Maths Questions for CTET Exam : 31st August 2018 (Solutions)_00.1
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Maths Questions for CTET Exam : 31st August 2018 (Solutions)

Maths Questions for CTET Exam : 31st August 2018 (Solutions)_40.1

Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB Exam & STET Exam.
Q1. The volume of the largest right circular cone that can be cut out of a cube, whose edge is 9 cm, is
(a) 2673/14 cm³
(b) 2683/14 cm³
(c) 2693/14 cm³
(d) None of the above
Q2. A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in the water completely. The change in the level of water in the jar is
(a) 15/16 cm
(b) 16/25 cm
(c) 16/75 cm
(d) None of these
Q3. The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity.
(a) 8172.64 cm³
(b) 8171.43 cm³
(c) 8070 cm³
(d) None of the above
Q4. One cubic meter piece of copper is melted and recast into a square cross section bar, 36 m long. An exact cube is cut off from this bar. If 1 m³ of copper cost is Rs. 108, the cost of this cube is
(a) 30 paise
(b) 40 paise
(c) 50 paise
(d) None of the above
Q5. The dimensions of a field are 12 m × 10 m. A pit 5 m long, 4 m wide and 2 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field. The level of the field is raised by
(a) 30 cm
(b) 35 cm
(c) 38 cm
(d) 40 cm
Maths Questions for CTET Exam : 31st August 2018 (Solutions)_50.1
Q7. The perimeter of two squares are 40 cm and 32 cm. The perimeter of a third square whose area is the difference of the area of two square is
(a) 24 cm
(b) 42 cm
(c) 40 cm
(d) 20 cm
Q8. The sides of a triangle are 3cm, 4 cm and 5 cm. The area of the triangle formed by joining the mid points of this triangle is
(a) 3 cm²
(b) 6 cm²
(c) 3/2 cm²
(d) 3/4 cm²
Q9. The difference between the areas of two squares drawn on two line segment of different length is 32 cm², the length of the greater line segment, if one is longer than the other by 2 cm is
(a) 7 cm
(b) 9 cm
(c) 11 cm
(d) 16 cm
Q10. In the given figure, OABC is a rhombus where three vertices A, B and C lies on the circle of radius 10 cm. The area of rhombus is
(a) 50√3 cm²
(b) 100√3 cm²
(c) 75√3 cm²
(d) 125√3 cm²
Solutions:
Maths Questions for CTET Exam : 31st August 2018 (Solutions)_60.1
Maths Questions for CTET Exam : 31st August 2018 (Solutions)_70.1
S3. Ans.(b)
Sol. Capacity of bucket = Volume of frustum of cone
=πh/3 [R²+r²+Rr]
=22/7×24/3 [(15)² +5²+15×5]
=176/7(325)
= 8171.43 cm³
S4. Ans.(c)
Sol. Volume of bar = Volume of piece of copper = 1 m³
Length of bar = 36 m
∴ Area of square cross section of bar
=(Volume )/Length=1/36 m^2
∴ Side of square cross section = √(1/36)=1/6  m
Volume of exact cube which is cut off from bar
=(1/6)³=1/6×1/6×1/6=1/216  m³
Cost of cube =1/216×108=1/2= 50 paise
S5. Ans.(d)
Sol. Area of the field = Length × Breadth
= 12 × 10 = 120 m²
Area of the pit’s surface = 5 × 4 = 20 m²
Area of which the earth is to be spread
= 120 – 20 = 100 m²
Volume of earth dug out
= 5 × 4 × 2 = 40 m³
∴ Level of field raised = 40/100
= 2/5 m = 1/2 × 100 = 40 cm
S6. Ans.(a)
Sol. Volume of cone = 1/3 πr²h
=1/3×22/7×1×1×7
=22/3  cm³
Volume of conical block
= 10 × 5 × 2 = 100 cm²
∴ Wastage = 100 – 22/3 = 278/3 cm³
∴ Wastage percentage
=278/3×100/100=92 2/3%
S7. Ans.(a)
Sol. Let P₁ = 40 cm and P₂ = 32 cm
⇒ a₁ = 10 cm and a₂ = 8 cm
∴ A₁ = 100 cm² and A₂ = 64 cm²
∴ Area of third square
= 100 – 64 = 36 cm²
⇒ Side of third square = 6 cm
⇒ Perimeter of third square
= 4 × 6 = 24 cm
Maths Questions for CTET Exam : 31st August 2018 (Solutions)_80.1
S9. Ans.(b)
Sol. Let the side of one square be x cm.
∴ The side of another square
= (x + 2) cm
According to the question,
(x + 2) ² – x² = 32
⇒ (x + 2 – x) (x + 2 + x) = 32
⇒ 2 (2x + 2) = 32
⇒ 2 (x + 1) = 16 ⇒ x = 7 cm
∴ Length of longer side = 7 + 2 = 9 cm
S10. Ans.(a)
Sol. Here, OA = OB = OC = 10 cm
Also, AC = 2CP
and CP = √((OC)²-(OP)² )
=√75=5√3 cm
Also, AC = 2CP
∴ AC = 10√3 cm
∴ Area of the rhombus OABC
=1/2×OB×AC
=1/2×10×10√3=50√3  cm²
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