For a charity show, the total tickets sold were 420. Half of these tickets were sold at the rate of ₹ 5 each, one-third at the rate of ₹ 3 each and the rest for ₹ 2 each. What was the total amount received?
UPSC-2014
To find the total amount received, we need to calculate the number of tickets sold at each price point and then multiply by their respective prices. Finally, sum up these amounts to get the total.
Total number of tickets sold:
\[ \text{Total tickets} = 420 \]Tickets sold at ₹ 5 each:
Half of the total tickets were sold at ₹ 5 each.
Amount received from tickets sold at ₹ 5:
\[ \text{Amount}_1 = 210 \times 5 = \text{₹ } 1050 \]Tickets sold at ₹ 3 each:
One-third of the total tickets were sold at ₹ 3 each.
Amount received from tickets sold at ₹ 3:
\[ \text{Amount}_2 = 140 \times 3 = \text{₹ } 420 \]Tickets sold at ₹ 2 each (the rest):
First, find the number of tickets already accounted for:
Amount received from tickets sold at ₹ 2:
\[ \text{Amount}_3 = 70 \times 2 = \text{₹ } 140 \]Total amount received:
Sum the amounts from all three categories.
The total amount received was ₹ 1610.
C) ₹ 1,610
A gardener has 1000 plants. He wants to plant them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of plants that he needs more for this purpose?
UPSC-2013
The problem requires finding the minimum number of additional plants needed to arrange 1000 plants in a square formation (equal number of rows and columns). This means we need to find the smallest perfect square greater than or equal to 1000.
Let the number of rows be \(N\) and the number of columns be \(N\). The total number of plants will be \(N \times N = N^2\).
We are given 1000 plants. We need to find the smallest perfect square that is greater than or equal to 1000.
First, let's find the square root of 1000:
\[ \sqrt{1000} \approx 31.62 \]Since the number of rows and columns must be an integer, we need to consider the next whole number greater than 31.62, which is 32.
Now, calculate the square of 32:
\[ 32^2 = 32 \times 32 = 1024 \]This means that to arrange the plants in 32 rows and 32 columns, the gardener needs a total of 1024 plants.
The gardener currently has 1000 plants. The number of additional plants needed is the difference between the required number of plants and the available number of plants:
\[ \text{Additional plants} = 1024 - 1000 = 24 \]Therefore, the gardener needs 24 more plants.
B) 24 — As calculated, the smallest perfect square greater than 1000 is 1024. The difference between 1024 and 1000 is 24, which is the minimum number of plants needed.
Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds respectively. The number of times they would fire the bullets together at the target in an hour is
UPSC-2014
To find out when the five persons fire bullets together, we need to determine the least common multiple (LCM) of their individual firing intervals. This LCM will represent the time after which they will all fire simultaneously again. Once we have this time, we can calculate how many times this simultaneous firing occurs within one hour, including the initial firing at time zero.
Correct Option: B) 7
The LCM of 6, 7, 8, 9, 12 is 504 seconds. One hour is 3600 seconds. The number of times they fire together after the initial firing (at \(t=0\)) within one hour is given by \(\lfloor \frac{3600}{504} \rfloor = 7\). This interpretation is common in questions of this type where the initial event is often implicitly excluded from the count of events within a duration, or the question specifically asks for events after the first one. If the question intended to include the initial firing, the answer would be 8, which is not an option. Therefore, 7 is the most appropriate answer based on the given options.
What is the remainder when 91x92x93x94x95x96x97x98x99 is divided by 1261?
UPSC-2022
To find the remainder when a product of numbers is divided by another number, we can look for factors of the divisor within the numbers being multiplied. If the divisor is a factor of any of the numbers in the product, or if the divisor can be formed by multiplying factors from the numbers in the product, then the remainder will be 0.
D) 0 — The divisor 1261 can be factored into \(13 \times 97\). The product includes 91, which is \(7 \times 13\), and it also includes 97. Since both 13 and 97 are present as factors within the numbers being multiplied, the entire product is divisible by \(13 \times 97 = 1261\). Therefore, the remainder is 0.
What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?
UPSC-2022
To find the smallest number greater than 1000 that leaves a remainder of 3 when divided by 6, 9, 12, 15, or 18, we first need to find the Least Common Multiple (LCM) of these divisors. The number will be of the form \(\text{LCM} \times k + 3\), where \(k\) is an integer. We then find the smallest such number that is greater than 1000.
C) 1083 is the smallest number greater than 1000 that leaves a remainder of 3 when divided by 6, 9, 12, 15, or 18.
Let p be a two-digit number and q be the number consisting of same digits written in reverse order. If pxq = 2430, then what is the difference between p and q?
UPSC-2022
We are given a two-digit number \(p\) and another number \(q\) formed by reversing the digits of \(p\). Their product is given as \(p \times q = 2430\). We need to find the difference between \(p\) and \(q\).
Let the two-digit number \(p\) be represented as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Here, \(a \in \{1, 2, ..., 9\}\) and \(b \in \{0, 1, ..., 9\}\).
The number \(q\) formed by reversing the digits will be \(10b + a\).
We are given that \(p \times q = 2430\). So, we have:
\[ (10a + b)(10b + a) = 2430 \]Since \(p \times q = 2430\), and \(p\) and \(q\) are numbers with the same digits (just reversed), they should be relatively close to each other. If \(p \approx q\), then \(p^2 \approx 2430\).
Let's find the square root of 2430:
So, \(p\) and \(q\) must be around 40-50. Since \(50^2 = 2500\) is slightly greater than 2430, one number must be less than 50 and the other could be slightly more or less than 50.
We are looking for two numbers whose product is 2430 and whose digits are reversed. The units digit of 2430 is 0. This means that the product of the units digits of \(p\) and \(q\) must end in 0. Let the units digit of \(p\) be \(b\) and the units digit of \(q\) be \(a\). So \(b \times a\) must end in 0. This implies either \(a=0\), \(b=0\), or one of them is 5 and the other is an even number.
If \(a=0\), then \(p\) would be \(b\) (a single digit number), which contradicts \(p\) being a two-digit number. So \(a \neq 0\).
If \(b=0\), then \(p = 10a\) and \(q = a\). This also contradicts \(q\) being a two-digit number (unless \(a=0\), which makes \(p=0\), not a two-digit number). So \(b \neq 0\) if \(q\) is a two-digit number. However, if \(b=0\), then \(p = 10a\) and \(q = a\). For example, if \(p=40\), then \(q=04=4\). \(40 \times 4 = 160 \neq 2430\).
Let's consider the case where one digit is 5 and the other is even. Since \(p\) and \(q\) are around 40-50, let's try numbers in that range.
So, \(p=45\) and \(q=54\) (or vice-versa, which doesn't affect the difference).
The difference between \(p\) and \(q\) is \(|q - p|\) or \(|p - q|\).
\[ |54 - 45| = 9 \]D) 9
