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SSC CHSL 2017 question paper 21 Jan 2017: Quantitative Aptitude

Apr 17, 2018 19:24 IST
    ssc chsl question paper
    ssc chsl question paper

    As there are a plenty of topics to cover in Quantitative Aptitude section, so you have to practice much harder. The questions from this section may or may not be difficult, but lengthy and require time. There are 25 questions in aptitude section. Therefore, to solve all of them, you have to be very fast and allocate time accordingly-

    Please find SSC CHSL 2017 quantitative aptitude questions of question paper held on 21 Jan 2017. 

    Online Quiz

    Question 1. Triangle ABC is right angled at B. BD is the altitude. AD is 8cm and DC is 18 cm. Find length of BD?

    (a) 6 cm

    (b) 9 cm

    (c) 12 cm

    (d) 15 cm

    Ans.:-c.

    Explanation:-

    Suppose, the length of BD = x cm.

    The line of altitude is always perpendicular to line AC.

    In triangle ABD,

    AB2 = AD2 + BD2;

    AB2 = 82 + x2; ……………(i)

    In triangle BCD,

    BC2 = BD2 + CD2;

    BC2 = x2 + 182; ……………(ii)

    In triangle ABC,

    AB2 + BC2 = AC2;

    Put the value from eq. (i) and eq. (ii) in it.

    64 + 324 + 2x2 = 262;

    388 + 2x2 = 676;

    x2 = 144;  => x= 12 cm.

    Question 2. Painter ‘A’ can paint a house in 50 days and ‘B’ can do it in 25 days. With help of ‘C’, they did the job in 10 days only. Then, ‘C’ alone can do the job in

    (a) 8 days

    (b) 16 days

    (c) 25 days

    (d) 15 days

    Ans.:-c.

    Explanation:-  

    1 day work of A= (1/50);

    1 day work of B = (1/25);

    1 day work of (A +B + C) = (1/10);

    Hence, the 1 day work of C alone = 0.1 – 0.04 – 0.02 = 0.04.

    Therefore, C will take = 1/0.04= 25days to get work done.

    Question 3. If 4 + 2x ≤ 6 + x and 2x + 5 < 2 + 4x; then x can take which of the following values?

    (a) 3

    (b) 1

    (c) 0

    (d) 2

    Ans.:-d.

    Explanation:-

    4 + 2x <=6  ----------- (i.)

    2x + 5 < 2 + 4x  ------------- (ii.)

    Subtracting eq. (i) from eq. (ii);

    1 < 4x -4;

    4x > 5; => x > 1.25;

    Question 4. What are the roots of the quadratic equation 4x2 + 6x – 18 = 0?

    (a) 3, -3

    (b) 3, 6

    (c) 3/2, -3

    (d) 3, 3

    Ans.:-c.

    Explanation:-

    4x2 + 6x – 18 = 0;

    2x2 + 3x – 9 = 0;

    2x2 + 6x – 3x – 9 = 0;

    2x (x + 3) -3 (x + 3) =0;

    (x +3)* (2x – 3)=0;

    x= -3, 3/2;

    Question 5. The price of an article is cut by 21%, to restore to its original value, the new price must be increased by

    (a) 21 percent

    (b) 26.58 percent

    (c) 17.36 perent

    (d) 26.25 percent

    Ans.:-b.

    Explanation:- Suppose the price of an article = Rs. x;

    New price = 79% of x; => 0.79x;

    % increase = (0.21/0.79)*100;

    = 26.58%.

    Question 6. If Gafur’s salary is 4/3 times of Haashim’s and Satish’s is 5/4 times of Haashim’s, what is the ratio of Gafur’s salary to Satish’s?

    (a) 16 : 15

    (b) 3 : 5

    (c) 5 : 3

    (d) 15 : 16

    Ans.:-a.

    Explanation:-

    Salary of Gafur = 4/3 * Salary of Haashim;

    Salary of Satish = 5/4 * Salary of Haashim;

    Salary (Gafur) : Salary (Satish) = [4/3]/[5/4] = 16/15.

    Question 7. If cosecA/(cosecA – 1) + cosecA/(cosecA + 1) = x, then x is

    (a) 2cosec2A

    (b) 2cosecA

    (c) 2secA

    (d) 2sec2A

    Ans.:-d.

    Explanation:-


    Question 8. Which of the following is correct?

