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Mathematics — Old Questions (TPSC)
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Today (March 23, 2026) Daily MCQs: 60 questions — Showing page 3 of 6
Q21. If an angle is 30° more than one-half of its complement, the value of the angle is :
Correct Option:
A [ 50° ]
Explanation: Explanation:
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°.
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°.
Q22. 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 :
Correct Option:
C [ 38.5 sq.m ]
Explanation: Explanation:
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
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
Q23. The sum of 5 successive odd numbers is 1075. What is the largest of these numbers ?
Correct Option:
C [ 219 ]
Explanation: Explanation:
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).
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).
Q24. Which is the correct order to show units from largest to smallest ?
Correct Option:
B [ Litre, decilitre, millilitre ]
Explanation: Explanation:
Units of capacity in descending order (largest to smallest) are:
1 litre (L) = 10 decilitres (dL)
1 decilitre (dL) = 100 millilitres (mL)
So,
Litre > Decilitre > Millilitre
Checking the options:
(A) Millilitres, decilitre, litre ❌ (small to large)
(B) Litre, decilitre, millilitre ✔️
(C) Decilitre, millilitre, litre ❌ (incorrect order)
(D) Litre, millilitre, decilitre ❌ (wrong middle order)
Correct Option: (B) Litre, decilitre, millilitre
📌 Exam tip: Remember the ladder:
L → dL → mL (each step down is ×10 or ×100).
Units of capacity in descending order (largest to smallest) are:
1 litre (L) = 10 decilitres (dL)
1 decilitre (dL) = 100 millilitres (mL)
So,
Litre > Decilitre > Millilitre
Checking the options:
(A) Millilitres, decilitre, litre ❌ (small to large)
(B) Litre, decilitre, millilitre ✔️
(C) Decilitre, millilitre, litre ❌ (incorrect order)
(D) Litre, millilitre, decilitre ❌ (wrong middle order)
Correct Option: (B) Litre, decilitre, millilitre
📌 Exam tip: Remember the ladder:
L → dL → mL (each step down is ×10 or ×100).
Q25. The difference between 78 % of a number and 59 % of the number is 323. What is 62% of that number ?
Correct Option:
A [ 1054 ]
Explanation: Explanation:
Let the number be x
Given,
Difference between 78% and 59% of the number = 323
So,
(78% − 59%) of x = 323
19% of x = 323
x = 323 × 100 ÷ 19
x = 1700
Now find 62% of the number:
62% of 1700 = (62 ÷ 100) × 1700
= 1054
Answer: 62% of the number is 1054.
📌 Exam tip: When the difference of percentages is given, first find the base number, then apply the required percentage.
Let the number be x
Given,
Difference between 78% and 59% of the number = 323
So,
(78% − 59%) of x = 323
19% of x = 323
x = 323 × 100 ÷ 19
x = 1700
Now find 62% of the number:
62% of 1700 = (62 ÷ 100) × 1700
= 1054
Answer: 62% of the number is 1054.
📌 Exam tip: When the difference of percentages is given, first find the base number, then apply the required percentage.
Q26. 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.
Correct Option:
B [ 2.56 ]
Explanation: Explanation:
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)%
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)%
Q27. 48 % of the first number is 60 % of the second number. What is the ratio of the first number to the second number ?
Correct Option:
C [ 5:4 ]
Explanation: Explanation:
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.
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.
Q28. The product of two consecutive odd numbers is 6723. What is the greater number?
Correct Option:
D [ 83 ]
Explanation: Explanation:
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
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
Q29. What is the value of two-fifth of 60 % of 320 ?
Correct Option:
C [ 76.8 ]
Explanation: Explanation:
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.
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.
Q30. Average of 5 numbers is 50. If every number is multiplied by 2, find the average.
Correct Option:
C [ 100 ]
Explanation:
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