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3. The ratio of the present ages of Lata and her mother is 2:9, mother's age at the time of Lata's birth was 28 years , what is saritha's present age? Let us assume that their ages are 2X and 9X 9X-2X = 28 7X = 28 x=4 thus their ages are : 2X = 8 and 9 X = 36 so Lata is 8 years old and her mother is 36 years old. Answer

4. The incomes of A and B are in the ratio of 4:3 and their expenditure are in the ratio of 2:1 . if each one saves Rs 1000,what are their incomes? Let us assume that their incomes are 4X and 3X their expenditure are : 2Y and 1 Y 4X – 2Y = 1000 -- - 1 3X-1Y = 1000 - - - -2 solving it we get : 2X = 1000 or X = 500 by putting this value in 1 st equation we get Y = 1000. thus their incomes are (4*500) = 2000 and (3*500) = 1500 respectively .

5. A certain number of one rupee,fifty parse and twenty five paise coins are in the ratio of 2.5:3:4, add up to Rs 210. How many 50 paise coins were there? The numbers are in the ratio 5:6:8 let us assume that they are in this number only : so their value is : 5*1 = 5, 6 * .5 =3 and 8 *.25 = 2 thus their value is : 5+3+2 value of 50 paise coins : 3/10 * 210 = 63 total number of 50 paise coins : 63*2 = 126 answer

6. Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together. The total of 1:5 is 6 and total of 3:5 is 8 and total of 5:7 is 12 their LCM is 24. Let us assume that the quantities are : (24*5) (24* 4) (24*5) respectively. Thus milk from first container is : 1/(1+5) * (24*5) = 20 milk from 2 nd container : 3/(3+5) *(24* 4)=36 milk from 3 rd container=5/(5+7) *(24* 5)=50 total milk =106 total water = 230 thus their ratio is : 53:115 answer

7. A sum of Rs 1162 is divided among A,B and C. Such that 4 times A's share share is equal to 5 times B's share and is also equal to 7 times C's share . What is the share of C? 4A=5B=7C = X Let X be LCM of 4,5,7 = 140 4A = 140 or A = 35, 5B = 140 or B = 28 7C=140 or C = 20 A gets : 35/(35+28+20) *1162 =35/83 * 1162 = 490 answer

8. Two cylindrical jars have their diameters in the ratio 3:1, their heights are in the ratio 1:3 the ratio of their capacities are ? Volume of cylinder : = pi * radius * radius * height volume of 1 st =pi * 3r*3r*1h volume of 2 nd = pi * 1r*1r*3h let us remove common elements from both these : (remove Pi, r^2, & h ) 9: 3 or 3:1 answer

9. If a:b =5:9 and b:c=4:7 Find a:b:c? We can see that B is common in both the ratios. So take is for our calculations. B is 9 in first ratio and 4 in second ratio. We shall multiply the first ratios by 4 and second ratios by 9 (reverse method) thus new ratio is available : A:B = 20: 36 B:C : 36: 63 now we notice that B has same value in both the ratios. A:B:C = 20:36:63 answer

10. Find the fourth proportion to 4,9,12 We can write it as : 4:9:12:X as per the rule, we can multiple the outer two terms and the inner two terms and they will be equal. So : 4X = 9 *12 OR 4X = 108 or X = 27 answer

11. Find third proportion to 16,36? We can write it as : 16: 36 :: 36:X (as the third proportion has the same ratio to 36 as 36 has to 16) apply the rule : we can multiple the outer two terms and the inner two terms and they will be equal.outer terms are 16 & X and inner terms are 36 & 36. 16X = 36*36 X = 1296/16 = 81 answer

12. Find mean proportion between 0.08 and 0.18 ? We can write it as : .08:X :: X:.18 (as we want to find a mean proportion – say X - s it has the same ratio to .08 as .18 has to it.) apply the rule : we can multiple the outer two terms and the inner two terms and they will be equal.outer terms are .08 & .18 and inner terms are X & X X^2 = .0144 X = .12 answer

