## SSC GD Constable Elementary Mathematics Syllabus

The SSC GD Elementary Mathematics section carries 1/4th of the weightage of the marks and questions asked in the Tier 1 exam. Hence, it is crucial to acquaint yourself with SSC GD Constable Elementary Mathematics Syllabus. When prepared meticulously, you can easily pass this section. So, to help you become a pro in SSC GD Constable Elementary Mathematics, here we have listed all the topics that are covered in SSC GD Syllabus of Elementary mathematics.

- Number Systems
- Computation of Whole Numbers
- Decimals and Fractions and the relationship between Numbers
- Fundamental arithmetical operations
- Percentages
- Ratio and Proportion
- Averages
- Interest
- Profit and Loss
- Discount
- Mensuration
- Time and Distance
- Ratio and Time
- Time and Work, etc.

## SSC GD Elementary Mathematics Questions

As per the SSC GD exam pattern, around 80 questions are asked out of which 20 questions are from the elementary mathematics section and the remaining questions are asked from the General Intelligence and Reasoning, General Knowledge and General Awareness and English/Hindi. SSC GD Maths questions are of matriculation or 10th Class examination standard and of easy-moderate difficulty level. With the application of formulae and short calculation tricks, candidates can attempt Elementary mathematics Questions from the topic number system easily. Check some of the examples of the questions that are being asked in the exam in the next segment.

## Previous Year SSC GD Maths Questions

To help you get hold of the question trends and patterns of the SSC GD Maths Questions that have been asked in the tier 1 CBT exam over the years, we have provided you with some of the frequently asked questions from the elementary mathematics questions.

1. A alone can complete a work in 6 days and B alone can complete the same work in 8 days. In how many days both A and B together can complete the same work?

A. daysB. days

C. days

D. days

Answer ||| D

Solution |||

Let the total work be 24x (LCM of 6x and 8x).

Then, efficiency of A = = 4x

And efficiency of B = = 3x

Efficiency of A and B together = 4x + 3x = 7x

Time taken by A and B together to complete the work = = days

2. Anil, Deepak and Dinesh together can complete a work in 35 days. Anil and Dinesh together can complete the same work in 60 days. In how many days Deepak alone can complete the same work?

A. 105 daysB. 84 days

C. 96 days

D. 110 days

Answer ||| B

Solution |||

Let the total work be 420x (LCM of 35x and 60x).

Then, the efficiency of Anil, Deepak and Dinesh together = = 12x

And efficiency of Anil and Dinesh together = = 7x

Therefore, efficiency of Deepak = 12x – 7x = 5x

Time taken by Deepak to complete the work = = 84 days

3. A car covers a certain distance at a speed of 48 km/hr in 14 hours. How much time it will take to cover the same distance at the speed of 84 km/hr?

B. 8 hr

C. 9 hr

D. 12 h

Answer ||| B

Solution |||

Given, Speed of car = 48 km/hr

Time = 14 Hours

Distance = Speed × Time

= 48 × 14 = 672 km

Time required in covering the same distance at the speed of 84 km/hr

= = = 8 hr.

4. If x : y = 5 : 4, then what will be the ratio of ?

B. 16 : 25

C. 4 : 5

D. 5 : 4

Answer ||| A

Solution |||

Let x is equal to ‘5a’.

Then, y = 5a × = 4a

Now, required

= = 25 : 16

5. If the length of a rectangle increases by 50% and the breadth decreases by 25%, then what will be the percent increase in its area?

B. 17.5%

C. 12.5%

D. 25%

Answer ||| C

Solution |||

Let the length and breadth of the rectangle are 100L and 100B respectively.

So, area of the rectangle = 100L × 100B = 10000LB

Now, new length(l1) = 100L + 50% of 100L

= 100L + = 100L + 50L = 150L

Similarly, new breadth = (100 – 25)% of 100B

⇒ 75% of 100B = = 75B

∴ New area of the rectangle will be

= 150L × 75B = 11250LB

Increase in area = 11250LB – 10000LB = 1250LB

Required percentage increase

= = 1250/100 = 12.5%

6. A sum of Rs. 4500 is lent at compound interest. If the rate of interest is 10% per annum (interest is compounded annually), then what will be the amount after 3 years?

B. Rs. 5689.5

C. Rs. 5686.5

D. Rs. 5889.5

Answer ||| A

Solution |||

Given:

Principal = ₹ 4500, Rate of interest = 10% p.a., Time = 3 years

Compounded Amount = Principal ×

= 4500 × = 4500 × (1.1)^{3} = 4500 × 1.331

= ₹ 5989.5

7. M can complete a work in 14 days less than the time taken by L. If both M and L together can complete the same work in 24 days, then in how many days M alone can complete the same work?

B. 56 days

C. 21 days

D. 42 days

Answer ||| D

Solution ||| Let time taken by L = x days

Time taken by M = (x−14) days

Now, L’s one day work + M’s one day work =

Both M and L together can complete the same work in 24 days.

(M+N)’s one day work = 1/24

According to the question,

⇒

⇒

⇒ 48x^{2} – 336 = x^{2} – 14x

⇒ x^{2} – 62x + 336 = 0

⇒ x^{2} – 56x –6x + 336 = 0

⇒ x(x – 56) –6(x + 56) = 0

⇒ (x – 56)(x –6) = 0

⇒ x = 56 or 6

(i) If x = 56, Time taken by L = 56 days

Time taken by M = (56-14) days = 42 days

(ii) If x = 6, Time taken by L = 6 days

Time taken by M = (6−14) days = −8 days, which is not possible.