    (a) (2x – y)2 = 4x2 – 4xy + y2

    (b) (2x – y)2 = x2 – 4xy + 4y2

    (c) (2x + y)2 = x2 – 4xy + 4y2

    (d) (2x + y)2 = 4x2 – 4xy + y2

    Ans.:-a.

    Explanation:-

    (2x-y)2= 4x2+y2-4xy;

    (2x – y)2 = 4x2 – 4xy + y2;

    Question 9. Of the 3 numbers whose average is 77, the first is 3/4 times the sum of other 2. The first number is

    (a) 148

    (b) 66

    (c) 99

    (d) 198

    Ans.:- c.

    Explanation:- Let these numbers are a, b, and c.

    a + b + c = 77*3 = 231; --------(i)

    a = &frac34;* (b + c);  ------------ (ii)

    replacing b + c value in eq.(i);

    a + 4/3*a = 231;

    7a/3 = 231;  => a = 33 * 3=99.

    Question 10. If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 34992, and Rs 37791.36 respectively, what is the rate of interest?

    (a) 4 percent

    (b) 16 percent

    (c) 8 percent

    (d) 13 percent

    Ans.:-c.

    Explanation:- By the compound interest formula-

    Question 11. Slope of the side DA of the rectangle ABCD is -3/4. What is the slope of the side AB?

    (a) -4/3

    (b) 3/4

    (c) -3/4

    (d) 4/3

    Ans.:-d.

    Explanation:- In a rectangle ABCD, line DA and AB will be perpendicular to each other.

    Slope (DA) * Slope (AB) = -1;

    Slope (AB) = -1/(-3/4)=4/3;

    Question 12. If cot 30 – cos 45 = x, then x is

    (a) √3 + 2

    (b) (√6-1)/√2

    (c) (√3-2√2)/√6

    (d) (1+√2)/2

    Ans.:-b.

    Explanation:- cot 30 – cos 45 = x;

    Replacing the value of these trigonometric measures-

    √3 – 1/√2 =x;

    x = (√6-1)/√2;

    Question 13. The length of the diagonal of a rectangle is 10 cm and that of one side is 8 cm. What is the area of this rectangle?

    (a) 80 sq cm

    (b) 48 sq cm

    (c) 60 sq cm

    (d) 32 sq cm

    Ans.:-b.

    Explanation:-

    Suppose that AC = 10 cm and AB = 8cm.

    Since, triangle ABC is a right-angled triangle.

    AC2 = AB2 + BC2;

    BC2 = 102 – 82;

    BC2 = 36;

    BC = 6;

    The Area of Rectangle = AB x BC;

    =8*6

    =48 sq. cm.

    Question 14. The two numbers are 55 and 99, HCF is 11, What is their LCM?

    (a) 486

    (b) 479

    (c) 476

    (d) 495

    Ans.:-d.

    Explanation:-

    HCF x LCM = first number x second number;

    LCM = 55 x 99 /11 = 55 * 9 = 495.

    Question 15. A dishonest milkman buys milk at Rs 25 per litre and adds 1/5 of water to it and sells the mixture at Rs 29 per litre. His gain is

    (a) 16 percent

    (b) 39.2 percent

    (c) 24 percent

    (d) 32 percent

    Ans.:-b.

    Explanation:- Suppose that he bought 10 litre of Milk and add 2 litre water into it.

    Then, the price of 10 litre milk = Rs. 250;

    The price of Milk and water mixture = 12 *29 = Rs. 348;

    Realized gain = (348-250/250)*100 = 39.2%.

    Question 16. If 9x – 6y = 15 and x + 6y = 15, then the value of x and y.

    (a) 3, 4

    (b) 3, 2

    (c) 4, 5

    (d) 5, 6

    Ans.:-b.

    Explanation:-

    9x -6y = 15 ---------(i)

    x + 6y = 15  ---------------(ii)

    Add both the equations,

    10x=30; => x=3 and y = 2.

    Question 17. A cone of height 25 cm and diameter 21 cm is mounted on a hemisphere of the same diameter. Find the volume of the solid thus formed? (Take π = 22/7)

    (a) 3155 cubic cms

    (b) 4158 cubic cms

    (c) 3315 cubic cms

    (d) 5313 cubic cms

    Ans.:-b.

    Explanation:-  

    Question 18. To cover a distance of 297 km in 4.4 hours what should be the average speed of the car in meters/second?

    (a) 67.5 m/s

    (b) 33.75 m/s

    (c) 37.5 m/s

    (d) 18.75 m/s

    Ans.:-d.

