In this article, we have shared all answers and proper explanations of the quantitative aptitude questions asked in SSC CHSL tier-1 exam held on 22 Jan 2017 (evening session). Go to the questions, which you find difficult and understand the logic behind it. Let us go-
Question 1 P and Q can do a project in 12 and 12 days respectively. In how many days can they complete 50% of the project if they work together?
(a) 6 days
(b) 3 days
(c) 18 days
(d) 1.5 days
Ans.:-b.
Explanation:- P’s 1 day work = 1/12; Q’s 1 day work = 1/12;
[P + Q]’s 1 day work = [1/12 +1/12]=1/6;
Suppose, it takes n days to complete half work;
n*1/6 = ½; = > n =3 days.
Question 2 If 4(x + 5) – 3 > 6 – 4x ≥ x – 5; then the value of x is
(a) 2
(b) 1
(c) 3
(d) 4
Ans.:-a.
Explanation:-
6 – 4x ≥ x - 5; it implies , x < = 2.2
Hence, 2 will be the correct answer.
Question 3 The point R (a, b) is first reflected in origin to R1 and R1 is reflected in x- axis to (-4, 2). The coordinates of point R are
(a) (-4, -2)
(b) (-4, 2)
(c) (4, 2)
(d) (4, -2)
Ans.:-c.
Explanation:- if point R (a, b) is reflected across the x-axis, then its co-ordinate will be R2 (a, -b) in fourth quadrant & R2 ( -a, -b) in third quadrant and vice-versa.
Since, point (-4 , 2) is in second quadrant and reflected one, therefore the original point will be (4, 2).
Question 4 For triangle PQR, find equation of altitude PS if co-ordinates of P, Q and R are (1, 2), (2, -1) and (0,5) respectively?
(a) x – 3y = -7
(b) x + 3y = -5
(c) x – 3y = -5
(d) x + 3y = -7
Ans.:-c.
Explanation:- The perpendicular line will pass through P(1, 2) and makes 90 degree angle with the line QR.
The slope of QR= [5 – (-1)]/[0-2]= -3.
Hence, the slope of the required line = (-1/slope of QR) = 3.
The perpendicular line that passes through P (1, 2)-
y - 2 = 3 (x – 1);
x – 3y =-5;
Question 5 If sec 300° = x, then value of x is
(a) -√2
(b) 1/√3
(c) 2
(d) -1
Ans.:-c.
Explanation:- sec 300 = sec(360-60) = sec60 = 2.
Question 6 To cover a distance of 225 km in 2.5 hours what should be the average speed of the car in meters/second?
(a) 90 m/s
(b) 45 m/s
(c) 50 m/s
(d) 25 m/s
Ans.:-d.
Explanation:- use formula, speed = distance/time;
Hence, speed = 225/2.5 = 90 kmph = 90*5/18 = 25 m/s.
Question 7 An angle is greater than its complementary angle by 30°. Find the measure of the angle?
(a) 30°
(b) 90°
(c) 60°
(d) 120°
Ans.:-c.
Explanation:- suppose the angle = x degrees; then its complementary angle will be (90 – x);
Following the question-
x = 90 –x +30;
2x = 120; = > x = 60;
Question 8 A rice trader buys 8 quintals of rice for Rs 3,600. 10% rice is lost in transportation. At what rate should he sell to earn 15% profit?
(a) Rs 352.1 per quintal
(b) Rs 517.5 per quintal
(c) Rs 569.25 per quintal
(d) Rs 582.3 per quintal
Ans.:-c.
Explanation:- The price of 8 quintals rice = Rs. 3600;
Price of 1 quintal rice = Rs. 450.
The cost of transportation = 10% on each quintal = Rs. 45;
The total price of 1 quintal of rice = Rs. 495.
Hence, the Selling price of 1 quintal rice to gain 15% profit= 495 * 1.15 = Rs. 569.25
Question 9 What is the value of (3/5 + 7/9)?
(a) 62/45
(b) 31/28
(c) 5/7
(d) 1/7
Ans.:-a.
Explanation:- After taking LCM of the giving expression, we will get-
= (3*9 + 7*5)/5*9 = (27 + 35)/45;
=62/45
Question 10 Marked price of an item is Rs 200. On purchase of 1 item discount is 5%, on purchase of 2 items discount is 14%. Rajeshri buys 3 items, what is the effective discount?
(a) 37 percent
(b) 26.25 percent
(c) 11 percent
(d) 30.2 percent
Ans.:-c.
