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# CTET/ UPTET Exam | Practice Mathematics Questions | 21st October 2019

CTET/ UPTET Exam Practice Mathematics Questions-
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.With proper system, Study Notes, Quizzes, Vocabulary one can quiet his/her nerves and exceed expectations in the blink of an eye. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2019DSSSB ,KVS,STET Exam.

Q1. The perimeter of a rectangle is 60 metres. If its length is twice its breadth, then its area is

(a) 160 m²
(b) 180 m²
(c) 200 m²
(d) 220 m²
Q2. A man is walking in a rectangular field whose perimeter is 6 km. If the area of the rectangular field be 2 sq. km, then what is the difference between the length and breadth of the rectangle? Q3. The area of a rectangle is 252 cm² and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter?

(a) 64 cm
(b) 68 cm
(c) 96 cm
(d) 128 cm
Q4. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
(a) 60
(b) 50
(c) 120
Q5. A carpenter is designing a table. The table will be in the form of a rectangle whose length is 4 feet more than its width. How long should the table be if the carpenter wants the area of the table to be 45 sq ft?
(a) 6 ft
(b) 9 ft
(c) 11 ft
(d) 13 ft
Q6. The perimeter of a rectangular field is 480 metres and the ratio between the length and the breadth is 5 : 3. The area is
(a) 1350 sq. m
(b) 1550 sq. m
(c) 13500 sq. m
(d) 15500 sq. m
Q7. A rectangular farm has to be fenced on one long side, one short side and the diagonal. If the cost of fencing is Rs. 100 per m, the area of the farm is 1200 m² and the short side is 30 m long, how much would the job cost?
(a) Rs. 7000
(b) Rs. 12000
(c) Rs. 14000
(d) Rs. 15000
Q8. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, what is the area of the field?
(a) 18750 sq. m
(b) 37500 sq. m
(c) 40000 sq. m
(d) 48000 sq. m
Q9. The ratio between the length and the perimeter of a rectangular plot is 1 : 3. What is the ratio between the length and breadth of the plot?
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
Q10. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is
(a) 15360 sq. m
(b) 153600 sq. m
(c) 30720 sq. m
(d)307200 sq. m
Solutions
S1. Ans.(c)
Sol. Let the breadth of the rectangle be x metres.
Then, length of the rectangle = (2x) metres.
2 (2x + x) = 60 ⇒ 6x = 60 ⇒ x = 10.
So, length = 20 m, breadth = 10 m.
∴ Area = (20 × 10) m² = 200 m².
S2. Ans.(b)
Sol. Let the length and breadth of the field be l and b km respectively.
Then, 2 (l + b) = 6 or l + b = 3 and lb = 2.
(l – b)² = (l + b)² – 4 lb = 3² – 4 × 2 = 1 ⇒ (l – b) = 1 m
S3. Ans.(a)
Sol. Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively.
Then, 9x × 7x = 252 ⇒ 63 x² = 252 ⇒ x² = 4 ⇒ x = 2
So, length = 18 cm, breadth = 14 cm.
∴ Perimeter = 2 (18 + 14) cm = 64 cm S5. Ans.(b)
Sol. Let the width of the table be x feet. Then, length of the table = (x + 4) ft.
∴ x (x + 4) = 45 ⇒ x² + 4x – 45 = 0
⇒ x² + 9x – 5x – 45 = 0
⇒ x (x + 9) – 5 (x + 9) = 0
⇒ (x + 9) (x – 5) = 0 ⇒ x = 5
Hence, length of the table = (5 + 4) ft = 9 ft.
S6. Ans.(c)
Sol. Let the length and breadth of the field be (5x) metres and (3x) metres respectively.
Then, 2 (5x + 3x) = 480 ⇒ 8x = 240 ⇒ x = 30.
So, length = 150 m, breadth = 90 m.
∴ Area of the field = (150 × 90) sq. m = 13500 sq. m. Sharing is caring!

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