Mathematics — Old Questions (TPSC)
Q51. A man bought a cow for Rs. 850.00 and sold it at a gain of 8%. For how much did he sell the cow?
Cost Price = Rs. 850.
Gain = 8%.
Selling Price = 850 × (108/100) = Rs. 918
Q52. A car is running at an average speed of 42 kilometres per hour. What time will it take to cover a distance of 350 metres?
Convert speed into m/s:
42 km/hr = (42 × 1000) / 3600 = 35/3 m/s.
Time = Distance / Speed = 350 ÷ (35/3) = 30 seconds.
Q53. Ninety six percent of the cost of a TV is Rs. 10464.00. What is its total cost?
Let total cost = Rs. x.
96% of x = 10464.
x = (10464 × 100) / 96 = Rs. 10900
Q54. A sum will be double in 10 year if the rate of simple interest per annum is
For simple interest, when a sum doubles, the interest earned equals the principal.
Using SI formula: (P × R × T) / 100 = P.
So (R × 10) / 100 = 1 ⇒ R = 10% per annum.
Q55. In a family, the age of father, mother, son and grandson are A, B, C, and D respectively. Given that A-B = 3, B+C = 78, C+D = 33 and the average age of the family is 34 year, then B-C is:
Average age = 34, so total age = 34 × 4 = 136.
A = B + 3.
So (B+3) + B + C + D = 136 ⇒ 2B + C + D = 133.
Given C + D = 33 ⇒ 2B + 33 = 133 ⇒ B = 50.
From B + C = 78 ⇒ C = 28.
Hence B − C = 22
Q56. 5% of 30% of 100 is
30% of 100 = 30.
5% of 30 = (5/100) × 30 = 1.5
Q57. In an examination, a student score marks for every correct answer and losses 1 mark every wrong answer. If he attempts all 75 questions and secures 125 marks, the total number of questions he attempted correctly is:
If all answers were wrong, marks = −75.
Actual marks = 125, so extra marks gained = 125 + 75 = 200.
Each correct answer increases marks by 5 (4 for correct + 1 saved).
Correct answers = 200 ÷ 5 = 40
Q58. A shopkeeper make the price of his goods at 20% higher than the original price. After that, he allows discount of 10% on market price. What will be his profit or loss percentage?
Assume Cost Price = 100.
Marked Price = 120 (20% increase).
Discount = 10% of 120 = 12.
Selling Price = 120 − 12 = 108.
Profit = 108 − 100 = 8%
Q59. P can do a piece of work in 10 days and Q can do the same work in 20 days, with the help of R they can finish the work in 5 days. How many days will it take for R alone to finish the work?
Let the total work be 1 unit.
P can do the work in 10 days,
so P’s one-day work = 1/10.
Q can do the work in 20 days,
so Q’s one-day work = 1/20.
Work done by P and Q together in one day
= 1/10 + 1/20 = 3/20.
Given that P, Q and R together finish the work in 5 days,
so their one-day work = 1/5 = 4/20.
Therefore, R’s one-day work
= (P + Q + R) − (P + Q)
= 4/20 − 3/20 = 1/20.
So, R alone can finish the work in 20 days.
Q60. A train, moving with uniform velocity, crosses a pole in 15 second, while it crosses 100 metre long platform in 25 seconds. The length of the train is:
Let the length of the train be x metres.
Speed of train = x / 15 m/s (crossing a pole).
While crossing a 100 m platform, distance = x + 100 and time = 25 s.
So speed = (x + 100) / 25 m/s.
Since speed is same:
x / 15 = (x + 100) / 25
⇒ 25x = 15x + 1500
⇒ 10x = 1500
⇒ x = 150 metres