Mathematics — Old Questions (TPSC)
Q51. 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).
Q52. 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.
Q53. 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.
Q54. 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.
Q55. 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).
Q56. 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.
Q57. 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.
Q58. 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 = 527 + 2 = 529.
Hence x2 + 1/x2 = 23.
Now (x + 1/x)2 = x2 + 1/x2 + 2 = 23 + 2 = 25.
Therefore x + 1/x = 5.
Answer: 5
Q59. By selling a bicycle for Rs. 2200, a man gains 10%. At what price did he buy the bicycle?
Gain = 10%, Selling Price = Rs. 2200.
Cost Price = (Selling Price × 100) / (100 + Gain%)
= (2200 × 100) / 110 = Rs. 2000
Q60. A number when added to its half gives 63. Find the number.
Let the number be x.
x + x/2 = 63
⇒ (3x)/2 = 63
⇒ x = (63 × 2)/3 = 42