Nagarro aptitude test

Total Capacity = I cm (5,4) = 20

Total time to empty the full tank = Total Capacity/ Total Efficiency = 20/(4-5) = - 20 Hours

Here, Negative sign indicates emptying efficiency/time.

• Total capacity = LCM (10, 20) = 20

• Efficiency of A \( = \frac{20}{10} = 2 \) and Efficiency of B \( = \frac{10}{10} = 1 \)

• Time taken by (A+B) to fill the tank =\( \frac{Total Capacity}{Total Efficiency } \)= \( \frac{20}{2+1} \) = \( \frac{20}{3} \) min

Let the ages of Ram and Rahim 10 yr ago be x and 3x respectively.

After 5 yr from now ( 10 for present + 5 for next)

\(\frac{x + 15}{3x + 15} = \frac{2}{3}\)

\(\Rightarrow 6x + 30 = 3x + 45\)

\(\Rightarrow\) 3x = 45 - 30 = 15

\(\therefore\) x = 5.

Ratio of their present ages = (x +10) : (3x + 10)

\(= 15 : 25 = 3 : 5\).

• Given, A and B together can complete a work in \( \frac{24}{5} \) days.
• B and C together can complete a work in \( \frac{36}{5} \) days.
• A and C together can complete a work in \( \frac{72}{13} \)days.
• Work = Efficiency\(\times\)Time
• Then total work is LCM of \(\frac{24}{5}, \frac{36}{5}, \frac{72}{13}\) = 72.
• Therefore the efficiency of A and B or A+B = \(\frac{5}{24}\times 72\) = 15.
• Efficiency of B and C or B+C = \(\frac{5}{36}\times 72\) = 10.
• Efficiency of A and C or A+C = \(\frac{13}{72}\times 72\) = 13.

• That is,

  \(\Rightarrow 2A + 2B + 2C = 15 + 10 + 13 \)

  \(\Rightarrow 2A + 2B + 2C = 38 \)

  \(\therefore A + B+ C = \frac{38}{2}\)

  = 19.

• Then the efficiency of B is,

  \(\Rightarrow A + C + B = 19\)

  A+C = 13

  \(\therefore B = 6\)

• The efficiency of B is = 6.

• B alone can complete the work in \(\frac{72}{6}\)

  = 12 days.

Average = \(\frac{n+1}{2}\)

= \(\frac{187+1}{2}\) = \(\frac{188}{2}\) = 94

• Average = \(\frac{n+1}{2}\)

• \(\frac{111+1}{2}\) = \(\frac{112}{2}\) = 56

• Answer is 56.

• We know that,

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

• \(\sqrt{105\sqrt{105\sqrt{105.......}}}\) =105

• Answer is 105

• We know that,

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

• Therefore,

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

• Answer is 15

Let age of father & son are \( 7x, 3x \)

4 year hence, son‘s age = \( 3x+4 \)

Ratio of ages of son & mother is \( 3x+4 : M = 1:2 \)

M = \( 6x+8 \)

\( F + M = 7x : 6x+4 \)

We can‘t find value of \( x \)

• Speed = 20 Km / hr

• Time = 7 hrs

• Distance = speed x time

  = \(20 \times 7 \)

  = 140 Km

• The speed required to reach 140km in 4 hours = Distance / time =\(\frac{140 }{4 }\)

  =35 Km /hr

• Speed to be increased =35 -20 = 15 Km/hr.

Cost price =100%

12% profit =100%+12%=112%

23% profit =100%+23%=123%

123%=61500

1%=500

112%=\(500\times112\)=56000.

Cost Price=100%

Difference in given profits = 18-13 = 5%

5% = 960

1% = \( \frac{960}{5}=192 \)

\( 100 \)% = \( 19200 \)

Cost Price = 100%\( = 19200 \).

• \((x+y)^2=x^2+y^2+2xy \)

• \(21^2=x^2+y^2+(2\times104)\)

  \(441=x^2+y^2+208 \)

• \(x^2+y^2=441-208=233 \)

• Amount=2500+ 827.5=3327.5

  Then,\(2500[1+\frac{10}{100}]^{n}=3327.5\)

• \([1+\frac{10}{100}]^{n}=\frac{33275}{25000}\)

  \([\frac{110}{100}]^{n}=\frac{1331}{1000}=[\frac{11}{10}]^{3}\)

• n =3.

A = \(P(1+\frac{R}{100})^{n} \)

2197 = \( P(1+\frac{30}{100})^{2} \)

2197 = \( P\times\frac{130}{100}\times\frac{130}{100} \)

P = \( \frac{2197\times100\times100}{130\times130} \)

P = 1300

\(A = P(1+\frac{R}{100})^N\)

\(1440 = P (1+\frac{20}{100})^2\)

\(1440 = P (\frac{6}{5})^2\)

P = 1440 x \(\frac{25}{36}\) = 1000.

Let the present age of \(A\) be \( a \) years and the present age of \(B\) be \(b \) years.

According to the problem, the ratio of their present ages is given as: \[ \frac{a}{b} = \frac{3}{4} \]

\[ a = \frac{3}{4}b \]

Ten years ago, A was half of B in age, which gives us the equation: \[ a - 10 = \frac{1}{2}(b - 10) \]

\[ \frac{3}{4}b - 10 = \frac{1}{2}(b - 10) \]

\[ \frac{3}{4}b - 10 = \frac{1}{2}b - 5 \]

\[ 3b - 40 = 2b - 20 \]

\[ 3b - 2b = 40 - 20 \] \[ b = 20 \]

Now substitute \(b = 20 \) into \(a = \frac{3}{4}b \) to find \(a\): \[ a = \frac{3}{4} \times 20 = 15 \]

Therefore, \(A's\) present age is \(15\) years and \(B's\) present age is \(20\) years. The sum of their present ages is: \[ a + b = 15 + 20 = 35 \]

Thus, the sum of their present ages is \(\boxed{35} \) years.

• My brother was born 6 years after my sister was born and 3 years before I was born.
• Age of father during my brother’s birth = (30 + 6) = 36 years.
• Age of mother during my brother’s birth = (38 – 3) = 35 years.
• Hence, during the birth of my brother, my father was 36 years old and my mother was 35 years old.

• The speed of the boat in still water=

  \( \frac{1}{2} \) \(\times\)(speed against the stream+speed along the stream)

  \( \frac{1}{2} \)\( \times \)(\( V_{A} \)+\( V_{B} \))

  \( \frac{1}{2} \)\( \times \)(11+5)

  =8 km/hr.

• Yearly basis Compound interest=\(3000\times8\%=240\)

• Half yearly basis compound interest\((3000\times4\%=120)+(3120\times4\%=124.8)\)=244.8

• Difference=244.8-240=4.8