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There are infinitely many rational numbers between any two rational numbers. So, two rational numbers between -2 and 2 may be 0 and 1

-1,03, 3 it's a answer

take the L.C.M of both the denominatorsthat is, 63now make the unlike rational numbers into like rational numbers1x9/7x9 and 1x7/9x79/63 and 7/63

multiply the numerator and denominator of both the rational numbers by 2.9x2/63x2 and 7x2/63x218/126 and 14x126

18/126>17/126>16/126>15/126>14/126therefore, two rational numbers between 1/7 and 1/9 can be16/126 and 15/126you can also show them in simplest form 8/63 and 5/42

1/2 = 12/243/4 = 18/24

Now, Rationals between 1/2 & 3/4 = 13/24 , 14/24 , 15/24 , 16/24 , 17/24

1/2 × 3 /3 =3/63/4 × 3/3 ÷9/12Lasa9/12. ;. 6/12 Between number 7/12. And 8/12

(1/2)(5/6+7/8)1/2(20+21/24)1/2(41/24)41/48

1kg 20g= 1000g+20g=1020g

No. of pouches of 20g which can be made=1020/20=51

1 man can dig it in 9×10 days=90 daysand 15 men can dig it in 90/15 days=6 days

10 men dig the well in 9 days.Let's suppose the total work done by these men is one.So if 10 men take 9 days then 10 men did 1/9 work each day.Assuming efficiency for each man equal.Therefore one man does 1/90 work each day.So 15 men can do 15/90 work each day.Let no of days taken by 15 men is x.So, (15/90 )* x= 1Therefore x = 6So 15 men can dig well in 6 days.

if 9 pens cost rs 45.90. what Will be the cost of 6 such pen

iii. December 14In India, the National Energy Conservation Day is observed every year onDecember 14. The day is observed to raise awareness about the importance of energy and its conservation.

(i) Area of ΔABC = 1/2 × Base × Height = 1/2 × BC × AC = 1/2 ab Area of ΔABC = 1/2 × Base × Height = 1/2 × AB × CD = 1/2 cp ⇒ 1/2 ab = 1/2 cp ⇒ ab = cp Hence proved.(ii) In right angled triangle ABC,AB2= BC2+ AC2 c2= a2+ b2(ab/p)2= a2+ b2a2b2/p2= a2+ b2-------- From proof (1)1/p2= (a2+ b2) / a2b21/p2= (a2/ a2b2+ b2/ a2b2) 1/p2= (1/b2+ 1/a2)1/p2= (1/a2+ 1/b2)Hence proved.

Given ΔABC is an isosceles triangle in which ∠B = 90°⇒ AB = BCBy Pythagoras theorem,

we have AC^2= AB^2+ BC^2⇒AC^2= AB^2+ AB^2[Since AB = BC]∴ AC^2= 2AB^2→ (1)It is also given that ΔABE ~ ΔACD

Recall that ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides

Hence ar(ABE)/ar(ACD)=AB^2/AC^2ar(ABE)/ar(ACD)=AB^2/2AB^2 (From1)=1/2The ratio is 1:2

Mean = Summation of (f*x) / Summation of f= ( 10*3 + 12 * 10 + 20*15 + 25*7 + 35*5 ) / ( 3+10+15+7+5)= 800 / 40= 20

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x= 0.34

Multiply both sides by 10 (Since number of digits without bar is 1)

10x = 3.4

10x = 3.43434… ---------- (i)

10 ×10x = …… × 10 (Since number of digits having bars is 1)

100x = 34.3434….. ------------- (ii)

Subtracting (i) from (ii)

100x - 10x = 30.9

90x = 30.9

x = 30.9/90=309/900

Therefore, the vulgar fraction = 309/900

34/99 two numbers bar and two 99

thanks

Product of two numbers= product of their HCF and LCMi.e,AB= HLB=HL/Aoption a

6555444555+54654444447765

a.-2/3b.7/19c.5d.0e.1f.1/19

a) 2/3b)7/19c)5d)0e)1f)1/9

105 7/8 dry fruits are used for 1kg of cookiesthus, 1 dry fruit is needed for 1/ (105 7/8) kg of cookies cookies made from 338 4/5 dry fruit would be = 338 4/5 * (1/ (105 7/8) ) = 3.2 kgs

Mean of marks of boys=Sum/1533=Sum/15 Sum=33*15=495

Mean of marks of girls=Sum/2535*25=Sumsum=875

Total sum=875+495=1370Mean=1370/40=34.25

Distance covered by the car by using 1-liter petrol = 16 kmDistance covered by the car by using 11/4 liters of petrol = 16*11/4 =44 km

in 1 litre petrol car runs = 16 kmin 2(3/4) litre petrol that is 11/4 litre petrol car runs = (11/4)16 = 44 km ANS..

the answer is 44 km.

44 km Is the right answer

3(7)+4/7=21+4/7=25/7

12/5 = 2 (2/5)

Use the longdivisionsolutionto convert the originalfraction(135)(135)to amixed number. Thewhole numberportion of themixed numberwill be the number of times thedenominatorof the originalfractiondividesevenly into thenumeratorof the originalfraction(2)(2), and thefractionportion of themixed numberwill be theremainderof the originalfractiondivision(3)(3)over thedenominatorof the originalfraction(5)(5)

235

Use the longdivisionsolutionto convert the originalfraction(135)(135)to amixed number. Thewhole numberportion of themixed numberwill be the number of times thedenominatorof the originalfractiondividesevenly into thenumeratorof the originalfraction(2)(2), and thefractionportion of themixed numberwill be theremainderof the originalfractiondivision(3)(3)over thedenominatorof the originalfraction(5)(5)

235

2 mixed 3/ 5 is the answer of this question.

235 is the answer of this question

5whole2/3 is the right answer

1^3 = 12^2 = 43^3 = 274^2 = 165^3 = 1256^2 = 367^3 = 343

So, the missing number is 125.the series is 1, 4, 27, 16, 125, 36, 343.

7x/15 = 200/5

7x/15 = 40

x = 600/7

speed = 8*600/7*15 = 320/7 km/hr