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
- A is 40% more than B, then A is less than B by?
- In a PQR right angle triangle, the value of angle R=90o then find the value of:
- Find
- Mass = 25 Kg, Height = 5 m, then find the value of potential energy?
- In a right angle triangle DEF, having right angle at E & sec D = 25/7. Find cosec F?
- If the total surface area of the cube is 337.5. Find it’s radius?
- If circumference is 22, then area of circle is?
- 39>?<75
- If
- 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
- Name the highest peak.
- Name the largest city of Myanmar.
- In UMANG App, G stand for?
- Which country has used helicopter for its coast security?
- Who decides MSP?
- Who imposed Jizya Tax for the 1st time in Gujarat?
- Which state Chief Minister received Peacock Award?
- India’s 1st solar mission to be launched in 2019.
- Valency depend on?
a) Electron gain b) Electron donated c) Electron shared d) All a, b, c
- Name the part of cell which takes in/out (Based on biology)?
- 1 Qs from Photosynthesis?
- Which river flows from south to north?
- Indian Navy purchased the ship from which country on 31st October 2017?
- The article 352, 356, 360 is related to?
- Burzil pass is located in?
- Ancient cultivation of Sri Lanka is called?
- Kautilya was Prime Minister of which emperor?
SSC CHSL Questions Asked 15th March 2018 – English
- Antonym – Candid, Evince
- Synonym – Glib, Indissoluble
- Similar meaning – Debonair, Rivet
- Spelling Test – Vacuum
SSC CHSL Questions Asked 15th March 2018 – Reasoning
- 18@12 = 3
2@14 = -6
4@4 = 0
2@16 = ?
- 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