Maths 20 May 1

Maths for competitive exam Dr. T.K. Jain. AFTERSCHOOOL M: 9414430763 [email protected] www.afterschool.tk www.afterschoool.tk

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What is the least number by which 72 must be multipled so that it is divisible by 112?

• • • • •

Take factors of 72 These are: 2*2*2*3*3 Take factors of 112, these are: 2*2*2*2*7 Thus, we have to multiply 72 with one 2 and one 7, so ans: 72*14. ans.

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What is the sum of all even numbers between 1 and 31? • • • • • • • •

The first no. is 2, Difference between numbers is 2 The Arithmetic progression is 2,4,6,8, … There are 15 terms (30/2) Formulae is n/2*(2a+(n-1)d), A= first term D = difference, and n = no. of terms which gives us: 15/2(4+(14)2) = 240 Ans.

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What is the sum of the following series: 1,2,4,8,16 .. For first 5 terms? • It is geometric series – because ratio is fixed. • Ratio is second term divided by first term. • Formulae: (A*(r^n – 1))/(r-1) • A = first term, r = ratio • [1*(2^5 -1)]/(2-1) = 31 Ans. www.afterschoool.tk

Sumit secured a speed of 10 on three occasion and 12 on 1 occasion on 100 meters races around a square field. What is his average speed in the entire race?

• Remember, use harmonic mean always for such questions. • Formulae for this is : • inverse of ¼ {3/10 + 1/12} • = 10.43 Ans.

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21 mango trees, 42 banana trees and 56 date trees have to be planted in such a way that each row contains same no. of trees of one variety only. How many rows will be there minimum? • • • •

It is a question of HCF HCF = heighest common factor Factors of 21: 3*7, 42:2*3*7 and 56: 2*2*2*7 HCF of 21, 42 and 56 is 7, so each row will contain 7 trees of one variety only. There will be 3 rows of mango, 7 rows of banana and 8 rows of date. Thus there will be minimum 18 rows.

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What is the sum of the series: 1^2, 2^2, 3^2, 4^2, 5^2, 6^2? • Formulae for addition of such series is : {n(n+1)(2n+1)}/6 • N is 6. • {6(7)(13)}/6 • 91 is the answer. • We can verify here : 1+4+9+16+25+36=91 www.afterschoool.tk

What is the sum of the series: 1^3, 2^3, 3^3, 4^3, 5^3? • Formulae for addition of such series is : {n^2(n+1) ^2}/4 • N is 5 • {(25)(36)}/4 • 225 Ans. • We can verify here : 1+8+27+64+125=225 www.afterschoool.tk

What will be remainder when 188^276- 112^276 is divided by 300? • The remainder will always be zero. The fundamental laws are: • x^n- y^n is always divisible by x+y for all even values of n. • x^n +y^n is always divisible by x+y for all odd values of n. • x^n- y^n is always divisible by x-y for all values of n. • x^n+ y^n is never divisible by x- y.

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What will be remainder when 188^275+ 112^275 is divided by 300? • The remainder will always be zero. The fundamental laws are: • x^n- y^n is always divisible by x+y for all even values of n. • x^n +y^n is always divisible by x+y for all odd values of n. • x^n- y^n is always divisible by x-y for all values of n. • x^n+ y^n is never divisible by x- y.

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get, when you divide 188^3+ 112^3 by (188^2+ 112^2 188*112)? • • • •

300 Ans. x^3+ y^3 = (x+ y) (x^2+ y^2 -xy )\ Other important forumulae is x^3- y^3 = (x- y) (x^2+ y^2 +xy )

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Put the sets of real number, whole numbers and natural numbers, integers, rational numbers in ascending order. • The smallest set here is that of natural numbers (natural numbers don’t include 0), then whole number (0,1,2,…) set is bigger, integers is bigger set then whole numbers, then rational number set is bigger and the biggest is real number set. So the order is : • Natural – whole – integers - rational – real www.afterschoool.tk

What will be remainder when 188^276- 112^276 is divided by 76? • The remainder will always be zero. The fundamental laws are: • x^n- y^n is always divisible by x+y for all even values of n. • x^n +y^n is always divisible by x+y for all odd values of n. • x^n- y^n is always divisible by x-y for all values of n. • x^n+ y^n is never divisible by x- y.

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