Target SSC CGL | 10,000+ Questions | Number System For SSC CGL : Day 11

Dear aspirants,

This is the new year with new goals, new experiences and lots of exams to be scheduled soon. SSC CGL has recently released the exam dates so now it is time to gear up your preparations and try hard to get the success. ADDA247 never fails to deliver something new and fruitful for you all. This time also we are providing you the best study plan as well as a study material. We are here going to prepare your Quantitative Aptitude section for the SSC CGL. In this article, we are providing you the details that how we are going to help you to clear the examination this year. We ADDA247 is going to provide you daily tests for all the subjects. The topic-wise quiz will be done from January till April. This will help you to get a deeper knowledge of all the topics and will prepare you thoroughly.

Q2. Two different prime numbers X and Y, both are greater than 2, then which of the following must be true?
(a) X – Y = 23
(b) X + Y ≠ 87
(c) Both (a) and (b)
(d) None of these

Q3. What is the remainder when 1! + 2! + 3! ……… + 100! Is divided by 7?
(a) 0
(b) 5
(c) 6
(d) 3

Q4. On dividing 2272 as well as 875 by 3–digit number N, we get the same remainder. The sum of the digits of N is:
(a) 10
(b) 11
(c) 12
(d) 13

Q6. There are two integers 34041 and 32506, when divided by a three–digit integer n, leave the same remainder. What is the value of n?
(a) 298
(b) 307
(c) 461
(d) Can’t be determined

Q7. The LCM of two numbers is 40 times their HCF. The sum of the LCM and HCF is 1,476. If one of the numbers is 288, find the other numbers?
(a) 169
(b) 180
(c) 240
(d) 260

Q9. In a problem involving division, the divisor is eight times the quotient and four times the remainder. If the remainder is 12, then the dividend is ?
(a) 300
(b) 288
(c) 512
(d) 524

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Solutions

   

S2. Ans.(b)
Sol.
Two prime numbers greater than 2 must be odd.
Sum of two odd numbers must always be even, thus,
X + Y = 87 is not possible.

S3. Ans.(b)
Sol.
7! + 8! + 9! + 10! + . . . . . . . . . + 100 = 7.6! + 8.7.6! + 9.8.7.6! + . . . . . . . . + 100! Is completely divisible by 7 as each of the terms contain at least one 7 in it.
Now, 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873
which leaves a remainder of 5 when divided by 7.

S4. Ans.(a)
Sol.
Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now, 1397 = 11 × 127
∴ The required 3-digit number is 127, the sum of whose digits is 10.

S6. Ans.(b)
Sol.
Let the common remainder be x. Then numbers (34041 – x) and (32506 – x) would be completely divisible by n.
Hence the difference of the numbers (34041 – x) and (32506 – x) will also be divisible by n
or (34041 – x – 32506 + x) = 1535 will also be divisible by n.
Now, using options we find that 1535 is divisible by 307.

In the last month i.e. in May daily we will provide you with a test of the previous years’ question papers, this will increase your confidence in solving the real exam and will make you familiar with the real-time exam.

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