SSC CGL solved question paper Tier-I exam held on 2 September 2016: Quantitative Aptitude

In this article, you will get 25 solved questions of Quantitative Aptitude asked in SSC CGL tier-1 exam held on 2^ndSeptember, 2016 (Morning shift). Please look at topic wise questions distribution for the respective examination in the table given below-

 From the above described table, we can observe that more questions were asked from Algebra, Geometry, Data Interpretation, Profit & Loss, and Trigonometry. The level of these questions was difficult and very time-consuming. Hence, we can recommend you to allocate more time on such topics to score more in the upcoming SSC CGL exams. Let us go through all these questions-

Question 1. If 20 men working 8 hours per day can complete a piece of work in 21 days. How many hours per day must 48 men work to complete the same job in 7 days?
a. 12
b. 20
c. 10
d. 15

Ans. 10

Explanation: Suppose the hours to complete the same job by 48 men in 7 days = t1;

Hence, 20 x 8 x 21 = 48 x t1 x 7;

t1 = 10 hours.
Question 2.

a. 83°
b. 80°
c. 75°
d. 60°

Ans. 83°

Explanation: angle BAD = 180 –(33 + 50) = 97;

Since, the sum of opposite angle in cyclic quadrilateral is 180 degrees, hence-

Angle BCD = 180 -97 = 83;

Question 3. A shop-keeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of cost price to the printed price of the book is

a. 45:56
b. 50: 61
c. 99: 125
d. None of these

Ans. 45:56

Explanation: suppose, the printed price of the book= Rs. x;

Selling price of book = 0.90x;

Hence, the cost price of book =  0.90x/(1+ 0.12);

= 0.9x/1.12 = 0.8035x;

Required Ratio= 0.8035x / x =0.8035 = 45: 56.

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Question 4.The number of pupils of a class is 55. The ratio of the number of male pupils to the number of female pupils is 5: 6. The number of female pupils is

a. 11
b. 25
c. 30
d. 35

Ans. 30

Explanation: The number of Female pupils = [6/(6 +5) ]* total population of class;

= 6 * 55/11 = 30;

Question 5. 5% more is gained by selling a watch for Rs. 350 than by selling it for Rs. 340. The cost price of the watch is

a. Rs. 110
b. Rs. 140
c. Rs. 200
d. Rs. 250

Ans. Rs. 200

Explanation: 5% more gain in selling the watch = 350 – 340 = Rs. 10;

Hence, the 100% gain on the watch = 10 * 20 = Rs. 200;

100% gain on any item = Cost price of the item= Rs. 200;

Question 6.If 60% of the students in a school are boys and number of girls is 812, how many boys are there in the school?

a. 1128
b. 1218
c. 1821
d. 1281

Ans. 1218

Explanation: No. of girls in school = 0.40 * total population of school;

Hence, Total population of school = 812/0.40 = 2030;

Therefore, the no. of boys in the school = 0.60 * 2030= 1218;

Question 7.It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is:

a. 3:5
b. 3:4
c. 4: 3
d. 4: 5

Ans. 3:4

Explanation: Let the speed of train = x kmph and the speed of car = y kmph;

Now following the question, we get the following two equations-

120/x  +  480/y = 8;  => 15/x + 60/y = 1------------(i.)

200/x + 400/y = 25/3;  => 8/x + 16/y = 1/3 ----------(ii.)

By solving both the equation , we will get-

x = 60 kmph and y = 80 kmph;

hence, the required ratio = 3: 4;

Question 9.If m + n = 1, then the value of m³ + n³ + 3mn is equal to

a. 0
b. 1
c. 2
d. 3

Ans. 1

Explanation:

= m³ + n³ + 3mn;

= (m + n) (m² + n² –mn) + 3mn;

=1*(m² + n² –mn) + 3mn;

=(m² + n² –mn) + 3mn;

= m² + n² + 2mn = (m +n)² =1.

Question 10.The maximum number of common tangents that can be drawn to two disjoint circles is
a. 1
b. 2
c. 4
d. Infinitely many

Ans. 4

Explanation:

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Question 11.In figure, DE || BC. If DE = 3 cm, BC = 6 cm and area of ΔADE = 15 sq cm, then the area of ΔABC is

a. 75 sq cm
b. 45 sq cm
c. 30 sq cm
d. 60 sq cm

Ans. 60 sq cm.

Explanation:

If DE|| BC, and D and E bisects the sides AB and AC. (DE = ½ BC).

Hence, in the given figure, area of ABC, DEF, BDF and FEC will be equal.

Therefore, The Area of ABC = 4*15 = 60 sq. cm.

Question 12.If sin4θ - cos4θ = 1/5, then the value of tan2θ is
a. 1/2
b. 1/3
c. 1/4
d. 1/5

Ans. ½

Explanation:

Question 13.If a perfect square, not divisible by 6, be divided by 6, the remainder will be

a. 1, 3 or 5
b. 1, 2 or 5
c. 1, 3 or 4

d. 1, 2 or 4

Ans. 1, 3 or 4

Explanation: Whenever a perfect square is divided by 6, the possible remainders will be 0, 1, 3 and 4.

