SSC GD Math Questions: Download GD Math Question PDF in Hindi/English

1.Simplify: [7.8 − 0.4 of (5.1 − 3.8) + 9.3 × 1.5]

A. 23.12
B. 12.23
C. 21.23
D. 23.21

Answer ||| C

Solution |||

[7.8 − 0.4 of (5.1 − 3.8) + 9.3 × 1.5]

= [7.8 – (0.4 × 1.3) + (9.3 × 1.5)]

= [7.8 – 0.52 + 13.95]

= (21.75 – 0.52)

= 21.23

2.A is twice as good as a workman as B. And together, they finish a piece of work in 20 days. In how many days, will A alone finish the work?

A. 30 days
B. 25 days
C. 26 days
D. 28 days

Answer ||| A

Solution |||

Let the efficiency of B is x.

∴ Efficiency of A is 2x.

Efficiency of A and B together = x + 2x = 3x

We know that,

Work = Time × Efficiency

So, work = 20 × 3x = 60x

Thus, A will alone finish the work in = 60x/2x = 30 days

3.If a shopkeeper gives two successive discounts of 20% and 15% on a book whose marked price is Rs. 850, then what is the selling price of the book?

A. Rs. 758
B. Rs. 587
C. Rs. 785
D. Rs. 578

Answer ||| D

Solution |||

Since

∴ Required selling price of the book is =

= 5780000/10000

= Rs. 578

4.Study the bar chart and answer the question based on it.

The average production of 1999 and 2000 was less than the average production of which of the following pairs of years?

Figure

A. 1998 and 2000
B. 1996 and 1997
C. 2000 and 2001
D. 1995 and 2001

Answer ||| C

Solution |||

Average production of 1999 and 2000 = (65+50)/2 = 57.5

Average production of 1998 and 2000 = (45+50)/2 = 47.5

Average production of 1996 and 1997 = (40+60)/2 = 50

Average production of 2000 and 2001 = (50+75)/2 = 62.5

Average production of 1995 and 2001 = (25+75)/2 = 50

Thus, it is clear that in average production of 2000 and 2001 is more than that of 1999 and 2000.

5.A cloth was 50cm broad and 8 cm long. When washed, it was found to have lost 25% of its length and 14% of its breadth, then the percentage decreased in its area is ……….

A. 34.5%
B. 35.5%
C. 36%
D. 35%

Answer ||| B

Solution |||

Area of the cloth = length × breadth = 50 × 8 = 400 cm²

After getting washed,

New length = 50 – 25% of 50 = 50 – 25/2 = 75/2 cm

New breadth = 8 – 14% of 8

= 172/25 cm

New area = 

= 3 × 86 = 258 cm²

Net change in area = (400 – 258) = 142 cm²

∴Percentage change in its area =

= 142/4 = 35.5%

6.A batsman scored 120 runs which included 15 fours and 2 sixes. What percent of runs scored by him running between the wickets?

A. 40%
B. 35%
C. 37.50%
D. 25%

Answer ||| A

Solution |||

Let the runs taken between the wickets by the batsman is x.

According to the question,

Total Runs = (15 × 4) + (2 × 6) + x = 120

⇒ 60 + 12 + x = 120

⇒ 72 + x = 120

⇒ x = 120 – 72 = 48

∴ Required percentage will be = (48/120)×100 = 40%

7.The value of [4 ÷ 2 + 3 x 6 – 7] is:

A. 13
B. 14
C. 11
D. 12

Answer ||| A

Solution |||

[4÷2+3 x 6–7]

= [(4/2) + (3 × 6) – 7]

= (2 + 18) – 7

= 20 – 7 = 13

8.A and together have Rs 1800. If 1/3 of A’s amount is equal to 2/3 of B’s amount, how much does B have?

A. Rs. 1200
B. Rs. 900
C. Rs. 750
D. Rs. 600

Answer ||| D

Solution |||

Let A and B have a and b amount respectively.

