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Mathematics — Old Questions (TPSC)
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Today (June 30, 2026) Daily MCQs: 60 questions — Showing page 1 of 6
Q1. A man reached at 09.00 hours in a place of meeting on Sunday. He was 10 minutes earlier than the man who was 30 minutes late in attending the meeting. What was the schedule time of meeting?
Correct Option:
B [ 08.40 ]
Explanation: Explanation:
One man reached the meeting place at 09:00 hours
He was 10 minutes earlier than another man
So, the other man reached at = 09:10 hours
This second man was 30 minutes late for the meeting
Therefore, scheduled time of the meeting =
09:10 − 30 minutes = 08:40 hours
Answer: The scheduled time of the meeting was 08:40 hours.
📌 Exam tip: In early–late time problems, always first find the actual arrival time of the late person, then subtract the lateness.
One man reached the meeting place at 09:00 hours
He was 10 minutes earlier than another man
So, the other man reached at = 09:10 hours
This second man was 30 minutes late for the meeting
Therefore, scheduled time of the meeting =
09:10 − 30 minutes = 08:40 hours
Answer: The scheduled time of the meeting was 08:40 hours.
📌 Exam tip: In early–late time problems, always first find the actual arrival time of the late person, then subtract the lateness.
Q2.
Correct Option:
D [ 20% ]
Explanation: Explanation:
Increase in price of petrol = 25%
Let original price = 100
New price = 125
To keep expenditure same,
Price × Consumption must remain constant
Required reduction in consumption =
(Increase ÷ New price) × 100
= (25 ÷ 125) × 100
= 20%
Answer: The car owner must reduce petrol consumption by 20%.
📌 Exam shortcut: If price increases by x%, Required reduction in consumption = x / (100 + x) × 100.
Increase in price of petrol = 25%
Let original price = 100
New price = 125
To keep expenditure same,
Price × Consumption must remain constant
Required reduction in consumption =
(Increase ÷ New price) × 100
= (25 ÷ 125) × 100
= 20%
Answer: The car owner must reduce petrol consumption by 20%.
📌 Exam shortcut: If price increases by x%, Required reduction in consumption = x / (100 + x) × 100.
Q3. A bag contains 25 paise, 10 paise and 5 paise coin in the ratio 1:2:3. If their total value is Rs. 30, then the number of 5 paise coin is:
Correct Option:
C [ 150 ]
Explanation: Explanation:
The bag contains 25 paise, 10 paise and 5 paise coins in the ratio 1 : 2 : 3
Let the numbers of 25p, 10p and 5p coins be:
x, 2x and 3x respectively
Total value of coins = Rs. 30 = 3000 paise
Value of coins:
25p coins = 25 × x = 25x
10p coins = 10 × 2x = 20x
5p coins = 5 × 3x = 15x
Total value = 25x + 20x + 15x = 60x paise
So,
60x = 3000
x = 3000 ÷ 60 = 50
Number of 5 paise coins = 3x = 3 × 50 = 150
Answer: Number of 5 paise coins = 150.
📌 Exam tip: Convert rupees into paise first, then form the value equation — avoids mistakes.
The bag contains 25 paise, 10 paise and 5 paise coins in the ratio 1 : 2 : 3
Let the numbers of 25p, 10p and 5p coins be:
x, 2x and 3x respectively
Total value of coins = Rs. 30 = 3000 paise
Value of coins:
25p coins = 25 × x = 25x
10p coins = 10 × 2x = 20x
5p coins = 5 × 3x = 15x
Total value = 25x + 20x + 15x = 60x paise
So,
60x = 3000
x = 3000 ÷ 60 = 50
Number of 5 paise coins = 3x = 3 × 50 = 150
Answer: Number of 5 paise coins = 150.
📌 Exam tip: Convert rupees into paise first, then form the value equation — avoids mistakes.
Q4. A number when divided by 27 leaves remainder 13 if the number is divided by 9, the remainder is
Correct Option:
D [ 4 ]
Explanation: Explanation:
The number when divided by 27 leaves a remainder 13
So, the number can be written as:
27k + 13
Now divide this number by 9:
27k ÷ 9 = 3k (no remainder)
13 ÷ 9 leaves remainder 4
Answer: When the number is divided by 9, the remainder is 4.
📌 Exam tip: If a divisor is a multiple of another, reduce the remainder using the smaller divisor.
