This is the sample paper for the SSC CGL pre-examination of the Aptitude section. It comprises 25 questions as per the latest SSC CGL pattern with shortcut solutions. 25 questions are asked in each section of the SSC CGL pre-examination. Aspirants usually face problems in solving the Quant section with the right speed & accuracy and hence lose on some sure shot marks. This sample paper will make you ready for the real exam by improving your speed & accuracy.
Que 1. There are 3 numbers such that A is 50% more than B and B is 25% less than C. If the difference between A & C is 216, then find B.
A)1212
B)1296
C)1248
D)1272
Answer: B
Explanation:
Let C be 200x.
Then ATQ, B will be 150x and A will be 225x.
Now, the difference between A & C is 216
This means, 25x = 216, x = 216/25
Now, B is 150x
B = 216/25 × 150
B = 216 × 6 = 1296.
Hence the correct answer is option B
Que 2. Find the circumradius of a triangle with sides of 88cm, 105cm, and 137cm.
A) 62.5
B) 67.5
C) 68.5
D) 72.5
Answer: C
Explanation
(88,105,137) is a Pythagorean triplet (Hence these sides will make a right-angle triangle).
Then, circumradius(R) = hypotenuse/2 (Biggest Side/2)
So, R= 137/2 = 68.5 cm
Hence the correct answer is option C
Que 3. Find the unit digit of 234 + 472 – 159
A) 4
B) 5
C) 6
D) 8
Answer: B
Explanation:
In finding the unit digit of a number we must check the last digit of every term.
234, we must check it unit digit by multiplying 3 four times the answer comes to 1.
Then, 472: we must multiply 7 two time and then we get 9 as its unit digit.
Again, 159: unit digit of a number ending with 5 is always 5.
Now, we simply put the values in the given equation
= 1+ 9 – 5
=5
Hence the correct answer is option B.
Que 4. The CP of an article is Rs360. It is sold at a certain price. If it is sold at one third of the original selling price, there is a loss of 50/3%. What is the original SP.
A) 900
B) 700
C) 500
D) 600
Answer: A
Explanation:
Let the original selling price be 3x.
If it is sold at one-third of 3x then there will be a loss of 50/3% of CP i.e. 1/6 of Rs360 = 60
SP in second case = CP – Loss
= Rs 360-60
= Rs 300
Original SP = Rs 3 × 300 = Rs 900
Hence the correct answer is option A.
Que 5. A, B and C started partnership by investing Rs10000, Rs25000 and Rs15000 respectively. A is a working partner and gets 20% of the profit & the remaining will be distributed in the ratio of their capital. If total profit is Rs60000, then find the share of A.
A) 18000
B) 21000
C) 21600
D) 19200
Answer: C
Explanation:
Profit will be divided among A,B & C in the following ratio.
A gets 20% of total profit for being an active partner, i.e 20% of 60000 = 12000.
Now, remaining amount of 48000 will be divided in the ratio of their investment.
Ratio of their investment is, 10000: 25000: 15000 = 2: 5: 3
So, complete share of A = 12000 (Share for active partner from total Profit) + 48000 × (2/10) {Share for capital invested}
A= Rs12000+9600
A= Rs 21600
Hence the correct answer is option C.
Que 6. ABCD is a trapezium in which AB is parallel to CD. AB= 8cm & CD=12cm. If the midpoints of the diagonal AC & BD are joined then find its length.
A) 2
B) 4
C) 6
D) 1
Answer: A
Explanation:
Formula = 1/2 × difference between the length of parallel sides ( As line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides are equal to half of their difference)
= 1/2 × (12 – 8) = 2 cm
Hence the correct answer is option A.
Que 7. If the sum of 1 + 3 + 5 + 7……….. till the nth term is 144, then what is n here?
A) 13
B) 10
C) 12
D) 11
Answer: C
Explanation:
The sum of the first n odd numbers is equal to n2
144 = n2
n = 12
Hence the correct answer is option C
Que 8. If x2 – 7x + 1 = 0, Find x3 + 1/x3?
A) 364
B) 322
C) 384
D) 326
Answer: B
Explanation:
Given, x2 – 7x + 1 = 0
On divide by x we get,
We get, x + 1x = 7
On Cubing both sides we get,
x3 + 1/x3 + 3 (x) × (1/x) (x + 1/x) = 343
x3 + 1/x3 + 3 × 7 = 343
It gives, x3 + 1/x3 = 322
Hence the correct answer is option B.
