The natural numbers below 660 which are divisible by 5 and 11 but not by 3 are?
Answer: A
If the number is divisible by 5 and 11 it must be divisible by 55.
The numbers are less than 660.
Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).
The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.
The numbers of such, multiples is the quotient of 11 by 3.
Quotient of 11/3=3.
Out of 11 multiples of 55, 3 are multiples of 3.
Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8
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If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:
Answer: A
Putting x=2, we get \(2^{2}(2^{2}-1) = 12\). So, \(x^{2}(x^{2}-1)\) is always divisible by \(12\)
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The number 3 divides ‘a’ with a result of ‘b’ and a remainder of 2.
The number 3 divides ‘b’ with a result of 2 and a remainder of 1.
What is the value of a?
Answer: D
Since 3 divides b with a result of 2 and a remainder of 1, b=2×3+1=7.
Since number 3 divides a with a result of b (which we now know equals 7) and a remainder of 2, a=b×3+2=7×3+2= 23.
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The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is:
Answer: D
\(x+(x+36)+y=100 \\ \Rightarrow 2x+y=64\)
Therefore \(y\) must be even prime, which is 2.
Therefore,
\(2x+2=64 \\ \Rightarrow x=31\)
Third prime number
\(= (x+36)\\= (31+36)\\= 67\)
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A set of S consists of,
i). All odd numbers from 1 to 55.
ii). All even numbers from 56 to 150.
What is the index of the highest power of 3 in the product of all the elements of the set SS?
Answer: A
No answer description available for this question.
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A bank issued credit card numbers and the corresponding PIN (Personal Identification Number). Both are 3-digit numbers up to 996. Pinaki was the last to get the credit card and so he had the last possible credit card number.
He was afraid of forgetting his PIN. He wrote down the number 123 in his diary to remember his PIN. He also wrote out the way to calculate 123 : "Multiply the card number by PIN. Divide the product by 997. The remainder is 123".
Once, Prafull saw his diary in which Pinaki wrote this number 123. Prafull did a lot of purchasing, as he now knows Pinaki's PIN. What is Pinaki's PIN?
Answer: A
No answer description available for this question.
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A chain smoker had spent all the money he had. He had no money to buy his cigarettes. Hence, he resorted to join the stubs and to smoke them.
He needed 4 stubs to make a single cigarette. If he got a pack of 10 cigarettes as a gift, then how many cigarettes could he smoke in all?
Answer: D
Ten cigarettes give 10 stub.
From 10 stubs 3 more cigarettes can be made (2 stubs would be obtained from 2 cigarettes formed by joining 8 stubs.)
So he could smoke 13 cigarettes in all.
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In a meet, people from five different places have assembled in Bangalore High School. From the five places the persons come to represent are 42,60,210,90 and 84.
What is the minimum number of rooms that would be required to accommodate so that each room has the same number of occupants and occupants are all from the same places?
Answer: C
All the students from each have to be accommodated in a certain number of rooms.
There should be no person left over (remainder) from any places who can be clubbed together with the persons left over from other places.
To have the minimum number of rooms, the capacity of each room is HCF of all the numbers.
HCF(42,60,210,90,84)=6HCF(42,60,210,90,84)=6
Thus Min number of rooms =(42+60+210+90+84) / 6=81
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