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MathsMathsDr. T.K. Jain.Dr. T.K. Jain.

AFTERSCHOOOLAFTERSCHOOOLM: 9414430763M: 9414430763

tkjainbkn@yahoo.co.intkjainbkn@yahoo.co.in

www.afterschool.tkwww.afterschool.tk

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(1+3+5+ 3983)/1992? This is an example of AP. Sum of AP = n/2(1st + last) N=1991+1 =1992

1992/2(1+3983) =996*3984=3968064 Divide it by 1992 =1992 Or

1992*1992 / 1992 =1992 Ans.

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Sue of 1 + + 9/16 .. .. To

infinite? This is an example of infinite series. Sum of an infinite series =

a/ (1-r) A = 1, R =

1 / (1-3/4) = 1/(1/4) = 4 Ans.

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.3111 +.6666.. +

.02222..? These can be written as (31-3)/90 +6/9 + 2/90 Rule: deduct the number which is not

repeated and divide by 10 for this number.Divide by 9 for the number which isrepeated.

=28/90+60/90+2/90

=90/90 =1 ans.

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Find the greatest number of threedigits, which when divided by 6,9,12

gives remainder of 3 in each case? Now first find LCM of 6,9,12, the

LCM is 36. Now find the greatest

number of 3 digits divisible by 36.Let us divide 1000 by 36. theremainder is 28. Deduct 28 from

1000 = 972. Now add 3 in this = 975 Ans. = 975

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(1^3 + 2^3 + 3^3 15^3)

(1+2+3..15) ? The sum of the first series is: (N+1)^2*N^2 / 4

=(15+1)^2

*15*15 / 4 =256*225/4 = 14400.

The sum of the next series is : n *(n+1) /2

=15*16/2 = 120 Answer = 14400 - 120 = 14280 ans.

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K-2, 2K+1, 6K+3 are in

GP, what is K? In GP, the square of middle (2nd ) term is equal tomultiple of first and 3rd term.

(K-2)*( 6K+3 ) = (2k+1)^2

6k^2 -9K 6 = 4k^2 +4k+1 2k^2-13k-7 = 0 2k^2 -14k + k -7 =0 2K(k-7)+1(k-7) = 0

K = 7 or -1/2 Ans.

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. + . 7 + . 7.006^2 + .047^2 +

.0079^2 ? Taking common values and cutting

these, we are left with:

(10/1)^2=100 Ans.

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A garrison of 2000 men had enough food tolast for 30 days. After 10 days 500 more men

joined then how long will the food last? Total food = 2000*30 = 60000

Food now left: 2000*20 = 40000

Number of persons now = 2000+500= 2500

No, of days left: 40000 / 2500 = 16

ans

k l

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A contractor undertook to complete awork in 120 days. He employed 200 menin 72 days. Only half of the work was

done. How many more men should beemployed to complete the remainingwork in time?

Half of the work is 200*72 = 14400 This work is to be done in (120-72)

=48 days. No. of persons requred 14400/48

=300 The contractor has to employ (300-

200) 100 more workers. Ans.

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20 men can complete a work in 40 days.When should 4 men leave so that the

work is completed in 48 days? 16 people will work for all 48 days.

16*48 = 768

Remaining work = 800 768 = 32 No. of days when 4 workers will work

= 32/4 = 8

They will leave after 8 days. Ans.

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A & B can do a work in 30 days, B+C cando it in 24 days, A + C can do it in 20days. They all work for 12 days. B +C

leave. How much more time will berequired? One days work =(1/30+1/24+1/20)*1/2 =15/240, = 1/16

12 days work (12*15) = 180/240 = 3/4 Remaining work is = As one days work is 1/16-1/24 = 1/48 48*1/4 = 12 Remaining work will be done by A in 12

days. Ans.

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A man goes to a garden and runs in thefollowing way:

from start, 25 m. west, then 60 mnorth, 80 m. east and finally 12 m.

south. What is the distance betweenstarting and finishing points?

Difference of east and west = 80-25 = 55 East

Difference of north and south= 60-12 = 48 north

Applying pythogorus rule = 55^2 +48^2 =

=3025+ 2304 = 5329

Square root of 5329 = 73 ans.

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Median of a triangle passes through thepoint which divides each median in the

ratio? 2:1 ans.

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What is the point called which is atequal perpendicular distance from the

sides of the triangle? Incentre it has equal distance from

sides.

A 25 l dd i t ll h it f t

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A 25 m. ladder against wall has its foot7 m. far from foot of wall. If top ofladder slides by 4m downwards, how

much will the foot slide? Assume the height of ladder = x X^2 +7^2 = 625

X= square root of 576 = 24.

Now height is reduced by 4, new height is20.

625-400 = 225, square root of 225 is 15.

thus the ladder has slided by 15-7 = 8 m.ans.

ladder

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