In this article, we are providing 25 solved questions of Aptitude section asked in SSC CGL tier-1 exam held on 7^{th}September, 2016 (Morning shift). Please find the topic wise questions’ distribution in the following table-
Sub-topics |
No. of questions |
Number System |
1 |
Algebra |
5 |
Percentages |
1 |
Averages |
1 |
Simple and Compound Interest |
1 |
Profit, Loss, and Discount |
2 |
Time and Distance |
1 |
Time and Work |
1 |
Geometry |
4 |
Mensuration |
1 |
Trigonometry |
3 |
Data Interpretation |
4 |
From the above table, we can find that the major questions were asked from Algebra, Geometry, Data Interpretation, Profit/loss, and Trigonometry. The level of these questions was very difficult and time-consuming. Hence, we recommend you to give more time on these topics to score high in the upcoming SSC CGL exam. Let us go through these questions-
Question 1.Find the least number, which must be subtracted from 18265 to make it a perfect square
a. 30
b. 38
c. 40
d. 45
Ans. 40
Explanation: Let us find the square root of the given number-
Hence, option(c.) is the correct answer.
Question 2.A and B together can do a piece of work in 36 days, B and C together can do it in 24 days. A and C together can do it in 18 days. The three working together can finish the work in
a. 8 days
b. 16 days
c. 30 days
d. 32 days
Ans. 16 days
Explanation: The efficiency of (A + B) = (100/36)%;
The efficiency of (B + C)= (100/24)%;
The efficiency of (A + C)=(100/18)%;
Adding above equations-
The efficiency of 2*(A + B + C)= (100/36) + (100/24) + (100/18)= (25/2)%;
The efficiency of (A + B + C) = (25/4)%.
The number of days to complete the task by all of threes = 100/(25/4) = 16 days;
Question 3.Two adjacent sides of a parallelogram are 21 cms and 20 cms. The diagonal joining the endpoints of these two sides is 29 cms. The area of the parallelogram (in sq. cms) is
a. 240
b. 120
c. 210
d. 420
Ans. 420
Explanation: Area of Rectangle ABCD = 2 * ( Area of triangle ABC);
Hence, The area of triangle ABC = √s(s-a)(s-b)(s-c); where s = (a + b + c)/2;
s = (20 + 21 + 29)/2 = 35;
The area of triangle ABC = √35(35-20)(35-21)(35-29) = √35*15*14* 6 = 7*5*3*2 = 210 sq. cms.
Hence, the area of parallelogram= 210*2 = 420 sq. cms.
SSC CGL 2018 Quantitative Aptitude Preparation Strategy: Detailed Chapterwise and Yearwise Analysis
Question 4.A dealer marks his goods 20% above cost price. He then allows some discount on it and makes a profit of 8%. The rate of discount is
a. 4%
b. 6%
c. 10%
d. 12%
Ans. 10%
Explanation: Marked Price =1.2* Cost price; ----- eq.(i.)
Let the discount percentage = x%;
(100 – x)% of Marked price = 1.08 * Cost price; -------eq.(ii.)
Divide eq.(ii.) by eq.(i.)-
(100-x)/100 = 1.08/1.2; => 100 –x = 0.9*100 => x= 10%.
