## Seventh SSC CGL level Solution Set, topic Arithmetic Number system

This is the seventh solution set of 10 practice problem exercise for SSC CGL exam on topic Arithmetic Number system. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set.

It is emphasized here that answering in MCQ test is not the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

- must have complete understanding of the basic concepts of the topics
- is adequately fast in mental math calculation
- should try to solve each problem using the most basic concepts in the specific topic area and
- does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving happens in the last step. But how to do that?

You need to use **your problem solving abilities** only. There is no other recourse.

If you have not taken the corresponding test yet, take the test by referring to * SSC CGL level Question Set 7 on Number system* and come back to go through the solution.

### Seventh solution set- 10 problems for SSC CGL exam: topic Arithmetic Number system - time 12 mins

**Q1. **The smallest number by which if 392 is multiplied, the result becomes a perfect cube, is,

- 3
- 6
- 7
- 9

** Solution:** By mental factorization we find, $392=8\times{49}=2^3\times{7^2}$. This will be transformed to the next perfect cube if multiplied by 7.

**Answer:** Option c: 7.

**Key concepts used:** Factorization -- concept on powers or indices expressing the factors in terms of powers of factors -- finding the missing factor to make all powers 3.

**Q2.** If the numbers 1, 2, 3, 4, 5, .......998, 999, 1000 are multiplied together, how many 0s on the right the product will have?

- 211
- 200
- 249
- 30

**Solution:** Let's analyze the mechanism of how a 0 is formed on the right of a number. One 0 is created by $5\times{2}=10$. Two 0s are created by $25\times{4}=100$. Three 0s are created by $125\times{8}=1000$ and four zeros are created by $625\times{16}=10000$. These are all the mechanisms by which 0s are created in the product of our problem.

There is only one 625 and it actually contributes to 4 numbers of 0s.

There are in total 8 numbers of multiples of 125, out of which 625 we have already considered. So rest 7 numbers of 125s, contribute $7\times{3}=21$ numbers of 0s.

In the next step we find the number of multiples of 25 to be 40 out of which 8 already have been consumed by 625 and 125s. Rest 32 numbers of 25s then contribute $32\times{2}=64$ numbers of zeros.

At the last step we have, 200 numbers of multiples of 5 out of which again we have already consumed 40. So the rest 160 each contributes one 0.

The total number of 0s is then, $4 + 21 + 64 + 160 = 249$.

**Answer:** Option c : 249 .

**Key concepts used:** Basic mechanisms of 0 creation -- relation between these mechanisms -- actual counting of each and calculating excluded sum resulting from each mechanism starting from the four 0s mechanism first.

**Q3.** In a two digit number, the unit's digit exceeds ten's digit by 2 and the product of the sum of the digits and the number is 144, then the number is,

- 46
- 26
- 24
- 42

**Solution:** 144 being a small number, its 2 digit factors can be enumerated directly as 12, 16, 18, 24, 36, 48, and 72. Only 24 satisfies both the criteria.

**Answer:** Option c: 24.

**Key concepts used:** Quick enumeration of the suitable factors -- examination with respect to given condition.

**Q4. **The sum of 37 consecutive positive integers is 851. The largest of the numbers is,

- 54
- 41
- 37
- 72

**Solution:** By the nature of sum of natural numbers we know, sum of first 9 natural numbers as $9\times{\frac{9 +1}{2}} = 9\times{5} = 45$ and this pattern will hold for all odd number of consecutive integers.

Essentially if the sum is divided by the the odd number of the set of integers, we get the average which is exactly the middle number of the series of the numbers.

Dividing 851 by 37 we get 23. We do it mentally. So there are 18 numbers below 23 and 18 above it in the series. Thus the largest number in the series is $23 + 18 = 41$.

**Answer:** Option b: 41.

**Key concepts used:** Concept on sum of odd number of consecutive integers -- concept on position of average of the series with respect to other numbers below and above it.

**Q5. **The unit's digit of $459\times{459} + 77\times{77} + 2785\times{2789}$ is,

- 7
- 9
- 0
- 5

**Solution:** The unit's digit of $459\times{459} + 77\times{77} + 2785\times{2789}$ is the sum of unit's digit of each of the three terms. So the resultant unit's digit is, the unit's digit of,

$9\times{9} + 7\times{7} + 5\times{9} = 81 + 49 + 45$.

Again, unit's digit of a sum of numbers is the sum of the unit's digit. So the desired unit's digit is,

$1 + 9 + 5 = 15$, that is, 5.