13. If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is We can see that B is common in the first 2 ratios. So take is for our calculations. B is 3 in first ratio and 4 in second ratio. We shall multiply the first ratios by 4 (the value of B in 2 nd ratio) and second ratios by 3 (the value of B in 1 st ratio) (reverse method) = A:B = 8:12 and B:C = 12:15 the new A:B:C = 8:12:15 Now we compare it to 3 rd ratio : C:D. C is common in both these ratio, it is : 15 & 6 respectively. So mulitply in reverse : by 6 to (8:12:15) and by 15 to (6:7) = A:B:C = 48:72:90 AND C:D=90:105 ANSWER = A:B:C:D =48:72:90:105 but all the values are divisible by 3, so reduce them : 16:24:30:35 answer

14. 2A=3B=4C then A:B:C? Let us assume that 2A=3B=4C= X let us take LCM of 2,3,4 LCM of 2,3,4 = 12 , let us assume X=12 so : 2A=3B=4C = 12 2A=12 or A = 6, 3B=12 or B=4, 4C=12 or C = 3 we can write as : A:B:C = 6:4:3 answer

15. 15% of x=20% of y then x:y is Here we can start by assuming any value for X or Y. Let us assume Y = (15 * 100/20) = 75 we have assumed Y = 75, (you can also start from X, if you wish, the answer will be same) so what is 20% of Y = 15 so what is X = 15 * 100/15 = 100 So we get X = 100 The ratio of X :Y = 100:75 or 4:3 answer

16. a/3=b/4=c/7 then (a+b+c)/c=? Let us assume that .a/3=b/4=c/7 =1 thus A= 3, B = 4, C = 7 now put the values of A,B,C in the formula and get the answer : (3+4+7) / 7 = 14/7 = 2 answer

17. Rs 3650 is divided among 4 engineers, 3 MBAs and 5 CAs such that 3 CAs get as much as 2 MBAs and 3 Engineers as much as 2 CAs .Find the share of an MBA. Let us assume CA=x,MBA = Y, ENG=Z 3X= 2 Y and 3 Z= 2 X here X is common. Let us make them equal. LCM of 3 and 2 is 6. Let us multiply the first ratios by 2(from 2 nd ratio) and let us multiply the 2 nd ratio by 3( from the first ratio). We get : 6X=4Y & 9Z=6X or 6X=4Y=9Z let us assume that it is equal to (LCM of 6,4,9 = 36) so 6X=36 or X=6, 4Y=36, Y=9, 9Z=36 or Z=4. Share of all engineers : (4*4=16), share of all MBAs (3*9=27), share of all Cas(5*6=30) total for MBA = 27/73 * 3650 = 1350 or 1MBA=450 total for all Eng : 16/73 * 3650 = 800 or 1 eng=200 share of all Cas : 30/73*3650 =1500 or 1CA=300

18. What is log 10 100000? = 5 answer

19. What is natural log ? Here the base is e, which is 2.71828 (not 10, which we had discussed earlier).

20. What is log 2 8? = 3 (here base is 2, and we know that 2^3 = 8)

21. What is log 1982 1? Answer = 0 We know that in any base b, the logarithm of 1 is 0.

22. What is log b b^x = ? Answer = x it is a rule that : log b b^x = x

23. What is : 5 = log 2 n ( what is n?) 32 (answer ) 2^5 = 32

24. What is log 10 .001 = -3

25. Find log of 1728 to the base 2*sqrt(3) ? Let us take factors of 1728 : 2*2*2*2*2*2*3*3*3 thus we know that 2*sqrt(3) ^ 6 = 1728 so answer is 6.

26. Change the base of log 200 to 2? = log 2 200 / log 2 10

27. Given: log 2 = .3010 log 3 = .4771 log 5 = .6990, what is log 6? We know that log 6 = log 2 * 3 we know that log 2*3 = .3010+.4771 =.7781

28. Given: log 2 = .3010, log 3 = .4771 Evaluate log 18. Log 2 * log 3*log 3 =.3010+.4771+.4771 =1.2552

29. Given: log 6 = .7781 Use the laws of logarithms to evaluate log 600 =log 6 * 100 = .7781 + 2 = 2.7781 answer