Important Quantitative Aptitude Questions for SSC CHSL Tier-1 2018

Dear students, you know that QUANT is a part of getting points and every chapter is important. Therefore, we are providing 15 questions of quant. Solve all these quizzes every day so that you can improve your accuracy and speed. We also provide lots of quant questions. So you can practice that chapter which takes more time to solve the questions.

प्रिय पाठकों, आप सभी जानते हैं कि संख्याताम्क अभियोग्यता का भाग बहुत ही महत्वपूर्ण है. इसलिए हम आपको संख्यात्मक अभियोग्यता कि 15 प्रश्नों कि प्रश्नोत्तरी प्रदान कर रहे हैं. इन सभी प्रश्नोत्तरी को दैनिक रूप से हल कीजिये ताकि आप अपनी गति और सटीकता में वृद्धि कर सकें. हम आपको अन्य कई संख्यात्मक अभियोग्यता के प्रश्न प्रदान करेंगे. ताकि आप पाठ्यक्रम अनुसार उन्हें हल कर पायें.


Q1. If 3/4 of a number is 7 more than 1/6 of the number then 5/3 of the number is?
(a) 12
(b) 18
(c) 15
(d) 20

S1. Ans.(d)
Sol. Let the number be x
⇒ According to the question,
⇒ 3x/4-x/6 = 7
⇒ (9x – 2x)/12 = 7
⇒ 7x = 7 × 12
⇒ x = 12
⇒ Then 5/3 of the number will be
= (x ×5)/3 ⇒ (12 × 5)/3 = 20

Q2. If x = 1/√(2 + 1) then (x + 1) equal to?
(a) √2+1
(b) √2-1
(c) √2
(d) 2

S2. Ans.(c)
Sol. Given
x=1/(√2+1)
Find (x + 1) = ?
⇒ x + 1 = 1/(√2  + 1)+1
⇒(√2+1+1)/(√2+1)
⇒(√2+2)/(√2+1)
⇒(√2 (√2+1))/((√2+1) )
⇒ √2

Q3.  If a * b = a + b + a/b, then the value of 12 * 4 is?
(a) 20
(b) 21
(c) 48
(d) 19

S3. Ans.(d)
Sol. 12 * 4 = 12 + 4 + 12/4
= 12 + 4 +3 = 19

Q4. Find the maximum number of trees which can be planted, 20 meters apart, on two sides of a straight road 1760 meters long?
(a) 180
(b) 178
(c) 174
(d) 176

S4. Ans.(b)
Sol. No of trees planted on one side of Road
= (Length of road)/(Distance between trees) + 1 ← first tree
=1760/20+1
= 88 + 1 = 89
Total trees (both sides) = 89 × 2 = 178

Q5. Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is?
(a) 15
(b) 20
(c) 25
(d) 10

S5. Ans.(b)
Sol. Mohan does ‘x’ correct sums
According to question

⇒ 3x – 60 + 2x = 40
5x = 100
x = 20

Q6. Two trains 108 m and 112 m in length are running towards each other on the parallel lines at a speed of 45 km/hr and 54 km/hr respectively. To cross each other after they meet, it will take?
(a) 12 sec
(b) 9 sec
(c) 8 sec
(d) 10 sec

S6. Ans.(c)
Sol. Time taken by them to cross each other
=(l1+l2)/(Relative speed in opposite direction)
Time = ((108 + 112))/((45 + 54)×5/18)=(220 × 18)/(99 × 5)
Time = 8 second

Q7. Two trains 150 m and 120 m long respectively moving from opposite directions cross each other in 10 seconds. If the speed of the second train is 43.2 km/hr, then the speed of the first train is?
(a) 54 km/hr
(b) 50 km/hr
(c) 52 km/hr
(d) 51 km/hr

S7. Ans.(a)
Sol. Let the speed of second trains = x km/hr
⇒ Their  relative speed in opposite direction
= (43.2 + x)km/hr
⇒ According to question,
⇒ time = (l1  + l2)/speed
⇒ 10 sec = ((150 + 120))/((43.2 + x)×5/18 m/s)
⇒ 10 = (270 × 18)/((43.2 + x) × 5)
43.2 × 5 + 5x = 486
⇒ x = (486 – 216)/5
⇒ Speed of second train = 54 km/hr

Q8. A bus moving at a speed of 45 km/hr overtakes a truck 150 meters ahead going in the same direction in 30 seconds. The speed of the truck is?
(a) 27 km/hr
(b) 24 km/hr
(c) 25 km/hr
(d) 28 km/hr

