SSC CHSL Questions Asked 15th March 2018 – Tier I

SSC CHSL Questions Asked 15th March 2018 – Tier I (All Shifts)

Here are the subject-wise questions asked for SSC CHSL Questions Asked 15th March 2018 all shift:

SSC CHSL Questions Asked 15th March 2018 – Quantitative Aptitude

1. A is 40% more than B, then A is less than B by?
2.  In a PQR right angle triangle, the value of angle R=90o then find the value of:
   
3. Find
   
4. Mass = 25 Kg, Height = 5 m, then find the value of potential energy?
5. In a right angle triangle DEF, having right angle at E & sec D = 25/7. Find cosec F?
6. If the total surface area of the cube is 337.5. Find it’s radius?
7. If circumference is 22, then area of circle is?
8. 39>?<75
9. If
   
10. In PQR traingle with 90 degree at Q, tan P = 3/4 and PQ = 2.5, find PR?

SSC CHSL Questions Asked 15th March 2018 – General Awareness

1. Name the highest peak.
2. Name the largest city of Myanmar.
3. In UMANG App, G stand for?
4. Which country has used helicopter for its coast security?
5. Who decides MSP?
6. Who imposed Jizya Tax for the 1st time in Gujarat?
7. Which state Chief Minister received Peacock Award?
8. India’s 1st solar mission to be launched in 2019.
9. Valency depend on?
   a) Electron gain             b) Electron donated                c) Electron shared            d) All a, b, c
10. Name the part of cell which takes in/out (Based on biology)?
11. 1 Qs from Photosynthesis?
12. Which river flows from south to north?
13. Indian Navy purchased the ship from which country on 31st October 2017?
14. The article 352, 356, 360 is related to?
15. Burzil pass is located in?
16. Ancient cultivation of Sri Lanka is called?
17. Kautilya was Prime Minister of which emperor?

SSC CHSL Questions Asked 15th March 2018 – English

1. Antonym – Candid, Evince
2. Synonym – Glib, Indissoluble
3. Similar meaning – Debonair, Rivet
4. Spelling Test – Vacuum

SSC CHSL Questions Asked 15th March 2018 – Reasoning 

1. 18@12 = 3
   2@14 = -6
   4@4 = 0
   2@16 = ?
2. WARD = 44, LOAD = 30, then BRIT = ?

SSC CHSL Questions Asked 15th March 2018 in 3rd Shift – Quantitative Aptitude

Q1) A product of Rs. 1200 purchased at 25% discount if sold at Rs. 1200, then how much is the profit?
A) 33%
B) 25%
C) 40%
D) 35%
Ans: A
Solution: Cost Price of product after 25% discount on marked price = 1200 – (25/100) × 1200 = 900.
Selling Price = 1200
Therefore, Profit = [(Selling Price – Cost Price)/Cost Price] × 100 = [(1200 – 900)/900] × 100 = 33.33%

Q2) Find the area of circle whose circumference is 44 cm? (Given π =22/7)
A) 108 sq. cm
B) 196 sq. cm
C) 154 sq. cm
D) 144 sq. cm
Ans: C
Solution: Circumference of circle = 2πr = 44
2 × 22/7 × r = 44
r = 7 cm
Area of circle = πr2, where r is the radius of circle.
Therefore, area of circle = 22/7 × (7)2 = 154 sq. cm

Q3) The ratio of X: Y: Z is 15: 3: 8. If the share of Z is 3680 more than Y, then find the share of X?
A) 10840
B) 11820
C) 11040
D) 12120
Ans: C
Solution: X: Y: Z = 15: 3: 8
Z = Y + 3680
Y/Z = 3/8
Y/(Y + 3680) = 3/8
8Y = 3Y + 11040
Y = 2208
X/Y = 15/3 = 5
Therefore, X = 2208 × 5 = 11040

Q4) If the volume of hemisphere is 5743.33 cubic cm. Find diameter of hemisphere?
A) 18 cm
B) 26 cm
C) 28 cm
D) 12 cm
Ans: C
Solution: Volume of Hemisphere = 2/3(πr3), where r is the radius of hemisphere.
2/3(πr3) = 5743.33
(2/3) × (3.14) × r3 = 5743.33
r3 = 2743.62
r = 14 cm
Therefore, diameter of sphere = 28 cm.

