In this article, you will find 25 solved questions of Quantitative Aptitude asked in SSC CGL tier-1 exam held on 28^{th}August, 2016 (Morning shift). Please find the questions distribution in this examination in the table given below-
Sub-topics |
No. of questions |
Algebra |
4 |
Percentages |
2 |
Averages |
1 |
Simple and Compound Interest |
1 |
Profit, Loss, and Discount |
2 |
Time and Distance |
1 |
Time and Work |
1 |
Partnership |
1 |
Geometry |
4 |
Mensuration |
1 |
Trigonometry |
3 |
Data Interpretation |
4 |
From the above table, we can conclude that SSC has inquired more questions from Algebra, Geometry, Data Interpretation, and Trigonometry. The levels of questions were very typical and time-consuming. Hence, we recommend you to spend more time on these topics to excel in upcoming SSC CGL exams. Let us go through the questions-
1.A father can do job as fast as 2 sons working together. If one son does the job in 3 hours and the other in 6 hours, the number of hours taken by the father, to do the job alone is
a. 1
b. 2
c. 3
d. 4
Ans. 2
Explanation:- The efficiency of first son=1*100/3=33.33%.
The efficiency of second son= 100/6= 16.67%.
Hence, the efficiency of father= 33.33 + 16.67= 50%.
And, the number of hours taken by father to complete the same job= 100/50=2 hours.
2.The perimeter of a rhombus is 240 m and the distance between any two parallel sides is 20 m. The area of the rhombus in sq.m. is
a. 600
b. 1200
c. 2400
d. 4800
Ans. 1200
Explanation:- the side of rhombus= 240/4= 60 m.
Altitude= distance between any parallel sides= 20 m.
Hence, the area of rhombus= 60*20=1200 sq. m.
3.A man sold an article for Rs. 450, after allowing a discount of 16 2/3 % on the printed price. What is that printed price?
a. Rs.525
b. Rs.530
c. Rs.535
d. Rs.540
Ans. Rs.540
Explanation:-Let the printed price= Rs. x;
If the discount is (50/3)%, then the sold percentage will be (250/3)%.
x*[250/(3*100)]=450;
x= 540;
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4. A sum of Rs. 770 has been divided among A, B, C in such a way that A receives 2/9th of what B and C together receive. Then A's share is
a. Rs.140
b. Rs.154
c. Rs.165
d. Rs.170
Ans. Rs.140
Explanation:- A=(2/9)(B+C);
Hence, B + C=9A/2;
Since, A + B + C= 770;
11A/2=770; = > A= Rs. 140.
5. A man bought 4 dozen eggs at Rs. 24 per dozen and 2 dozen eggs at Rs. 32 per dozen. To gain 20% on the whole, he should sell the eggs at
a. 16 per dozen
b. 21 per dozen
c. 32 per dozen
d. 35 per dozen
Ans. 32 per dozen
Explanation:- Total price paid for the eggs= 4*24 + 2* 32= Rs. 160.
The total price after 20% gain= 160*1.2= Rs. 192;
Hence, the selling price per dozen= 192/6= Rs. 32;
6. P's salary is 25% higher than Q, what percentage is Q's salary lower than that of P?
a. 20
b. 29
c. 31
d. 331/3
Ans. 20
Explanation:- Let the salary of Q= Rs. x; then, the salary of P= Rs. 1.25x.
% change in salary= (0.25x/1.25x)*100= 20%.
7.A and B start running at the same time and from the same point around a circle. If A can complete one round in 40 seconds and B in50 seconds, how many seconds will they take to reach the starting point simultaneously?
a. 10
b. 200
c. 90
d. 2000
Ans. 200
Explanation:- time taken to reach the starting point simultaneously= LCM(40, 50)= 200 seconds.
8.
a. -1
b. 1
c. -2
d. 2
Ans. 1
Explanation:- The answer is obvious= 3*1/3=1.
9.
a. 81
b. 27
c. 120
d. 119
Ans. 119
Explanation:- p-1/p=3;
Squaring both sides-
P^{2}+1/p^{2}-2=9; => p^{2}+1/p^{2}=11;
Square again of both sides-
P^{4}+1/p^{4}+2=121; => p^{4}+1/p^{4}= 119;
10.∆ABC is an isosceles triangle with AB = AC = 15 cm and altitude from A on BC is 12 cm. Length of side BC is
a. 9 cm
b. 12 cm
c. 18 cm
d. 20 cm
Ans. 18 cm
Explanation:- Since, BL=CL; Hence, using Pythagoras theorem;
15^{2}=12^{2} + BL^{2}; BL=9 cm. Hence, the length of base= 9*2= 18 cm.
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11.The mid points of AB and AC of a triangle ABC are respectively X & Y. If BC + XY = 12 units, then the value of BC - XY is:
a. 2 units
b. 6 units
c. 8 units
d. 4 units
Ans. 4 units
Explanation:- If the mid points of two sides of a triangle is merged, then BC||XY and XY= ½ BC.
By putting this value in the given equation, BC + ½ BC=12; => BC= 8 units and XY = 4 units.
Hence, BC –XY = 8-4= 4 units.