Explanation:- MP of item = Rs.200;
On purchase of 1 item, discount= 200 * 5% = Rs. 10.
On purchase of 2 items, discount = 2*200*14% =Rs. 56.
On purchase of 3 items, discount = 10 + 56 = Rs. 66.
% discount = 66*100/600 = 11%.
Question 11 If the radius of a circle is increased by 16%, its area increases by
(a) 34.56 percent
(b) 32 percent
(c) 16 percent
(d) 17.28 percent
Ans.:-a.
Explanation:- Let the initial radius = r; then, increased radius = 1.16r;
Change in area= pi*[(1.16r)2- r2] = pi*0.3456*r2;
% increase in area= [pi*0.3456*r2/pi*r2]*100 = 34.56%.
Question 12 Expand and simplify (x + 4)2 + (x – 2)2
(a) 2(x2 + 2x + 10)
(b) 2(x2 + 2x - 10)
(c) 2(x2 – 2x + 10)
(d) 2(x2 – 2x – 10)
Ans.:-a.
Explanation:- (x + 4)2 + (x – 2)2 = x2 + 16 + 2*x*4 + x2 + 4 – 2*x*2;
=2 x2 + 20 +4x;
= 2(x2 +2x + 10);
Question 13 A cone is hollowed out of a solid wooden cube of side 6 cm. The diameter and height of the cone is same as the side of the cube. What is the volume of the remaining cube?
(a) 159.42 cubic cms
(b) 323.15 cubic cms
(c) 106.33 cubic cms
(d) 210.66 cubic cms
Ans.:-a.
Explanation:- the side of cube =6 cms = diameter of cone = height of cone;
Hence, r = 3 cms; and h = 6cms;
The volume of the cube = 1/3*pi* r2*h;
= 1/3 * pi * 9 * 6 = 18* pi = 56.57 cm3;
The volume of remaining cube = total volume of cube – cone volume;
= 63 – 56.57 = 216 – 56.57 = 159.43 cm3.
Question 14 In Δ ABC, the median AD is 7 cm and CB is 14 cm, measure of ∠CAB is
(a) 30°
(b) 60°
(c) 90°
(d) 120°
Ans.:-c.
Explanation:- 
AD = 7cms; BD = CD = 14/2= 7 cms.
Since, AD is the median of CB; therefore, it will be also divides the angle CAB into half.
Suppose, angle BAD = CAD = x;
Since, AD = BD = CD = 7cms.
Therefore, ABD = BAD = x;
CAD = ACD =x;
Since, the sum of all angles in triangle is equal to 180 degrees.
2x + x + x= 180;
2x = 90 degree; = > Angle CAD = 90 degree.
Question 15 Among three numbers, the first is twice the second and thrice the third. If the average of three numbers is 198, then what is the difference between the first and the third number?
(a) 216
(b) 297
(c) 661
(d) 431
Ans.:-a.
Explanation:- Let the first number =a; the second number = b; and the third number = c;
Given: a = 2b and a = 3c;
(a + b + c ) /3 = 198; = > a + b + c = 594;
a + a/2 + a/3 = 594; = > (11/6)*a = 594;
a = 594*6/11 = 54*6 = 324.
Hence, The first number = 324;
The third number = 108;
The difference between them = 324-108= 216
Question 16 If secA/√(sec2A – 1) = x, then value of x is
(a) tanA
(b) cosecA
(c) sinA
(d) cosA
Ans.:-b.
Explanation:- x = secA/√(sec2A – 1);
Apply, sec2A = 1 + tan2A; => sec2A - 1 = tan2A;
x= sec A/tan A = cosec A.
Question 17 If Gangadhar's salary is 3/11 times of Hari's and Shambhu's is 3/4 times of Hari's, what is the ratio of Gangadhar's salary to Shambhu's?
(a) 44:9
(b) 4:11
(c) 9:44
(d) 11:4
Ans.:-b.
Explanation:- Gangadhar’s salary = 3/11* Hari’s salary;
Shambhu’s salary= ¾*Hari’s salary;
Divide both statements-
Salary (Gangadhar : Hari) = 4:11
Question 18 If a + b = 8 and ab =15, then a3 + b3 is
(a) 224
(b) 244
(c) 152
(d) 128
Ans.:-c.
