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Most Important topics related Maths Questions | 4th December 2019

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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. If two adjacent sides of a square paper are decreased by 20% and 40% respectively, by what percentage does the new area decrease? 
यदि एक वर्गाकार कागज की दो आसन्न भुजाएं क्रमशः 20% और 40% घटा दी जाती हैं तो नया क्षेत्रफल कितने प्रतिशत कम होगा?   
(a) 50
(b) 52
(c) 60
(d) 43

Q2. In the given figure, PS = SQ = SR and ∠SPQ = 54°. Find the measure of ∠x. 
दी गई आकृति में, PS = SQ = SR और ∠SPQ = 54°. ∠x का माप ज्ञात करें.

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(a) 72°
(b)108°
(c) 36°
(d) 54°

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(a) 225
(b) 226
(c) 216
(d) 215

Q4. If xy = 6 and x²y + xy² + x +y = 63, then the value of x² + y² is 
यदि xy = 6 और x²y + xy² + x +y = 63, तो x² + y² का मान होगा
(a) 61
(b) 69
(c) 23
(d) 55

Q5. Two positive numbers x and y are inversely proportional. If x increase by 10%, then y decreases by
दो धनात्मक संख्याएं x और y व्युत्क्रमानुपाती हैं. यदि x में 10% कि वृद्धि होती है, तो y में कमी होगी  
(a) 100/11%
(b) 10/11%
(c) 10%
(d) 2/11%

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(a) 3
(b) 4
(c) 1
(d) 2

Q8. The sum of two positive number is 63. if one number x is double the other, then the equation is 
दो धनात्मक संख्या का योग 63 है. यदि एक संख्या x दूसरे से दोगुनी है, तो समीकरण होगा
(a) (x-63)/x=2
(b) x/(63-x)=2
(c)  x/(x-63)=2
(d)  (63-x)/x=2

Q9. Which of the numbers -20,-3/4,1/2  ,10 is greater than its square?
-20,-3/4,1/2  ,10 में से कौन सी संख्या इनके वर्ग से बड़ी है?
(a) 10
(b) -20
(c) 1/2
(d) -3/4

Q10. Which of the following fractions 3/8,4/5,31/40,9/20,7/10 is greater than 1/2 and less than 3/4? 
 निम्नलिखित में से कौन सा भिन्न 3/8,4/5,31/40,9/20,7/(10 ) 1/2 से अधिक है और  3/4 से कम है?
(a) 4/5
(b) 7/10
(c) 31/40
(d) 9/20

Q11. In the figure, ABC is a triangle. Measure of ∠ ABC, (in degrees) is 
दी गई आकृति में, ABC एक त्रिभुज है. ∠ ABC का माप (डिग्री में) है

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(a) 72
(b)80
(c) 57
(d) 61

Q12. The distance between two places is 12 km. A map scale is 1:25000. The distance between the two places on the map, (in cm) is 
दो स्थानों के बीच की दूरी 12 किमी है. एक मानचित्र पैमाना 1: 25000 है. मानचित्र पर दो स्थानों के बीच की दूरी, (सेमी में) है
(a) 24
(b) 36
(c) 48
(d) 60

Q13. The number of vertices in a polyhedron which has 30 edges and 12 faces is 
एक बहुफलक जिसमें 30 किनारे और 12 पृष्ठ हैं, में शीर्षों की संख्या है 

(a) 12
(b) 15
(c) 20
(d) 24

Q14. The mean of 10 numbers is 0. If 72 and -12 are included in these numbers, the new mean will be 
10 संख्याओं का माध्य 0 है. यदि 72 और -12 इन संख्याओं में शामिल हैं, तो नया माध्य होगा
(a) 0
(b) 5
(c) 6
(d) 60

Q15. The circumference of the base of a right circular cylinder is 44 cm and its height is 15 cm. The volume (in cm³) of the cylinder is
 (use π=22/7)
एक लंब वृतीय बेलन के आधार की परिधि 44 सेमी है और इसकी ऊंचाई है 15 सेमी. बेलन का आयतन (सेमी³ में) है(use π=22/7)
(a) 770
(b) 1155
(c) 1540
(d) 2310

Solutions
हल

S1. Ans.(b)
Sol. Let us consider one side of a square be 100 cm.
∴ Area = 100 × 100 =10000cm²
Reducing the sides by 20% and 40%, we get the sides as 80 cm and 60 cm.
∴ Area = 80 × 60 = 4800cm²
∵ Reduction in area = 10000 – 48000 = 5200 cm²
∴ Percentage reduction =( 5200×100)/10000=52%
Shortcut Method
 (20+40)-(20×40)/100=60-8=52%

