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Mathematics — Old Questions (TPSC)
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Today (February 05, 2026) Daily MCQs: 60 questions — Showing page 4 of 6
Q31. 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.
Q32. 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.
Q33. 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.
Q34. 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.
Q35. 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 %
Q36. 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
Q37. 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.
Q38. A trader will loss 12 % if he sells an article for Rs. 2640.00. For how much should he sell it to gain 12 % ?
Correct Option:
C [ Rs. 3360.00 ]
Explanation: Explanation:
Selling price at 12% loss = Rs. 2640
Let Cost Price = CP
At 12% loss,
Selling Price = 88% of CP
So,
0.88 × CP = 2640
CP = 2640 ÷ 0.88
CP = 3000
Now, to gain 12%,
Required Selling Price = 112% of CP
= 1.12 × 3000
= Rs. 3360
Answer: To gain 12%, the article should be sold for Rs. 3360.
📌 Exam shortcut: If loss = x%, CP = SP × 100 / (100 − x) If gain = x%, SP = CP × (100 + x) / 100
Selling price at 12% loss = Rs. 2640
Let Cost Price = CP
At 12% loss,
Selling Price = 88% of CP
So,
0.88 × CP = 2640
CP = 2640 ÷ 0.88
CP = 3000
Now, to gain 12%,
Required Selling Price = 112% of CP
= 1.12 × 3000
= Rs. 3360
Answer: To gain 12%, the article should be sold for Rs. 3360.
📌 Exam shortcut: If loss = x%, CP = SP × 100 / (100 − x) If gain = x%, SP = CP × (100 + x) / 100
Q39. What is the sum of numbers from 1 to 100?
Correct Option:
C [ 5050 ]
Explanation: Explanation:
The sum of numbers from 1 to 100 is the sum of first 100 natural numbers.
Formula for sum of first n natural numbers =
n(n + 1) ÷ 2
Here, n = 100
Sum = 100 × 101 ÷ 2
= 5050
Answer: The sum of numbers from 1 to 100 is 5050.
📌 Exam tip: Always use the formula n(n + 1)/2 for fast calculation.
The sum of numbers from 1 to 100 is the sum of first 100 natural numbers.
Formula for sum of first n natural numbers =
n(n + 1) ÷ 2
Here, n = 100
Sum = 100 × 101 ÷ 2
= 5050
Answer: The sum of numbers from 1 to 100 is 5050.
📌 Exam tip: Always use the formula n(n + 1)/2 for fast calculation.
Q40. The product of two numbers is 7200. If their HCF is 20, then what will be the LCM?
Correct Option:
B [ 360 ]
Explanation: Explanation:
Product of two numbers = 7200
HCF of the two numbers = 20
Relationship between two numbers:
Product of numbers = HCF × LCM
So,
LCM = 7200 ÷ 20
= 360
Answer: The LCM of the two numbers is 360.
📌 Exam tip: Always remember: Number₁ × Number₂ = HCF × LCM
Product of two numbers = 7200
HCF of the two numbers = 20
Relationship between two numbers:
Product of numbers = HCF × LCM
So,
LCM = 7200 ÷ 20
= 360
Answer: The LCM of the two numbers is 360.
📌 Exam tip: Always remember: Number₁ × Number₂ = HCF × LCM
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