Que 9. A sum of Rs 5000 is borrowed at a rate of 10% p.a. compounded annually. At the end of each he return Rs1000. How much money will he pay at the end of 3rd year to clear his debt?
A) 3225
B) 3475
C) 3345
D) 4345
Answer: D
Explanation:
Principle= Rs 5000
Interest for the first year= (10/100) × 5000 = 500
Amount = Rs 5000 + 500 = Rs 5500
Amount paid at the end of the first year = Rs 1000
Now new principle for second year = 5500 – 1000 = Rs 4500
Interest for the second year = (10/100) × 4500 = Rs 450
Amount at the end of 2nd year = 45000 + 4500 = Rs 4950
Amount paid at the end of 2nd year = Rs1000
Now, the principle for the 3rd year = 4950 – 1000 = Rs 3950
Interest for the 3rd year = (10/100) × 3950 = Rs 395
The amount he has to pay at the end of 3rd year to clear off his debt = 3950 + 395 = Rs4345
Hence the correct answer is option D.
Que 10. (3√3x3 + 5√5y3)/(√3x + √5y) = Ax2 + By2 + Cxy, Find A2 + B2 + C2?
A) 25
B) 36
C) 49
D) 64
Answer: C
Explanation:
We know, x3 + y3 = (x + y) (x2 + y2 – xy)
Comparing it with the given equation, we get
Ax2 + By2 + Cxy = 3x2 + 5y2 – √15xy.
Hence, A = 3 , B = 5 , and C = -√15.
A2 + B2 + C2 = 9 + 25 + 15 = 49.
Hence the correct answer is option C.
Que 11. Find the area of the sector of a circle with an angle of 90 degrees and the radius of the circle is 14 cm.
A) 154 cm sq
B) 144 cm sq
C) 516 cm sq
D) 576 cm sq
Answer: A
Explanation:
Area of sector = (θ/360) × πr2
= (90/360) × 22/7 × 14 × 14
=154 cm sq.
Hence the correct answer is option A.
Que 12. Two vessels A & B contain milk & water in the ratio 3: 2 and 7 : 3. They are mixed in the ratio of 2:3. Find the ratio of milk & water in the resulting mixture.
A) 23 : 33
B) 33 : 17
C) 15 : 17
D) 15 : 33
Answer: B
Explanation:
In first vessel ratio of M : W = 3 : 2 ( Total = 5 units)
In, Second vessel ratio of M : W = 7 : 3 ( Total= 10 units)
First, make the total units of both vessels the same, multiply vessel 1 by 2 for that.
Now we have, M : W = 6 : 4 (First vessel)
And, M : W = 7 : 3 ( Second vessel)
ATQ, they are mixed in the ratio of 2 : 3, so multiply vessel 1 by 2 & vessel 2 by 3 & simply add to get the answer.
M : W = 12 : 8 (First vessel)
M : W = 21 : 9 (Second vessel)
Adding we get, M : W = 33 : 17.
Hence the correct answer is option B
Que 13. A trader sold the article at 5% loss. Had he sold it at 20% profit then he gains Rs 100 more. Find CP.
A) Rs 300
B) Rs 400
C) Rs 500
D) Rs 600
Answer: B
Explanation:
Let CP= 100x, then SP1 = 95x
ATQ, SP2 = 120x, then he gains 100 Rs more.
So, 120x – 95x = 25x = 100
It gives, x = 4
So, CP = 100 × 4 = 400.
Hence the correct answer is option B
Que 14. A train starts from point A towards B at a speed of 54km/hr and another train starts from B towards A at a speed of 72 km/hr. They meet after 5 hours. What is the distance between A & B ?
A) 540
B) 620
C) 630
D) 720
Answer: C
Explanation:
Let the distance between AB be x.
Distance between points / Relative speed=Time taken
Relative Speed = Speed of train1 + speed of train 2 (When traveled in the opposite direction)
Relative Speed= 54 + 72 =126 km/hr.
So x = 5 hours × 126 km/hr
x = 630 Km.
Hence the correct answer is option C
Que 15. A can do the work in 6 days. B can do it in 12 days & C can do it in 18 days. If they started work together then in how many days the whole work will we completed?
A) 35/11
B) 34/11
C) 36/11
D) 42/11
Answer: C
Explanation:
Let the total work be 36 units (LCM of 6, 12 & 18)
Then, A does 6 units, B does 3 units & C does 2 units per day respectively.
Time is taken to complete the work by A, B & C = Total work / Total efficiencies of A, B & C
= 36 / (6+3+2)
= 36/11 days.
Hence the correct answer is option C.
