Mathematics — Old Questions (TPSC)
Q21. The value of √156 . 25
Given expression = √156.25
Write 156.25 as a fraction:
156.25 = 15625 ÷ 100
So,
√156.25 = √(15625 ÷ 100)
= √15625 ÷ √100
√15625 = 125
√100 = 10
Therefore,
√156.25 = 125 ÷ 10 = 12.5
Answer: The value of √156.25 is 12.5.
📌 Exam tip:
For square roots of decimals, convert into fractions first—calculation becomes easy and accurate.
Q22. What is 50 % of 40 % of Rs. 3,450 ?
We have to find 50% of 40% of Rs. 3,450
First find 40% of Rs. 3,450:
40% of 3450 = (40 ÷ 100) × 3450 = 1380
Now find 50% of 1380:
50% of 1380 = (50 ÷ 100) × 1380 = 690
Answer: 50% of 40% of Rs. 3,450 is Rs. 690.
📌 Exam tip:
When multiple percentages are involved, apply them step by step or multiply their decimal forms directly.
Q23. If a : b = 4 : 5 and b : c = 6 : 7, then a : c will be equal to :
Q24. A student was asked to add 16 and subtract 10 from a number. He by mistake added 10 and subtracted 16. if his answer is 14, what is the correct answer ?
Q25. A train crosses a pole in 15 seconds, while it crosses 100 metre long platform in 25 seconds. The length of the train is :
Q26. One-Fourth percent of 18 is :
One-fourth percent = 1 ÷ 4 % = 0.25%
0.25% of 18 = (0.25 ÷ 100) × 18
= 0.045
Answer: One-fourth percent of 18 is 0.045.
📌 Exam tip:
To find a fraction of a percent, convert it into decimal first, then multiply.
Q27. A sum of Rs. 427 is to be divided among A, B and C in such a way that 3 times A's share, 4 times B's share and 7 times C's share are all equal. The share of C is then :
Given that,
3 times A's share = 4 times B's share = 7 times C's share
Let the common value be x
So,
A's share = x ÷ 3
B's share = x ÷ 4
C's share = x ÷ 7
Total sum = 427
So,
(x ÷ 3) + (x ÷ 4) + (x ÷ 7) = 427
Taking LCM of 3, 4 and 7 = 84
(28x + 21x + 12x) ÷ 84 = 427
61x ÷ 84 = 427
61x = 427 × 84
x = (427 × 84) ÷ 61
x = 588
C's share = x ÷ 7
= 588 ÷ 7
= 84
Answer: The share of C is Rs. 84.
📌 Exam tip: When multiples of shares are equal, express each share in terms of a common variable first, then form the total-sum equation.
Q28. If an angle is 30° more than one-half of its complement, the value of the angle is :
Let the angle be x degrees
The complement of the angle = (90 − x)°
One-half of its complement = (90 − x) ÷ 2
Given that the angle is 30° more than one-half of its complement:
x = (90 − x) ÷ 2 + 30
Multiply both sides by 2:
2x = 90 − x + 60
2x = 150 − x
3x = 150
x = 50
Answer: The value of the angle is 50°.
📌 Exam tip:
For angle problems involving complements, always remember:
Complementary angles sum to 90°.
Q29. A goat is tied to one corner of a square plot of side 12 m by a rope 7 m long. The area it can graze is :
Side of the square plot = 12 m
Length of rope = 7 m
The goat is tied at one corner of the square.
Since the rope length (7 m) is less than the side of the square (12 m),
the goat can graze only in a quarter circle of radius 7 m.
Area grazed = (1 ÷ 4) × π × r2
= (1 ÷ 4) × π × 72
= (49π) ÷ 4
Using π = 22 ÷ 7,
Area = (49 × 22) ÷ (7 × 4)
= 154 ÷ 4
= 38.5 m2
Answer: The area the goat can graze is 38.5 square metres.
📌 Exam tip:
Goat tied at a corner → area = ¼ circle
Goat tied at midpoint of side → ½ circle
Q30. The sum of 5 successive odd numbers is 1075. What is the largest of these numbers ?
Let the 5 successive odd numbers be:
(x − 4), (x − 2), x, (x + 2), (x + 4)
Sum of these numbers = 1075
So,
(x − 4) + (x − 2) + x + (x + 2) + (x + 4) = 1075
5x = 1075
x = 215
Thus, the numbers are:
211, 213, 215, 217, 219
Largest number = 219
Answer: The largest of the numbers is 219. 📌 Exam tip:
For consecutive odd/even numbers,
the average = middle number = (sum ÷ count).