Math Quiz for CTET Exam 2016_00.1
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Math Quiz for CTET Exam 2016

Math Quiz for CTET Exam 2016_40.1
Q1. If x=2/3 and y=3/4, then a rational number between (x-y)^(-1) and (x^(-1)-y^(-1) ) is
(a) 2/3
(b) -71/2
(c) -71/12
(d) 1/6
Sol. 
Given, x=2/3 and y=3/4
∴ x^(-1) = 1/x= 3/2
And y^(-1) = 1/y= 4/3
Now, (x-y)^(-1) = (2/3-3/4)^(-1)
=(8/12-9/12)^(-1)= ((-1)/12)^(-1)
= -12 = (-12 × 12)/12 = (-144)/12
and x^(-1)-y^(-1) = 3/2-4/3 = (3 × 3)/(2 × 3)-(4 × 2 )/(3 × 2)
= 9/6-8/6 = 1/6 = (1 × 2)/(6 × 2) = 2/12
Here, (-71)/12 lie between -144/12 and 2/12

Q2. If 3(5x-7)-4(8x-13)=2(9x-11)-17, then the value of (7x – 5)/(11x –  9) is
(a) -14/31
(b) -9/13
(c) 9/13
(d) 14/31
Sol. 
3(5x-7)-4(8x-13) = 2(9x-11)-17
⇒ (15x-21)-(32x-52) = (18x-22)-17
⇒ (15x-32x-18x)-21+52+22+17=0
⇒ -35x+70=0
⇒ -35x=-70
⇒ x = 70/35 = 2
∴ (7x – 5)/(11x – 9) = (7 (2)- 5)/(11(2)-9)
=(14 – 5)/(22 – 9) = 9/13

Q3. If (√1058  × √648)/x=18, then the value of x is
(a) 36
(b) 46
(c) 58
(d) 23
Sol.
 (√1058  × √648)/x=18
⇒ x=(√1058  × √648)/18 = (√1058  × √648)/(18)^2
=(√1058  × √648)/√324 = √1058 = √2 = √(1058×2)
=√2116 = 46  [∴√a×√b=√(a×b)]
∴ x = 46

Q4. The value of m such that
(2/9)^3×(4/81)^(-6)=(2/9)^(2m-1) is
(a) –6
(b) 0
(c) –4
(d) 3
Sol.
 (2/9)^3×(4/81)^(-6) = (2/9)^(2m-1)
 ⇒ (2/9)^3×(81/4)^6 = (2/9)^(2m-1) [As (a/b)^(-1)=b/a]  
⇒ (2/9)^3×[(81/4)^2 ]^3 = (2/9)^(2m-1)
⇒ [2/9×(81/4)^2 ]^3 = (2/9)^(2m-1)
⇒ (729/8)^3 = ( 2/9)^(2m-1)
⇒ [(9/2)^3 ]^3 = (2/9)^(2m-1)⇒[(2/9)^(-3) ]^3 = (2/9)^(2m-1)
⇒ (2/9)^(-9) = (2/9)^(2m-1)
⇒ 2m – 1 = – 9 ⇒ 2m=-8
∴ m = (-8)/2=-4

Q5. Let x = 597×〖10〗^22+73.5×〖10〗^21. When x is expressed in standard from as 6.0435 × 〖10〗^m, then the value of m is
(a) 22
(b) 23
(c) 24
(d) 21
Sol. 
x = 594×〖10〗^22+73.5×〖10〗^21
= 5970×〖10〗^21+73.5×〖10〗^21
=(5970+73.5)×〖10〗^21=6043.5×〖10〗^21
On multiply numerator and denominator by 〖10〗^3, we get
=6043.5×〖10〗^21×〖10〗^3=6.0435×〖10〗^24
On compare this with 6.0435×〖10〗^m, we get m = 24

Q6. If 2m-1/2m=2, when m≠0, then the value of m^2+1/(16m^2 ) is
(a) 2 1/2
(b) 2
(c) 1 1/2
(d) 4
Sol. 
Given, 2m-1/2m = 2
On squaring both side (2m-1/2m)^2 = (2)^2
⇒ 4m^2+1/(4m^2 )-2.(2m).(1/2m) = 4
⇒ 4m^2+1/(4m^2 )-2 = 4
⇒ 4m^2+1/(4m^2 ) = 6
On divide both side by 4, we get
∴ m^2+1/(16m^2 ) = 6/4 = 1 1/2

Q7. The mean of the five observations x, x + 2, x + 4, x + 6, x + 8 is 11. Then, the mean of the first three observations is
(a) 12
(b) 2
(c) 20
(d) 9
Sol. 
Given, the mean and the 5 observations
x, x + 2, x + 4, x + 6, x + 8 is 11
∴(x+(x+2)+(x+4)+(x+6)+(x+8))/5 = 11
 ⇒ 5x + 20 = 55 ⇒ 5x = 55 – 20 = 35
x=35/5=7
∴Mean of the first three observations
=(x+(x+2)+(x+4))/3=(7+(7+2)+(7+4))/3  
=27/3=9

Q8. A bag has 5 red marbles, 4 green marbles and 3 blue marbles. All marbles are identical in all respects other than colour. A marble is taken out from the bag without looking into it. What is the probability that it is non-green marble?
(a) 1/3
(b) 2/3
(c) 7/12
(d) 5/12
Sol. 
Given, a bag contains 5 red marbles, 4 green marbles and 3 blue marbles.
Total number of marbles = 12
Total number of non-green marbles = 8
Here, total number of cases = 12
And number of favourable casses = 8
∴ Probability of getting non-green marbles
=(Number of favourable cases )/(Total number of cases )
=8/12=2/3

Q9. If x, y and z respect the number of faces, number of vertices and number of edges respectively of a polyhedral, then which of the following is true?
(a) x + 2 = z – y
(b) z – 2 = x – y
(c) y – 2 = z + x
(d) x – 2 = z – y
Sol.
 Let the given polyhedral is tetrahedron in which number of faces, number of vertices and number or edges are 4, 4 and 6, respectively.
∴ x = 4, y = 4 and z = 6
In option (d), x – 2 = z – y
x – 2 = 4 – 2 = 2
and z – y = 6 – 4 = 2
∴ x – 2 = z – y (True)

Q10. Ravi purchased two articles for Rs. 1500 each. He sold them, gaining 6% on one and losing 4% on the other. His gain/loss per cent in the whole transaction is
(a) Gain, 1%
(b) Loss, 1 1/2%
(c) Gain, 2%
(d) Loss, 1%
Sol. 
Given, Ravi purchase two articles for Rs. 1500 each.
∴ Total purchase cost = Rs. 3000
Ravi get 6% profit on first article, then he got
Amount after selling of 1st article
= 1500 + 1500 ×6/10 = Rs. 1590
After selling, 2nd article, he lose 4%, Then, his amount of 2nd article after selling it
=1500-1500×4/100 = Rs. 1440
∴ Total amount after selling of both article
= 1590 + 1440 = Rs. 3030
Which is 1% more than total purchase cost of both article.  

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