WeCP Numerical Ability Questions

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required largest number.

9995 - 7652 = 2343

9995 - 8504 = 1491

8504 - 7652 = 852

 

Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder

= HCF of 2343, 1491, 852

= 213

 

Now we need to find out the common remainder.

 

Take any of the given numbers from 7652, 8504 and 9995, say 7652

 

7652 ÷ 213 = 35, remainder = 197

Let maximum marks of the examination = x

Marks that Karan got = 35% of x = 35 x 100

 

Given that Karan failed by 10 marks.

pass mark = \frac{35x}{100} + 10     …..(1)

 

Marks that Sujit got = 15 marks more than the passing marks.

Passing marks = \frac{45x}{100} - 15   ……(2)

 

From (1) and (2), 

\frac{35x}{100} + 10  =  \frac{45x}{100} - 15

\frac{10x}{100} = 25

Hence, x = 10 x 25 = 250

Let the present age of Abhishek= x

Then the present age of Amitabh= 3x+3

 

Given that, three years hence, Amitabh’s age will be 10 years more than twice the age of the son

(3x + 3 + 3) = 2(x+3) + 10

x = 10

Amitabh’s present age

= 3x + 3 = 3 x 10 = 33

Let’s take speed of Air India be x km/hr and speed of Indigo be y km/hr,

\Rightarrow  7200/x = 7200/y + 1        ----(1)

Now after engine trouble, speed of the Indigo will be 5y/6 and it takes 36 minutes more that Air India,

\Rightarrow \frac{7200}{x} = \frac{7200 x 6}{5y} - \frac{36}{60}     ----(2)

Equating both equations,

\Rightarrow \frac{7200}{y} + 1 = \frac{7200 x 6}{5y} - \frac{36}{60}

\Rightarrow 8640/y - 7200/y = 1.6

\Rightarrow 1440/y = 1.6

\Rightarrow y = 900 km/hr

From equation 1

\Rightarrow x = 800 km/hr

∴ Speed of Air India in m/s  is 800 x \frac{5}{18} m/s.

= 222.22 m/s.

21 is the only number which is multiple of both 3 and 7,

so while calculating probability separately for 3 and 7, this number 21 will be taken twice.

 

So to avoid confusion we take it once here and will exclude 21 while calculating probability separately.

 

Also, number 21 can only be taken once.

 Probability of getting number multiple of both 3 and 7 (number 21) =1/28

 

\Rightarrow Probability of getting numbers which are multiple of 3 ( 3,6,9,12,15,18,24,27) = 9 / 28

 

\Rightarrow  Probability of getting numbers which are multiple of 7 (7,14,28) =3/28

 

\Rightarrow Probability of getting a number either multiple of 3 or 7 = 3/28 + 9/28 + 1/ 28 = 12/28

Hence, Probability of getting a number either multiple of 3 or 7 = \frac{3}{7}

Cost price of 1 pencil = 16/9

 

Selling price of 1 pencil = 22/11

 

Gain percentage = \frac{2-\frac{16}{9}\times 100}{\frac{16}{9}}

= 25/2 % = 12.5%

 

Therefore, Cost price of ice-cream = 16/9 + 16/9 * 12.5%

= Rs. 1.125

In 24 days, they both would have done 1/3 and 5/8 of the work.

Therefore, the remaining work is 1 – (\frac{1}{3} + \frac{5}{8}) = 1/24.

This means Raina has done 1/24th of the work

So he should be paid \frac{1}{3}th of the amount i.e. 480 × 1/24 = Rs. 20

Let us take H=Length of the head

                    T=Length of the tail

                    B=length of the body

Given that H=12 cms.

                        T=H+\frac{1}{3} B  => T=12+\frac{1}{3}B 

Let this be the first equation

                          B=H+T

  substituting the value of H and T here we will get

                          B=12+12+\frac{1}{3}B  

                           => B=24+\frac{1}{3}B  

Therefore, B=36  cm

Substituting this in T we get

                           T=12+\frac{1}{3}(36)

                           T=12+12   =24

So now that you got all the values add 

                  H=12 cm

                  B=36 cm

                  T= 24 cm

Hence, total length of cat  = H+B+T

    = 72 cm.

Weight of zinc in the alloy= 6 gm

weight of copper in the alloy = 9 gms.

 

So, Weight of zinc in the alloy=6 + 10= 16.(4 parts value)

 

Therefore one part = 4 gm

 

So, 5 gm should be subtracted from 9 gm of copper to make 4 gm

A chess board consists of 8 x 8 square and 

9 parallel lines x 9 parallel lines.

 

For a rectangle we need to select 2 parallel lines and the other two parallel lines that are perpendicular to the first set.