Question 5.Sum of two numbers is thrice their difference. Their ratio is
a. 1:2
b. 2:1
c. 3:1
d. 1:3
Ans. 2:1
Explanation: Let these numbers are x and y;
(x + y) = 3*(x – y); => 2x = 4y;
Hence, x: y = 2: 1;
Question 6.The average age of 36 students in a group is 14 years. When the teacher's age is included in it, the average increases by one. The teacher's age in years is
a. 31
b. 51
c. 36
d. 50
Ans. 51
Explanation: Suppose S_{1}, S_{2}, S_{3,………., }S_{36 }are the students in the group-
(S_{1} + S_{2} + S_{3………., }+S_{36})/36 = 14; => (S_{1} + S_{2} + S_{3………., }+S_{36}) = 504; ------eq.(i)
[(S_{1} + S_{2} + S_{3………., }+S_{36}) + T]/37 = 15;
Put the value of equation(i) in the above equation-
504 + T = 37*15;
T = 555 – 504 = 51 years;
Question 7.A dishonest dealer professes to sell his goods at cost price but uses a weight of 875 grams for the kilogram weight. His gain in percentage is
a. 17%
b. (103/7) %
c. (100/7) %
d. 14%
Ans. (100/7)%
Explanation: Suppose the selling price of 1 kg goods = Rs. x;
The selling price of 875 grams goods = Rs. (875x/100);
% profit in selling good = [x – (875x/1000)]/[875x/100] = (100/7)%;
Question 8.A's salary is 50% more than that of B. Then, B's salary is less than that of A by
a. 50%
b. (100/3)%
c. (133/4)%
d. (89/2)%
Ans. (100/3)%
Explanation: Salary of A = 1.5 * B’s salary;
%change in B’s salary= 0.5* B’s salary*100/1.5* B’s salary = (100/3)%
Question 9.Speed of a boat along and against the current are 14 kms/hr and 8 kms/hr respectively. The speed of the current is
a. 11 kms/hr
b. 6 kms/hr
c. 5.5 kms/hr
d. 3 kms/hr
Ans. 3 kms/hr
Explanation: The speed of boat = ½ [downstream speed + Upstream Speed] = ½ [14 + 8] = 11 kmph;
Hence, the speed of current = 14 – 11 = 11-8 = 3 kmph;
Question 10.If the simple interest on Rs. 1 for 1 month is 1 paisa, then the rate percent per annum will be
a. 10%
b. 8%
c. 12%
d. 6%
Ans. 12%
Explanation: Use I =PRT/100; let the annum interest rate = r%;
1 = [100 * r * 1/12 ]/100 ; => r = 12%;
SSC CGL 2018 Tier-I Exam: Preparation Tips and Strategy
Question 11.If a/b + b/a = 1, then the value of a^{3} + b^{3} will be
a. 1
b. 0
c. -1
d. 2
Ans. 0
Explanation:
Question 12.If p = 99, then the value of p(p^{2} +3p+3) will be
a. 999999
b. 1000000
c. 1000001
d. 999998
Ans. 999999
Explanation: p(p^{2} +3p+3)= p(p^{2} +2p+p+1 + 2) = p[(p+1)^{2}+(p+2)];
= 99*[(99+1)^{2} + 99+2] = 99*[10000+101]= 990000+9999 = 999999;
Question 13.If a-b=1 and a^{3}-b^{3 }= 61, then the value of ab will be
a. -20
b. 20
c. 30
d. 60
Ans. 20
Explanation: (a-b)^{2} = a^{2}+b^{2}-2ab =1; ----eq.(i)
(a^{3}-b^{3})=(a-b)( a^{2}+b^{2}+ab) = a^{2}+b^{2}+ab =61;-------eq.(ii)
Subtract eq.(i) from eq.(ii)-
3ab = 60; =>ab=20;
Question 14.
a. 2a-b
b. a+b
c. a-b
d. 2a+b
Ans. a-b
Explanation: It can be simply observed with substituting the value x= a-b;
Question 15.The point where the 3 medians of a triangle meet is called
a. centroid
b. Incentre
c. Circumcentre
d. orthocentre
Ans. centroid
Explanation: The centroid of a triangle is the intersection of the three medians of the triangle (each median connecting a vertex with the midpoint of the opposite side). It lies on the triangle's Euler line, which also goes through various other key points including the orthocenter and the circumcenter.
