# SSC GD Constable Maths Questions PDF

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## SSC GD Constable Maths Questions 2019 PDF:

SSC GD Constable Question and Answers download PDF based on previous year question paper of SSC GD 2018-19 exam. Latest 20 Very important Maths questions in Quant for GD Constable.

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Question 1: A man spends 75% of his income. His income increases by 20% and his expenditure also increases by 10%. The percentage of increase in his savings is

a) 40%
b) 30%
c) 50%
d) 25%

Question 2: A fruit-seller buys some oranges and by selling 40% of them he realises the cost price of all the oranges. As the oranges being to grow over-ripe, he reduces the price and sells 80% of the remaining oranges at half the previous rate of profit. The rest of the oranges being rotten are thrown away. The overall percentage of profit is

a) 80
b) 84
c) 94
d) 96

Question 3: A dishonest dealer professes to sell his goods at the cost price but uses a false weight of 850 g instead of 1 kg. His gain percent is

a) $17\frac{12}{7}$%
b) $17\frac{11}{17}$%
c) $71\frac{11}{17}$%
d) $11\frac{11}{17}$%

Question 4: The ratio of present ages of Golu and Polu is 17 : 19. After 5 years the age of Polu will be 81 years. What is the present age of Golu?

a) 56 years
b) 68 years
c) 52 years
d) 74 years

Question 5: The ratio of present ages of Y and Z is 5 : 8. 9 years ago, the age of Y was 7/13th the age of Z. What is the present age of Z?

a) 46years
b) 52 years
c) 48 years
d) 40 years

Question 6: The ratio of the present age of Monu and Sonu is 13:11. The ratio of the age of Monu 4 years from now and the age of Sonu 4 years earlier is 5:3. What is the difference between the ages of Sonu and Monu?

a) 4 years
b) 6 years
c) 8 years
d) 10 years

Question 7: Two successive price increase of 10% and 10% of an article are equivalent to a single price increase of

a) 19%
b) 20%
c) 21%
d) 22%

Question 8: If each side of a square is increased by 10%. its area will be increased by:

a) 10%
b) 21%
c) 44%
d) 100%

Question 9: The marked price of an item is Rs. 480. The shopkeeper allows a discount at 10% and gains 8%. If no discount is allowed, his gain percent would be

a) 18%
b) 18.5%
c) 20.5%
d) 20%

Question 10: The sum of a fraction and its reciprocal is $\frac{113}{56}$. Find the fraction.

a) $\frac{7}{8}$
b) $\frac{5}{8}$
c) $\frac{8}{9}$
d) $\frac{3}{7}$

Question 11: Arrange the fractions $\frac{3}{4},\frac{5}{12},\frac{13}{16},\frac{16}{29},\frac{3}{8}$ in their descending order of magnitude.

a) $\frac{3}{4}>\frac{3}{8}>\frac{13}{16}>\frac{16}{29}>\frac{5}{12}$
b) $\frac{3}{8}>\frac{5}{12}>\frac{16}{29}>\frac{3}{4}>\frac{13}{16}$
c) $\frac{13}{16}>\frac{3}{4}>\frac{16}{29}>\frac{5}{12}>\frac{3}{8}$
d) $\frac{13}{16}>\frac{16}{29}>\frac{3}{4}>\frac{5}{12}>\frac{3}{8}$

Question 12: Manish can complete a work in 21 days and Karan can complete the same work in 28 days. If both together work for 7 days, then what fraction of total work is left?

a) $\frac{3}{5}$
b) $\frac{2}{3}$
c) $\frac{7}{12}$
d) $\frac{5}{12}$

Question 13: Arrangement of the factions $\frac{4}{3},-\frac{2}{9},-\frac{7}{8},\frac{5}{12}$ into ascending order are

a) $-\frac{7}{8},-\frac{2}{9},\frac{5}{12},\frac{4}{3}$
b) $-\frac{2}{9},-\frac{7}{8},\frac{5}{12},\frac{4}{3}$
c) $-\frac{2}{9},-\frac{7}{8},\frac{4}{3},\frac{5}{12}$
d) $-\frac{7}{8},-\frac{2}{9},\frac{4}{3},\frac{5}{12}$

Question 14: If a sum of money doubles itself in 8 yrs, then the interest rate in percentage is ?

a) 8 ½%
b) 10%
c) 10 ½ %
d) 12 ½ %

Question 15: A certain principal is invested in a scheme of compound interest. The amount obtained after 1 year is Rs 2400 and the amount obtained after 2 years is Rs 2880. What is the rate of interest (in percentage)?