Question 14.A batsman in his 12th innings makes a score of 120, and thereby increase his average by 5. The average score after 12th innings is

a. 60
b. 55
c. 65
d. 70

Ans. 65

Explanation: Suppose he played I₁, I₂,……, I₁₁ innings and the average is A₁₁.

I₁ + I₂ +…+ I₁₁ = 11*A₁₁;  ---------eq.(i.)

I₁ + I₂ +…+ I₁₁ + 120= 12*(A₁₁ + 5);  --------------eq.(ii.)

Subtract eq.(i) from eq.(ii.),

120 = A₁₁ + 60;

A₁₁= 60;

Hence, average after 12 innings = 60 + 5=65.

Question 15.

a. 2
b. 4
c. ±2
d. -2

Ans. 2

Explanation: Put √3 = 1.732 in the given expression-

7 + 4√3 = 13.928;  => √(7 + 4√3) = 3.73;

√(3 + 8√ (7 + 4√3) = 5.73;

√(-√3+(√(3 + 8√ (7 + 4√3) = 1.999 ~ 2.

Question 16.

a. 6
b. 12
c. 11
d. 3

Ans. 3

Explanation:

Question 17.The side BC of the ΔABC is extended to the point D. If ∠ ACD = 112° and ∠B = 3/4 ∠A, then the value of ∠B is

a. 64°
b. 48°
c. 46°
d. 50°

Ans. 48°

Explanation: Angle ACB = 180-112 = 68.

4/3 B +B + 68 =180;  => B = 48;

Question 18.ΔABC is a right angled triangle, the radius of its circumcircle is 3 cm and the length of its altitude drawn from the opposite vertex to the hypotenuse is 2 cm. Then the area of the triangle is

a. 12 sq cm
b. 3 sq cm
c. 6 sq cm
d. 5 sq cm

Ans. 6 sq cm

Explanation: The hypotenuse of the triangle will be the diameter of the circumcircle and altitude of the triangle will be the 2 cms as per the diagram given below-

Hence, The area of the right-angled triangle = ½ * 6*2 = 6 sq. cms.

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Question 19.The height of a tower is 50√3 m. The angle of elevation of a tower from a distance 50 m from its feet is

a. 30°
b. 45°
c. 60°
d. 90°

Ans. 60°

Explanation: from the figure, tanC= AB/BC = √3;

Hence, C = 60;

Question 20.The amount of Rs. 10,000 after 2 years, compounded annually with the rate of interest being 10% per annum during the first year and 12% per annum during the second year, would be (in rupees)

a. 11,320
b. 12,000
c. 12,320
d. 12,500

Ans. 12,320

Question 21.The value of tan80° tan10° + sin²70° + sin²20° is 

a. 0
b. 1
c. 2
d. √3/2

Ans. 2

Explanation:

= tan80° tan10° + sin²70° + sin²20°;

= tan80° tan(90^o -80^o) + sin²(90⁰-20°) + sin²20°;

= tan80° cot80^o + cos²20 + sin²20° = 1 + 1=2;

Question 22.The bar graph given below shows the per acre yield (in kg) of different countries. Study the graph carefully and answer the questions

The average yield of the given countries is

a. 1321/3
b. 1331/3
c. 1341/3
d. 1351/3

Ans. 1351/3

Explanation: Average yield = (200 + 160 + 120 + 132 + 120 + 80)/6 = 812/6 =406/3.

Question 23.The bar graph given below shows the per acre yield (in kg) of different countries. Study the graph carefully and answer the questions

By how much percentage is India's per acre yield more than that of Pakistan's?

a. 20%
b. 25%
c. 33 1/3%
d. 35%

Ans. 25%

Explanation: % change in yield for India and Pakistan = (200 – 160)*100/160 = 4000/160 = 25%.

Question 24.The bar graph given below shows the per acre yield (in kg) of different countries. Study the graph carefully and answer the questions

Srilanka's yield (approximately) is what percentof total yield of all the countries?

a. 17.8%
b. 16.2%
c. 18.2%
d. 15.4%

Ans. 16.2%

Explanation: Total yield of six countries = = (200 + 160 + 120 + 132 + 120 + 80) = 812.

The required percentage = 132*100/812 =16.25%.

Question 25.The bar graph given below shows the per acre yield (in kg) of different countries. Study the graph carefully and answer the questions

Writing the yields of all countries in ascending order, the difference between the sum of yields of first three countries to that of last three countries is

a. 200 kg
b. 212 kg
c. 172 kg
d. 162 kg

Ans. 172 kg

Explanation: the required difference = (200 + 160 + 132) – ( 120 + 120 + 80) =492  - 320=172 kg.

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