    Explanation:-

    Speed = Distance/time;

    = 297/4.4 = 67.5 kmph;

    = > (67.5*5)/18 = 18.75 m/sec.

    Question 19. The diameter of a circle is equal to the side of the square. What is the area of the square if the area of the circle is 36π sq cm?

    (a) 72 sq cm

    (b) 144 sq cm

    (c) 36 sq cm

    (d) 18 sq cm

    Ans.:-b.

    Explanation:- Suppose, the diameter of circle = side of square =2a cms.

    Area of Circle = ��a2 = 36π; => a = 6;

    Hence, the area of square = 4a2=4*36 = 144 sq. cm.

    Question 20. Marked price of an item is Rs 100. On purchase of 2 items discount is 25%, on purchase of 4 items discount is 43%. Rajasi buys 6 items, what is the effective discount?

    (a) 37 percent

    (b) 26.25 percent

    (c) 9.6 percent

    (d) 24.6 percent

    Ans.:-a.

    Explanation:-

    2* 100 ------(-25%)------> Rs. 150 (After discount);

    4*100 --------(-43%) ------> Rs. 228 (After discount);

    If she purchased 6 items;

    6*100 ------- > Rs. 378 (after discount);

    Discount percentage = [(600-378)/600]*100 = 37%.

    Question 21. If tan2A = x, then x is

    (a) 2tanA/(1 – tan2A)

    (b) (1 – tan2A)/2tanA

    (c) 2tanA/(1 + tan2A)

    (d) (tan2A – 1)/2tanA

    Ans.:-a.

    Explanation:-

    tan2A =(2*tanA)/(1-tan2A);

    Question 22. Refer the below data table and answer the following Question.

     

    Quantity of stock

    Average Cost (Rs)

    Mobile Phones

    40

    10000

    Cameras

    93

    20000

    TVs

    97

    52000

    Refrigerators

    21

    22000

    ACs

    39

    23000

    What is the value of the total stock (in lakh rupees)?

    (a) 866.3

    (b) 127

    (c) 290

    (d) 86.63

    Ans.:-d.

    Explanation:-

    Value of total stock = 40*10000 + 93*20000 + 97*52000 + 21*22000 + 39*23000;

    = 4 lacs + 18.6 lacs + 50.44 lacs + 4.62 lacs + 8.97 lacs

    =86.63 lacs.

    Question 23. Refer the below data table and answer the following Question.

    Year

    Ratio Import/Export

    2011

    0.8

    2012

    1.2

    2013

    0.8

    2014

    1.4

    2015

    0.8

    If the imports in 2012 was Rs. 1000 crores and the total exports in the years 2012 and 2013 together was Rs. 3600 crores, then the imports in 2013 was?

    (a) 2767

    (b) 2213

    (c) 833

    (d) 3458

    Ans.:-b.

    Explanation:- Suppose the imports in 2013 was = Rs. x crores.

    Export in 2012= 1000/1.2 = 833.33 crores.

    Export in 2013 = 3600 – 833.33 = 2766.67 crores.

    Import in 2013 = 2766.67*0.8=2213.336 crores.

    Question 24. Refer the below data table and answer the following Question.

    Measured on Birthday

    Height of the child (in cms)

    4

    100

    5

    110

    6

    115

    7

    120

    8

    125

    9

    135

    10

    140

    11

    150

    12

    160

    13

    170

    14

    175

    15

    185

    16

    195

    What was the increase in the height of the child from the 7th Birthday to the 15th Birthday?

    (a) 75 cms

    (b) 70 cms

    (c) 65 cms

    (d) 60 cms

    Ans.:-c.

    Explanation:-

    Increased height = 185-120 = 65 cms.

    Question 25. Refer the below data table and answer the following Question.

    Deep Sleep

    5 min

    Dreaming

    10 min

    Light sleep

    10 min

    Extremely light sleep

    20 min

    Awake

    55 min

    Between 10 pm to 6am, a fitness band records the following data. How long was the user dreaming or was awake?

    (a) 3.7 hours

    (b) 4.7 hours

    (c) 5.2 hours

    (d) 5.7 hours

    Ans.:-

    Explanation:-

    Duration for which user is dreaming or awake = 55 + (20 +55) + (10 + 20 + 55) + (10 + 10 + 20 +55)

    = 55 + 75 + 85 + 95

    = 310 min  5.2 hours.

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