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251.

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Answer» (i) `13/3125 = 0.00416`
(ii) `17/8 = 2.125`
(iii) `15/1600 = 0.009375`
(iv) `23/(2^3*5^2) = 23/200`
`= 0.115`
(ix) `35/50 = 7/10 = 0.7`
answer
Discussion
252.

Ravin and Skha drive around a circualr sports field. Ravi takes 16 minutes to take one round while sikha completes the round in 20 minutes. If both start at the same point at the same time and go in same direction, after how much ltime will they meet at the starting point ? [

Answer» Required number of minutes = LCM(n 16,20)
Now , ` 16 =- 2^(4) and 20 = ( 2^(2) xx 5)`
LCM ( 16,20) = ( ` 2^(4) xx 5) = ( 16xx 5) = 80`
Hence , both will meet at the starting point after 80 minutes.
Discussion
253.

Find the simplest form of ` 148/185`

Answer» Correct Answer - `sqrt9, root3(6), pi, 0.232332333`
Discussion
254.

The area bounded by the curves y=lnx, y=ln|x|, y=|lnx| and y=|ln||x| is

Answer» We can draw graphs for these `4` curves.
Please refer to the video to see the graphs.
The common intersection area is the required area.
From the graph, we can see that,
Area of shaded region `(A)= 4 int_(-oo)^0 xdy`
Now, `y = lnx => x = e^y``:. A = 4 int_(-oo)^0 e^y dy`
`=>A = 4[e^y]_(-oo)^0`
`=>A = 4[e^0 - e^(-oo)]`
`=>A = 4[1-0]`
`=>A = 4`
So, area bounded by the given curves is `4` square units.
Discussion
255.

Examine whether `17/30` is a terminating decimal

Answer» Correct Answer - `4/5`
Discussion
256.

Given that HCF ( 252, 594) = 18 find LCM ( 252 , 594)

Answer» we have
` LCM = ( " product of two given numbers")/(" their HCF")`
` (( 252 xx 594))/18= 8316`
Hence , LCM ( 252 , 594) = 8316.
Discussion
257.

The HCF of two numbers is 18 and their product is 12960. find their LCM.

Answer» Correct Answer - 720
Discussion
258.

The HCF of two numbers is `27` and their LCM is `162`. if one of numbers is `81`, find the other.

Answer» Correct Answer - No
Discussion
259.

The LCM of two numbers is 1200. show that the HCF of these numbers cannot be 500. why ?

Answer» Correct Answer - since 500 is not a factor of 1200
Discussion
260.

Find the HCF and LCM of `12, 15, 18, 27`

Answer» Correct Answer - `(2+sqrt3) and ( 2-sqrt3)`
Discussion
261.

If the product of two numbers is 1050 and their HCF is 25, find their LCM.

Answer» Correct Answer - 42
Discussion
262.

The HCF of `2472, 1284` and a third number `N` is `12`. If their LCM is `2^(3)xx3^(2)xx5xx103xx107`. Find the number `N`.

Answer» `2472=2^(3)xx3xx103`
`1284=2^(2)xx3xx107`
LCM`=2^(3)xx3^(2)xx5xx103xx107` (given)
`HCF=2^(2)xx3`
We know that HCF of the numbers is the highest number which divides all the numbers i.e, HCF is common to all the numbers.
So, N should be a multiple of `2^(2)xx3`.
Also, we known that LCM is the largest number that is divide by the given numbers. As LCM contains `3^(2)` so `3^(2)` must be in any of the 3 number necessarily.
So, it is only N which will contain `3^(2)`.
Also, LCM contains 5 which will necessarily be in any of the 3 numbers. So, it is only N which will contain 5. Since LCM contains 103 and 2472 contains it so no need to having it in N. Similarly LCM contains 107 and it is in 1284. So, no need to having it in N....(2)
So, from above (1) and (2) cases of HCF and LCM , we are sure that
`N=2^(2)xx3^(2)xx5=180`
Note
Although N can also be `2^(2)xx3^(2)xx5xx103,2^(2)xx3^(2)xx5xx107 " and " 2^(2)xx3^(2)xx5xx103xx107 ` i.e., 18540,19260 and 1983780
Caution:
Some students solve this question as,
Product of numbers= Product of theri LCM and HCF
`implies (2^(3)xx3xx103)xx(2^(2)xx3xx107)xxN=(2^(3)xx3^(2)xx5xx103xx107)xx(2^(2)xx3)`
`:. N=(2^(3)xx3^(2)xx5xx103xx107xx2^(2)xx3)/(2^(3)xx3xx103xx2^(2)xx3xx107)=15`
This method (also the answer) is WRONG because formula used in this method is applicable only for two numbers.
Discussion
263.

