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.

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.

If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.

Answer = 1800 

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

72 --- 9
? ---- 1

=> Down Stream = 8

45 ---- 9 
? ---- 1

=> Up Stream = 5

Speed od current S = ?

S = (8 - 5)/2 = 1.5 kmph.

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.