Mathematics — Old Questions (TPSC)
Q11. How many multiples of 3 are there from 1 to 100 which are not multiples of 2?
Multiples of 3 from 1 to 100 = floor(100/3) = 33.
Multiples of both 3 and 2 are multiples of 6: floor(100/6) = 16.
Required = 33 - 16 = 17.
Q12. In a group of cows and hens, the number of legs are 14 more than twice the number of heads. The number of cows is
Let cows = c, hens = h.
Heads = c + h.
Legs = 4c + 2h.
Given: 4c + 2h = 2(c + h) + 14
=> 4c + 2h = 2c + 2h + 14
=> 2c = 14
=> c = 7.
Q13. A sum of money tripled itself at compound interest in 10 years. In how many years will it become 27 times?
If it becomes 3 times in 10 years, then every 10 years it multiplies by 3 (same rate).
27 = 33 = 3 × 3 × 3.
So time = 3 × 10 = 30 years.
Q14. The sum of 6 consecutive integers is 9. What is the lowest integer?
Let the integers be n, n+1, n+2, n+3, n+4, n+5.
Sum = 6n + 15 = 9
=> 6n = -6
=> n = -1 (lowest integer).
Q15. A number consists of two digits whose sum is 11. If 27 is added to the number, then the digits change their places. What is the number?
Let number = 10x + y, with x + y = 11.
Reversed = 10y + x.
Given: 10x + y + 27 = 10y + x
=> 9x - 9y = -27
=> x - y = -3
Solving with x+y=11 gives x=4, y=7? (Check) Actually x-y=-3 and x+y=11 => x=4, y=7.
Number = 47, and 47 + 27 = 74 (reversed).
Answer: 47.
Q16. A number when divided by 80 leaves remainder 20. What is the remainder when the same number is divided by 16?
Number N = 80k + 20.
Divide by 16: 80k is divisible by 16 (since 80 = 16×5).
So remainder = 20 mod 16 = 4.
Q17. When the price of sugar decreases by 10%, a man could buy 1 kg more for Rs. 270. Then, the original price of sugar per kg is
Let original price = Rs. P per kg.
New price after 10% decrease = 0.9P.
With Rs.270: quantities are 270/P and 270/(0.9P).
Given difference = 1 kg:
270/(0.9P) - 270/P = 1
270(1/0.9 - 1)/P = 1
270((10/9) - 1)/P = 1
270(1/9)/P = 1
30/P = 1 => P = 30.
Q18. The least number which must be added to 3931 to make it a perfect square, is
622 = 3844 and 632 = 3969.
3931 lies between them.
3969 − 3931 = 38, so 3931 + 38 = 3969 (a perfect square).
Q19. If x4 + 1/x4 = 527 , then the value of x + 1/x is
x4 + 1/x4 = (x2 + 1/x2)2 − 2.
So (x2 + 1/x2)2 = 529 ⇒ x2 + 1/x2 = 23.
(x + 1/x)2 = 23 + 2 = 25 ⇒ x + 1/x = 5.
Q20. There are five types of envelope and four types of stamps in a Post office. How many ways are there to buy an envelope and a stamp?
Number of ways = (choices of envelope) × (choices of stamp)
= 5 × 4 = 20.