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Divisibility by 2 or 5: A number is divisible by 2 or 5 if the last digit is divisible by 2 or 5.

Divisibility by 3 (or 9): All such numbers the sum of whose digits are divisible by 3 (or 9) are divisibleby 3 (or 9).

Divisibility by 4: A number is divisible by 4 if the last 2 digits are divisible by 4.

Divisibility by 6: A number is divisible by 6 if it is simultaneously divisible by 2 and 3.

Divisibility by 8: A number is divisible by 8 if the last 3 digits of the number are divisible by 8

Divisibility by 11: A number is divisible by 11 if the difference of the sum of the digits in the odd placesand the sum of the digits in the even places is zero or is divisible by 11.

Divisibility by 12: All numbers divisible by 3 and 4 are divisible by 12.

Divisibility by 7, 11 or 13: The integer n is divisible by 7, 11 or 13 if and only if the difference of thenumber of its thousands and the remainder of its division by 1000 is divisible by 7, 11 or 13.

For Example: 473312 is divisible by 7 since the difference between 473-312 = 161 is divisible by 7.



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If the present age is y, then n times the present age = ny.

If the present age is x, then age n years later/hence = x + n.

If the present age is x, then age n years ago = x - n

The ages in a ratio a: b will be ax and bx.

If the current age is y, then 1/n of the age is y/n.



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The Proportionalities Implicit in the Equation S x T= D The above equation has three implicit proportionality dimensions each of which has its own critical bearing on the solving of time, speed and distance problems. 1.Direct proportionality between time and distance (when the speed is constant) time ╬╝ distance. A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the same speed. Find the ratio of distances covered by the two cars. Solution: Since, the speed is constant, we can directly conclude that time ╬╝ distance. Hence ta/tb=Da/Db Since, the times of travel are 2 and 3 hours respectively, the ratio of distances covered is also 2/3. 2.Direct Proportionality between speed and distance (when the time is constant) speed ╬╝Distance. d1/d2 = s1/s2 3.Inverse proportionality between speed and time . S1/S2 = t2/t1



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Profit, P = SP-CP; SP>CP Loss, L = CP-SP; CP>SP P% = (P/CP) x 100 L% = (L/CP) x 100 SP = {(100 + P)/100} x CP SP = {(100-L)/100} x CP CP = {100/(100 + P)} x SP CP = {100/(100-L)} x SP Discount = MP-SP SP = MP-Discount For false weight, profit percentage will be P% = (True weight-false weight/ false weight) x 100. When there are two successful profits say m% and n%, then the net percentage profit equals to (m+n+mn)/100 When the profit is m% and loss is n%, then the net % profit or loss will be: (m-n-mn)/100 If a product is sold at m% profit and then again sold at n% profit then the actual cost price of the product will be: CP = [100 x 100 x P/(100+m)(100+n)]. In case of loss, CP = [100 x 100 x P/(100-m)(100-n)] If P% and L% are equal then, P = L and %loss = P2/100

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