Arithmetical Reasoning Questions

Let 'n' number of days it take to complete.

we know that sum of n number= n/2[2a+(n-1)d]
where a=20,d=1;
n/2[2x20+(n-1)1]=275
n[n+39]=550
n2 + 39n - 550 = 0
 n2+ 50n - 11n - 550 = 0
(n-11) (n-50) = 0
n-11 = 0;
n = 11(Days).

The Multiples of 18 are 18, 36, 54, 72, 90, 108, 126, ....

• Multiples of 18 are nothing but numbers which are divisible by 18 leaving remainder 0.

• An infinite number of multiples of 18 can be achieved by adding 18 to each subsequent multiple.

We know that, to be in the form that the number of rows and the number of columns to be equal the number should be a perfect square (a2).

Given number is 1200

The perfect number which is above and near to 1200 is 1225 which is 35  35.

Therefore, The minimum number of coins he need is 25. 

Here asked for the value of 28 tens

=> 28 x 10 = 280.

No. of students will be 98 x 12 = 1176
let us consider n circles then 1 + 2 + 3 + ....+ n = 1176
n(n+1)/2 = 1176
n = 48 circles.

15 girls are absent => b = 2(g-15)...... eqn...(1)

45 boys are absent => g - 15 = 5(b-45).....eqn...(2)

=> g = 5(b-45) + 15

=> b = 2((5(b-45)+15)-15)

=> b = 2(5b - 225 + 15 - 15)

=> b = 2(5b-225)

=> b = 10b - 450

=> 9b = 450

=> b = 50

so from equation(2)

g - 15 = 5(b - 45)

=> g - 15 = 5(50-45)

=> g - 15 = 5(5)

=> g = 25 + 15

=> g = 40

Therefore, the number of boys = 50 and girls = 40

Given that 50 paise and 25 paise coins are in the ratio of 7:3, they can be either 7 & 3 or 14 & 6.
As total 20 coins also includes Rs. 1 coins, so 50 paise & 25 paise coins are 7 & 3 respectively & Rs. 1 coins are 10.
Total amount = (100x10) + (50x7) + (25x3)=1425 paise or Rs.14.25.

'0' is called OFF State in Binary...