Q5) The side of a square is increased by 20%. Find the percentage increase in circumference of square?
A) 16%
B) 20%
C) 18%
D) 25%
Ans: B
Solution: Let the side of square be a.
Circumference of square = 4a
New length of the side after 20% increment = a + 20a/100 = 1.2a
New Circumference = 4 × 1.2a = 4.8a
Therefore, percentage increase = (.8a/4a) × 100 = 20%

Q6) A and B completes the work in 20 days when working together. If B finished the work alone in 36 days, then in how many days will A finish the work alone?
A) 30 days
B) 35 days
C) 40 days
D) 45 days
Ans: D
Solution: Work completed by A and B in 1 day = 1/20
Work completed by B alone in 1 day = 1/36
Therefore, work completed by A alone in 1 day = 1/20 – 1/36 = 1/45
So, A will finish the work alone in 45 days.

Q7) Find the principal if the difference between compound and simple interest is Rs. 8 at 4% per annum for 2 years?
A) Rs. 4500
B) Rs. 5400
C) Rs. 5000
D) Rs. 4800
Ans: C
Solution: Simple Interest = (P × R × T)/100, where P is the principal, R is the rate of interest and T is the time duration.
Simple Interest = (P × 4 × 2)/100 = 2P/25 = 0.08P
Compound Interest = P (1 + R/100)n – P, where P is the principal, R is the rate of interest and n is the time duration
Compound Interest = P (1 + 4/100)2 – P = 1.0816P – P = .0816P
Given, .0816P – .08P = 8
Therefore, P = 5000

Q8) If y/x = 1/3, then find the value of (x2 + y2)/(x2 – y2)?
A) 4/3
B) 5/4
C) 5/7
D) 3/2
Ans: B
Solution: (x2 + y2)/(x2 – y2) = [1 + (y/x)2 ]/ [1 – (y/x)2] Given, y/x = 1/3
Therefore, [1 + (y/x)2 ]/ [1 – (y/x)2] = [1 + (1/3)2 ]/[1 – (1/3)2 ] = 5/4

Q9) The parallel sides of a trapezium are 17 cm and 8 cm respectively. If its area is 25 square cm, then what is the height of trapezium?
A) 2 cm
B) 1 cm
C) 3 cm
D) 4 cm
Ans: A
Solution: The area of a trapezium is the product of half of sum of parallel sides [(L1 + L2)/2] and the distance (h) between them.
Here the parallel sides are 17 cm and 8 cm and the area is 25.
Hence, [(17 + 8)/2] × h = 25
on solving, we get height of trapezium is 2 cm.

Q10) A train running at a speed of 45m/s crossed a bridge 550 m longer in 25 seconds. Find the length of train?
A) 575 m
B) 525 m
C) 560 m
D) 570 m
Ans: A
Solution: Speed = Distance/Time
S = (L1 + L2)/T
Speed = 45m/s
Length of bridge, L1 = 550 m and Length of Train, L2 = x
Time = 25 seconds
45 = (550 + x)/25
x = 1125 – 550 = 575 m
Therefore, length of train = 575 m.