a) 20
b) 15
c) 25
d) 10

Question 16: A person borrows some money for 8 years at a rate of simple interest. If the ratio of principal and total interest is 5: 8, then what is the rate ( in percentage) of interest?

a) 10
b) 20
c) 25
d) 30

Question 17: The marked price of a pen is Rs 3000. The shopkeeper gives two successive discounts of 15% and a% to the customer. If the customer pays Rs 2142 for the pen, then what is the value (in percentage) of a?

a) 16
b) 14
c) 18
d) 17

Question 18: The marked price of a book is Rs 4200. The shopkeeper gives two successive discounts of 25% and y% to the customer. If the customer pays Rs 2898 for the book, then what is the value (in percentage) of y?

a) 7
b) 8
c) 6
d) 5

Question 19: A man purchased an article for ₹1500 and sold it at 25% above the cost price. If he has to pay ₹75 as tax on it, his net profit percentage will be:

a) 25%
b) 30%
c) 15%
d) 20%

Question 20: A shopkeeper marks his goods 20% higher than the cost price and allows a discount of 5%. The percentage of his profit is.

a) 14%
b) 15%
c) 10%
d) 20%

let the income of person be Rs y So his expenditure be 75% of income which is = 0.75y Saving = y- 0.75y = 0.25yNow after increment of 20 % in income the new income becomes = 1.2y Given that bew expenditure is 10% more than previous one.So , new expenditure is = 1.1 × 0.75y = 0.825y New savings = 1.2y – 0.825y = 0.375yPercentage increase in savings =$\frac {0.375y-0.25y}{0.25y}$×100 = 50%

Let us assume that the fruit seller buys 100 oranges for Rs. 100
He sells 40 oranges for Rs. 100
Profit obtained = 100 – 40 = Rs. 60
% Profit = $\frac{60}{40} \times 100 = 150$%
Now, he sells 80% of the remaining oranges at half the profit
i.e., he sells 48 oranges at 75% profit.
Selling Price of 48 oranges = 48 + 75% of 48 = Rs. 84
Rest of them are thrown away.
Total SP = 100 + 84 = 184
Profit = 184 – 100 = 84
% Profit = 84%

Let assume that he buys 1000 gram of quantity at 1000 Rs So, Cost Price of 1 gram = 1 Rs Now as it is mentioned that he is selling only 850 gram in 1000 Rs So , selling Price of 1 gram = $\frac{1000}{850}$ = Rs 1.1764 So profit percentage = $\frac{SellingPrice-CostPrice}{CostPrice}$ = $\frac{1.1764-1}{1}\times100$ = 17.64 % = $17\frac{11}{17}$%

Let the present age of Golu and Polu be ‘x’ and ‘y’ years
Thus, $\frac{x}{y} = \frac{17}{19}$
After 5 years the age of Polu will be 81 years.
So, y+5 = 81
y = 76
So, x = $\frac{17*76}{19}$ = 68 years
Hence, option B is the right answer.

Let the present age of Y and Z be y and z respectively.
So, y:z = 5:8
Also,
y-9 = 7*(z-9)/13
13y-117 = 7z – 63
13y-7z = 54
13*5z/8 – 7z = 54
9z/8 = 54
z = 48 years
Hence, option C is the right answer.

Let the present age of Monu and Sonu be ‘x’ and ‘y’ respectively.
So, x:y = 13:11
It is given that,
$\frac{x+4}{y-4} = \frac{5}{3}$
3x + 12 = 5y – 20
3x-5y = -32
3*$\frac{13y}{11}$ – 5y = -32
-16y = -32*11
y = 22
So, x = 26
Thus, y-x = – 22 + 26 = 4 years
Hence, option A is the right answer.