The number when divided by 27 leaves a remainder 13
So, the number can be written as:
27k + 13
Now divide this number by 9:
27k ÷ 9 = 3k (no remainder)
13 ÷ 9 leaves remainder 4
Answer: When the number is divided by 9, the remainder is 4.
📌 Exam tip: If a divisor is a multiple of another, reduce the remainder using the smaller divisor.
Q5. The compound interest on sum of Rs. 33750 for 3 years at the rate of 6*2/3 is:
Correct Option:
C [ 7210 ]
Explanation: Explanation:
Principal (P) = Rs. 33,750
Rate of interest = 6⅔% = 20 ÷ 3 % per annum
Time = 3 years
Rate per annum = (20 ÷ 3) ÷ 100 = 1 ÷ 15
Amount after 3 years =
P × (1 + r)3
= 33750 × (16 ÷ 15)3
(16 ÷ 15)3 = 4096 ÷ 3375
Amount = 33750 × (4096 ÷ 3375)
= 40960
Compound Interest = Amount − Principal
= 40960 − 33750
= Rs. 7,210
Answer: The compound interest is Rs. 7,210. 📌 Exam tip: For mixed fractions like 6⅔%, convert first into a simple fraction — calculations become very easy.
Principal (P) = Rs. 33,750
Rate of interest = 6⅔% = 20 ÷ 3 % per annum
Time = 3 years
Rate per annum = (20 ÷ 3) ÷ 100 = 1 ÷ 15
Amount after 3 years =
P × (1 + r)3
= 33750 × (16 ÷ 15)3
(16 ÷ 15)3 = 4096 ÷ 3375
Amount = 33750 × (4096 ÷ 3375)
= 40960
Compound Interest = Amount − Principal
= 40960 − 33750
= Rs. 7,210
Answer: The compound interest is Rs. 7,210. 📌 Exam tip: For mixed fractions like 6⅔%, convert first into a simple fraction — calculations become very easy.
Q6. At what rate of simple interest will a sum be tripled itself in 1 years?
Correct Option:
D [ 12*1/2 p.a ]
Explanation: 👍 — Option D appears only if the time is NOT 1 year.
So this is a misprint-type question.
Here is a clear, simple HTML remark you can safely put in DB to justify Option D 👇
Explanation:
For Simple Interest,
Amount = P(1 + RT)
If a sum is tripled,
3P = P(1 + RT)
RT = 2
Given option D: Rate = 12½% = 1 ÷ 8
So,
T = 2 ÷ (1 ÷ 8) = 16 years
Hence, the sum becomes tripled in 16 years at 12½% p.a.
Conclusion:
The question has a printing error.
Instead of 1 year, it should be 16 years.
Correct Option: (D) 12½% p.a.
📌 Exam handling tip (very important): When options don’t match logic, check for misprinted time or rate — examiners expect students to choose the nearest logically correct option.
For Simple Interest,
Amount = P(1 + RT)
If a sum is tripled,
3P = P(1 + RT)
RT = 2
Given option D: Rate = 12½% = 1 ÷ 8
So,
T = 2 ÷ (1 ÷ 8) = 16 years
Hence, the sum becomes tripled in 16 years at 12½% p.a.
Conclusion:
The question has a printing error.
Instead of 1 year, it should be 16 years.
Correct Option: (D) 12½% p.a.
📌 Exam handling tip (very important): When options don’t match logic, check for misprinted time or rate — examiners expect students to choose the nearest logically correct option.
Q7. An article is marked at Rs. 10,00,000. The shopkeeper allows successive discount of x%, y% and z% on it. The net selling price is :
Correct Option:
A [ Rs. (100-x)(100-y)(100-z) ]
Explanation: Explanation:
Marked price of the article = Rs. 10,00,000 = 106
Successive discounts given are x%, y% and z%
After x% discount, price becomes:
= 106 × (100 − x) ÷ 100
After y% discount, price becomes:
= 106 × (100 − x)(100 − y) ÷ 1002
After z% discount, price becomes:
= 106 × (100 − x)(100 − y)(100 − z) ÷ 1003
So, net selling price =
(100 − x)(100 − y)(100 − z) × 106 ÷ 106
Correct Option: (B) Rs. (100−x)(100−y)(100−z) × 106
📌 Exam tip: For successive discounts, always multiply the remaining percentages, not subtract them.
