Problems on Numbers Questions

Q:

In a two-digit number, the digit in the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged,difference between the newly formed number and the oiginal number is less than the original number by 1. What is the original number ?

 A) 35 B) 36 C) 37 D) 39

Explanation:

Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

15 7652
Q:

In a two-digit number, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :

 A) 12 B) 24 C) 36 D) 48

Explanation:

Let the ten's digit be x.

Then, unit's digit = x + 2.

$\therefore$Number = 10x + (x + 2) = 11x + 2

Sum of digits = x + (x + 2) = 2x + 2.

$\therefore$(11x + 2) (2x + 2) = 144

=>$\inline \fn_cm 22x^{2}+26x-140=0$

=> (x - 2)(11x + 35) = 0

=> x = 2

Hence, Required Number = 11x + 2 = 24

10 4726
Q:

Three numbers are in the ratio 4 : 5 : 6 and their average is 25. The largest number is :

 A) 30 B) 40 C) 50 D) 60

Explanation:

Let the numbers be 4x, 5x and 6x, Then, (4x + 5x + 6x ) / 3 = 25

=> 5x = 25

=> x = 5.

Largest number  6x = 30.

8 4331
Q:

The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :

 A) 100 B) 200 C) 300 D) 400

Explanation:

Let the numbers be x and y.

Then, xy = 9375 and  x/y = 15.

$\inline \fn_cm \frac{xy}{(\frac{x}{y})}=\frac{9375}{15}$

=>$\inline \fn_cm y^{2}=625$

=> y = 25

=> x = 15y = 15 x 25 = 375.

Sum of the numbers = 375 + 25 = 400.

7 3838
Q:

The sum of three numbers is 136. If the ratio between first and second be 2 : 3 and that between second and third is 5 : 3, then the second number is

 A) 30 B) 40 C) 50 D) 60

A : B = 2 : 3 and B : C = 5 : 3 =  $\inline \fn_cm \left ( 5\times \frac{3}{5}: \right 3\times \frac{3}{5})$$\inline \fn_cm 3:\frac{9}{5}$
=>A : B : C = $\inline \fn_cm 2:3:\frac{9}{5}$= 10 : 15 : 9.
Second number = $\inline \fn_cm 136\times \frac{15}{34}$= 60.