Important Mathematics Questions for CTET 2020 : 4th February 2020 |_00.1
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Important Mathematics Questions for CTET 2020 : 4th February 2020

Important Mathematics Questions for CTET 2020 : 4th February 2020 |_20.1

Q1. Find the number of ways expressing 48 as a product of two factors?
निम्न में से कितने तरीकों से, दो गुणनखंड़ों के प्रयोग से गुणनफल के रूप में 48 प्राप्त किया जा सकता है?
(a) 6
(b) 5
(c) 4
(d) 8

Q2. What is the sum of the divisors of 120?
120 के भाजक का योग क्या है?
(a) 120
(b) 360
(c) 240
(d) 480

Q3. What is the remainder when the product 2658×9437×2246 is divided by 8?
2658 × 9437 × 2246 को 8 से विभाजित करने पर शेषफल क्या होगा?
(a) 4
(b) 6
(c) 2
(d) 7

Q4. What would be the sum of 1+4+7+10+—– up to the 20th term?
20 वें पद तक 1+4+7+10+—–का योग क्या होगा?
(a) 560
(b) 540
(c) 590
(d) 500

Important Mathematics Questions for CTET 2020 : 4th February 2020 |_30.1(a) 9
(b) 6
(c) 8
(d) 7

Q6. The difference between the place value and the face value of 4 in the numeral 643296
संख्यांक 643296 में 4 के स्थानीय मान और अंकित मान के बीच का अंतर:
(a) 39999
(b) 39996
(c) 40000
(d) 40004

Important Mathematics Questions for CTET 2020 : 4th February 2020 |_40.1(a) y
(b) x
(c) z
(d) None of these/ इनमें से कोई नहीं

Important Mathematics Questions for CTET 2020 : 4th February 2020 |_50.1

(a) 1
(b) 3
(c) 45
(d) None of these/ इनमें से कोई नहीं

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Important Mathematics Questions for CTET 2020 : 4th February 2020 |_70.1

Solutions

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Important Mathematics Questions for CTET 2020 : 4th February 2020 |_90.1

S3. Ans.(a)
Sol. 2658÷8 ⇒ remainder = 2
9437÷8 ⇒ remainder = 5
2246÷8 ⇒ remainder = 6
= (2×5×6)÷8
= 60÷8
Remainder = 4

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S6. Ans.(b)
Sol. Place value of 4 = 40000
Face value of 4 = 4
Difference = 40000-4 ⇒ 39996

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