The HCF of two numbers is 23 and their LCM is 1449. If one number is 207, find the other number.

Answer» Here, HCF=23
LCM =1449
Now firs no. `xx` second no. `=HCFxxLCM`
`implies` second no. `=(HCFxxLCM)/("first no. ")`
`= (23xx1449)/(207)=161`
Discussion
264.

The HCF of two numbers is 23 and their LCM is 1449. If one of the number is 161, find the other .

Answer» Correct Answer - 207
Discussion
265.

A vertical tower Stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of Elevation of the bottom and the top of the flag staff are `alpha and beta` respectively Prove that the height of the tower is `(htanalpha)/(tanbeta - tanalpha)`

Answer» We can draw a diagram with the given details.
Please refer to video for the diagram.
Here, height of the tower ` = BC = x`
`AB = h, CD = y`
`/_CDB = alpha, /_CDA = beta`
`:.tan alpha = (BC)/(CD) `
`=> tan alpha = x/y`
`y = x/ tan alpha->(1)`
Now, `tan beta = (AC)/(CD) `
`=> tan beta = (h+x)/(y)`
`=> y tan beta = h+x`
Putting value of `y` from (1),
`=>x/tan alpha tan beta= h+x`
`=>x(tanbeta/tanalpha-1) = h`
`=>x((tanbeta-tanalpha)/tanalpha) = h`
`=>x = (htanalpha)/(tanbeta-tanalpha)`
So, height of tower is `(htanalpha)/(tanbeta-tanalpha)`.
Discussion
266.

One card is taken out from a well- shuffled deck of 52 cards. Find the probability that it is :1.) a face card2.) the jack of hearts3.) queen of diamonds

Answer» Here, Total number of outcomes, `n(S) = 52`
1.)Number of face cards in a well shuffled deck of cards `= 12`
`:.` Probability of drawing a face card `= 12/52 = 3/13`
2.)Number of the jack of hearts in a well shuffled deck of cards `= 1`
`:.` Probability of drawing a jack of hearts `= 1/52`
3.)Number of the queen of diamonds in a well shuffled deck of cards `= 1`
`:.` Probability of drawing a queen of diamonds`= 1/52`
Discussion
267.

Use Euclids divisionLemma to show that the cube of any positive integer is either of the form `9m , 9m+1`or, `9m+8`for some integer `m`.

Answer» Let a be any positive integer .
Let `" " b = 3`
`therefore " " a = 3q +r`
where , `0 le r lt 3` i.e. , r = 0 , 1, 2,
(i) When r = 0 , then
a =3q
`implies " " a^(3) = (3q)^(3) = 27q^(3) = 9(3q^(3)) = 9m`
where , `m = 3q^(3)` is an integer .
(ii) When , r =1 , then
a = 3q +1
`implies " " a^(3) = (3q +1) ^(3) = 27q^(3) + 27q^(2) + 9q + 1 = 9(3q^(3) + 3q^(2) + 1) + 1 = 9m +1`
where , ` m = 3q^(3) + 3q^(2) +1 ` is an integer .
(iii) When r = 2 , then
a = 3q + 2
`implies " " a^(3) = (3q +1)^(3)`
=` 27q^(3) + 54q^(2) + 36 q + 8 = 9(3q^(3) + 6q^(2) + 4q) + 8 = 9m + 8`
where , `m = 3q^(3) + 6q^(2) + 4q` is an integer .
Hence , the cube of any positive integer is of the form 9m or 9m +1 or 9m + 8. Hence Proved .
Discussion