Aptitude and Reasoning Questions

Let present Anirudh's age be 'A' and Bhavana's age be 'B'

Given seven years, their ratio is 7:9

=> A-7 : B-7 = 7 : 9

=> 9A - 7B = 14 .......(1)

And given,

Chandhu is 12 years older than A nad 12 years younger than B

=> C = A + 12....(2)

=> B = C + 12

=> B = A + 12 + 12 .....(From (2))

=> B = A + 24 ....(3)

Put (3) in (1)

9A - 7(A+24) = 14

9A - 7A - 168 = 14

2A = 182

=> A = 91 years

=> C = A + 12 = 91 +12 = 103 years.

Let the Cost of Production of the article = 100

Then, Labour Cost = 20

Raw Material = 10

Other Expenditure = 100 - 10 – 20 = 70

Selling Price of the article  = 120

After increasing Labour and Raw material cost by 40% & 20% respectively,

New Labour cost = 28

New Raw material cost = 12

New Cost of Production = 70 + 28 + 12 = 110

Then, New SP = 110% of 120 = 120 x 110/100 = 132

=> New Gain = 132 - 110 = 22

=> New Profit % = 22 x 100/110 = 20%

Prashanth sold a album for $ 735 and made a profit of 22.5% .
CP = 735 x 100/122.5 = Rs. 600
New CP = 600 + 120 = Rs. 720
New SP = 735 + 120 = Rs. 855.

Profit = Rs. 135.

profit % = 100 x 135/720 = 18.75%.

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Let the cost price of 1 article = Rs. 1

From the given data,

Then, the selling price of 1 article = 22/18 = 11/9

Then, Profit = SP - CP = 11/9 - 1 = 2/9

Required, profit % = Profit/CP x 100

= [(2/9)/1] x 100

= 200/9

= 22.222%.

In these type of fractions, the decimal moves left the number of zeroes in the divisor times. Herein the given question 18/1000, there are 3 zeros in 1000 (divisor). Hence, the quotient is 0.018.

The quotient of 18/1000 = 0.018.

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 E3 =E1∩E2={6, 12, 18} 

=> n(E3) = 3

∴E=E1 ∪ E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

∴P(E)=n(E)n(S)=1218=23