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Most Important topics related Maths Questions | 5th 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 a 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.

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(a) 77
(b) 79
(c) 58
(d) 68

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(a) 8
(b) 16
(c) 2
(d) 4

Q3. The scale of a map is given as 1:10000. On the map, a forest occupies a rectangular region measuring 10cm ×100cm. The actual area of the forest (in km²) is 
प्र3. मानचित्र का पैमाना 1: 10000 दिया गया है. मानचित्र पर, एक आयताकार वन का माप है 10सेमी ×100सेमी. वन का वास्तविक क्षेत्रफल (किमी² में) होगा
(a) 10
(b) 1
(c) 1000
(d) 100

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(a) 90.903
(b) 16.2903
(c) 15.2903
(d) 91.903

Q6. The value of a machine depreciates at the rate of 10% per year. It was purchased 3 yr ago. If its present value is Rs. 145800 for how much was it purchased?
प्र6. एक मशीन के मूल्य में प्रति वर्ष 10% की दर से अवमूल्यन होता है. यह 3 वर्ष पहले खरीदी गई थी. यदि इसका वर्तमान मूल्य रू 145800 है तो इसे कितने में खरीदा गया था?
(a) Rs. 175800
(b) Rs. 180000
(c) Rs. 200000
(d) Rs. 210000

Q7. The base of an isosceles ∆ABC is 48 cm and its area is 168 cm². The length of one of its equal sides is 
प्र7. एक समद्विबाहू ∆ABC का अधार है 48 सेमी और इसका क्षेत्रफल है 168 सेमी². इसकी समान भुजाओं की लंबाई होगी
(a) 8 cm
(b) 15 cm
(c) 17 cm
(d) 25 cm

Q8. Four times the area of the curved surface of a cylinder is equal to 6 times the sum of the areas of its bases. If its height is 12 cm, then its volume (in cm³) is 
प्र8. एक बेलन का 4 गुना वक्र पृष्ठीय क्षेत्रफल इसके आधार के क्षेत्रफल के योग के 6 गुना के समान है. यदि इसकी ऊंचाई 12 सेमी है, तो इसका आयतन (सेमी³ में) होगा
(a) 48π
(b) 384π
(c) 546π
(d) 768π

Q9. LCM of 22, 54, 135 and 198 is
प्र9. 22, 54, 135 और 198 का लघुत्तम समापवर्त्य होगा 

(a) 2² × 3³ × 5 × 11
(b) 2³ × 3² × 5 × 11
(c) 2 × 3³ × 5 × 11
(d) 2² × 3² × 5 × 11

Q10. The value of a machine which was purchased 2 yr ago, depreciates at 12% per annum. If its present value is Rs. 9680 for how much was it purchased?
प्र10. एक मशीन जो 2 वर्ष पहले खरीदी गई थी, के मूल्य में प्रति वर्ष 12% की दर से अवमूल्यन होता है. यदि इसका वर्तमान मूल्य रु 9680 है तो यह कितने में खरीदी गई थी?
(a) Rs. 12142.60
(b) Rs. 11350.50
(c) Rs. 12500
(d) Rs. 10200

Q11. In the product (x² – 2) (1 – 3x + 2x²) the sum of coefficients of x² and x is 
प्र11. (x² – 2) (1 – 3x + 2x²) के गुणनफल में x² और x के गुणांक का योग होगा
(a) 5
(b) 3
(c) 6
(d) 2

Q12. The internal base of a rectangular box is 15 cm long and 12 1/2 cm wide and its height is 7 1/2 cm. The number of cubes will be 
प्र12. एक आयताकार डिब्बे का आंतरिक आधार 15 सेमी लंबा और 12 1/2 सेमी चौड़ा है और इसकी ऊंचाई 7 1/2 सेमी है. घनों की संख्या होगी
(a) 90
(b) 120
(c) 45
(d) 60

Q13. The additive inverse of S, where S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + … + 49 – 50, is 
प्र13. S का योगज प्रतिलोम, जिसमें S= 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + … + 49 – 50, होगा
(a) –25
(b) 1
(c) 0
(d) 25

Q14. The value of ∛(-2300)×∛5290 is
प्र14. ∛(-2300)×∛5290 का मान होगा
(a) –270
(b) 230
(c) –230
(d) –529

Q15. A sum of money amounts to Rs. 4818 after 3 yr and Rs. 7227 after 6 yr on compound interest. The sum is 
प्र15. चक्रवृद्धि ब्याज पर एक राशि 3 वर्ष बाद रु 4818 और 6 वर्ष बाद रु 7227 हो जाती है. राशि होगी 
(a) Rs. 3122
(b) Rs. 3212
(c) Rs. 2409
(d) Rs. 2490

Solutions

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S3. Ans.(a)
Sol. On a map scale is 1: 10000
i.e, 1 cm = 10000cm
or 1 cm² = 10000cm × 10000cm
Now, area of forest on map
= 10 cm × 100 cm, = 1000cm²
Now, actual area of forest
=1000 × (10000) ² cm²
[∵ 1 km² = (100000) ² cm²]
Area in km² = (1000×10000×10000)/(1000×1000×100×100)
= 10km²

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S5. Ans.(b)
Sol.
We have,
829030000
In standard form, k × 10n
∴ 829030000 = 8.2903 × 10⁸
On comparing, we get k = 8.2903, n = 8
∴ k + n = 8.2903 + 8 = 16.2903

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S7. Ans.(d)
Sol. Let the length of equal sides of an isosceles triangle be a cm.
Given, base = 48 cm and area = 168 cm²
Let the altitude (height) be x.
Area of an isosceles ∆ABC
= 1/2×Base×Height
⇒ 168=1/2×48×x⇒ x=168/24=7cm

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S8. Ans.(d)
Sol. Curved surface area of a cylinder = 2πrh
Area of the base = πr²
∴ Area of the both base = 2πr²
According to the question,
4 × 2πrh = 6 × 2πr²
 ⇒ 4 × 2πr × 12 = 6 × 2 × πr × r
⇒ r = 8
∴ Volume of a cylinder = πr²h
= π × 8 × 8 × 12
= 768π cm³

S9. Ans.(c)
Sol. ∵22=2× 11
 54=3×3 × 3×2
 135=3×3 × 3×5
 198=2×3 × 3×11
So, LCM of 22, 54, 135 and 198
 =2×3×3 × 3×5×11
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 So, option (c) is correct.

S10. Ans.(c)
Sol. Let purchases price of machine was Rs. x.
∵ Price is decrease at the rate of 12% per annum
So, price of machine after one year
= x-(12×x)/100=88x/100=22x/25
Now, price of machine after 2nd year
= 22x/25-  22x/25×12/100=  22x/25-  66x/(25×25 )
According to question,
 22x/25-  66x/(25×25 )=9680 [Present price of machine]
⟹ 25×22x-66x=9680×25×25
⟹ 484x=6050000
⟹ x=6050000/484
⟹ x=12500
So, purchases price of machine was Rs. 12500.
So, option (c) is correct.

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From Eq. (i) co-efficient of x² and are (-3) and 6 respectively
So, their sum = -3 + 6 = 3
Hence, option (b) is correct.

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S13. Ans.(d)
Sol.
S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8…+ 49 – 50
S = (1 + 3 + 5 + 7 + …+ 49) + (–2 – 4 – 6 –… – 50)
⇒ sum of first n odd number = n²
Sum of first n even numbers = n( n + 1)
∴ S = (25)² – (25)(26), S = 625 – 650, S = –25
∴ Additive inverse of (–25) is – (–25) = 25

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