So,

(a + b) = 1800 —- (i)

Also,

a/3 = 2b/3

⇒ a = 2b —-(ii)

Putting the value of (ii) in (i)-

⇒ 2b + b = 1800

⇒ 3b = 1800

⇒ b = 1800/3 = Rs. 600

9.The ratio between the perimeter and the breadth of a rectangle is 3 : 1. If the area of the rectangle is 310 sq. cm, the length of the rectangle is nearly:

A. 11.45 cm
B. 10.45 cm
C. 12.45 cm
D. 13.45 cm

Answer ||| C

Solution ||| The ratio between the perimeter and the breadth of a rectangle is 3 : 1.

Let perimeter of a rectangle = 3x cm

Breadth of a rectangle = x cm

We know that, Perimeter of a rectangle = 2(L + B)

Hence, 3x = 2(Length + x)

⇒2 × Length = x

⇒Length = x/2 cm

Now, Area of the rectangle is 310 sq. cm.

So,

Length × Breadth = 310

⇒(x/2) ×(x) = 310

⇒x²/2 = 310

⇒x²= 620

⇒x = 24.9 cm

Hence, Length of rectangle = x/2 = 24.9 = 12.45 cm

10.The volume of a solid sphere is 4851 m³. What is the surface area of the sphere? (Take Π = 22/7)

A. 1386 m²
B. 1364 m²
C. 1260 m²
D. 1408 m²

Answer ||| A

Solution ||| Volume of a solid sphere = 4851 m³

⇒4ϖr³/3 = 4851

⇒r³ = (4851 × 3)/4ϖ

⇒r³ = (4851 × 3)/4ϖ

⇒ r³ = 1157.625

⇒ r = 10.5 m

Surface area of the sphere = 4ϖr² =

= 1386 m²

11.Find the value of:

[(7×6 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

A. 12
B. 14
C. 16
D. 18

Answer ||| A

Solution |||

Consider: [(7×6 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

Using BODMAS:

⇒ [(42 + 5 − 15)÷4 + 6÷3 – 4 + 18÷3]

⇒ [32÷4 + 6÷3 – 4 + 18÷3]

⇒ 8 + 2 – 4 + 6 = 12

12.If A is equal to 20% of C and B is equal to 40% of C, then which one of the following is 150% of B?

A. 50% of C
B. 250% of A
C. 65% of C
D. 300% of A

Answer ||| D

Solution |||

Given,

A = 20% of C

A = 20C/100 = C/5

C = 5A ……………….(i)

And B = 40% of C

B = 40C/100 = 2C/5 …………..(2)

Also, 150% of B = 150B/100

Using (1) and (2):

= 300% of A

13.If the selling price is tripled, the profit becomes 5 times. What is the profit percent?

A. 100%
B. 200%
C. 50%
D. 150%

Answer ||| A

Solution ||| Let cost price = x unit

And Selling price = y unit

Let Profit = (y−x) unit

If the selling price is tripled, the profit becomes 5 times.

New Selling price = 3y unit

New profit = 5(y – x) unit ………(i)

But

if S.P. = 3y, then Profit = (3y – x) unit ………….(ii)

From (i) and (ii):

⇒ 3y − x = 5y – 5x

⇒ 2y = 4x

⇒ y = 2x

Hence Original Profit = y – x = 2x – x = x

Hence, Profit percentage = (x/x)×100 = 100%

14.If 15 apples cost as much as 6 strawberries, 2 strawberries cost as much as 16 bananas, 6 bananas cost as much as 15 potatoes, then what is the cost of 1 potato if an apple costs ₹20?

A. ₹25
B. ₹22.50
C. ₹2.50
D. ₹2.25

Answer ||| C

Solution |||

ATQ, an apple costs ₹20.

So, Cost price of 15 apples = 15 × 20 = ₹300

Also, 15 apples cost as much as 6 strawberries.

