Arithmetic Problems (CAT PYQs) Part-1
Q1–10 of 50The speed of a railway engine is 42 kmph when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 kmph when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is
The equation that will fit the situation is: \( S = 42 - K\sqrt{n} \). At n = 9, S = 24. Thus, putting these values in the equation we get: 24 = 42-3K. Hence, K = 6. So the equation becomes \( 42 - 6\sqrt{n} \). At n = 49, the value of S would become equal to 0, thus, the railway engine can carry a maximum of 48 compartments (option (b)).
Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is ₹700 when there are 25 boarders and ₹600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
Alternately, you can solve this through equations as follows:
Let 'a' be the fixed cost and 'b' the variable cost.
According to the question:
\( 700 \times 25 = a + 25b \) (1)
\( 600 \times 50 = a + 50b \) (2)
Solving the equation (1) and (2), we get
a = 5000, b = 500
Let the average expense of 100 boarders be 'X'. Then
\( 100 \times X = 5000 + (500 \times 100) \)
\( \therefore X = 550 \)
Forty per cent of the employees of a certain company are men, and 75% of the men earn more than ₹25,000 per year. If 45% of the company's employees earn more than ₹25,000 per year, what fraction of the women employed by the company earn less than or equal to ₹25,000 per year?
Forty per cent are men and 60 per cent are women. Out of the men category, 75 per cent earn more than 25000 per year. Thus, a total of 30 per cent of the total employees of the company are males who earn more than 25000 per year. Since, there are a total of 45 per cent of the employees who earn more than 25000 per year, it means that out of the 60 per cent who are women 15 per cent earn more than 25000 and 45 per cent earn ₹25000 or less than that per year. Thus, the required ratio is 3/4 (option (d)).
Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6.30 a.m. and travels at 50 kmph towards Baroda situated 100 km away. At 7.00 a.m. Howrah-Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kmph. At 7.30 a.m. Mr Shah, the traffic controller at Baroda realizes that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?
The distance between Ahmedabad and Baroda being 100 kms, it is evident that by 7:30 am, Navjivan Express would have covered 50 kms (travelling @ 50kmph for 1hour), while the Howrah-Ahmedabad Express would have covered 20 kms (travelling 40 kmph for 30 minutes). Thus, the distance between the two trains would be 30 kms at 7:30 am. Since their relative speed is 90 kmph, the remaining distance of 30 kms would be covered in 1/3rd of an hour - or 20 minutes. Option (b)is correct.
Directions for question: The following table presents the sweetness of different forms relative to sucrose, whose sweetness is taken to be 1.00.
Substance / Sweetness
Lactose 0.16
Maltose 0.32
Glucose 0.74
Sucrose 1.00
Fructose 1.70
Saccharin 675.00
What is the minimum amount of sucrose (to the nearest gram) that must be added to one gram of saccharin to make a mixture that will be at least 100 times as sweet as glucose?
If we mix 8 grams of sucrose to 1 gram of saccharine, we would have 9 grams with a sweetness quotient of 683. The average sweetness would be 683/9 = 75.88 which is greater than 100 times the sweetness of sucrose. If we mix 9 grams of sucrose to 1 gram of saccharine, the average would be below 74. Hence, option (b) is the required answer.
Directions for question: The following table presents the sweetness of different forms relative to sucrose, whose sweetness is taken to be 1.00.
Substance / Sweetness
Lactose 0.16
Maltose 0.32
Glucose 0.74
Sucrose 1.00
Fructose 1.70
Saccharin 675.00
Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1:2:3?
The average sweetness of the mixture as defined would be: \( (1\times0.74 + 2\times1 + 3\times1.7)/6 = 7.84/6 = 1.306 \). Thus, option (a) is correct.
Directions for question: These questions are based on the situation given below. A road network connects cities A, B, C and D. All road segments are straight lines. D is the midpoint on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long. Mr. X and Mr. Y leave A at 8:00 am and take different routes to city C, reaching at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 km per hour. Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.
What is the average speed of Y in km per hour?
Y covers two equal distances AD and CD with speeds 45 kmph and 55 kmph respectively. Therefore the average speed of y must be \( \dfrac{2 \times 45 \times 55}{45+55} = 49.5 \) kmph.
Alternately, you could also think of this as:
The average speed for Y would be the weighted average of 45 and 55 in the ratio 55:45 (as the distance on both the segments AD and DC are equal). The value would be 49.5 (option (b)).
Directions for question: These questions are based on the situation given below. A road network connects cities A, B, C and D. All road segments are straight lines. D is the midpoint on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long. Mr. X and Mr. Y leave A at 8:00 am and take different routes to city C, reaching at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 km per hour. Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.
The total distance traveled by Y during the journey is approximately
\( \dfrac{100+BC}{61.875} = \dfrac{AC}{49.5} \)
\( BC = \sqrt{AC^2-100^2} \)
\( \dfrac{100+\sqrt{AC^2-100^2}}{61.875} = \dfrac{AC}{49.5} \)
Now put the value of AC from the option and check.
We get that for AC = 105, LHS = RHS.
Alternately, you could think as follows:
Since ABC is a right triangle, and D is the midpoint of the hypotenuse BD, it would be half the length of the hypotenuse. Question 8 is asking for the length of AC, while question 9 is asking for the length of BD. Looking at the information contained in the question, it is evident that these distances would not come under the cannot be determined category. Thus, we can solve questions 8 and 9 simultaneously by looking at a value for the answer to 9, which should be half the answer to question 8. Thus, the answer to question 8 is :105 km (as that is the only value that fits).
Directions for question: These questions are based on the situation given below. A road network connects cities A, B, C and D. All road segments are straight lines. D is the midpoint on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long. Mr. X and Mr. Y leave A at 8:00 am and take different routes to city C, reaching at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 km per hour. Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.
What is the length of the road segment BD?
AC = 105 km and D is the midpoint of AC. So, AD = DC = BD = 105/2 = 52.5 km
Directions for question: These questions are based on the situation given below.
Rajiv reaches city B from city A in 4 hours, driving at 35 km per hour for the first 2 hours and at 45 km per hour for the next two hours. Aditi follows the same route, but drives at three different speeds: 30, 40 and 50 km per hour, covering an equal distance in each speed segment. The two cars have similar petrol consumption characteristics (km per liter) as shown in the below.
The amount of petrol consumed by Aditi for the journey is
The distance between City A and City B would be \( 45 \times 2 + 35 \times 2 = 160 \) kms (as per Rajiv's movement plan). Aditi would cover this distance in three equal parts of 53.33 kms @ of 30 kmph, 40 kmph and 50 kmph respectively.
Petrol consumed by Aditi = \( 53.33/16 + 53.33/24 + 53.33/16 = 160/48 + 160/72 + 160/48 = 640/72 = 8.9 \) liters (option (c)).