 

Hence, ^{9}C_{2} \times ^{9}C_{2} = 1296 rectangles

Therefore, 2 \times N - 5 = 2 x 1296 - 5 = 2587

Given number is 97215p6

(6+5+2+9) + (p+1+7) = (14-p)

 

This (14-p) must be divisible by 11

 

Therefore, p = 3

Average of 5, 7, 14 and Y = \frac{( 5 + 7 + 14 + Y )}{4}

 

Therefore, X = 80% of \frac{( 5 + 7 + 14 + Y )}{4}

 \Rightarrow \frac{(80 )}{100}\times\frac{26+Y}{4}

\Rightarrow X = (26 + Y)/5

5X - Y = 26          ----- (i)

 

Also, (X + Y)/2 = 26             ----- (ii)

 

From (i)and (ii)

X = 13

Y = 39. 

 

Hence,

X+Y = 52

We have-

6^{7} \times 35^{3} \times 11^{10}

\Rightarrow {(2\times 3)}^{7} \times {(5\times 7)}^{3} \times 11^{10}

\Rightarrow {2}^{7} \times {3}^{7} \times {5}^{3} \times  {7}^{3} \times 11^{10}

 

Thus, there are (7 + 7 + 3 + 3 + 10) = 30 prime number

The total angle traced by the hour hand is the angle traced in 6 hours and 20 minutes.

 

From the geometry of a clock, We know that the angle traced by the hour hand in one hour is 30º and in one minute is \frac{1}{2}^{\circ}.

 

Therefore, (30^{\circ} x 6) + (20 x \frac{1}{2}^{\circ}) = 190^{\circ} is the angle traced by the hour hand.

Suppose virat moves 4 km downstream in x hours. Then,

Speed downstream = \frac{4}{x} km/h

Speed upstream = \frac{3}{x} km/h

 

\therefore \frac{48}{4/x} + \frac{48}{3/x} = 14

\therefore x = \frac{1}{2}

 

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

 

Rate of the stream = \frac{1}{2} \times (8-6) km/h = 1 km/h

 

Mode = 3 median - 2 mean

 

\therefore median = \frac{mode + 2 mean}{3}

\Rightarrow median = \frac{<span style="font-weight: 400">(26.1 + 2 \times 20.6)</span>}{3}

\Rightarrow median = \frac{<span style="font-weight: 400">67.3</span>}{3}

\therefore median = 22.43

Let the side of the square be of Length = a

For Inner Circle,

the diameter of the circle = length of the side of the square = a

Diameter (inner circle) = a

 

For Outer Circle,

the diameter of the circle = length of the Diagonal of the square = a\sqrt{2}

Dia (outer) = a\sqrt{2}

 

Area of a circle = \frac{\pi d^{2}}{4}

 

Ratio of Area of 2 Circles

Area Outer / Area Inner = \frac{d_{outer}^{2}}{d_{inner}^{2}}  =\frac{(a\sqrt{2})^{2}}{(a)^{2}} = \frac{2}{1}

Area Outer : Area Inner = 2 : 1

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required largest number.

9995 - 7652 = 2343

9995 - 8504 = 1491

8504 - 7652 = 852

 

Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder

= HCF of 2343, 1491, 852

= 213

 

Now we need to find out the common remainder.

 

Take any of the given numbers from 7652, 8504 and 9995, say 7652

 

7652 ÷ 213 = 35, remainder = 197

Let maximum marks of the examination = x

Marks that Karan got = 35% of x = 35 x 100

 

Given that Karan failed by 10 marks.

pass mark = \frac{35x}{100} + 10     …..(1)

 

Marks that Sujit got = 15 marks more than the passing marks.

Passing marks = \frac{45x}{100} - 15   ……(2)

 

From (1) and (2), 

\frac{35x}{100} + 10  =  \frac{45x}{100} - 15

\frac{10x}{100} = 25

Hence, x = 10 x 25 = 250

Let the present age of Abhishek= x

Then the present age of Amitabh= 3x+3

 

Given that, three years hence, Amitabh’s age will be 10 years more than twice the age of the son

(3x + 3 + 3) = 2(x+3) + 10

x = 10

Amitabh’s present age

= 3x + 3 = 3 x 10 = 33

Let’s take speed of Air India be x km/hr and speed of Indigo be y km/hr,

\Rightarrow  7200/x = 7200/y + 1        ----(1)

Now after engine trouble, speed of the Indigo will be 5y/6 and it takes 36 minutes more that Air India,

\Rightarrow \frac{7200}{x} = \frac{7200 x 6}{5y} - \frac{36}{60}     ----(2)

Equating both equations,

\Rightarrow \frac{7200}{y} + 1 = \frac{7200 x 6}{5y} - \frac{36}{60}

\Rightarrow 8640/y - 7200/y = 1.6

\Rightarrow 1440/y = 1.6

\Rightarrow y = 900 km/hr

From equation 1

\Rightarrow x = 800 km/hr

∴ Speed of Air India in m/s  is 800 x \frac{5}{18} m/s.

