SSC CGL solved question paper Tier-I exam held on 1 September, 2016: Quantitative Aptitude
Find all solved questions of Quantitative Aptitude asked in SSC CGL tier-1 exam held on 1^{st}September, 2016 (Morning shift) with prescriptive analysis. Read full paper here-
In this article, you will get 25 solved questions of Quantitative Aptitude asked in SSC CGL tier-1 exam held on 1^{st}September, 2016 (Morning shift). Please look at topic wise questions distribution for the respective examination in the table given below-
Sub-topics |
No. of questions |
Number system |
1 |
Algebra |
4 |
Percentages |
1 |
Averages |
1 |
Simple and Compound Interest |
1 |
Ratio and Proportion |
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 described table, we can see that more questions were asked from Algebra, Geometry, Data Interpretation, Profit & Loss, and Trigonometry. The level of these questions was difficult and time-consuming. Hence, we recommend you to allocate more time on such topics to score in the upcoming SSC CGL exams. Let us go through all questions-
Question 1. A and B together can finish a work in 30 days. They worked for it for 20 days and then B left the work. The remaining work was done by A alone in 20 days more. In how many days can A alone finish the work?
a. 48 days
b. 50 days
c. 54 days
d. 60 days
Ans. (d.)
Explanation: (A + B)’s combined efficiency = 100/30 = (10/3)%.
The work done by both in 20 days= 20*(10/3) = (200/3)%.
Remaining work = 100 – (200/3) = (100/3)%.
If A does (100/3)% of work in 20 days, then A’s one day work = (100/60)%.
Hence, A will alone finish the work in = 100/(100/60) = 60 days.
Question 2. The centroid of an equilateral triangle ABC is G. If AB is 6 cms, the length of AG is
a. √3 cm
b. 2√3 cm
c. 3√2 cm
d. 2√2 cm
Ans. (b.)
Explanation: When the medians of a triangle converges at a point, that point is known as centroid. In equilateral triangle, every median is perpendicular to the base line and centroid divides the median into 2: 1.
Hence, in triangle AEC,
AE = √6^{2}-3^{2} = 3√3 cms.
=> AG = (2/3)(AE)= 2√3 cms.
Question 3. A merchant changed his trade discount from 25% to 15%. This would increase selling price by
a. 3⅓%
b. 6⅙%
c. 13⅓%
d. 16⅓%
Ans. (c.)
Explanation: For discount of 25% to 15%, selling price moves from 75% to 85%.
Hence, % change in SP= (85-75)*100/75 = 13⅓%.
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Question 4. If 177 is divided into 3 parts in the ratio 1/2 : 2/3: 4/5, then the second part is
a. 75
b. 45
c. 72
d. 60
Ans. (d.)
Explanation: ½ : 2/3 : 4/5 == 15 : 20 : 24.
15x + 20x + 24x = 177; => 59x = 177;
x = 3;
Hence, the second part = 3*20 = 60.
Question 5. If percentage of profit made,when an article is sold for Rs.78, is twice as when it is sold for Rs.69, the cost price of the article is
a. Rs. 49
b. Rs. 51
c. Rs. 57
d. Rs. 60
Ans. (d.)
Explanation: Suppose, the cost price = Rs. x;
(78 – x)/x = 2* (69 – x) /x;
78 –x = 2*(69 – x);
78 –x = 138 – 2x;
x = 138 – 78 = 60.
Question 7.Gautam travels 160 kms at 32 kmph and returns at 40 kmph. Then average speed is
a. 72 kmph
b. 71.11 kmph
c. 36 kmph
d. 35.55 kmph
Ans. (d.)
Explanation: The total traveled distance = 320 kms.
Total time taken in up and down = (160/32) + (160/40) = 5 + 4 = 9 hrs.
Hence, the average speed = 320/9 = 35.55 Kmh.
Question 8.If x=3/2, then the value of 27x^{3}-54x^{2}+36x-11 is
a. 11⅜
b. 11⅝
c. 12⅜
d. 12⅝
Ans. (d.)
Explanation:
Question 9.If a+b+c = 6 and ab+bc+ca = 11, then the value of bc(b+c) + ca(c+a) +ab(a+b) +3abc is
a. 33
b. 66
c. 55
d. 23
Ans. (b.)
Explanation: bc(b+c) + ca(c+a) +ab(a+b) +3abc = bc(b+c) + abc + ca(c+a) + abc + ab(a+b) +abc;
= bc (a + b + c) + ca (a + b + c) + ab ( a + b + c);
= (a + b + c ) (ab + bc + ca) = 6 * 11= 66.
