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AS in first coloumb8+3+2=13now in second3+6-4=5now in third4+1=5now 5-1=4so 1 will be there

1

E = 55 × 1 = 53 × 7 = 2121 - 5 = 6

Set 1M = 132 × 13 = 265 × 9 = 4545 - 26 = 19(Multiply numbers on opposite side of the square and subtract)Set 2B = 29 × 2 = 1811 × 2 = 2222 - 18 = 4

Given: ΔABC ~ ΔPQR. &

ar ΔABC =ar ΔPQR

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQRar ΔABC =ar ΔPQR. (given)

ΔABC / ar ΔPQR = 1

⇒ AB²/PQ² = BC²/QR² = CA²/PR² = 1

[ USING THEOREM OF AREA OF SIMILAR TRIANGLES]

⇒ AB= PQ , BC= QR & CA= PR

Thus, ΔABC ≅ ΔPQR[BY SSS criterion of congruence]

thanxx for solution

If interior angle=40°then, exterior angle=180°-40°=140°So, angle BOY= 140°/2 = 70°

Hope it helps.Mark as best if I am correct.

if interior angle =40°then, exterior angle =180°-40°=140°so,angle boy=140°/2=70°

let x and y be the runs scored by two batsman

total runs scored=164

equation: x+y=164this equation can have many solutions like

x. y 100. 64 50. 114 64. 100and many more solutions

The Quadrilateral is cyclic , so, angle C + angle A = 180°

=> angle DAB = 180°-100° = 80°

now in ∆ADB , sum of angles are 180°

so , angle ADB = 180° -50-80 = 50°

y=2/10 {(50)2+10 (50)}y=2/10{100+500}y=2/10 {600}y=2*60=120

Surface Area of 2 cuboids = 2*2*(1*1.5 + 1.5*0.5 + 0.5*1)= 4*(1.5 + 0.75 + 0.5)= 4* (2.75)= 11 sq metres. Cost = 120*11= Rs. 1320

Where is the figure?

please send me answer

please give me like😊

He has Rs 8696 in his bank account

8696 is the correct answer

he has rs 8696 in his bank account

0 is the right answer

12,-3 the correct answer of the given question

A rhombus is kind of a parallelogram , henceAngle.QPS=AngleQRS=50°Now, in ∆PQSanQRS +RSQ+RQS =180°[Angle sum property of triangle]

50°+x+x=180°(Opposite sides of rhombus are equal, therefore angles opposite to equal sides are equal)

50°+2x=180°2x=130°x=65°I.e. RQS=65°

I have already solve it bt thanks for cooperation.....

Step. 1 find angle PQR ( QPS and PQR are supplementary)

step 2 divide it by 2 to get angle RQS

this is because the 2 triangles SPQ and QRS are congruenttherefore Angle RQS = PQS = 1/2 PQR

the answer is -> 65°

thank you😊

Thanks

let the height be h

and the vertical distance be x

now,

tan 45=1

h/(x+80)=1

h=x+80 ___(i)

tan 60=root 3

h/x=root 3

h=x root 3 ------(II)

PUtting the valu of h in eq (i)

x root3=80+x

1.73x-x=80

0.73x=80

x=80/0.73

therefore,x=109.5

putting the value of x in eq(i)

h=109.5*1.73=189.43 m

give full figure please

Given that :the diameter of wheel of cycle = 1.6m

we have to find:how many revolution the wheel of cycle make covering a distance of 352 m

solution:-the Radius of wheel = 1.6/2 = 0.8 m

and the circumference of wheel = 2πr = 2π×0.8 => 1.6 × 22/7 = 5.02857143 m

now, the number of revolution by wheel covering a distance 352 m = 352/ 5.02857143= 70 rounds Answer

area of garden =side^2=50^2=2500m^2hence total cost=rate*area=2500*10=25000rupees

option b is the correct answer of the given question

b) 9340537 is the right answer

option (b) is a correct answer

option b is correct answer.

9340537p=9r=3o=4c=0u=5r=3e=7

P=9, R=3, O=4, C=0, U=5R=3, E=7So, PROCURE=9340537

9340537 is the option is 2nd is answer

arithmetic mean=a+b/2 =5+7/2 =12/2 =6

A+B-------- 25+7------- 2

12----- 2 =6

that is a+b/2 so 5+7/2=6

So2(7/9)== 18+7/9 25/9

and 5 4/11= 55+4/11= 59/11

tan30=qr/pq; 1/V3=h/30; hV3=30; h=30/V3 = 30/ v3 x V3/ V3=30V3/3=10V3

let, AB=30m BC=X <C=30° ABC got to be the same as AB by BC=tan 30° =)30 by x= 1 by wrute3 =)x=30wrute3

tan30=pq/qr; 1/V3=30/x; x=30V3

ans: (√3 - 1)cuz tanα=perpendicular / basegiven perpendicular is 100m. α1=30°,α2=45° .put the values in above formula get 2 equations and solve them. Then difference between the two values give the the distance.

ln ∆ABC,

h/x = tan 60°

⇒ h = x√3

ln ∆BCD,

50/x = tan 30°

⇒ x = 50√3

∴ h = 150

∴ Height of hill = 150 m

Let AB is the Tower of height = h = 50 m.And, let the Height of Hill CD = H m.Distance between The root of the tower and hill = BCNow,In ΔABC∠C = 30° TAN(C) = AB/BC⇒ TAN(30) = 50/BC⇒ 1/√3 = 50 /BC⇒ BC = 50√3 m.Now,InΔBCD,∠B = 60° Tan(B) = CD/BC⇒ Tan(60) = H/BC⇒ BC√3 = H⇒ H = 50√3*√3 = 150 m.