Let’s say price of article is 100
After first increase its price will be 100(1+ $\frac{10}{100}$) = 110
Now second increment will be applied on 110
Hence new price will be 110(1+ $\frac{10}{100}$ ) = 121
Which is 21 more than before any increment
Hence total percentage increment = 21

Let’s say side of square is 100 unit
its area will be $10^{4} unit^{2}$
after 10% increment its value will be 110 unit
and area will become $1.21 \times 10^{4} unit^{2}$
change in area $.21 \times 10^{4} unit^{2}$
percentage change will be 21

Marked price = 480
discount = 10%
Selling price = 480 – $\frac{480 \times 10}{100}$ = 432
Gain = 8%
Cost price = $432 \times (\frac{100}{108})$ = 400
After no discount ,gain will be = 480-400 = 80
Percentage gain = $\frac{80}{400} \times 100$ = 20%

Let the fraction be $x$ According to ques, => $x+\frac{1}{x}=\frac{113}{56}$ => $\frac{x^2+1}{x}=\frac{113}{56}$ => $56x^2-113x+56=0$ => $56x^2-49x-64x+56=0$=> $7x(8x-7)-8(8x-7)=0$=> $(7x-8)(8x-7)=0$=> $x=\frac{8}{7},\frac{7}{8}$=> Ans – (A)
$\frac{3}{4}=0.75$$\frac{5}{12}=0.42 \frac{13}{16}=0.81 \frac{16}{29}=0.55 \frac{3}{8}=0.37Descending order = \frac{13}{16}>\frac{3}{4}>\frac{16}{29}>\frac{5}{12}>\frac{3}{8} => Ans – (C) 12) Answer (D) Let total work to be done = L.C.M. (21,28) = 84Manish can complete the work in 21 days, => Manish’s efficiency = \frac{84}{21}=4 units/day Similarly, Karan’s efficiency = \frac{84}{28}=3 units/dayNow, (Manish+Karan)’s 7 day’s work = (4+3)7=49 unitsFraction of work that is left = \frac{(84-49)}{84}=\frac{35}{84}=\frac{5}{12}=> Ans – (D) 13) Answer (A) Given factions \frac{4}{3},-\frac{2}{9},-\frac{7}{8},\frac{5}{12} Multiply 72(LCM) with each fraction, then we get96, -16, -63, 30Arrange them in ascending order i.e -63 < -16 < 30 < 96 (or) -\frac{7}{8}<-\frac{2}{9}<\frac{5}{12}<\frac{4}{3}Hence, option A is the correct answer. 14) Answer (D) Let the sum = Rs. x and rate of interest = r\%=> Amount after 8 years = Rs. 2x=> Simple interest = (2x-x)=Rs. xAlso, SI = \frac{P \times r \times t}{100}=> \frac{x \times r \times 8}{100}=x=> r=\frac{100}{8}=12\frac{1}{2}\%=> Ans – (D) 15) Answer (A) Amount obtained after 1 year = Rs. 2400Amount obtained after 2 years = Rs. 2880=> Rate of interest = \frac{(2880-2400)}{2400}\times100= \frac{480}{24}=20\%=> Ans – (A) 16) Answer (B) Let Principal amount = Rs. 5x and interest = Rs. 8xLet rate of interest = r\% and time period = 8 years=> Simple interest = \frac{P\times R\times T}{100} => \frac{5x\times r \times 8}{100}=8x=> \frac{5r}{100}=1=> r=\frac{100}{5}=20\%=> Ans – (B) 17) Answer (A) Marked price = Rs. 3000After first discount of 15%, price of pen = 3000-(\frac{15}{100}\times3000)= 3000-450=Rs. 2550After 2nd discount of a\%, price = 2550-(\frac{a}{100}\times2550)=2142=> 25.5a=2550-2142=> 25.5a=408=> a=\frac{408}{25.5}=16\%=> Ans – (A) 18) Answer (B) Marked price = Rs. 4200 After first discount of 25%, price of pen = 4200-(\frac{25}{100}\times4200) = 4200-1050=Rs. 3150 After 2nd discount of y\%, price = 3150-(\frac{y}{100}\times3150)=2898 => 31.5y=3150-2898 => 31.5y=252 => y=\frac{252}{31.5}=8\% => Ans – (B) 19) Answer (D) Cost price = Rs. 1500Markup % = 25%=> Selling price = 1500+(\frac{25}{100}\times1500)= 1500+375=Rs. 1875Total cost price (including tax) = 1500+75=Rs. 1575$$\therefore$ Profit % = $\frac{(1875-1575)}{1575}\times100$= $\frac{300}{15.75}\approx 20\%$=> Ans – (D)
Let cost price = Rs. $100$Markup % = 20%=> Marked price = $100+(\frac{20}{100}\times100)$= $100+20=Rs.$ $120$After allowing discount of 5%, => Selling price = $120-(\frac{5}{100}\times120)$ = $120-6=Rs.$ $114$$\therefore$ Profit % = $\frac{(114-100)}{100}\times100=14\%$=> Ans – (A)