = 50 Γ ?
βΉ 3.6x = 5 Γ ?
βΉ ? = 0.72x or she sells at 72% of the printed price.
Therefore the discount to be given = 100 β 72 = 28%
Let us consider the two ratios given to us.
From (1), we can say that, a + 3 = Β± 3b, So
a = 3b β 3 ------- (3) Or,
a = -3b β 3 ------- (4)
Sub (3) in (2)
βΉ (a β 1) = 2(b β 1) or (a β 1) = -2(b β 1)
βΉ (3b β 4) = 2b β 2 or (3b β 4) = -2b + 2
Both these cases are not possible since a and b are said to be of opposite signs.
Letβs try condition (4).
Sub (4) in (2), we get
βΉ (a β 1) = 2(b β 1) or (a β 1) = -2(b β 1)
βΉ (-3b β 4) = 2b β 2 or (-3b β 4) = -2b + 2
Here a = 15 and b = β 6 are possible.
A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is:
Let the average marks scored by boys during the mid semester exam be n.
Then, the girlsβ average mark will be n + 5.
On calculation, the average of the entire class will be n + 3.
This is in the ratio of 2 : 3 since there are 20 boys and 30 girls in the class.
During the final exam, the average score of the girls dropped by 3.
So, n + 5 becomes n + 2 while the average score of the entire class increased by 2 or it becomes n + 5.
Using alligation we can say that the difference between the average mark of entire class and average mark of girls is 3 which is
Thus the average of boys = (n + x) β (n + 5) = 4.5,
On solving we get x = 9.5.
The area of the closed region bounded is given by the equation,
| x | + | y | = 2.
We can substitute x = 0 or y = 0.
The coordinates we obtain are as follows;
(2,2) , (-2,2) , (2,-2) and (-2,-2)
On joining these points you will get a square whose diagonal is 4 units. Therefore, the sides of the square will be 2β(2) and its area will be 2β(2) Γ 2β(2) = 8
The area of the closed region is 8 sq. units.
From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is:
Given that from a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. Here GBC is the one third of the area of the triangle.
We can join AG and GD which is the median. Each of the shaded triangle has the same area and therefore the remaining area is two-thirds of ABC
The ratio of the sides are 8 : 5 : 7
Area of triangle = β(s(sβa)(sβb)(sβc))
where, semi perimeter (s) = = 10
Area = β(10(10 β 8)(10 β 5)(10 β 7))
Area = β(10(2)(5)(3))
Area = 10β3
Area of ABC = 25 Γ 10 β3 (As 8 : 5 : 7 multiplied by 5 gives the sides of triangle ABC)
Therefore area of the remaining triangle
Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq. cm, of the region enclosed by BPC and BQC is:
Given that ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC and let BPC be an arc of a circle centred at A and lying between BC and BQC.
If AB has length 6 cm then the area, in sq. cm, of the region enclosed by BPC and BQC has to be found.
i.e. the shaded region is,
Area of semicircle BQC = Ο
= 9Ο - 9Ο - 18
= 18
A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to:
Given that a solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. We have to find the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube.
Let us take the volume of the 5 cubes to be
Volume of the original cube =
Volume of the original cube =
Sides of the original cube =
Similarly, sides of the 5 smaller cubes = x , x , 2x , 3x , 3x .
Surface Area of a cube =
Surface Area of the original cube = (4x) Γ (4x) =
Surface area of the smaller cubes =
Sum of the surface areas of the smaller cubes =
Change in surface area =
%change =
Hence the percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to 50%.
A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is: (TITA)
Given that the height of the cylinder is 3 cm, while its volume is
A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically.
Since OPQ is the right angled triangle
We can find the OP = 1 cm.
We have to find the vertical distance of the topmost point of the ball from the base of the cylinder.
Since OP = 1, to reach the topmost point still it has to go 2 cm from the point O.
The vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is 2 + 1 + 3 = 6 cm
is:
= 10
Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is: (TITA)
