Mathematics — Old Questions (TPSC)
Q1. Simantika sold two radios for Rs. 2000 each. On one she gains 16 % and on the other she losses 16 %. Find her gain or loss per cent in the whole transaction.
Selling price of each radio = Rs. 2000
First radio (16% gain):
Cost Price = 2000 × 100 ÷ 116
= 2000 ÷ 1.16
= 1724.14 (approx.)
Second radio (16% loss):
Cost Price = 2000 × 100 ÷ 84
= 2000 ÷ 0.84
= 2380.95 (approx.)
Total Cost Price:
= 1724.14 + 2380.95
= 4105.09
Total Selling Price:
= 2000 + 2000 = 4000
Loss = 4105.09 − 4000 = 105.09
Loss % = (105.09 ÷ 4105.09) × 100
≈ 2.56%
Answer: There is a 2.56% loss in the whole transaction.
📌 Exam shortcut (very important):
When two articles are sold at the same selling price with equal gain and loss % = x,
Net result = loss of (x² / 100)%
Q2. 48 % of the first number is 60 % of the second number. What is the ratio of the first number to the second number ?
Let the first number be A and the second number be B
Given,
48% of A = 60% of B
So,
(48 ÷ 100)A = (60 ÷ 100)B
Cancel 100 from both sides:
48A = 60B
Divide both sides by 12:
4A = 5B
Therefore,
A : B = 5 : 4
Answer: The ratio of the first number to the second number is 5 : 4.
📌 Exam tip: In percentage–ratio problems, remove % first, then simplify the equation.
Q3. The product of two consecutive odd numbers is 6723. What is the greater number?
Let the two consecutive odd numbers be x and (x + 2)
Given product = 6723
So,
x(x + 2) = 6723
x2 + 2x − 6723 = 0
Check nearby odd numbers:
81 × 83 = 6723
So, the two odd numbers are 81 and 83
Greater number = 83
Answer: The greater number is 83.
📌 Exam tip: For products of consecutive odd numbers, try factorisation by inspection
Q4. What is the value of two-fifth of 60 % of 320 ?
We have to find two-fifth of 60% of 320
First find 60% of 320:
60% of 320 = (60 ÷ 100) × 320 = 192
Now find two-fifth of 192:
(2 ÷ 5) × 192 = 384 ÷ 5 = 76.8
Answer: The required value is 76.8.
📌 Exam tip: Always calculate percentage first, then apply the fraction — faster and safer.
Q5. Average of 5 numbers is 50. If every number is multiplied by 2, find the average.
Q6. A boat is rowed down stream at 35 Km./hr. and up stream at 17 km./hr. Find the speed of the stream.
Speed of stream = (Downstream speed − Upstream speed) / 2
= (35 − 17) / 2
= 18 / 2
= 9 km/hr
Q7. The sum of the two digit number and 40% of the number obtained by reversing its digits is 87.4. If the sum of digits of original number is 11, then find 60% of original number.
Let the number be 10x + y, so the reversed number is 10y + x.
Given: x + y = 11 ... (1)
Also: (10x + y) + 40% of (10y + x) = 87.4
=> (10x + y) + 0.4(10y + x) = 87.4
=> 10x + y + 4y + 0.4x = 87.4
=> 10.4x + 5y = 87.4
Multiply by 5: 52x + 25y = 437 ... (2)
From (1), y = 11 - x. Put in (2):
52x + 25(11 - x) = 437
=> 52x + 275 - 25x = 437
=> 27x = 162 => x = 6, so y = 5
Original number = 65
60% of 65 = (60/100) × 65 = 39
Answer: 39
Q8. A total of 324 coins of 20 paise and 25 paise make a sum of Rs. 71. The number of 25-paise coins is
Let the number of 25-paise coins be x.
Then number of 20-paise coins = 324 − x.
Total value = Rs. 71 = 7100 paise
25x + 20(324 − x) = 7100
=> 25x + 6480 − 20x = 7100
=> 5x = 620
=> x = 124
Since 124 is not in the options, the correct answer is None of these.
Q9. The ratio of cost price and selling price of an article is 20 : 21. Then the profit per cent on it is
Let Cost Price = 20 and Selling Price = 21.
Profit = 21 − 20 = 1.
Profit % = (Profit / Cost Price) × 100
= (1 / 20) × 100
= 5%.
Q10. Rahul is 4 years younger than his elder brother Shyam. Rahul also has a sister who is 3 years younger to him. What is the difference between the age of Rahul and the average age of the three siblings?
Let Rahul = R years.
Shyam = R+4, Sister = R-3.
Average age = (R + (R+4) + (R-3)) / 3 = (3R+1)/3 = R + 1/3.
Difference = 1/3 year = 4 months.