Explanation:- consider (a + b)3 = a3 + b3 + 3ab (a + b);
a3 + b3 = (a + b)3 - 3ab (a + b);
= 83 – 3*15*8 = 512 -360 = 152;
Question 19 What is the value of (1 + cotA)2 + (1- cotA)2?
(a) 2cosec2A
(b) 2sec2A
(c) 1 – 2cosec2A
(d) 1 – 2sec2A
Ans.:-a.
Explanation:- apply (a + b)2 = a2 + b2 + 2ab;
(1+ cot A)2 = 12 + cot 2A + 2*1*cot A; ------------eq.(i)
(1 - cot A)2 = 12 + cot 2A - 2*1*cot A; -------------eq.(ii)
Add the above equations-
(1 + cotA)2 + (1- cotA)2 = 2 + 2 cot2A = 2* ( 1+ cot2A) = 2 cosec2 A.
Question 20 The sum of twice a number and thrice its reciprocal is 25/2. What is the number?
(a) 7
(b) 6
(c) 5
(d) 4
Ans.:-b.
Explanation:- suppose, the number is x.
2x + 3/x = 25/2; => 4x2 -25x + 6=0;
4x2 -24x –x + 6=0;
4x (x -6) -1 (x-6)=0;
(x-6)(4x-1)=0; = > x= 6, ¼;
Question 21 A sum fetched a total simple interest of Rs. 7728 at the rate of 7% per year in 8 years. What is the sum?
(a) Rs 13800
(b) Rs 16560
(c) Rs 11040
(d) Rs 8280
Ans.:-a.
Explanation:- SI = P*R*T/100;
P = SI*100/R*T;
P = 7728*100/7*8 = Rs. 13800
Question 22 Refer the below data table and answer the following Question.
|
|
Number of employees |
Annual salary (in lakhs) |
Bonus as percent of annual salary |
| Manager |
3 |
60 |
30% |
| Excecutive |
6 |
12 |
30% |
| Trainee |
4 |
4 |
20% |
What is the average bonus (in rupees)?
(a) 606154
(b) 7880002
(c) 253333
(d) 224000
Ans.:-a.
Explanation:- Average bonus = Total Bonus/total no. of employees;
Average bonus = (3*60*30% + 6*12*30% + 4*4*20%)*100000/13;
= (54 + 21.6 + 3.2)*100000/13 = Rs. 606153.84.
Question 23 Refer the below data table and answer the following Question.
|
|
2011 |
2012 |
2013 |
2014 |
2015 |
| Company A |
3000 |
3000 |
1000 |
2000 |
2000 |
| Company B |
5000 |
4000 |
2000 |
3000 |
4000 |
| Company C |
3000 |
5000 |
3000 |
4000 |
5000 |
For which of the following pairs of years the total exports from the three Companies together are equal?
(a) 2011 & 2013
(b) 2013 & 2015
(c) 2011 & 2015
(d) 2014 & 2015
Ans.:-c.
Explanation:-
Total exports of all three companies in 2011 = 3000 + 5000 + 3000 = 11,000;
Total exports of all three companies in 2012 = 3000 + 4000 + 5000 = 12,000;
Total exports of all three companies in 2013 = 1000 + 2000 + 3000 = 6,000;
Total exports of all three companies in 2014 =2000 + 3000 + 4000 = 9,000;
Total exports of all three companies in 2015 = 2000 + 4000 + 5000 = 11,000;
Question 24 Refer the below data table and answer the following Question.
| Year |
Profit or (-Loss) in Rs Crore |
| 2011 |
5 |
| 2012 |
15 |
| 2013 |
-20 |
| 2014 |
-15 |
| 2015 |
-5 |
What was the total Profit or loss of the company in last 5 years?
(a) Profit of Rs 20 crores
(b) Loss of Rs 30 crores
(c) Loss of Rs 20 crores
(d) Profit of Rs 30 crores
Ans.:-c.
Explanation:- Total profit/loss of company is all years = 5 + 15 -20 -15 -5 = -20 crores.
Question 25 Refer the below data table and answer the following Question.
| India’s exports in 2015 |
Value in million US$ |
| Jewellery |
550 |
| Software |
700 |
| Cotton |
875 |
| Steel |
775 |
| Electronics |
550 |
Software was what percent of total exports?
(a) 20.29 percent
(b) 22.79 percent
(c) 25.29 percent
(d) 17.79 percent
Ans.:-a.
Explanation:- % total exports of software = Software exports*100/ total exports;
= 700*100/ 3450 =20.289%.
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