= 222.22 m/s.

21 is the only number which is multiple of both 3 and 7,

so while calculating probability separately for 3 and 7, this number 21 will be taken twice.

 

So to avoid confusion we take it once here and will exclude 21 while calculating probability separately.

 

Also, number 21 can only be taken once.

 Probability of getting number multiple of both 3 and 7 (number 21) =1/28

 

\Rightarrow Probability of getting numbers which are multiple of 3 ( 3,6,9,12,15,18,24,27) = 9 / 28

 

\Rightarrow  Probability of getting numbers which are multiple of 7 (7,14,28) =3/28

 

\Rightarrow Probability of getting a number either multiple of 3 or 7 = 3/28 + 9/28 + 1/ 28 = 12/28

Hence, Probability of getting a number either multiple of 3 or 7 = \frac{3}{7}

Cost price of 1 pencil = 16/9

 

Selling price of 1 pencil = 22/11

 

Gain percentage = \frac{2-\frac{16}{9}\times 100}{\frac{16}{9}}

= 25/2 % = 12.5%

 

Therefore, Cost price of ice-cream = 16/9 + 16/9 * 12.5%

= Rs. 1.125

In 24 days, they both would have done 1/3 and 5/8 of the work.

Therefore, the remaining work is 1 – (\frac{1}{3} + \frac{5}{8}) = 1/24.

This means Raina has done 1/24th of the work

So he should be paid \frac{1}{3}th of the amount i.e. 480 × 1/24 = Rs. 20

Let us take H=Length of the head

                    T=Length of the tail

                    B=length of the body

Given that H=12 cms.

                        T=H+\frac{1}{3} B  => T=12+\frac{1}{3}B 

Let this be the first equation

                          B=H+T

  substituting the value of H and T here we will get

                          B=12+12+\frac{1}{3}B  

                           => B=24+\frac{1}{3}B  

Therefore, B=36  cm

Substituting this in T we get

                           T=12+\frac{1}{3}(36)

                           T=12+12   =24

So now that you got all the values add 

                  H=12 cm

                  B=36 cm

                  T= 24 cm

Hence, total length of cat  = H+B+T

    = 72 cm.

Weight of zinc in the alloy= 6 gm

weight of copper in the alloy = 9 gms.

 

So, Weight of zinc in the alloy=6 + 10= 16.(4 parts value)

 

Therefore one part = 4 gm

 

So, 5 gm should be subtracted from 9 gm of copper to make 4 gm

A chess board consists of 8 x 8 square and 

9 parallel lines x 9 parallel lines.

 

For a rectangle we need to select 2 parallel lines and the other two parallel lines that are perpendicular to the first set.

 

Hence, ^{9}C_{2} \times ^{9}C_{2} = 1296 rectangles

Therefore, 2 \times N - 5 = 2 x 1296 - 5 = 2587

Given number is 97215p6

(6+5+2+9) + (p+1+7) = (14-p)

 

This (14-p) must be divisible by 11

 

Therefore, p = 3

Average of 5, 7, 14 and Y = \frac{( 5 + 7 + 14 + Y )}{4}

 

Therefore, X = 80% of \frac{( 5 + 7 + 14 + Y )}{4}

 \Rightarrow \frac{(80 )}{100}\times\frac{26+Y}{4}

\Rightarrow X = (26 + Y)/5

5X - Y = 26          ----- (i)

 

Also, (X + Y)/2 = 26             ----- (ii)

 

From (i)and (ii)

X = 13

Y = 39. 

 

Hence,

X+Y = 52

We have-

6^{7} \times 35^{3} \times 11^{10}

\Rightarrow {(2\times 3)}^{7} \times {(5\times 7)}^{3} \times 11^{10}

\Rightarrow {2}^{7} \times {3}^{7} \times {5}^{3} \times  {7}^{3} \times 11^{10}

 

Thus, there are (7 + 7 + 3 + 3 + 10) = 30 prime number

The total angle traced by the hour hand is the angle traced in 6 hours and 20 minutes.

 

From the geometry of a clock, We know that the angle traced by the hour hand in one hour is 30º and in one minute is \frac{1}{2}^{\circ}.

 

Therefore, (30^{\circ} x 6) + (20 x \frac{1}{2}^{\circ}) = 190^{\circ} is the angle traced by the hour hand.

Suppose virat moves 4 km downstream in x hours. Then,

Speed downstream = \frac{4}{x} km/h

Speed upstream = \frac{3}{x} km/h

 

\therefore \frac{48}{4/x} + \frac{48}{3/x} = 14

\therefore x = \frac{1}{2}

 

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

 

Rate of the stream = \frac{1}{2} \times (8-6) km/h = 1 km/h