Cost price of 6 strawberries = ₹300

Cost price of 2 strawberries = (300/6)×2 = ₹100

Now, 2 strawberries cost as much as 16 bananas.

Cost price of 16 bananas = ₹100

Cost price of 6 bananas = (100/16)×6 = ₹37.5

And 6 bananas cost as much as 15 potatoes.

Cost price of 15 potatoes = ₹37.5

Cost price of a potato = 37.5/15 = ₹2.5

15.A wheel makes 4000 revolution is covering a distance of 60 km. The radius of the wheel is:

A. 8 m
B. 8.25 m
C. 4.68 m
D. 2.39 m

Answer ||| D

Solution |||

A wheel makes 4000 revolution is covering a distance of 60 km.

60 km = 60000 m

Distance covered in 1 revolution = 60000/4000 = 15 meter

We know, Distance covered in one revolution = Perimeter of Wheel

⇒ 2Πr = 15

⇒ r = 15/2Π

⇒ r = (15×7)/(2×22)

⇒ r = 2.39 m

Hence, radius of the wheel = 2.39 m

16.Shopkeeper selling on article for Rs.46 loses 8%. In order to gain of 6%, what should be the selling price of the article?

A. Rs.65
B. Rs.56
C. Rs.53
D. Rs.85

Answer ||| C

Solution |||

Given, If Shopkeeper sells an article for Rs.46, he loses 8%.

Let cost price of article = 100 unit

Loss% = 8%

So, 8% of 100 unit = 8 unit

Selling price = 100 unit – 8 unit = 92 unit

According to the question,

92 unit → Rs. 46

1 unit → Rs. 0.5

100 unit → Rs. 50

Hence, Cost price of article = Rs. 50

If Profit% = 6%

Then Selling price of article = 50 + 6% of 50 = 50 + 3 = 53

17.The areas of two squares are 16:9 . The ratio of their perimeter is:

A. 9:16
B. 9:12
C. 12:16
D. 16:12

Answer ||| D

Solution |||

Given, the areas of two squares are 16:9

Let area of first square = 16x²

Area of second square = 9x²

Hence,

Side of first square = √16x² = 4x

Side of second square = √9x² = 3x

Perimeter of first square = 16x

Perimeter of second square = 12x

Ratio of their perimeter = 16 : 12

18.The difference between upper limit and lower limit of the class interval is called as:

A. Range
B. Mode
C. Frequency distribution
D. Class interval

Answer ||| D

Solution ||| We know that difference between upper limit and lower limit of the class interval is called class size.

19.In how many years, a sum will be thrice of it at the rate of interest 5% per annum?

A. 25 years
B. 40 years
C. 30 years
D. 20 years

Answer ||| B

Solution |||

Given, Rate of interest = 5%

Let the number of years required = t

Let Principle = P

Amount = 3P

Interest = Amount – principle =3P – P = 2P

Now using formula,

⇒ t = 200/5 = 40 years

20.In a university, the number of students studying science, mathematics and language are in the ratio of 2:4:9. If the number of students in Science, Mathematics and Language be increased 10%, 20% and 40% respectively. What will be the new ratio?

A. 12:23:63
B. 11:24:63
C. 24:11:63
D. 63:11:24

Answer ||| B

Solution |||

Let number of students studying science = 2x

Number of students studying mathematics = 4x

Number of students studying language = 9x

The number of students in Science, Mathematics and Language be increased 10%, 20% and 40% respectively. 