S8. Ans.(a)
Sol. Let the speed of truck is = x km/hr
Their relative speed in same direction
= (45 – x) km/h (Here (45 – x) has been written because bus crosses the truck which is running 150 metres ahead from it. i.e. Truck speed will be lower than that of bus)
According to the question,
Time = (total distance)/(total speed)
=150/((45-x)×5/18)=30
=(150×18)/((45-x)×5)=30
= x = 27 km/h
So speed of the truck is 27 km/h

Q9. A passenger train 150m long is traveling with a speed of 36 km/hr. If a man is cycling in the direction of the train at 9 km/hr., the time taken by the train to pass the man is?
(a) 10 sec
(b) 15 sec
(c) 18 sec
(d) 20 sec

S9. Ans.(d)
Sol. Their relative speed in same direction = 36 – 9 = 27 km/h
⇒ Time, taken by train to cross the man = 150/27[Distance/Speed=time]
⇒ Time=150/(27×5/18)   [1 km/hr  =5/18 m/s]
⇒ Time = (150 × 18)/(27 × 5)
⇒ Time = (30 × 2)/3
  Time = 20 second

Q10. The speed of two trains is in the ratio 6 : 7. If the second train runs 364 km is 4 hours, then the speed of the first train is?
(a) 60 km/hr
(b) 72 km/hr
(c) 78 km/hr
(d) 84 km/hr

S10. Ans.(c)
Sol. Given
1st  : 2nd train
Ratio of speed of trains =  7
⇒ Second train covers 364 kms in 4 hours then its speed = 364/4 = 91km/hr
⇒ In the question it is given that speed of the second train = 7 Ratio
but actual speed = 91 km/hr
i.e. 7 ratio → 91
⇒ 1 ratio → 13 km.
Therefore,
Speed of the first train is
⇒ 6R ⇒ 6 × 13 = 78 km/hr

Q11. On mixing two classes A and B of students having average marks 25 and 40 respectively. The over all average obtained is 30. Find the ratio of the students in the class A and B?
(a) 2 : 1
(b) 5 : 8
(c) 5 : 6
(d) 3 : 4

S11. Ans.(a)
Sol. By alligation,


Let, no. of students in class A be x
No. of students in class B be y
⇒ Total marks of class A = 25x
Total marks of class B = 40y
∴ Total marks of A + B = (25x + 40y)
⇒ Now on mixing the two class no. of students become (x + y) and Average = 30
Total marks = 30 (x + y)
⇒ ∴ 25x + 40y = 30 (x + y)
25x + 40y = 30x + 30y
x = 2y
x : y = 2 : 1

Q12. At present the ratio of the age of Maya and Chhaya is 6 : 5 and fifteen years from now, the ratio will get changed to 9 : 8. Maya’s present age is?
(a) 21 years
(b) 24 years
(c) 30 years
(d) 40 years

S12. Ans.(c)
Sol.

3 units = 15 years
1 unit = 5 years
∴ present age
Maya → 6 × 5 = 30 years
Chhaya → 5 × 5 = 25 years

Q13. What must be added to each term of the ratio 7 : 11, so as to make it equal to 3 : 4?
(a) 8
(b) 7.5
(c) 6.5
(d) 5

S13. Ans.(d)
Sol. A/B=7/11 Given
Let x be added to both A & B
⇒(7+x)/(11+x)=3/4
Cross multiply the equation
28 + 4x = 33 + 3x
x = 5

Q14. The sum of three numbers is 68. If the ratio of the first to the second be 2 : 3 and that of the second to the third be 5 : 3, then the second number is?
(a) 30
(b) 58
(c) 20
(d) 48

S14. Ans.(a)
Sol. A + B + C = 68
A : B : C
2 : 3
      5 : 3

A : B : C = 10: 15 : 9

10x + 15x + 9x = 34x
34x = 68
x = 2
∴ A = 2 × 10 = 20
B = 2 × 15 = 30
C = 9 × 2 = 18

Q15. Two number are in the ratio 4 : 5 and their L.C.M. is 180. The smaller number is?
(a) 9
(b) 15
(c) 36
(d) 45


S15. Ans.(c)
Sol.
A : B
4 : 5
4x : 5x
LCM = 4 × 5 × x = 20x
20x = 180
x = 9
∴ Smallest number is = 4 × 9 = 36
Largest number is = 5 × 9 = 45

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