Real Numbers - Olympiad Level MCQ, Class 10 Mathematics - Class 10 MCQ

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Let the common remainder be  10

From the given numbers we now subtract the common remainder

i.e 38 − 10 = 28,66 −10 = 56, 80 −10 = 70

Now  we will find the HCF of 28, 56, 70

28 = 2 x 7 x 2

56 = 2 x 2 x 2 x 7

70 = 2 x 5 x 7

So HCF of 28, 56 and 70 = 2 x 7 

So the required number is 14

LCM of 24 , 32 & 42 = 672

The least possible number for your given case = 672 +5 = 677

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.

N= 18a+2= 35b+19=42c+26

=> N+16= 18a=35b=42c

=> N+16 is divisible by LCM (18,35,42)

=> N+16 is divisible by 630 (=3*3*2*5*7)

=> N= 630k-16

=> Least possible number= 630*1-16= 614

Here,
2 - 1 = 1
3 - 2 = 1
4 - 3 = 1
5 - 4 = 1
6 - 5 = 1
That is, if we subtract 1 from the number divisible by 2, 3, 4, 5, 6, we get the required number.

The smallest number divisible by 2, 3, 4, 5, 6 will be the LCM of 2, 3, 4, 5, 6.

LCM of 2,3, 4, 5, 6 is 60
So, the required number which leaves a remainder of 1,2,3,4,5= L.C.M of (2,3,4,5,6)-1=59

The least common multiple of 2, 3, 4, 5, 6 is their LCM=60.
So, the answer is the least number such that it is a multiple of 7 and a multiple of 60 leaving remainder 1.
∴The required answer is 301.

Number lies between 11 and 1111 When divide by 9 a remainder of 6 and when divided by 21 leave a remainder of 12 = 33, 96, 159 (all separated by 63)

a =33  d=63
1111=a+(n-1)63

1111=33+(n-1)63

1078=(n-1)63

1078/63=n-1

n=(1078+63)/63

n= 1141/63

n = 18.11

then total number=18

Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70-5),
117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65,117                 [for the largest number]
For this, 117 = 65 x 1 + 52 [∵ dividend = divisor x quotient + remainder]
⇒ 65 = 52 x 1 + 13
⇒ 52 = 13 x 4 + 0
∴ HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.

Let a|b and the GCD(a,b) = m, then b=aq for some integer q and the GCD(a,b) can be expressed as a linear combination with some integer x and y,

       ax + by = m .

Substituting in b we get,

      ax + (aq)y = m

      a(x + qy) = m

But x + qy = GCD(1,q) =1. Thus,

      a(1) = m

Hence a = GCD(a,b), which is what we needed to show.

According to problem
(10x+y)(10a+b)=100p + 10q + r
By substituting the values
r=2y
q=(2x+y)
p=2x
(10x+y)(10a+b)=100*2x+10*2(x+y)+2y
(10x+y)(10a+b)=220x+22y
10a+b=22

Consider,

a² + b² + c² – ab – bc – ca = 0

Multiply both sides with 2, we get
2( a² + b² + c² – ab – bc – ca) = 0
⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ a² + a² + b² + b² + c² + c²– 2ab – 2bc – 2ca = 0
⇒ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0
⇒ (a – b)² + (b – c)² + (c – a)² = 0
Since the sum of squares is zero, it means each term should be zero.
⇒ (a – b)² = 0,  (b – c)² = 0, (c – a)² = 0
⇒ a – b = 0,  b – c = 0,  c – a = 0
⇒ a = b,  b = c, c = a
⇒ a = b = c

π (Pi) and e are famous irrational numbers

π = 3.1415926535897932384626433832795... (and more)

e = 2.7182818284590452353602874713527 (and more ...)

We cannot write down a simple fraction that equals Pi.

The popular approximation of  ²²/₇ = 3.1428571428571... is close but not accurate.

The first few terms of e add up to: 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 = 2.71666...

The product of a rational and irrational number is irrational.

2 × √5 = 2√5 is an irrational number.

Product of a rational and irrational number always irrational if rational number is not zero.

Take an example of 2 and 3.
When you multiply,
√2×√3=√6(Irrational).
When we add,
√2+√3=Irrational.

Given that, a = x³y² = x x x x x x y x y
and b = xy³ = x x y x y x y
∴ HCF of a and b = HCF (x³y², xy³) = x x y x y = xy²
[Since, HCF is the porduct of the smallest power of each common prime facter involved in the numbers]

Correct Answer :- D

Explanation : 15/(10)^1/2 + (20)^1/2+ (40)^1/2- (125)^1/2

(3*5)/[(2)^1/2(5)^1/2 + 2(5)^1/2 + 2(2)^1/2(5)^1/2 - 5(5)^1/2]

= (3*5)/[3(2)^1/2(5)^1/2 - 3(5)^1/2]

= (3*5)/3(5)^1/2[(2)^1/2 - 1]

= (5)^1/2/[(2)^1/2 - 1]

Rationalise it, we get

= (5)^1/2/[(2)^1/2 - 1] * [(2)^1/2 + 1]/[(2)^1/2 + 1]

= (5)^1/2[(2)^1/2 + 1]/(2 - 1)

= (5)^1/2/[(2)^1/2 + 1]