New number of students studying science = 2x + 10% of 2x

= 2x + (10/100)×2x = 2x + x/5 = 11x/5

New number of students studying mathematics = 4x + 20% of 4x

= 4x + (20/100)×4x = 4x + 4x/5 = 24x/5

New number of students studying language = 9x + 40% of 9x

= 9x + (40/100)×9x = 9x + 18x/5 = 63x/5

New ratio = 11x/5 : 24x/5 : 63x/5 = 11 : 24 : 63

21.Rama and Hari can together finish a piece of work in 15 day. Rama works twice as fast as Hari, then Hari alone can finish work in:

A. 45 days
B. 30 days
C. 25 days
D. 20 days

Answer ||| A

Solution |||

Let Number of days taken by Hari alone to finish work = 2x

Hari’s one day work = 1/2x unit

Number of days taken by Rama alone to finish work = x

Rama’s one day work = 1/x unit

It is given that Rama and Hari can together finish a piece of work in 15 day.

According to the question,

One day Work of Hari + One day work of Rama = One day work of Hari and Rama Together

⇒ 15 + 30 = 2x

⇒ 2x = 45

Number of days taken by Hari alone to finish work = 45 days

22.The distance between two stations is 500 km. A train starts from station ‘A’ at 9 am and travels towards station ‘B’ at 60 km/hrs. Another train starts from station ‘B’ at 10 am and travels towards ‘A’ at 40 km/hrs. At what time both the train meet to each other.

A. 12:45 PM
B. 1:36 PM
C. 3:36 PM
D. 2:24 PM

Answer ||| D

Solution |||

Distance between two stations = 500 km.

A train starts from station ‘A’ at 9 am and travels towards station ‘B’ at 60 km/hrs. Another train starts from station ‘B’ at 10 am and travels towards ‘A’ at 40 km/hrs.

Let train P starts from station ‘A’ and travels towards station ‘B’ and train Q starts from station ‘B’ and travels towards station ‘A’

Distance covered by train P from 9 am to 10 am = 60 km /hours × 1 hour = 60 km

Remaining distance between station A and station B = 500 km – 60 km = 440 km

After 10 am both trains are moving towards each other.

So, Relative speed = 60 + 40 = 100 km/hrs

Time taken to cover 440 km at the relative speed of 100 km/hr = 440/100 = 4 hour 24 minutes

Both trains will meet after 4 hours and 24 minutes post 10 am.

Hence, Both trains will meet at 2.24 PM.

23.A trader allows two successive discounts of 5% each on the marked price of a sofa set for Rs.24,500. The selling price of sofa set is:

A. Rs.23,274.75
B. Rs.23,275
C. Rs.1,163.75
D. Rs.22,111.25

Answer ||| D

Solution ||| Marked price of a sofa set = Rs.24,500.

Trader allows two successive discounts of 5% each on the marked price.

Selling price of sofa set =

= Rs. 22111.25

24.The captain of a cricket team of 11 members is 28 years old and the wicket keeper is 4 years older than him. If the ages of these two are excluded, the average age of the remaining players is two years less than the average age of the whole team. What is the average age of the team?

A. 15 years
B. 22 years
C. 21 years
D. 29 years

Answer ||| C

Solution |||

Let the average age of 11 team members is x.

So, total age of the team = 11x.

Also, the age of the captain = 28 years

Age of the wicket-keeper = 28 + 4 = 32 years

According to the question,

Age of team – Age of captain and Wicket-keeper = Age of other 9 members

⇒ 11x – (28 + 32) = 9×(x – 2)

⇒ 11x – 60 = 9x – 18

⇒ 11x – 9x = 60 – 18

⇒ 2x = 42

⇒ x = 42/2 = 21

∴ Average age of the team is 21 years.

25.Two numbers are respectively 25% and 50% more than a third number. The ratio of the first number to second number is:

A. 1 : 2
B. 2 : 1
C. 6 : 5
D. 5 : 6

Answer ||| D

Solution |||

Let the third number is 100x.

∴ First number = 25% of 100x + 100x

First number = (25/100)×100x + 100x = 25x + 100x = 125x

Second number = 50% of 100x

Second number = (50/100)×100x + 100x = 50x + 100x = 150x

So, the required ratio will be –

= 125x : 150x

= 5 : 6