Quantitative Aptitude - Arithmetic Ability Questions

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

Therefore, the number of boys in the class = 8.

Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600

Find a number which exactly divide all these numbers 

That is HCF of 18000, 9600& 3600 

All the value has 00 at end so the factor will also have 00.

HCF for 180, 96 & 36.

Factors of  

180 = 3 x 3 x 5 x 2 x 2

96 = 2 x 2 x 2 x 2 x 2 x 3 

36 = 2 x 2 x 3 x 3 

Common factors are 2x2×3=12

 Therefore, Actual HCF is 1200

Gold Coins 18000/1200 will be in 15 rooms

Silver Coins 9600/1200 will be in 8 rooms

Bronze Coins 3600/1200 will be in 3 rooms

Total rooms will be (15+8+3)  =  26 rooms.

Volume required in the tank = (200 x 150 x 2) cu.m = 60000 cu.m 

Length of water column flown in1 min =(20 x 1000)/60 m =1000/3 m 

Volume flown per minute = 1.5 x 1.25 x (1000/3) cu.m= 625 cu.m. 

Required time = (60000/625)min = 96min

The mutiplicative inverse of a number is nothing but a reciprocal of a number.

Now, the product of a number and its multiplicative inverse is always equal to 1.

For example :

Let the number be 15

Multiplicative inverse of 15 = 1/15

The product of a number and its multiplicative inverse is = 15 x 1/15 = 1.

Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 x 8 x 1) / (100 x 2) = Rs. 4
S.I. for last 6 months = (104 x 8 x 1) / (100 x 2) = Rs. 4.16
So, amount at the end of 1 year = (100 + 4 + 4.16) = Rs. 108.16
Effective rate = (108.16 - 100) = 8.16%.

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7!/2!= 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways.

Required number of ways = (2520 x 20) = 50400.

If we consider the third term to be ‘x”
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d).
Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
x + 9d will be the 12th term.

Thus, 12th term of the A.P will be zero.

20 large cakes will require the equivalent of 10 helpers working for one hour. 700 small cakes will require the equivalent of 20 helpers working for one hour. This means if only one hour were available we would need 30 helpers. But since three hours are available we can use 10 helpers.