Skip to main content

SSC Question

SSC Question
D1: Question 1. A & B together finish a job in 24 days, while A, B & C together can finish the same job in 8 days. C alone will finish the job in
a. 12 days
b. 14 days
c. 16 days
d. 24 days

Ans. 12 days
Explanation: The efficiency of (A + B) = 100/24 = (25/6)%;
The efficiency of (A + B + C) = 100/8 = (25/2)%;
The efficiency of C = 25/2 – 25/6 = (50/6)%;
Hence, C can alone finish this job in = 100/(50/6) = 12 days;
D2: Question 2.Area of the circle inscribed in a square of diagonal 6√2 cm (in sq. cms) is 
a. 9 Π
b. 6 Π
c. 3 Π
d. 9√2Π
Ans. 9 Π
Explanation: Diagonal of square = side* √2; => Side = 6 cms;
Side = Diameter = 6 cms; => radius = 3 cms;
Hence, the area of circle = 9π sq. cms
D1: Question 3.The original price of a TV set is Rs. 6,000. If the price is discounted by 20% and then raised by 10% for service contract, the price charged by the shopkeeper is
a. Rs. 5400
b. Rs. 5280
c. Rs. 5100
d. Rs.4200
Ans. Rs. 5280
Explanation: Original price = Rs. 6000;
Price after discount = 6000 – 1200 = Rs. 4800;
Price after raising service contract = 4800 + 480 = Rs. 5280
D1: Question 4.A certain sum of money was divided between A, B and C in the ratio 5:6:9. If A received Rs. 450 the sum divided was
a. 2000
b. 1800
c. 2250
d. 1000
Ans. 1800
Explanation: Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x.
5x = 450; => x = 90;
Hence, the total money = 20x = 20*90 = Rs. 1800;
D1: Question 5.By selling a bag at Rs. 230, profit of 15% is made. The selling price of the bag, when it is sold at 20% profit would be
a. Rs.250
b. Rs.205
c. Rs.240
d. Rs.200
Ans. Rs.240
Explanation: Suppose the Cost price of the bag= Rs. x;
Hence, x + 0.15x = 230; => x = Rs. 200;
Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240
D1: Question 6.The weights of two iron balls are 3.5 kg and 7.5 kg. What is the percentage weight of the 1st ball with respect to 2nd ball.
Explanation: The required percentage = 3.5*100/7.5 = (140/3)%.
D1: Question 7.A Bus travels at the speed of 36 km/h, then the distance covered by it in one second is
a. 10 m
b. 15 m
c. 12.5 m
d. 13.5 m
Ans. 10 m
Explanation: Distance traveled in one second = (36* 1000 meter)/(60 * 60 seconds); => 10 m
Question 8.
a. (a+b)/(a-b)
b. -1
c. 2ab
d. 1
Ans. 1
Explanation:
= a/(a-b) – b/(a-b);
=(a-b)/(a-b) =1;
D1: Question 9.The value of (1 - √2 + (√2 - √3 + (√3 - √4 + ..... + (√15 - √16) is 
a. 0
b. 1
c. -3
d. 4
Ans. -3
Explanation:
= (1 - √2 + (√2 - √3 + (√3 - √4 + ..... + (√15 - √16);
= 1-√16 = 1 – 4= -3;
D2: Question 10.∆ABC and ∆DEF are two similar triangles and the perimeter of ∆ABC and ∆DEF are 30 cm and 18 cm respectively. If length of DE = 36 cm, then length of AB is 
a. 60 cm
b. 40 cm
c. 45 cm
d. 50 cm
Ans. 60 cm
Explanation: As per the property of similar triangle,
perimeter of ∆ABC/ perimeter of ∆DEF = AB/DE;]
30/18 = AB/36; => AB = 60 cms
D4: Question 11.If the length of a chord of a circle is equal to that of the radius of the circle, then the angle subtended, in radians, at the center of the circle by chord is
a. 1
b. ∏/2
c. ∏/3
d. ∏/4
Ans. ∏/3
Explanation: From the given figure below, we can see that the triangle OAB is a equilateral triangle. Hence, each angle of this triangle will of 60 degrees, which in radian will be equal to ∏/3.
D2: Question 12.The value of (sec245 - cot245) - (sin230 + sin260) is 
a. 1
b. 2√3
c. 0
d. 1/√2
Ans. 0
Explanation: Put the numeric values of these trigonometric ratios-
= (2)2 – 1 –[(1/2)2 + (√3/2)2];
After simplifying the above expression,
= 0;
D1: Question 13.The average salary of male employees in a firm was Rs. 5200 and that of females was Rs. 4200. The mean salary of all the employees was Rs. 5000. What is the % of female employees?
a. 80%
b. 20%
c. 40%
d. 30%
Ans. 20%
Explanation: Suppose the number of male employees = x; and the number of female employees = y;
5200*x + 4200*y = 5000*(x + y);
200x = 800y; =>x = 4y;
Hence, % of female employees = y*100/(x+y) = 100/5 = 20%.
D1: Question 15.The cube of 105 is
a. 1157625
b. 1175625
c. 1185625
d. 1158625
Ans. 1157625
Explanation: (105)3= (100 + 5)3;
= (100)3+ (5)3 + 3*100*5(100 + 5);
= 1000000 + 125 + 150000 + 7500;
= 1157625;
D3: Question 16.In ∆ABC, B is right angle, D is the midpoint of the side AC. If AB = 6 cm, BC = 8cm, then the length of BD is 
a. 4 cm
b. 5 cm
c. 8 cm
d. 12 cm
Ans. 5 cm
Explanation: In right-angled triangle ABC, AC = AB2 + BC2√62 + 82= 10 cms.
Since, AD = BD = CD = AC/2 = 5 cms

