Mathematics — Old Questions (TPSC)
Q41. If 4/5 of an estate be worth Rs. 16800, then what is the value of 3/7 of the estate?
4/5 of estate = Rs. 16800.
Total estate = 16800 × (5/4) = Rs. 21000.
Value of 3/7 of estate = (3/7) × 21000 = Rs. 9000
Q42. A student has to secure 40% marks in a paper to pass. He gets 40 and fails by 40 marks. Find the maximum marks of the paper.
Let maximum marks be x.
Pass marks = 40% of x = 0.40x.
Student scored 40 and failed by 40 marks, so pass marks = 40 + 40 = 80.
Therefore 0.40x = 80 ⇒ x = 200.
Q43. In an examination, a student was asked for to find 3/14 of a certain number. By mistake, he found 3/4 of it. His answer was 150 more than the correct answer. The given number is:
Let the number be x.
Wrong value = (3/4)x and correct value = (3/14)x.
Given (3/4)x − (3/14)x = 150.
⇒ (21 − 6)x / 28 = 150.
⇒ (15/28)x = 150.
⇒ x = 280
Q44. A, B, and C working together can finish a piece of work in 8 days. A alone can do it in 20 days and B alone can do it in 24 days. In how many days will C alone can do the same work?
Let total work = 1 unit.
A’s rate = 1/20 and B’s rate = 1/24.
So (A + B) rate = 11/120.
Given (A + B + C) rate = 1/8 = 15/120.
Therefore C’s rate = 15/120 − 11/120 = 4/120 = 1/30.
Hence C alone can do the work in 30 days
Q45. The average of 10 numbers is calculated as 25. It is discovered later on that while calculating the average, one number, namely 36 was wrongly read as 56. The correct average is:
Average of 10 numbers = 25
Wrong total = 25 × 10 = 250
One number 36 was wrongly taken as 56
Excess added = 56 − 36 = 20
Correct total = 250 − 20 = 230
Correct average = 230 ÷ 10 = 23
Q46. If selling price of 10 articles is the same as cost price 12 articles, find the gain percentage.
Selling price of 10 articles = Cost price of 12 articles
Assume cost price of 1 article = 1
Cost price of 12 articles = 12
So, selling price of 10 articles = 12
Selling price of 1 article = 12 ÷ 10 = 1.2
Gain per article = 1.2 − 1 = 0.2
Gain % = (0.2 ÷ 1) × 100 = 20%
Q47. The ratio between two numbers is 3:4. If their L.C.M is 600, then the numbers are:
Ratio between two numbers = 3 : 4
Let the numbers be 3x and 4x
Since 3 and 4 are co-prime,
L.C.M of the numbers = 3x × 4x = 12x
Given L.C.M = 600
So, 12x = 600
x = 600 ÷ 12 = 50
Therefore,
First number = 3 × 50 = 150
Second number = 4 × 50 = 200
Answer: The numbers are 150 and 200
Q48. A man sold two commodities for Rs.99 each. On one he gains 10% and the other he loses 10%. Find his gain or loss percent in the whole transaction.
Selling price of each commodity = 99
First commodity (10% gain):
Gain = 10%
So, Cost Price = 99 × 100 ÷ 110 = 90
Second commodity (10% loss):
Loss = 10%
So, Cost Price = 99 × 100 ÷ 90 = 110
Total Cost Price:
90 + 110 = 200
Total Selling Price:
99 + 99 = 198
Loss = 200 − 198 = 2
Loss % = (2 ÷ 200) × 100 = 1%
Answer: There is a 1% loss in the whole transaction.
Q49. A 150 metres long train takes 20 seconds to cross a 450 metres long platform. The speed of the train in kilometres per minute is
Length of train = 150 metres
Length of platform = 450 metres
Total distance covered to cross the platform
= 150 + 450 = 600 metres
Time taken = 20 seconds
Speed = Distance ÷ Time
= 600 ÷ 20 = 30 metres per second
Convert metres per second to kilometres per minute:
30 m/s = 30 × 60 = 1800 metres per minute
= 1.8 kilometres per minute
Answer: Speed of the train = 1.8 km per minute
Q50. 1/3 of Samir's money is equal to 1/2 of Rahul's money. If both of them together have Rs.500, what is the difference between their amounts?
Given,
(1 ÷ 3) of Samir's money = (1 ÷ 2) of Rahul's money
Let Samir's money = S
Let Rahul's money = R
So,
S ÷ 3 = R ÷ 2
∴ 2S = 3R
Ratio of Samir's and Rahul's money = 3 : 2
Total money = 500
Samir's money = (3 ÷ 5) × 500 = 300
Rahul's money = (2 ÷ 5) × 500 = 200
Difference between their amounts = 300 − 200 = 100
Answer: Difference between their amounts = Rs. 100 📌 Exam tip: Convert fractional relations into a ratio first — saves time and avoids mistakes.