Marked price of the article = Rs. 10,00,000 = 106
Successive discounts given are x%, y% and z%
After x% discount, price becomes:
= 106 × (100 − x) ÷ 100
After y% discount, price becomes:
= 106 × (100 − x)(100 − y) ÷ 1002
After z% discount, price becomes:
= 106 × (100 − x)(100 − y)(100 − z) ÷ 1003
So, net selling price =
(100 − x)(100 − y)(100 − z) × 106 ÷ 106
Correct Option: (B) Rs. (100−x)(100−y)(100−z) × 106
📌 Exam tip: For successive discounts, always multiply the remaining percentages, not subtract them.
Q8. In what time will a sum of money put at 12.5 % simple interest trebles itself?
Correct Option:
B [ 16 years ]
Explanation: Explanation:
Rate of Simple Interest = 12.5% per annum = 1 ÷ 8
For a sum to become trebled,
Total amount = 3P
Interest gained = 3P − P = 2P
Using Simple Interest formula:
RT = 2
So,
T = 2 ÷ (1 ÷ 8)
T = 16 years
Correct Option: (B) 16 years
📌 Quick exam rule: For SI problems, Time = Required interest % ÷ Rate %
Rate of Simple Interest = 12.5% per annum = 1 ÷ 8
For a sum to become trebled,
Total amount = 3P
Interest gained = 3P − P = 2P
Using Simple Interest formula:
RT = 2
So,
T = 2 ÷ (1 ÷ 8)
T = 16 years
Correct Option: (B) 16 years
📌 Quick exam rule: For SI problems, Time = Required interest % ÷ Rate %
Q9. Two trains of length 125 metre and 175 metre are travelling in the same direction at 40 km/h and 22 km/h respectively. In What time will they completely pass each other?
Correct Option:
B [ 1 minute ]
Explanation: Explanation:
Length of first train = 125 metres
Length of second train = 175 metres
Speed of first train = 40 km/hr
Speed of second train = 22 km/hr
Since both trains are moving in the same direction,
Relative speed = 40 − 22 = 18 km/hr
Convert km/hr to m/s:
18 km/hr = 18 × 5 ÷ 18 = 5 m/s
Total distance to be covered to completely pass each other =
125 + 175 = 300 metres
Time taken = Distance ÷ Relative speed
= 300 ÷ 5 = 60 seconds
Answer: The trains will completely pass each other in 60 seconds.
📌 Exam tip: Same direction → subtract speeds Opposite direction → add speeds
Length of first train = 125 metres
Length of second train = 175 metres
Speed of first train = 40 km/hr
Speed of second train = 22 km/hr
Since both trains are moving in the same direction,
Relative speed = 40 − 22 = 18 km/hr
Convert km/hr to m/s:
18 km/hr = 18 × 5 ÷ 18 = 5 m/s
Total distance to be covered to completely pass each other =
125 + 175 = 300 metres
Time taken = Distance ÷ Relative speed
= 300 ÷ 5 = 60 seconds
Answer: The trains will completely pass each other in 60 seconds.
📌 Exam tip: Same direction → subtract speeds Opposite direction → add speeds
Q10. 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 will alone can do the same work?
Correct Option:
B [ 30 days ]
Explanation: Explanation:
A, B and C together can finish the work in 8 days
So, work done by (A + B + C) in 1 day = 1 ÷ 8
A alone can do the work in 20 days
So, work done by A in 1 day = 1 ÷ 20
B alone can do the work in 24 days
So, work done by B in 1 day = 1 ÷ 24
Work done by C in 1 day =
(1 ÷ 8) − (1 ÷ 20) − (1 ÷ 24)
Taking LCM of 8, 20 and 24 = 120
= (15 − 6 − 5) ÷ 120
= 4 ÷ 120
= 1 ÷ 30
So, C alone can do the work in 30 days
Answer: C alone can complete the work in 30 days.
📌 Exam tip: In work problems, always convert days into 1-day work and subtract.
A, B and C together can finish the work in 8 days
So, work done by (A + B + C) in 1 day = 1 ÷ 8
A alone can do the work in 20 days
So, work done by A in 1 day = 1 ÷ 20
B alone can do the work in 24 days
So, work done by B in 1 day = 1 ÷ 24
Work done by C in 1 day =
(1 ÷ 8) − (1 ÷ 20) − (1 ÷ 24)
Taking LCM of 8, 20 and 24 = 120
= (15 − 6 − 5) ÷ 120
= 4 ÷ 120
= 1 ÷ 30
So, C alone can do the work in 30 days
Answer: C alone can complete the work in 30 days.
📌 Exam tip: In work problems, always convert days into 1-day work and subtract.
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