Q11) If a2 + b2 = 80 and ab = 32, then find the value of a + b?
A) 14
B) 12
C) 11
D) 13
Ans: B
Solution: a2 + b2 = 80 and ab = 32
(a + b)2 = a2 + b2 + 2ab = 80 + 64 = 144
Therefore, a + b = 12

Q12) If triangle ABC is similar to triangle DEF, AB = 4cm and DE = 9cm. Find the area of triangle ABC and DEF?
A)
B)
C)
D)
Ans: B

Q13) If √343 + 6√7 = X, then find the value of √7 + 3√7?
A) 4X/13
B) 3X/5
C) 2X/13
D) 4X/5
Ans: A
Solution: X = √343 + 6√7
X = 7√7 + 6√7
X = 13√7
Therefore, √7 + 3√7 = 4√7 = 4X/13

Q14) If √20 + √125 = X, then find the value of √45 + √5?
A) 4X/9
B) 4X/7
C) 3X/8
D) 2X/5
Ans: B
Solution: X = √20 + √125
X = 2√5 + 5√5
X = 7√5
Therefore, √45 + √5 = 3√5 + √5 = 4√5 = 4X/7

Q15) Initially the ratio of water and milk is 4: 5. After adding 7 liter of milk in it, the new ratio will be 2: 3. Find the quantity of water initially?
A) 22 liter
B) 24 liter
C) 35 liter
D) 28 liter
Ans: D
Solution: Let the ratio of water and milk be 4x and 5x.
After adding 7 liter of milk, new ratio will be 2: 3.
So, 4x/ (5x + 7) = 2/3
12x = 10x + 14
x = 7
So, quantity of water = 4x = 28 liter

Q16) Find the smallest number among 3/4, √ (9/49), .43, .72?
A) 3/4
B) √ (9/49)
C) 0.43
D) 0.7^2
Ans: B
Solution: 3/4 = 0.75
√ (9/49) = 3/7 = 0.428 ~ 0.43
0.72 = 0.49
Therefore, the smallest number is √ (9/49).

Q17) Find the value of angle BAC, if angle C = 15̊
A) 60̊
B) 75̊
C) 105̊
D) 90̊

Ans: B

Q18) Find the value of Cosec C + 1/3, if angle A = 45̊ and angle B = 90̊?
A) √3 + 1/3
B) √2 + 1/3
C) 2/√3 + 1/3
D) √3/2 + 1/3

Ans: B
Solution: Cosec C = cosec 45̊ = √2
Therefore, Cosec C + 1/3 = √2 + 1/3

Q19) Find the value of 2A + B, if AC = BC and angle ACB = 45̊?
A) 150̊
B) 180̊
C) 135̊
D) 120̊

Ans: B
Solution: Since we know AC = BC (isosceles triangle), this implies angle ACB = angle BAC = 45̊.
So angle B = 180̊ – 45̊ – 45̊ = 90̊
Therefore, 2A + B = 180̊

Q20) In a triangle PQR, if cos P = 5/13 and PQ = 2.5 cm, then find the length of PR.
A) 4.5 cm
B) 6.5 cm
C) 5.5 cm
D) 7.5 cm

Ans: B
Solution: Based on diagram, cos P = PQ/PR = 5/13 and PQ = 2.5
Therefore, PR = 6.5 cm

SSC CHSL Questions Asked 15th March 2018 3rd shifts – Data Interpretation

Directions: (21 – 23): Below pie-chart represents the number of seats of MLA in a given state.

Q21) Find the state which has second highest number of seats?
A) E
B) F
C) C
D) A
Ans: B
Solution: State F has the second highest number of seats.

Q22) Find the total number of seats of MLA among all the given states?
A) 430
B) 350
C) 400
D) 380
Ans: C
Solution: Total number of seats = 100 + 20 + 60 + 50 + 80 + 90 = 400

Q23) Find the average number of seats of MLA in states A, E and F together?
A) 90
B) 60
C) 80
D) 70
Ans: A
Solution: Total number of MLA seats in A, E and F together = 100 + 80 + 90 = 270
Therefore, average number of seats of MLA in states A, E and F together = 270/3 = 90

Q24) Find the angle of state F for the seats of MLA among total seats?
A) 108̊
B) 90̊
C) 72̊
D) 81̊
Ans: D
Solution: Seats of MLA in state F = 90
Total seats of MLA in all states = 400
Therefore, required angle = (90/400) × 360̊ = 81̊