SSC CGL 2018 NEWS
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Staff Selection Council (SSC) conducts Combined Graduate Level (CGL) entrance exam each year in order to recruit candidates in group B and C posts in various departments of Central government.
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Questions for Quantitative aptitude comprise a substantial portion in SSC CGL exam. Just like logical reasoning, it has a weightage of 50 marks. Weightage of different topics in SSC CGL over the years has been tabulated below:
|Topic||Weightage (Number of questions)||Topic||Weightage (Number of questions)|
|Simplification||1-2||Profit and loss||1-2|
|Ratio and Proportion||1-2||Mensuration||2-3|
|Problems on ages||0-1||Time and Work||1-2|
|Speed time and distance||1-2||Algebra||2-3|
The table provides a fair idea regarding the weightage of quantitative aptitude questions in previous years and one can expect the similar weightage in SSC CGL exam 2017 as well.
Check SSC CGL Syllabus
In order to secure good marks in quantitative aptitude, candidates can follow the tips and tricks mentioned below.
As one knows, mensuration is all about calculations, so one should be extremely good with calculations and remember all tables, square roots, cube roots and avoid spending so much time on them.
It is important that one is aware of all the basics of different chapters. One should memorize all basic theories and formulas and solve questions using them. Avoid opting for shortcut methods.
One is advised to make concise notes with keywords of important things taught in exam and practice them regularly.
Students go through quite a lot many difficulties during an exam and the list of challenges faced by students while appearing for SSC CGL exam is as follows:
Sample questions of quantitative aptitude, mensuration have been mentioned below:
Question 1. The length of rectangular plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per meter is Rs. 5300, what is the length of the plot?
Sol. Cost = Rate × Perimeter
⇒ Perimeter = 5300/26.50 = 200
⇒ 2 (L + B) = 200
⇒ 2 (L + L – 20) = 200
⇒ L – 10 = 50
⇒ L = 60 m
Question 2. The length and breadth of the floor of the room are 20 by 10 feet respect. Square tiles of 2 feet are to be laid. Black tiles are laid in first row on all sides, white tiles on 1/3rd of remaining sides and blue tiles on the rest. How many blue tiles are required?
Sol:- Side of a tile = 2 feet ⇒ Area of 1 tile = 22 = 4 sq ft. ----- (1)
Length left after lying black tile on 4 sides = 20 – 4
⇒ Area left after black tiles = (20 – 4)×(10-4) = 96 sq. ft.
Area left after white tiles = 2/3×96 = 64 sq. ft.
⇒ Area for blue tiles = 64 sq. ft.
Number of blue tiles = 64/4 = 16 (using 1)
Question 3. A cow is tethered in the middle of the field with a 14 ft long rope. If the cow grazes 100 sq. ft. per day, then approximate time taken to graze the whole field?
Sol: Here, the rope of cow is like radius
Area = π (14)2
No. of days = (Area of field)/Rate of cow = π (14)2/100 = 6 days(approx)
Question 4. A circle and a rectangle have same perimeter. Sides of rectangle are 18 by 26 cm. What is the area of circle?
Sol:- 2πr = 2 (18 + 26) ⇒ r = 14 cm
Area = π r2 = 616 cm2
Question 5. What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle?
Area of triangle = ½ × L × B
Area of rectangle = L × B
Area of a rectangle: area of a triangle = L × B: ½ × L × B = 2: 1
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