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1 A + 14 = 26 A = 26 - 14A = 8

sorry the above solution is wrong

x + 15x/100 = 322

x = 322*100/115

x = 280

SP = 280 + 280*25/100

= 280+70 = 350

Reaming distance=Total distance- covered distance=(5x+3)-(3x-2)=5x+3-3x+2=2x+5

Let sanjay bro's present age= xSanjaya's age= x/2after 6 years, sanjaya bro's age will be = x+6, Sanjay will be = x/2+6as per questionx/2+6=3/5(x+6)=>x/2+6=3x/5+18/5=>6-3.6=3x/5-x/2=>2.4=x/10=>x=24Sanjay's bro is 24 yrs old and sanjay is 12 yrs old.

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Euclid’s Division Lemma:

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exists unique integersqandrwhich satisfies the conditiona = bq + rwhere 0≤ r ≤ b.

The basis of Euclidean division algorithm is Euclid’s division lemma. To calculate the Highest Common Factor (HCF) of two positive integersaandbwe use Euclid’s division algorithm. HCF is the largest number which exactly divides two or more positive integers. By exactly we mean that on dividing both the integersaandbthe remainder is zero.

Let us now get into the working of this euclidian algorithm.

Consider we have two numbers 78 and 980 and we need to find the HCF of both of these numbers. To do this, we choose the largest integer first, i.e. 980 and then according to Euclid DivisionLemma,a = bq + rwhere 0 ≤r≤b;

980 = 78 × 12 + 44

Now, herea= 980,b= 78,q= 12 andr= 44.

Now consider the divisor as 78 and the remainder 44 and apply the Euclid division method again, we get

78 = 44 × 1 + 34

Similarly, consider the divisor as 44 and the remainder 34 and apply the Euclid division method again, we get

44 = 34 × 1 + 10

Following the same procedure again,

34 = 10 × 3 + 4

10=4×2+2

4=2×2+0

As we see that the remainder has become zero therefore proceeding further is not possible and hence the HCF is the divisorbleft in the last step which in this case is 2. We can say that the HCF of 980 and 78 is 2.

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Let x be the cost price(100%-10%)×x=7290/100x=72x=72×100/90x=4×20x=80

let the percentage be x(100%+x%)×80=600100%+x%=15/2100/100+x/100=15/2x/100=15/2-100/100x/100=750/100-100/100x=/100=650/100x=65%

Let the population of the village at the beginning of 2013 be x•population at the end of 2013= x+(10/100)x= (11/10)x

•population at the end of 2014= (11/10)x+(11/100)x= (121/100)x

•population at the end of 20157986 = (121/100)x+(121/1000)x7986 = (1210+121/1000)xx = 7986/1.331 x = 6000

hence the population at the beginning of 2013 was 6000

Population in 2015=120000×(106/100)×(95/100)

=12×106×95

=120840

The population in 2015 is increased by 840

percentage=100×(840/120000)

=0.7%

so the population increase by 0.7%

from 3 and 3 we get

AE > AD

in the form of p/qq is the denominatorhence -3/5 5 is denominator

In 2days he can finish

2 days is the correct answer of the given question

2 days is the best answer I think

Volume of brick=(15×10×6)cm3=900 cm3Volume of wall=(1500×1000×600)cm3=900000000cm3

So,Bricks needed=900000000cm3/900cm3=1000000 bricks

see question

Let SP of each orange be xSp of 42 oranges = 42xLoss = 8xCP = Sp + Loss = 50x

Loss % = Loss ÷ Cp x 100

=8x/50x X 100 = 16%

Question you have submitted is incomplete. Please post a complete question.

X+y+z=6,x+2y+3z=14, x+4y+7z=30

*In the first part of the solution, in matrix B, the number is 30. It's not 3.*

28000 = 10000(1+ r x 10/100)2.8 = 1 + r/101.8 = r/10r = 18%

28000 = 10000(1+ r x 10/100)2.8 = 1 + r/101.8 = r/10r = 18%

the bank gave an interest of Rupes 500 on deposit

We can consider the increase in population as a problem of compound interest.

Here, P= 6809000 t= 4 year and r= 4.7%pa

Population at the end of 2015 = P(1+r/100)⁴

= 6809000(1+0.047)⁴=6809000(1.047)⁴= 6809000*1.2016=81,81,694.4≈8181694

Hence the population at the end of 2015 will be 8181694.

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1. three point zero one2. twelve point three three

thanx can you do a physics question

One day work of roshan 1/20therefore work done by him 1/20*5=1/4remaining work=1-1/4=3/4let ramesh one day work be 1/xtherefore 1/x*12=3/41/x=3/48=1/16ramesh one day work is 1/16ramesh can complete the work in 16 daysnow if they work togather then their one day work=1/20+1/16=4+5/80=9/80implies 80/9 days=8whole8/9 days

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please post clearly

osm answer

Let selling price (sp)=100

Loss=20

Loss=cp — sp

Cp = sp +loss = 100+20=120

Loss % on cost price(cp) = (20/120)*100% = 16.66% lossOption-D is right

8400/5 = 1680g

each packet weighs 1Kg 680g

8times 7×3-7×2+3=10meters

in one attempt net value of distance boy climb is 3–2 = 1min 6 attempts boy climbs = 6 min the 7th attempt, he climbs = 6+3–2 = 7 min the eighth attempt he reaches 7+3 = 10 m

Hence answer 8 attempts

(a) 14/6 - 1/4= (28 - 3)/12= 25/15

(b) 5/12 - 1/6= (5 - 2)/12 = 3/12= 1/4

(c) 1/10 - 1/15= (6 - 4)/60= 2/60 = 1/30

(d) 1/2 - 1/3= (3 - 2)/6 = 1/6

Solution:Let us assume, initial quantity of air = 1

Therefore, quantity removed in first step = ¼

Remaining quantity after first step

=1−14=34=1-14=34

Quantity removed in second step

=34×14=316=34×14=316

Remaining quantity after second step

=34−316=12−316=916=34-316=12-316=916

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

i) all sides are equal so it is equilateral 🔺ii) one angle is 90° so it right angle 🔺 iii) obtuse angleiv) one angle is greater than 90° so it is obtise angle 🔺