12. Which of the following relations is correct for 0< θ<90?
a. sinθ = sin2θ
b. sinθ < sin2θ
c. sinθ > sin2θ
d. sinθ = cosecθ
Ans. sinθ > sin2θ
Explanation:- substitute θ=30;
Then, sin30> sin60; hence, option(c.) holds true.
13.The difference between two numbers is 9 and the difference between their squares is 207. The numbers are
a. 17, 8
b. 16, 7
c. 15, 6
d. 23, 14
Ans. 16, 7
Explanation:- Let number be x and y. then –
x - y = 9; ---------(i.)
x^{2} – y^{2} = 207; ----------(ii.)
x + y= 23; ---------(iii.)
solving eq.(i) and (iii), we get-
x= 16 and y= 7;
14.The average of 20 numbers is calculated as 35. It is discovered later on that while calculating the average, one number, namely 85, was read as 45. The correct average is
a. 36
b. 36.5
c. 37
d. 37.5
Ans. 37
Explanation:- the sum of estimated numbers= 35*20= 700;
The correct sum of the numbers will be= 700-45+85= 740;
Hence, the correct average= 740/20= 37;
15.If x^{2} - xy + y^{2} =2 and x^{4} + x^{2}y^{2} + y^{4} = 6, then the value of x^{2} + xy + y^{2} is:
a. 1
b. 12
c. 3
d. 36
Ans. 3
Explanation:- x^{2} - xy + y^{2} =2 ------------(i.)
x^{4} + x^{2}y^{2} + y^{4} = 6 ------------------(ii.)
Divide eq.(ii) by eq.(i)-
16.If a^{2} + 13b^{2} + c^{2} - 4ab - 6bc = 0, then a: b: c is
a. 1:2:3
b. 2:3:1
c. 2:1:3
d. 1:3:2
Ans. 2: 1: 3
Explanation:-
17.The circumcenter of a triangle ABC is O. If ∠BAC = 85^{}, ∠BCA = 75^{}, then ∠OAC is of
a. 70°
b. 72°
c. 75°
d. 74°
Ans. 70°
Explanation:- Since, OA=OB=OC; hence, angle OAC= angle OCA;
Hence, angle OAC= (180-40)/2=70 degree.
18.Radius of the incircle of an equilateral ΔABC of sides 2√3 units is x cm. The value of x is
a. ⅓
b. ½
c. 1
d. √3
Ans. 1
Explanation:- The altitude of the triangle= √3/2*side= 3 cm.
If AD= 3 cm; OA=(3-x) cm.
Then, (3-x)^{2}= (√3)^{2}+ x^{2};
9 + x^{2} – 6x = 3 + x^{2}; x= 1 cm.
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19. If tan3θ. tan7θ = 1, then the value of tan(θ + 36°) is
a. 1/√3
b. 0
c. 1
d. √3
Ans. 1
Explanation:- 1- tan3θ. tan7θ=0;
(tan3θ + tan7θ)/( 1- tan3θ. tan7θ)= tan90;
tan(3θ + 7θ)=tan(90);
10θ=90; => θ=9;
Hence, tan(9+36)=tan45=1;
20.If the angle of elevation of a cloud from a point 200 m above a lake is 30 and the angle of depression of its reflection in the lake is 60. Then the height of the cloud above the lake is
a. 100 m
b. 200 m
c. 300 m
d. 400 m
Ans. 400 m
Explanation:- In triangle ACD, tan60=CD/AC; = > CD = 300*tan60;
Since, CD=AB;
Hence, tan30=MB/AB; = > MB = AB*tan30;
MB=200*tan60*tan30=200m.
The required height of cloud= MB+BD = 200+200=400 m.
21.The difference between the simple interest and compound interest (compounded annually) on Rs. 40,000 for 3 years at 8% per annum is
a. Rs.684.32
b. Rs.788.48
c. Rs.784.58
d. Rs.4000
Ans. Rs.788.48
Explanation:- SI= PRT/100; => SI = 40000*8*3/100 = Rs. 9600;
CI = A[(1+r/100)^{n}-1];
CI= 40000[(1+8/100)^{3}-1]= Rs. 10388.48;
Hence, the difference between the interests= 10388.48-9600=Rs. 788.48;
The pie-chart shows Distribution of Special Children Population during the year 1994-96. Study the pie-chart and answer the following questions.
22. Find the approximate percentage distribution of children with auditory disorder.
a. 43.7%
b. 42.7%
c. 41.7%
d. 40.7%
Ans. 41.7%
Explanation:- Percentage of children with auditory disorder= 871*100/(871+222+275+657+60)=87100/2085 = 41.77%;
23. What is the average number of cases in different types of special children during the year 1994-96.
a. 417
b. 413
c. 433
d. 465
Ans. 417
Explanation:-Average types of special children= total number of children/no. of categories;
= (871+222+275+657+60)/5=2085/5=417.
24. Find the ratio between articulatory disorder and speech disorder cases.
a. 21:55
b. 55:21
c. 55:12
d. 12:55
Ans. 12:55
Explanation:- Ratio = 60/275 = 12: 55;
25. What is the ratio between language disorder and the average of the remaining disorder cases.
a. 219:119
b. 119:219
c. 919:419
d. 729:529
Ans. 219:119
Explanation:- Required Ratio= 657/(2085-657)=657/1428= 219: 119;
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