D2: Question 17.The diagonals of two squares are in the ratio 5:2.The ratio of their area is
a. 5:6
b. 25:4
c. 5:4
d. 125:8
Ans. 25:4
Explanation: Diagonal of square = side*2;
Side of first square a1= d1/2;
Side of second square a2= d2/2;
Ratios of the areas=(a1)2/(a2)2 =  (d1/2)2 / (d2/2)2 = (d1)2/(d2)2 = 25: 4;
D2: Question 18.The angle of elevation of a ladder leaning against a wall is 60 degrees and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is 
a. 2.3 m
b. 4.6 m
c. 9.2 m
d. 7.8 m
Ans. 9.2 m
Explanation: From the figure given below,
cos60 =4.6/h; => h = 4.6 * 2 = 9.2 m

D1: Question 19.The product of two 2-digit numbers is 2160 and their H.C.F. is 12. The numbers are
a. (12, 60)
b. (72, 30)
c. (36, 60)
d. (60, 72)
Ans. (36, 60)
Explanation: only option (c.) follows the given condition because there product is equal to 2160 and HCF is 12.
D2: Question 20.The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs. 1. The sum (in Rs.) is:
a. 620
b. 630
c. 640
d. 625
Ans. 625
Explanation: Suppose the principal amount = Rs. P;
Simple Interest = P*4*2/100 = 2P/25;
Compound Interest = P(1 + 4/100)2 -P = 51P/625;
(51P/625) – (2P/25) =1;
(51P – 50P)/625 =1;
P = Rs. 625;
D2: Question 21.In a mixture of 25 liters, the ratio of milk to water is 4:1. Another 3 liters of water is added to the mixture. The ratio of milk to water in the new mixture is
a. 5:1
b. 5:2
c. 5:3
d. 5:4
Ans. 5:2
Explanation: The amount of milk in the mixture = (4/5) * 25 = 20 liters
The amount of water in the mixture = (1/5) * 25 = 5 liters
When added 3 liters of water, then total quantity of water = 8 liters;
The required ratio = 20: 8 = 5: 2.
D3: Question 22.A constituency is divided in four regions A, B, C and D. Two candidates X & Y contested the last election from that constituency. The adjoining graph gives the break-up of voting in the four regions. Study the graph and answer the following question. 

Approximately how much percent of voters voted in favor of X?
a. 45.4
b. 47.5
c. 50
d. 225
Ans. 47.5
Explanation: Total number of voters, who have voted in favor of X = 45 + 72 + 51 + 56 = 225;
Total Voters = (45 + 40 +1) + (73 + 88 + 9) + (51 + 47 + 5) + (56 + 51 + 8)=474;
The required percentage = 225*100/474 = 47.46%
D3: Question 23.A constituency is divided in four regions A, B, C and D. Two candidates X & Y contested the last election from that constituency. The adjoining graph gives the break-up of voting in the four regions. Study the graph and answer the following question. 
Approximately how much percent of voters did not caste their votes?
a. 4.9
b. 4.5
c. 0.23
d. 23
Ans. 4.9
Explanation: Total number of voters, who have not voted = 1 + 9 + 5 + 8 = 23;
Total Voters = (45 + 40 +1) + (73 + 88 + 9) + (51 + 47 + 5) + (56 + 51 + 8)=474;
The required percentage = 23*100/474 = 4.85%
D3: Question 24.A constituency is divided in four regions A, B, C and D. Two candidates X & Y contested the last election from that constituency. The adjoining graph gives the break-up of voting in the four regions. Study the graph and answer the following question. 