Que 16. If tanθ = 8/15, then (sinθ – cosθ) / (sinθ + cosθ) = ?
A) 7/23
B) 5/23
C) -7/23
D) -5/23
Answer: C
Explanation:
= (sinθ – cosθ)/(sinθ + cosθ)
Divide numerator and denominator by cosθ we get, ( tanθ – 1)/(tanθ + 1)
= (8/15 – 1)/(8/15 + 1)
= (-7/15)/(23/15)
= -7/23
Hence the correct answer is option C.
Que 17. (1 + Tanx)/(1 – Tanx) = Tan 73, Find the value of x
A) 23
B) 29
C) 27
D) 28
Answer: D
Explanation:
We can write, (Tan45 + Tanx)/(Tan45 – Tanx) = Tan73 ( as Tan 45 = 1)
It gives us, Tan(45 + x) = Tan73 { As, Tan (A+B) = (Tan A + Tan B) / (1 – Tan A × Tan B) }
Hence x = 73- 45 = 28
Hence the correct answer is option D
Que 18. Radius of cylinder is 5 cm & height is 15cm . Find the approx difference between its volume & CSA.
A) 700
B) 730
C) 670
D) 707
Answer: D
Explanation:
volume = πr2h
volume= (22/7) × 5 × 5 × 15 cm3
= 1178.5 cm3
CSA = 2πrh = 2 × (22/7) × 5 × 15
= 471.4 cm2
Required difference = Volume – CSA = 707.1 cm2.
Hence the correct answer is option D
Que 19. If the average of 21 numbers is 73 and two numbers are misread as 41 and 53 instead of 14 and 35. Find out the correct average (approx.)
Answer: C
Explanation:
Average of 21 numbers = 73
Total of 21 numbers = 21 × 73
= 1533
Correct total = 1533 – Incorrect numbers + correct numbers
= 1533 – (41 + 53) + (14 + 35)
= 1488
Correct Average = Correct total / total number
= 1488/21
= 71 (approximately).
Hence the correct answer is option C
Que 20. ABCD is a cyclic quadrilateral of which AB is the diameter. Diagonals AC & BD intersect at E. If angle DBC is 26 degree. Then find angle AED.
A) 26
B) 64
C) 116
D) 52
Answer: B
Explanation:
∠DAC = 26° ( angles on the same segment in a circle)
∠ADB = 90° (Angle formed by diameter on any point on circumference)
Now, in triangle AED
∠AED + 26° + 90° = 180°.
∠AED = 180° – 116° = 64.
Hence the correct answer is option B
Que 21. 33x381y is a seven-digit number which is divisible by 88. Then find the value of x + y + xy?
A) 13
B) 11
C) 15
D) 17
Answer: A
Explanation:
For a number to be divisible by 88 it must be divisible by 8 and 11
For a number to be divisible by 8 then its last 3 digits must be completely divisible by 8.
Hence, the possibility of y is 6 ( As 816 is divisible by 8)
Now, for a number to be divisible by 11. then the difference of the sum of odd digits and even digits must be 0 or 11.
Then, the possibility of x is 1 {( 3 + x + 8 + 6) – ( 3 + 3 + 1) = 10 + x should be 0 or 11)
10 + x = 0 then x = -10 not possible.
10 + x = 11
x = 1.
So, x + y + xy
=1+ 6 + 1 × 6
= 13.
Hence the correct answer is option A.
Directions (22-25). The following table shows the data of total numbers of students of 5 schools (A, B, C, D &E) and the respective ratio of boys and girls in the given schools.
Que 22. What is the ratio of boys of school B and girls of school E?
A) 22 : 27
B) 44 : 27
C) 22 : 29
D) 44 : 29
Answer: B
Explanation:
As shown in the table above, the ratio of boys in school B to Girls in school E
= 176 : 108
= 44 : 27
Hence the correct answer is option B
Que 23. Total numbers of boys in school E & girls in school B?
A) 278
B) 248
C) 256
D) 298
Answer: D
Explanation:
Again from the table, we have
Total number of boys in school E & girls in school B = 90 + 208
= 298.
Hence the correct answer is option D.
Que 24. Percentage of the total number of boys with respect to total number of students in all schools is? (approx.)
A) 47%
B) 53%
C) 49%
D) 57%
Answer: C
Explanation:
From the table we have,
Total number of boys = (484/996) × 100
= 48.59%
= 49%
Hence the correct answer is option C.
Que 25. Number of girls in school C?
A) 72
B) 76
C) 82
D) 84
Answer: D
Explanation:
The total number of girls in school C = 84.
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