So, x ≠ 6

Hence, Time taken by M = 42 days

8.A : B = 5 : 8 and B : C = 11 : 13. If A = 110, then what is the value of C?

A. 176B. 104

C. 208

D. 88

Answer ||| C

Solution |||

Given:

A = 110

Therefore, B = 110 × = 176

And C = 176 × = 208

9. A can complete a task in 8 days and B in 16 days respectively. If they work together for 3 days then the remaining part of the work left is:

B. 9/16

C. 7/16

D. 11/16

Answer ||| C

Solution |||

Total efficiency of A and B together = (2 + 1) = 3

Work done by both in 3 days = efficiency × time = 3 × 3 = 9

Remaining work = (16 – 9) = 7

∴ Part of the work left will be = 7/16

10. The height of a cone is 24 cm and radius of its base 10 cm. If the rate of painting it is ₹ 28 cm^{2}, then what will be the total cost in painting the cone from outside?

B. ₹20820

C. ₹17660

D. ₹22880

Answer ||| D

Solution |||

ATQ, Height of the Cone ‘h’ = 24 cm

Radius of the base of the Cone ‘r’ = 10 cm

We know, Slant Height of a Cone ‘l’ =

l = = = 26 cm

Now, the lateral surface area of a Cone- A_{LSA}= Πrl

A_{LSA} = ×10×26 cm^{2}

And the rate of painting = ₹28 cm^{2}

Therefore, Total cost of painting = ×10×26×28 = ₹22880

11. The diagonal of a square is 14 cm. What will be the length of the diagonal of the square whose area is double of the area of first square?

B.

C. 28 cm

D.

Answer ||| B

Solution |||

We know that D_{1} = √2×L_{1}; (d=diagonal and l= length)

∴ 14 = √2 × L_{1}

⇒ L_{1} = cm = 7√2 cm

Area of the square = (L_{1})^{2}

= (7√2)^{2} = 49 × 2 = 98 cm^{2}

Area of the second square

= 2× 98 = 196 cm^{2}

So, (L_{2})^{2} = 196

⇒ L_{2} = √196 = 14 cm

Thus, D_{2} = √2 × L_{2};

(D_{2}=diagonal of second square) = √2 × 14 = 14√2 cm

12. Three cubes each of sides 7 cm are joined end to end. Find the surface area of the resulting solid.

^{2}

B. 696 cm

^{2}

C. 882 cm

^{2}

D. 782 cm

^{2}

Answer ||| A

Solution |||

Since three cubes are joined it will become a cuboid.

So, length = 7 + 7 + 7 = 21 cm,

Breadth = 7 cm and height = 7 cm

Now, required surface area of the resulting solid

= 2(lb + bh + lh)

= 2(21 × 7 + 7 × 7 + 21 × 7)

= 2(147 + 49 + 147)

= 2 × 343 = 686 cm^{2}

13. Two numbers are in the ratio 5 : 6. If 16 is subtracted from each, the numbers will be in the ratio 3 : 4. If 8 is added to the first number and 3 is subtracted from the second, then they will be in the ratio:

B. 16 : 15

C. 15 : 16

D. 13 : 14

Answer ||| B

Solution |||

Let the two numbers are 5x and 6x respectively.

According to the question,

⇒ 4(5x – 16) = 3(6x – 16)

⇒ 20x – 64 = 18x – 48

⇒ 20x – 18x = 64 – 48

⇒ 2x = 16

⇒ x = 16/2 = 8

∴ Two numbers are

(5 × 8) = 40 and (6 × 8) = 48

Now, 8 is added to the first number and 3 is subtracted from the second number.

(40 + 8) = 48 and (48 – 3) = 45

So, the required ratio will be -

⇒ 48 : 45 = 16 : 15

14. If A : B = 7 : 8 and B : C = 7 : 9, then what is the ratio of A : B : C ?

A. 56 : 49 : 72B. 49 : 56 : 72

C. 56 : 72 : 49

D. 72 : 56 : 49

Answer ||| B

Solution |||

Therefore, A : B : C = 49 : 56 : 72

15. The value of [25 of 4 − 30 + (2^{2} + 18)] is:

B. 84

C. 92

D. 37

Answer ||| C

Solution |||

Consider [25 of 4 − 30 + (2^{2} + 18)]

Using BODMAS

⇒ 25 of 4 − 30 + (22)

⇒ 100 – 30 + 22

= 92

## SSC GD Elementary Mathematics Questions PDF

As the General Duty Constable exam can be held anytime soon, candidates are looking for important SSC GD Elementary Mathematics Questions to practice and improve their overall score. So, to help aspirants improve their number of good attempts in the mathematics section, we have compiled important SSC GD Elementary mathematics Questions in PDFs in the Hindi and English languages. Simply click on the links provided below to download SSC GD Maths Questions PDF.

**SSC GD Elementary Mathematics Questions PDF in English****SSC GD Elementary Mathematics Questions PDF in Hindi**

## Preparation Tips for SSC GD Constable Elementary Mathematics Syllabus

Math is the section that determines the final score and the merit list of candidates in the SSC GD Constable exam.Candidates that perform well in this section have a decent chance of performing well in the final result.

- Candidates must utilize their time efficiently to finish the Elementary mathematics section during the SSC GD Constable tier 1 exam.
- Candidates should do extra practice on specified topics in order to answer this section.
- Candidates should set a target for the number of questions they want to attempt while paying close attention to the rest of the sections.
- Candidates are advised to skip time-taking questions.
- Only by practising SSC GD Mock test and attempting more questions, candidates can develop the habit of recognising the nature of a question within seconds.

Hope this was useful!!

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