Question 10.If the angles of a triangle are in the ratio of 2:3:4, then the difference of the measure of greatest angle and smallest angle is
a. 20°
b. 30°
c. 40°
d. 50°
Ans. (c.)
Explanation: As per the given ratio, the angles will be 40, 60, and 80.
Hence, the difference between the smallest and largest number = 80 – 40= 40 degrees.
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Question 11.In ΔABC, ∠A = 90°, AD ┴ BC and AD = BD = 2 cm. The length of CD is
a. 3 cm
b. 3.5 cm
c. 3.2 cm
d. 2 cm
Ans. (d.)
Explanation: In a rectangular triangle, if the a perpendicular is drawn to the hypotenuse, then it will bisect it and length of all the three lines will be same. Hence, CD = 2 cms.
Question 12. If tan45 = cotθ, then the value of θ in radians in?
a. Π/4
b. Π/9
c. Π/2
d. Π/12
Ans. (a.)
Explanation: tan 45 = tan (90 – θ);
45 = 90 –θ; => θ = 45;
Θ = 45* pi/180 = pi/4;
Question 13.(2^{51} + 2^{52}+2^{53}+2^{54}+2^{55}) is divisible by
a. 23
b. 58
c. 124
d. 127
Ans. (c.)
Explanation: (2^{51} + 2^{52}+2^{53}+2^{54}+2^{55}) = 2^{51}*(1 + 2 + 4 + 8 + 16) = 31*2^{51}=124 * 2^{49;}
Hence, option (c.) is the correct one.
Question 14.The average of 12 numbers is 9. If each number is multiplied by 2 and added to 3, the average of the new set of numbers is
a. 9
b. 18
c. 21
d. 27
Ans. (b.)
Explanation: The sum of all 12 numbers = 12*9 = 108;
The required average = (108*2 +3 )/12 = 18.25;
Question 17.The perimeter of two similar triangles ABC and PQR are 36 cms and 24 cms respectively. If PQ = 10 cm then the length of AB is
a. 18 cm
b. 12 cm
c. 15 cm
d. 30 cm
Ans. (c.)
Explanation: perimeter(ABC)/perimeter(PQR) = AB/PQ;
AB = (36*10)/24 = 15 cms.
Question 18.In a triangle ABC, AB = 8 cm, AC = 10 cm and ∠B = 90°, then the area of ΔABC is
a. 49 sq.cm
b. 36 sq.cm
c. 25 sq.cm
d. 24 sq.cm
Ans. (d.)
Explanation: AB = √10^{2} – 8^{2} = 6 cms.
Areas of triangle ABC = ½ * BC*AB = ½ *6*8= 24 cms.
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Question 20.The compound interest on Rs. 64,000 for 3 years, compounded annually at 7.5% p.a. is
a. Rs. 14,400
b. Rs. 15,705
c. Rs. 15,507
d. Rs. 15,075
Ans. (c.)
Explanation: CI = A(1+r/100)^{n} –A;
= 64000(1+7.5/100)^{3} – 64000;
= 64000(1.242296-1) = 64000*0.242296=15507.
Question 21.The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60° and 30° respectively. Then height of the temple is
a. 50 m
b. 43 m
c. 40 m
d. 45 m
Ans. (d.)
Explanation: tan30 = (h-30)/BC;
tan 60= h/BC;
Divide the above statements-
tan30/tan60 = h-30/h;
solving this equation-
h= 45 meters.
Question 22.Study the pie-chart given and answer the following questions.
If the miscellaneous charges are Rs. 6000, then the advertisement charges are
a. Rs.12000
b. Rs.27000
c. Rs.90000
d. Rs.25000
Ans. (b.)
Explanation: Advertising is 4.5 times of miscellaneous.
Hence, advertisement charges = 4.5*6000 = Rs. 27,000
Question 23. The central angle of printing charge is x more than that of advertisement charge. Then the value of x is
a. 72°
b. 61.2°
c. 60°
d. 54.8°
Ans. (b.)
Explanation: 360*18% +x = 360*35%;
x= 360*17% = 61.2 degrees.
Question 24. What should be the central angle of the sector 'cost of paper'?
a. 22.5°
b. 54.8°
c. 36°
d. 16°
Ans. (c.)
Explanation: central angle of cost of paper =10% of 360 = 36 degrees.
Question 25. The ratio between royalty and binder's charges is
a. 5:6
b. 5:8
c. 6:5
d. 8:13
Ans. (a.)
Explanation: Ratio between royalty and binder’s charges = 15: 18 = 5: 6.
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