Cisco aptitude questions

• If the number is = \(x \),

  \(\frac{80}{100}\times x+80=x\)

  \(\frac{80x}{100}+80=x\)

  \(\frac{80x + 8000}{100}=x\)

  \(80x + 8000 = 100 x\)

  \(8000 = 100x - 80x\)

  \(8000 = 20x\)

• \(x\) = \( \frac{8000}{20}\)= 400.

Easy Method :

• Suppose the number is \(x \).

  Adding 80% of x with 80 gives the same number.

  That is, 20% of x is 80.

  Number = 100%

  20% = 80

  100% =?

• To find the number, simply multiply them crosswise and divide them as shown below.

  

• Number = \(\frac{80\times 100}{20}=400\).

• Given \(\sqrt{8} ×\sqrt{18} ×\sqrt{32} ×\sqrt{50}\)

  = \(\sqrt{4×2} ×\sqrt{9×2} ×\sqrt{16×2} ×\sqrt{25×2}\)

• \(2\sqrt{2} ×3\sqrt{2} ×4\sqrt{2} ×5\sqrt{2}\)

  =\( 120×2×2 \)

  = \( 480 \)

• Answer is 480.

• Total weight = \(Average \times Total number\)

  = \(29\times18\)

  = \( 522 \) kg.

ATQ

• \( \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} \)

• Required Time = \( \frac{xy}{y-x} \) Hour.
• Answer is \( \frac{xy}{y-x} \).

• If we are given a value in kmph , so now for converting it into mps

• just divide it by 3.6

  hence 30kmph=\( \frac{30}{3.6} \)mps = 8.33mps

• Answer is 8.33 mps.

• Mohit's 5 day's work = 5 \( \times \) \( \frac{1}{20} \) = \( \frac{1}{4} \)

• Remaining work = 1 - \( \frac{1}{4} \) = \( \frac{3}{4} \)

• Mohan cando the remaing work in =

  10 \( \times \)\( \frac{3}{4} \) =7\( \dfrac{1}{2} \) days

• Let the present ages of son and father be \(x\) and \((60 -x)\) years respectively.

  Then, \((60 - x) - 6 = 5(x - 6)\)

  \(54 + x = 5x - 30\)

  \(4x = 24\)

  \(x = 6\)

• Son's age after \(6\) years = \((x+ 6) = 12 \) years.

• Answer is 12.

• We know that,

  \(\sqrt{x\sqrt{x\sqrt{x.......}}}\) =x

• Therefore,

\(\sqrt{15\sqrt{15\sqrt{15.......}}}\) = 15

• Answer is 15

• Rate downstream=; \( \frac{32}{6} \) kmph

• Rate upstream= \( \frac{14}{6} \)kmph

• Velocity of current=\( \frac{1}{2} \) \( \times \)\(( \frac{32}{6} \) - \( \frac{14}{6} \))

  =1.5kmph

• Let A's age \(10\) years ago = \(x\) years, Then, B's age \(10\) years ago = \(2x\) years.
• \(\frac{x+20}{2x+10}= \frac{3}{4}\) , \(x=5\)
• So, the total of their present ages =\(( x + 10 + 2x + 10) = (3x + 20) \) = 35years.

• Let the speed of train on onward journey be X km/h.

  Then, speed of train on return journey = 0.8X km/h

  Total time = 46 - 1 = 45

  \(\frac{1000}{X} + \frac{ 1000}{0.8X}\) = 45

  X = 50 km/h

• Speed of the train on return journey = 0.8 × 50

  = 40km/h

• Answer is 40km/h.

• Let krishna leave the job after x days .

  \(\frac{x}{30}\) + \(\frac{x}{40}\) + \(\frac{5}{40}\) = 1

  4x + 3x + 15 = 120

  7x = 105

• x = 15

• Hence Krishna leave the job 15 days before

• Average =\(\frac{sum}{number}\)

  Sum = \(average \times number\)

  Sum of ages of A,BandC = \(40\times 3\) =120

  Sum of ages of BandC = \(50\times 2\) =100

• Difference in sum =120-100 =20

  A' s age =20

• Distance=\(\frac{s_{1}\times s_{2}}{s_{1}- s_{2}}\times t_{1}- t_{2}\)

• Convert time into hour

  =\(\frac{60\times30}{30}\times \frac{4-(-6)}{60}\)

  =\(\frac{60\times30}{30}\times \frac{10}{60}\)

  =10km

• Answer is 10 km.

• Difference=\(P (\frac{r}{100})^{2}\)

• 20=\(P (\frac{5}{100})^{2}\)

  p=\(\frac{20\times10000}{25}=8000\)

  Sum=8000

• Answer is 8000.

• Required ratio

  = 120% of Rs. 30 : 110% of Rs. 20

  = 36 : 22

  = 18 : 11

• Answer is 18 : 11.

• \((a+b)^{2}=a^{2}+b^{2}+2ab\),

• Question show the format,\(a^{2}+2ab+b^{2}\)

\(6.4\times6.4+2\times55.04+8.6\times8.6 =(6.4+8.6)^{2}\)

=\(15^{2}=225\)

• Answer is 15.

• Let the two positive numbers be 4x and 5x

• According to the given condition,

  4x X 5x = 180

  x\( \pm \)3

• Only x = 3 is considered because the numbers are positive. So, the numbers are 12, 15.

• The difference between the two numbers is,

  d = 15-12 = 3

• Therefore, the difference between two numbers is 3.

• According to BODMAS rule, the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition, and at the end Subtraction.

• \(3104\div97+5\times84-69+2\)

  \(=32+5\times84-69+2\)

  \(=32+420-69+2\)

  =385

• Answer is 385.

• Number = \(x \) \(x \times \frac{40}{100} +42=x\) \(x \times \frac{2}{5} +42=x\) i.e, \(\frac{2x}{5}+42=x\) \( 2x + 210 = 5x \) \(210 = 3x \) \(\therefore x = 70 \)

Easy Method :

• Suppose the number is \( x \).

• Adding 40% of \( x \) to 42 gives the same number.

  That is, 60% of \( x \) is 42

  Number = 100%

  60% = 42

  100% =?

• To find the number, simply multiply them crosswise and divide them as shown below.

• 

• Number = \(\frac{42\times 100}{60}=70\)