27 May Maths Ii

MATHS Dr. T.K. Jain. AFTERSCHOOOL M: 9414430763 [email protected] www.afterschool.tk www.afterschoool.tk

www.afterschoool.tk

What is the value of (248+(52+(144)^1/2)^ 1/2)^ 1/2 ? • First take square root of 144 – 12, • Now add 12 in 52, we get 64, its square root is 8, add it in 248, we get 256, • Square root of 256 is 16, so ans. is 16 ans.

www.afterschoool.tk

Express .17777… as fraction? • • • •

Here we have to solve it as under: 17-1 / 90 =16/ 90 8/45 ans.

www.afterschoool.tk

Express .127777… as fraction? • • • •

Here we have to solve it as under: 127-12 / 900 =115/900 23/180 ans.

www.afterschoool.tk

Express .173737373… as fraction? • • • • •

Here we have to solve it as under: 173-1 / 990 =172/990 =86/495 ans. Here we put 99 as there are two repeating numbers, and one 0 because we have one non repeating number.

www.afterschoool.tk

Find HCF and LCM of 1.75, 5.6 and 35? • Step 1: Multiply these numbers by 100, we get, 175,560,3500, now find their HCF and LCM. • HCF is 35, divide it by 100, we get .35 • LCM is 14000= divide it by 100, we get 140 Ans.

www.afterschoool.tk

What is the product of 2*.2*.02*.002*.0002? • • • •

There are ten digits after decimals. 2^5 = 32 So the answer is : .0000000032 Ans.

www.afterschoool.tk

What is the value of 1/(.1*.01*.001*.0001)? • There are 10 digits after decimal in denominator. • So we will have 11 digits in the answer. The answer is : 10,00,00,00,000 or 1*10^10 ans.

www.afterschoool.tk

What is the value of .0002/(.02*.002*.0002)? • We can cut .0002 • Now we have 1/ .00004 • Multiplying both sides by 100000, we get: 100000/4 = 25000 ans.

www.afterschoool.tk

HCF of two numbers is 16 and their LCM is 136, find the numbers? • There are no such numbers. • 136 does not have 16 as its factor. Every LCM will always have HCF as its factor. Therefore the answer is no such factor is possible.

www.afterschoool.tk

Which is the number, from which 1657 and 2037 when divided gives remainders 6 &5 respectively? • Deduct remainders from these: • 1651 and 2032, find their HCF now. • Divide 2032 by 1651, remainder is 381. Now divide 1651 by 381, we get remainder : 127. now divide 381 by 127, remainder is zero. This is HCF. Ans. www.afterschoool.tk

What is the value of 81^ -3/4? • • • • • • •

Let us first remove – (negative sign). To solve this, we will make it 1/81 ¾ can be solved in two steps = ¼ and 3. Now let solve the power ¼. This will give us 1/3 Now let us solve the power of 3, We get 1/27 ans. www.afterschoool.tk

What is the unit digit in 2467^153? • We know that 7 has in its power only four digits in this sequence–7,9,3,1. If the power is in multiple of 4, the unit digit is always 1. • 153 divided by 4 gives remainder of one. Whenever we have power of 1, the unit digit is 7. so the answer is 7 ans. •

www.afterschoool.tk

What is the unit digit in 2467^154? • We know that 7 has in its power only four digits in this sequence–7,9,3,1. If the power is in multiple of 4, the unit digit is always 1. • 154 divided by 4 gives remainder of 2. Whenever we have power of 2, the unit digit is 9. so the answer is 9 ans. •

www.afterschoool.tk

What is the unit digit in 2467^155? • We know that 7 has in its power only four digits in this sequence–7,9,3,1. If the power is in multiple of 4, the unit digit is always 1. • 155 divided by 4 gives remainder of 3. Whenever we have power of 3, the unit digit is 3. so the answer is 3 ans. •

www.afterschoool.tk

I engaged a servant on the condition that I shall pay him Rs. 200 and a set of clothes for 30 days of work. The servant left after working for just 15 days. I gave him Rs. 20 cash and clothes. What is the value of clothes? • • • • •

Let x = value of clothes; 30 days =2*15days 200 + x = 2(20+x) 200 +x = 40 +2x -x = -160 X = 160 Ans.

www.afterschoool.tk

Sum of squares of two numbers is 3341, and the difference of their squares is 891 what are the numbers?

• • • • • •

X^2+y^2 = 3341; X^2-y^2 = 891 Adding them we get 2x^2 = 4232 Subtracting from each other we get 2y^2 = 2450 y^2 = 1225 X = 46; y = 35 Ans. www.afterschoool.tk

What is the sum of the series: 1^2+2^2 … so on for first 10 terms? • • • •

Formulae for sum of this term is: N(n+1)(2n+1) / 6 10(11)(21) / 6 385 Ans.

www.afterschoool.tk

What is the sum of the series: 1^3+2^3 … so on for first 10 terms? • • • •

Formulae for sum of this term is: N^2(N+1)^2/4 100(121)/4 = 3025Ans.

www.afterschoool.tk

Thanks Afterschoool conducts three year integrated PGPSE (after class 12 along with IAS, CA & CS) and 18 month PGPSE (Post Graduate Programme in Social Entrepreneurship) along with preparation for CS / CFP / CFA /CMA / FRM. This course is also available online also. It also conducts workshops on social entrepreneurship in schools and colleges all over India – start social entrepreneurship club in your institution today with the help from afterschoool and help us in developing society.

www.afterschoool.tk

About AFTERSCHOOOL PGPSE – the best programme for developing great entrepreneurs • Most flexible, adaptive but rigorous programme • Available in distance learning mode • Case study focused- latest cases • Industry oriented practical curriculum • Designed to make you entrepreneurs – not just an employee • Option to take up part time job – so earn while you learn www.afterschoool.tk

Workshops from AFTERSCHOOOL • • • • • • • • •

IIF, Delhi CIPS, Jaipur ICSI Hyderabad Branch Gyan Vihar, Jaipur Apex Institute of Management, Jaipur Aravali Institute of Management, Jodhpur Xavier Institute of Management, Bhubaneshwar Pacific Institute, Udaipur Engineering College, Hyderabad

www.afterschoool.tk

www.afterschoool.tk social entrepreneurship for better society

www.afterschoool.tk