This post contains detailed solution for Maths questions that were asked in the Kerala PSC Secretary (LSGD), which was conducted on November 09, 2024.
1. A bag contain 50p, 25p and 10p coins in the ratio 5 : 9 : 4, amounting to Rs. 206. Find the number of 10p coins.
- A) 200 B) 40 C) 160 D) 360
Explanation:
Let 'x' be the common ratio of the no of coins, then
The no. of 50p coins = 5x
The no. of 25p coins = 9x
The no. of 10p coins = 4x
∴ Value of coins =
50p coins = .50 × 5x = 2.5x rupees
25p coins = .25 × 9x = 2.25x rupees
10p coins = .10 × 4x = 0.4x rupees
∴ Total Value of coins = 2.5x + 2.25x + 0.4x = 206 (∵ Total value of coins is given as 206)
= 5.15x = 206
= x = 2-6/5.15
= x = 40
∴ the no. of 10p coins is 4x = 4 × 40 = 160.
Solution
∴ There are 160 10p coins in the bag.
2. Arun can do a piece of work in 80 days. He works at it for 10 days and then Anil alone finishes the remaining work in 42 days. In how much time will Arun and Anil working together, finish the work?
Explanation:
Given that Arun can do a piece of work in 80 days.
∴ Work done by Arun in 1 day = 1/80.
∴ Work done by Arun in 10 days = 1/80 × 10
= 1/8
∴ Remaining work done by Anil = 1 - 1/8
= 7/8
Given that Anil alone completes the remaining work ie, 7/8th of the work in 42 days.
∴ Total Work is done by Anil in = 42 × 8/7
= 48 days.
In conclusion, we can say that Arun completes 1/8th of the work in 1 day, where as Anil completes 1/48th in 1 day.
∴ Total Work is done by Arun & Anil is = 1/8 + 1/48
= 8/240
= 1/30
Solution
∴ Arun & Anil working together will finish the work in 30 days.
3. The average of five consecutive odd numbers is 61. What is the difference between highest and lowest numbers?
Explanation:
Let the 5 consecutive odd numbers be: x, x+2, x+4, x+6, x+8
Given that the average of 5 consecutive odd numbers is 61.
∴ (x+ x+2+ x+4+x+6+x+8) / 5 = 61
(5x + 20) / 5 = 61
5x = 285
x = 57
∴ The 5 consecutive odd numbers are 57, 59, 61, 63, 65
∴ the difference between highest and lowest numbers = 65 -57 = 8
Solution
∴ The difference between highest and lowest numbers is 8.
4. Which of the following interchanges of signs would make the given equation correct?
100 + 100 × 50 – 2 ÷ 3 = 51
- A) – and × B) + and ÷ C) + and × D) + and –
Explanation:
In here, our goals is to get LHS (left hand side) & RHS right hand side) equal, ie by interchanging the signs we should get 51 in the LHS and 51 on the RHS.
For that we'll solve it by substituting the options in given the question,
Step 1: – and ×
100 + 100 × 50 – 2 ÷ 3 = 51
Substituting - & × and applying BODMAS rule,
100 + 100 × 50 × 2 ÷ 3 = 51
100 + 10000 ÷ 3 = 51
100 + 3333.33 = 51
3433.33 = 51
∴ LHS ≠ RHS
Step 2: + and ÷
100 + 100 × 50 – 2 ÷ 3 = 51
Substituting + & ÷ and applying BODMAS rule,
100 ÷ 100 × 50 – 2 ÷ 3 = 51
1 × 50 - 2 + 3 = 51
50 + 3 - 2 = 51
53 - 2 = 51
51 = 51
∴ LHS = RHS (we can stop here, as we got the answer or can continue checking the other options)
Solution
∴ Option B is the correct answer.
5. Identify the number that does not belong to the following series.
2, 6, 14, 30, 62, 126, 250
- A) 14 B) 62 C) 126 D) 250
Explanation:
2, 6, 14, 30, 62, 126, 250
3 × 2 = 6,
6 × 2 + 2 =14
14 × 2 + 2 = 30
30 × 2 + 2 = 62
62 × 2 + 2 = 126
126 × 2 + 2 = 254
Solution
∴ The number that doesn't fit into the series is 250.
6. What was the day of the week on 4th June 2002?
- A) Sunday B) Monday C) Tuesday D) Wednesday
Explanation:
By the formula,
Day of the week = (Date + Month Code + Century Code + Century Upper Year + Quotient of Upper Year) mod 7 ---------------------------------------------- I
Month Code:
| Month | Code |
|---|
| January | 1 |
| February | 4 |
| March | 4 |
| April | 0 |
| May | 2 |
| June | 5 |
| July | 0 |
| August | 3 |
| September | 6 |
| October | 1 |
| November | 4 |
| December | 6 |
Century Code:
| Century | Code |
|---|
| 1700s | 4 |
| 1800s | 2 |
| 1900s | 0 |
| 2000s | 6 |
The remainder of the division by 7 gives the day of the week:
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
Given,
Date = 4 (4th day of the month)
Month Code = 5 (June's code)
Century Code = 0 (for the 20th century)
Century Upper Year = 20 (the first two digits of the year)
Quotient of Upper Year = 20 / 4 = 5 (integer part)
Substituting the values in equation I,
Day of the week = ( 4+5+0+20+5) mod 7
= 34 mod 7
= 34 ÷ 7 and remainder is 2.
Day of the week = Tuesday (A remainder of 2 corresponds to Tuesday)
Solution
∴ June 4, 2002, was a Tuesday.
7. (25√5)x = (3√5)x + 1, what does x equal?
- A) 1/5 B) 2/13 C) 2/15 D) 5/3
Explanation:
(25√5)x = (5 2+1/2)x
(3√5)x + 1 = (51/3)x+1
(5 2+1/2)x = (51/3)x+1
5 5x/2 = 5x+1/3
5x/2 = x+1/3
15x = 2x+2
13x = 2
x = 2/13
8. 25% of 24 + 25% of 32 –14% of 350 = ?
- A) -35 B) 35 C) -30 D) 30
Explanation:
= (25/100 × 24) + (25/100 × 32) – (14/100 × 350)
= 6 + 8 – 49
= -35
Solution
∴ The answer is -35.
Post a Comment
Post a Comment