In region B, Y gets A% more votes than X. Find the value of A?
a. 24%
b. 21%
c. 19%
d. 15%
Ans. 21%
Explanation: % more voters for Y as compared to X= (88-73)*100/73 = 1500/73 =20.54%;
Hence, the required value of A = 21%.
D3: Question 25.A constituency is divided in four regions A, B, C and D. Two candidates X & Y contested the last election from that constituency. The adjoining graph gives the break-up of voting in the four regions. Study the graph and answer the following question. 
Nearly what percentage of his total votes did X receive from region B?
a. 30
b. 31
c. 32
d. 35
Ans. 32
Explanation: The required percentage = 73*100/(73 + 45 + 51 + 56) = 7300/225 = 32.44%

Comments

Popular posts from this blog

SSC Practice Problem Solution

SSC Practice Problem Solution In this post, we will discuss solution of  SSC Practice Problem . Question 1  The efficiency of (A + B) = 100/24 = (25/6)% The efficiency of (A + B + C) = 100/8 = (25/2)% The efficiency of C = 25/2 – 25/6 = (50/6)% Hence, C can alone finish this job in = 100/(50/6) = 12 days Answer: 12 days option (1) Question 2  Original price = Rs. 6000; Price after discount = 6000 – 1200 = Rs. 4800; Price after raising service contract = 4800 + 480 = Rs. 5280 Answer: Rs. 5280 option (2) Question 3  Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x. 5x = 450; => x = 90; Hence, the total money = 20x = 20*90 = Rs. 1800; Answer: Rs. 1800 option (2) Question 4 Suppose the Cost price of the bag= Rs. x; Hence, x + 0.15x = 230; => x = Rs. 200; Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240 Answer: Rs. 240 option (3) Question 5  The required percentage ...

Average

Average In this post, We will learn about average.  Average of any collection of data is the sum of data divide by number of data. Average = (Sum of observations /  Number of observations) Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.  problems The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is (CAT 2018) As specified in the question mean of x and y and z is 80 x+y+z=3*80 mean of x, y, z,u,v is 75 x+y+z+u+v=75*5 2u=x+y 2v=y+z x+y+z+((x+y+y+z)/2)=75*5 y+3(x+y+z)=75*10 y=750-720 y=30 since x+y+z=240 y=30 so x+z=210 since x>=z so minimum value of x=210/2 = 105 Problem2: In first 5 inings sehwag has scored at average of 40. In sixth ining He had scored 52 what is His average after 6th ining. one simple approach is getting total score divided by no of inin...

Complex Number

Complex Number While solving the equation of the form  a x 2 + b x + c = 0 , the roots of the equations can take three forms which are as follows: Two Distinct Real Roots Similar Root No Real roots (Complex Roots) The introduction of complex numbers in the 16th century made it possible to solve the equation  x 2 + 1 = 0 . The roots of the equation are of the form  x = ± − 1 ‾ ‾ ‾ √  and no real roots exist. Thus, with the introduction of complex numbers, we have Imaginary roots. We denote  − 1 ‾ ‾ ‾ √  with the symbol i, where i denotes Iota (Imaginary number). An equation of the form z= a+ib, where a and b are real numbers, is defined to be a complex number. The real part is denoted by Re z = a and the imaginary part is denoted by Im z = b. Algebraic Operation on Complex numbers: Addition of two complex numbers Subtraction of two complex number Multiplication of two complex number Division of two complex number Power of Iota (i) ...

Logarithm

Logarithm When a x   = N , then we say that x = logarithm of N to the base a and write it as x = log a N . In simple words, it represents the power to which a number must be raised. Let me expand on that by giving a simpler example. If we are asked, what would be the result if ‘a’ is multiplied with itself ‘b’ times; then your answer would be x = a*a*a*…. b (times). This can also be written as a^b. This is also known as ‘a raised to the power of b’ If we are asked, which number multiplied with itself ‘b’ times, will result in a; then you are asked for the value of x such that x*x*x*… b(times) = a => x^b = a => x = a^(1/b) This is also known as ‘bth root of a’ If swe are asked, how many times should you multiply ‘a’ with itself to get ‘b’, that is where the concept of logarithm comes into the picture. You are asked for the value of ‘x’ such that a*a*a…. x (times) = b => x = Log  a  b This is also known as ‘ Log b to the base a’ ...