**Answer:** Option d: 5.

**Key concepts used:** Unit's digit concepts.

**Q6.** In three three digit numbers each of which when divided by 5 leaves a remainder of 3. The difference between the largest and smallest is twice the difference between the first two. If the sum of the three numbers is 429, the second largest among the three is,

- 129
- 243
- 133
- 143

**Solution:** The second statement means the three numbers form a series of three numbers with same difference between each two adjacent numbers. Sum of such a set of three numbers will be three times the average middle number.

Dividing 429 by 3 we get, 143.

**Answer:** Option d : 143.

**Key concepts used:** Detecting the nature of the series from the problem statement and noting that the middle number, the average, is required. We can find out this number but not the other two.

**Note:** The first statement is superfluous.

** Q7.** Two numbers both larger than 29 have HCF 29 and LCM 4147. The sum of the numbers is,

- 669
- 896
- 996
- 696

**Solution: **We need to have an idea of the factors of 4147 as we know, the product of the numbers is the product of the HCF and LCM. Dividing 4147 first by 29 mentally we get, 143. Now we factorize the product of HCF and LCF easily,

$HCF\times{LCM} = 29\times{4147} = 29\times{29}\times{11}\times{13}$.

As both numbers are larger than 29, their sum must be,

$29(11 + 13)=29\times{24}=696$

**Answer:** Option d: 696.

** Key concepts used:** Concept of HCF and LCM and their relationship with the pair of numbers -- careful consideration of the given conditions to identify the factors forming the two numbers.

** Q8.** The ten's digit of a two digit number exceeds the unit's digit by 5. The second number formed by reversing the digits is less than the first number by five times the sum of the digits of the number. The sum of the digits is,

- 7
- 13
- 9
- 11

** Solution:** The two digit such numbers can be 16, 27, 38 and 49, that's all. Out of these 4 27 satisfies the given condition.

$72 - 5\times{(2 + 7)} = 72 - 45 = 27$.

**Answer:** Option c: 9.

**Key concepts used:** Identifying the 4 possible numbers -- enumeration technique -- inspection of each with respect to the given condition.

**Q9.** How many of the following numbers are divisible by 132?

396, 4, 264, 792, 2178, 5184, 6336, 968, 762

- 6
- 4
- 7
- 5

**Solution:** On inspection we find, $132 = 11\times{3}\times{2}\times{2}$. It means any multiple of 132 must have all these factors.

Starting from the left, 396 is the 3rd multiple of 132 - count 1.

3rd term 264 is 2nd multiple of 132 - count 2.

968 is not divisible by 3 (sum of digits is not divisible by 3). So we are left with 5 more terms.

792 is the sixth multiple of 132 - count 3. So 762 is out.

For 2178 we use 396 to reach near it $396\times{5}=1980$, left is198. So 2178 is not a multiple of 132.

For 5184 we do the same, $396\times{13}=5148$. So 5184 is also out.

For 6336, $396\times{16}=6336$, a perfect hit - count 4.

We have used 396 as it is very near to 400.

**Answer:** Option b: 4.

**Key concepts used:** Factorization of 132 gives an idea of its constituent factors. Using 3 and its simple rule of divisibility we could eliminate one number 968.

264, 396 and 792 directly are identified to be multiples of 132 as these are small numbers.

396 being near 400, for the other larger numbers, multiples of 396 were used for reaching as near to the target number as possible. All by mental maths.

** Q10.** The natural numbers are grouped in the sequence 1, (1 + 2), (1 + 2 + 3), (1 + 2 + 3 + 4)......Then the value of the 10th term of the sequence is,

- 50
- 57
- 55
- 60

**Solution:** From careful identification of the pattern we find the last term within the brackets of the 2nd term of the 10 term sequence is 2, last of the 3rd term of the 10 term sequence is 3 and similarly for the 4th term of the sequence the last term within brackets is 4. So by induction we know for the 10th term of the 10 term sequence the last term inside the brackets must be 10.

It means the 10th term will be the series, (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). The sum upto 9 terms is 45 and so upto 10th term sum is 55.

**Answer:** Option c: 55.

**Key concepts used:** By pattern identification and principle of induction formation of the 10th term -- sum of natural numbers.

### Other resources on Number system and related topic

You may refer to our other useful resources on number system and other related topics especially algebra.

The first should of course be the solution to this question. Carefully go through the solution absorbing the ideas explained in the process of reaching the solutions. You won't be able to retain this knowledge unless you apply this analytical deductive thinking on other problems repeatedly.

#### Valuable guidelines with links to resources on SSC CGL

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**Numbers and Number system and basic mathematical operations**

**Factorization or finding out factors**

**Basic and Rich Algebra concepts for elegant solutions of SSC CGL problems - **** though on Algebra, it should be useful.**

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