हल. आइए हम एक वर्ग की एक भुजा को 100 सेमी मानते हैं.
∴ क्षेत्रफल =100 × 100 =10000सेमी²
भुजाओं को 20% और 40% तक घटाने पर 80 सेमी और 60 सेमी की भुजाएं प्राप्त होती हैं
∴ क्षेत्रफल = 80 × 60 = 4800सेमी²
∵ क्षेत्रफल में कमी = 10000 – 48000 = 5200 सेमी²
∴ प्रतिशत में कमी =( 5200×100)/10000=52%
संक्षिप्त विधि
(20+40)-(20×40)/100=60-8=52%

S2. Ans.(c)
Sol. PS = SQ = SP
∴ PRQ would be isosceles ∆.
∵ ∠ SPQ = 54°
∴ ∠PRQ = 54° (Base angles of isosceles ∆ are equal)
∴ ∠ SQR = x
∴ ∠ SQR = x (∵ ∆ is isosceles)
∴ ∠ RQP = 2x
∵ in ∆ PQR, ∠ P + ∠ R ∠ Q = 180°
54° + 54° + 2x = 180
2x = 180° – 108°
 x=(72°)/2=36°

हल. PS = SQ = SP
∴ PRQ समद्विबाहु ∆ होगा.
∵ ∠ SPQ = 54°
∴ ∠PRQ = 54° (समद्विबाहु ∆  के आधारीय कोणों समान होते हैं)
∴ ∠ SQR = x
∴ ∠ SQR = x (∵ ∆ समद्विबाहु है)
∴ ∠ RQP = 2x
∵ ∆ PQR में, ∠ P + ∠ R ∠ Q = 180°
54° + 54° + 2x = 180
2x = 180° – 108°
 x=(72°)/2=36°

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S4. Ans.(b)
Sol. x²y + xy² + x + y = 63
⟹ xy(x + y)+(x + y) = 63
⟹ (x + y) + (xy + 1) = 63
⟹ (x + Y ) + (6 + 1) = 63 (∵xy = 6 is given)
⟹ x + y = 9
⟹ (x + y) ² = 9²
⟹ x² + y² + 2xy = 81
⟹ x² + y² + 2×6 = 81
⟹ x² + y² = 81-12
⟹ x² + y² = 69

S5. Ans.(a)
Sol. Required decrease = a/(a+100 )×100%
 =10/(10+100)×100%
 =100/11%

5. उत्तर.(a)
हल. आवश्यक कमी = a/(a+100 )×100%
 =10/(10+100)×100%
 =100/11%

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S8. Ans.(d)
Sol. Let one of the numbers be x.
∴ other number will be 2x.
Then, by given conditions, x + 2x = 63
63 – x = 2x = (63-x)/x=2
8. उत्तर.(d)
हल. एक संख्या को x मानें.
∴ अन्य संख्या को  2x.
फिर, दी गई स्थिति में, x + 2x = 63
63 – x = 2x = (63-x)/x=2

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S10. Ans.(b)
Sol. 1/2=0.5
 3/4= 0.75
∴ Only7/10=0.7 lies between 1/2 and 3/4

10. उत्तर.(b)
हल. 1/2=0.5
 3/4= 0.75
∴ केवल 7/10=0.7  1/2  और  3/4  इनके बीच में आता है

S11. Ans.(c)
Sol. ∠ ADB + ∠ BDC = 180
[∴ Sum of on a straight line is 180°]
∠ ADB = 180° – ∠ BDC
∠ ADB = 180° – 119° [∴ ∠BDC = 119° given]
= 61°
Now, in ∆ ABD
∠ A + ∠ABD + ∠ ADB = 180°
[∴ Sum of angles of triangles is 180°]
∠ ABD = 180° – ∠ ADB – ∠ A
= 180° – 61° – 62°
= 180° – 123° = 57°

11. उत्तर.(c)
हल. ∠ ADB + ∠ BDC = 180
[∴एक सीधी रेखा का योग है 180°]
∠ ADB = 180° – ∠ BDC
∠ ADB = 180° – 119° [∴ ∠BDC = 119° ज्ञात]
= 61°
अब,  ∆ ABD में
∠ A + ∠ABD + ∠ ADB = 180°
[∴ त्रिभुज के कोणों का योग है 180°]
∠ ABD = 180° – ∠ ADB – ∠ A
= 180° – 61° – 62°
= 180° – 123° = 57°

S12. Ans.(c)
Sol. Required difference = 12km/25000=1200000/25000
 = 48 cm

12. उत्तर.(c)
हल. आवश्यक अंतर= 12km/25000=1200000/25000
 = 48 सेमी

S13. Ans.(a)
Sol. Let the number be X
According to the question,
 x/2+15=39⟹x=48
 ∴Required sum=4+8=12

13. उत्तर.(a)
हल. एक संख्या मानें  X
प्रश्न के अनुसार,
 x/2+15=39⟹x=48
 ∴आवश्यक योग=4+8=12

S14. Ans.(b)
Sol. New mean = (10×0+72-12)/(10+2)=5

14. उत्तर.(b)
हल. नया माध्य = (10×0+72-12)/(10+2)=5

S15. Ans.(d)

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