Mathematics — Old Questions (TPSC)
Q31. A number when added to its half gives 63. Find the number.
Let the number be x.
x + x/2 = 63
⇒ (3x)/2 = 63
⇒ x = (63 × 2)/3 = 42
Q32. A sum of money lent at simple interest becomes Rs. 3224.00 in 2 years and Rs. 4160.00 in 5 years. Find out the sum.
Difference of amounts = 4160 − 3224 = 936, which is interest for 3 years.
Yearly interest = 936 ÷ 3 = 312.
Interest for 2 years = 312 × 2 = 624.
Principal = 3224 − 624 = Rs. 2600
Q33. A, B, and C working together can finish a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone can do the same work?
A + B + C work rate = 1/8.
A''s rate = 1/20 and B''s rate = 1/24.
C''s rate = 1/8 − (1/20 + 1/24).
= 15/120 − 11/120 = 4/120 = 1/30.
Hence, C alone can do the work in 30 hours
Q34. 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
Q35. 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.
Q36. 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
Q37. 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.
Q38. 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
Q39. 5% of 30% of 100 is
30% of 100 = 30.
5% of 30 = (5/100) × 30 = 1.5
Q40. 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