Question 16. ΔABC a right angled triangle has ∠B = 90° and AC is hypotenuse. D is its circumcentre and AB = 3 cms, BC = 4 cms. The value of BD is
a. 3 cms
b. 4 cms
c. 2.5 cms
d. 5.5 cms
Ans. 2.5 cms
Explanation: In case of circumcircle, AD=CD=BD;
Hence, BD = 2.5 cms;
Question 17. ΔABC is an equilateral triangle and D, E are midpoints of AB and BC respectively. Then the area of Δ ABC : the area of the trapezium ADEC is
a. 5:3
b. 4:1
c. 8:5
d. 4:3
Ans. 4:3
Explanation: By the mid-point theorem-
Area(ADF) = Are(BDF)=Area(DEF)=Area(CEF)= ¼*Area(ABC);
Hence, the required ratio = Area(ABC): ¾ * Area(ABC) = 4: 3;
Question 18.
a. 75°
b. 30°
c. 150°
d. 105°
Ans. 105°
Explanation: Angle 1 = Angle 2; (Isosceles triangle property)
Hence, Angle 1 = Angle 2 = (180 -30)/2 = 75;
SSC CGL Syllabus 2018: Tier I, II, III and IV with Exam Pattern
Question 20.If x=aCosθCosΦ, y = aCosθSinΦ and z= aSinθ, then the value of x^{2} + y^{2} + z^{2} is
a. 2a^{2}
b. 4a^{2}
c. 9a^{2}
d. a^{2}
Ans. a^{2}
Explanation:
x^{2} + y^{2} + z^{2} = (acosθcosΦ)^{2} +( aCosθSinΦ)^{2}+(asinθ)^{2};
= (acosθ)^{2}[cos^{2}Φ+sin^{2}Φ]+ (asinθ)^{2};
=a^{2}[cos^{2}θ+sin^{2}θ]=a^{2}
Question 21.A 1.6 m tall observer is 45 meters away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in meters is
(Take √3 = 1.732).
a. 25.98
b. 26.58
c. 27.58
d. 27.98
Ans. 27.58
Explanation: In triangle APC,
tan30=h/45; => h = 45/√3= 25.98 meters;
hence, the height of tower = 25.98 + 1.6 = 27.58 meters;
Question 22.The following pie chart shows proportion of population of seven villages in 2009. Study the pie chart and answer the questions that follows :
If the below poverty line population of the village 'X' is 12160, then the population of village 'S' is
a. 18500
b. 20500
c. 22000
d. 20000
Ans. 22000
Explanation: 38% of population of village ‘X’= 12160;
Population of village ‘X’ = 12160*100/38= 32000;
Since, X: S = 16: 11;
The population of S= 11/16 * 32000 = 22000;
Question 23.The following pie chart shows proportion of population of seven villages in 2009. Study the pie chart and answer the questions that follows :
The ratio of the below poverty line population of village 'T' to that of the below poverty line population of village 'Z' is
a. 11:23
b. 13:11
c. 23:11
d. 11:13
Ans. 23:11
Explanation: The population of village T below poverty line= 46% of (21% of total population);
The population of village Z below poverty line= 42% of (11% of total population);
The required ratio= (46*21)/(42*11) = 23: 11;
Question 24.The following pie chart shows proportion of population of seven villages in 2009. Study the pie chart and answer the questions that follows :
If the population of the village 'R' is 32000, then the below poverty line population of village 'Y' is
a. 14100
b. 15600
c. 16500
d. 17000
Ans. 15600
Explanation: The population of village ‘R’ = 32000;
Let the population of village ‘Y’ = y;
16: 15= 32000: y; => y = 32000*15/16 = 30000;
Population of village ‘Y’ below poverty line = 30000*52/100 = 15600;
Question 25.The following pie chart shows proportion of population of seven villages in 2009. Study the pie chart and answer the questions that follows :
In 2010, the population of 'Y' and 'V' increases by 10% each and the percentage of population below poverty line remains unchanged for all the villages. If in 2009, the population of village Y was 30,000, then the below poverty line population of village 'V' in 2010 is _____.
a. 11250
b. 12760
c. 13140
d. 13780
Ans. 12760
Explanation:
Population of village Y = 30000.
Let the population of village V in 1997 be x.
Then, 15 : 10 = 30000 : x =>x = (30000x10)/15= 20000.
Now, population of village V in 2010 = 20000 + (10% of 20000) = 22000.
Therefore, Population of village V below poverty line in 2010 = 58% of 22000=12760.
Comments