Most Important topics related Maths Questions | 6th December 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 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 2019, DSSSBKVS, STET Exam.

Q1. 4/6-1/8=3/8,6/7-2/9=4/2 The above represents the work of student. If this error pattern continues, the student’s answer to 5/11-2/7 will be
4/6-1/8=3/8,6/7-2/9=4/2  उपरोक्त छात्र के काम को दर्शाता है. यदि यह त्रुटि पैटर्न जारी रहता है, तो छात्र का उत्तर होगा 5/11-2/7
(a) 3/4
(b) 3/7
(c) 2/18
(d) 7/18

Q2. Find the value of 547527/82, if 
547.527/0.0082 =x.
547527/82 मान ज्ञात कीजिए, यदि 547.527/0.0082 =x
(a) 10x
(b) 100x
(c) x/100
(d) x/10

Q3. Shown here are expressions given to Seema, Anees, Asha and Tessy with their answers
सीमा, अनीस, आशा और टेसी को दी गई अभिव्यक्तियाँ उनके उत्तरों सहित यहाँ दर्शाई गई हैं
Seema 4×1+8÷2=8
Anees 6+4÷2-1=4
Asha 9+3×2-4÷2=10
Tessy 27÷3-2×3=21
सीमा 4×1+8÷2=8
अनीस 6+4÷2-1=4
आशा 9+3×2-4÷2=10
टेसी 27÷3-2×3=21
Who has got the correct answer?
किसका उत्तर सही है?
(a) Asha / आशा
(b) Tessy/ टेसी
(c) Seema/ सीमा
(d) Anees/ अनीस

Q4. If A=3/4÷5/6, B=3÷[(4÷5)÷6]
C = [3÷(4÷5)]÷6 and
D = 3÷4(5÷6), then
यदि  A=3/4÷5/6, B=3÷[(4÷5)÷6]
C = [3÷(4÷5)]÷6 और
D = 3÷4(5÷6), तो
(a) A and D are equal/ A और D समान हैं
(b) All are equal/ सभी समान हैं
(c) A and B are equal/ A और B समान हैं
(d) A and C are equal/ A और C समान हैं

(a) 4
(b) 10
(c) 2
(d) 3
Q6. If 2/3,23/30,9/10,11/15 and 4/5 are written in ascending order, then the fraction in the middle most will be 
यदि  2/3,23/30,9/10,11/15 और 4/5 आरोही क्रम में लिखे गए हैं, तो मध्य का भिन्न होगा
(a) 23/30
(b) 4/5
(c) 2/3
(d) 11/15
Q7. When half of a number is increased by 15, the result is 39. The sum of digits of the original number is 
जब किसी संख्या के आधे भाग को 15 से बढ़ाया जाता है, तो परिणाम 39 निकलता है. मूल संख्या के अंकों का योग होगा
(a) 6
(b) 7
(c) 9
(d) 12
a और b  अंकों का योग है
(a) 13
(b) 12
(c) 15
(d) 14
Q9. The value of .001+1.01+0.11 is 
.001+1.01+0.11 का मान होगा
(a) 1.013
(b) 1.121
(c) 1.111
(d) 1.101
(a) 6
(b) 8
(c) 9
(d) 12
Q11. What should be subtracted from (-5)/7  to get (-2)/3 ?
(-5)/7  में से क्या घटाने पर (-2)/3  प्राप्त होगा?
(a) 29/21
(b) (-1)/21
(c) (-29)/21
(d) 1/21
Q12. The difference between the greatest and smallest fraction amongst 6/7,8/9,9/10 and 7/8 is 
6/7,8/9,9/10 and 7/8 में से सबसे बड़ा और सबसे छोटे भिन्न का अंतर होगा
(a) 1/72
(b) 2/63
(c) 3/70
(d) 1/56
 (a) 1492.16×10^8
(b) 1.49216×10^11
(c) 149216×10^9
(d) 14.9216×10^10
Q14. The sum of all possible values of a, for which the 4-digit number 547a is divisible by 3, is 
a के सभी संभावित मानों का योग, जिसके लिए 4-अंक की 547a संख्या 3 से विभाज्य है
(a) 7
(b) 15
(c) 10
(d) 13
Q15. Which one of the following statements is correct?
निम्नलिखित में से कौन सा कथन सही है?
(a) sum of two prime numbers is always a prime number/ दो अभाज्य संख्याओं का योग हमेशा अभाज्य संख्या होती है 
(b) 1 is the smallest prime number/1 सबसे छोटी अभाज्य संख्या है
(c) a composite number can be odd / समग्र संख्या विषम हो सकती है
(d) there is no even prime number/ कोई अभाज्य संख्या सम नहीं होती
Solutions
S1. Ans.(a)
Sol. The student has done the following error
4/16-1/8=(4-1)/(16-8)=3/8
6/7-2/9=(6-2)/(7-9)=4/2
5/11-2/7=(5-2)/(11-7)=3/4
S2. Ans.(d)
Sol. Given, x=547.527/0.0082
x=10×(547527/82)
547527/82=x/10
S3. Ans.(c)
Sol. Seema 4×1+8÷2=4+4=8
Anees 6+4÷2-1=6+2-1
=8-1=7≠4
Asha 9+3×2-4÷2=9+6-2
=15-2=13≠10
Tessy 27÷3-2×3= 9-6=3≠21
Hence, answer of Seema is correct.
S4. Ans.(a)
Sol. A=3/4÷5/6=3/4×6/5=9/10
B=3÷[(4÷5)÷6]
=3÷[4/5÷6]=3÷[4/30]=3×30/4=45/2
C=[3÷(4÷5)]÷6
=[3÷4/5]÷6=(3×5/4)÷6 
=15/4÷6=15/24=5/8
D= 3÷4(5÷6)=3÷4×5/6=3÷20/6
=3×6/20=18/20=9/10
Hence , A and D are equal.
S6. Ans.(a)
2/3,23/30,9/10,11/15,4/5
⇒20/30,23/30,27/30,22/30,24/30
∵Required fraction=23/30
S7. Ans.(d)
Sol. Let number be x.
According to the question, 
x/2+15=39⇒x=48
∵Required sum of digits is 4+8=12
S8. Ans.(d)
Sol.  
Here, taking carry at units place 
i.e. 17-9=8
Now, at 10’s place taking carry
14-a=b⇒a+2=14
Now, at 100’s place 9 becomes 8 because of previous carry.
∵                     a+b=14
S9. Ans.(b)
Sol. The value of 0.001+1.01+0.11 On adding, 
S11. Ans.(b)
Sol. Let x should be subtracted from 
((-5)/7)  Then,(-5)/7-x=(-2)/3
⇒ x=(-5)/7+2/3=(-15+14)/21=(-1)/21
S12. Ans.(c)
Sol. The difference between the greatest and smallest fraction 
=9/10-6/7=(63-60)/70=3/70
S14. Ans.(b)
Sol. 5+4+7+a=16+a
For 16+a to be divisible by 1, the possible values of a are 2, 5 and 8.
∵ Sum of these values is 15.
S15. Ans.(c)
Sol. A composite number can be odd.
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