CAT 2017 Quant Question: Algebra
If a and b are integers of opposite signs such that 
is:
A. 9 : 4
B. 81 : 4
C. 1 : 4
D. 25 : 4
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Answer:
Let us consider the two ratios given to us.
From (1), we can say that, a + 3 = ± 3b, So
a = 3b – 3 ------- (3) Or,
a = -3b – 3 ------- (4)
Sub (3) in (2)
⟹ (a – 1) = 2(b – 1) or (a – 1) = -2(b – 1)
⟹ (3b – 4) = 2b – 2 or (3b – 4) = -2b + 2
Both these cases are not possible since a and b are said to be of opposite signs.
Let’s try condition (4).
Sub (4) in (2), we get
⟹ (a – 1) = 2(b – 1) or (a – 1) = -2(b – 1)
⟹ (-3b – 4) = 2b – 2 or (-3b – 4) = -2b + 2
Here a = 15 and b = – 6 are possible.
Hence, the answer is 25 : 4
Choice D is the correct answer.
CAT 2017 Quant Question: Mixture & Alligation
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:
A. 9.5
B. 10
C. 4.5
D. 6
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Answer:
Let the average marks scored by boys during the mid semester exam be n.
Then, the girls’ 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) – (n + 5) = 4.5,
On solving we get x = 9.5.
Hence, the answer is 9.5
Choice A is the correct answer.
CAT 2017 Quant Question: Co-ordinate Geometry
The area of the closed region bounded by the equation | x | + | y | = 2 in the two-dimensional plane is
A. 4π
B. 4
C. 8
D. 2π
Answer:
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 2√(2) and its area will be 2√(2) × 2√(2) = 8
The area of the closed region is 8 sq. units.
Hence, the answer is 8 sq. units
Choice C is the correct answer.
CAT 2017 Quant Question: Geometry
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:
Answer:
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 = √(s(s−a)(s−b)(s−c))
where, semi perimeter (s) = 
= 10
Area = √(10(10 − 8)(10 − 5)(10 − 7))
Area = √(10(2)(5)(3))
Area = 10√3
Area of ABC = 25 × 10 √3 (As 8 : 5 : 7 multiplied by 5 gives the sides of triangle ABC)
Therefore area of the remaining triangle 
Hence, the answer is 
Choice B is the correct answer.
CAT 2017 Quant Question: Geometry
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:
A. 9π - 18
B. 18
C. 9π
D. 9
Answer:
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 = π 
= 9π - 9π - 18
= 18
Hence, the answer is 18
Choice B is the correct answer.
CAT 2017 Quant Question: Mensuration & Geometry
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:
A. 10
B. 50
C. 60
D. 20
Answer:
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) × (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%.
Hence, the answer is 50%
Choice B is the correct answer.
CAT 2017 Quant Question: Mensuration & Geometry
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)
Answer:
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
Hence, the answer is 6 cm
CAT 2017 Quant Question: Geometry
Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is: (TITA)
Answer:
Given that ABC be a right-angled triangle with BC as the hypotenuse.
Lengths of AB and AC are 15 km and 20 km, respectively.
We have to find the minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour.
We should first find the minimum distance in order to find the minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour.
Therefore minimum distance AD has to be found and then it should be divided by the 30 km per hour.
Using the idea of similar triangles
Area of the triangle ABC
Hence 12 kms is travelled at 30km per hour 
The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is 
× 60 = 24 minutes
Key thing to be noted here is using Pythagoras theorem to find the altitude AD and then using Speed, Time and Distance formula to find the time.
Hence, the answer is 24
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