Real Numbers - Olympiad Level MCQ, Class 10 Mathematics
40 Questions MCQ Test | Real Numbers - Olympiad Level MCQ, Class 10 Mathematics
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Q. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
The largest possible number with which when 60 and 98 are divided, leaves the remainder 3 in each case, is –
The largest possible number with which when 38, 66 and 80 are divided,leaves the remainder same is –
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
What is the least possible number which when divided by 24, 32 or 42 in each case it leaves the remainder 5?
LCM of 24 , 32 & 42 = 672
The least possible number for your given case = 672 +5 = 677
What is the least possible number which when divided by 24, 32 or 42 in each case leaves remainder 5. How many such numbers are possible between 666 and 8888?
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
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.
What is the least possible number which when divided by 18, 35 or 42 leaves 2, 19, 26 as the remainders
respectively?
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
What is the least possible number which when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively?
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 number which when divided by 2, 3, 4, 5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is
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.
How many numbers lie between 11 and 1111 which when divided by 9 leaves a remainder of 6 and when divided
by 21 leaves a remainder of 12?
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
The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is
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.
If x|y, where x > 0, y > 0 (x, y ε z) then –
If a|b, then gcd of a and b is –
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.
A number 10x + y is multiplied by another number 10a + b and the result comes as 100p + 10q +r, where r = 2y, q = 2(x + y) and p = 2x; x, y < 5, q ≠ 0. The value of 10a + b may be:
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
If x, y ε R and |x| + |y|= 0, then –
If a, b, c ε R and a2 + b2 + c2 = ab + bc + ca, then –
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
If x, y ε R and x < y ⇒ x2 > y2 then –
If x, y ε R and x > y ⇒ |x| > |y|, then–
If x, y ε R and x > y ⇒ |x| < |y|, then–
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...
If a, b ε R and a < b, then
If x is a non-zero rational number and xy is irrational, then y must be –
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.
The arithmetical fraction that exceeds it's square by the greatest quantity is –
If x and y are rational numbers such that 
is irrational, then 
is
Take an example of 2 and 3.
When you multiply,
√2×√3=√6(Irrational).
When we add,
√2+√3=Irrational.
If x and y are positive real numbers, then –
If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is
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]
is equal to –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]
+ 
is equal to –The expression 
is equal to –
If x, y, z are real numbers such that 
then the values of x, y, z are respectively
if a, b, c ε R and a > b ⇒ ac < bc, then –
If a, b, c ε R and ac = bc ⇒ a = b, then –
Between any two distinct rational numbers –
Infinite number of rational numbers exist between any two distinct rational numbers. We know that a rational number is a number which can be written in the form of p/q where p and q are integers and q not equal to 0.
The total number of divisors of 10500 except 1 and itself is –
10,500 = 2^2 x 3^1 x 5^3 x 7^1
d(n) = (a + 1)(b + 1)(c + 1)(d + 1)
d(10500) = (2 + 1)(1 + 1)(3 + 1)(1 + 1)
d(10500) = (3)(2)(4)(2)
d(10500) = 48
Now 1 and 10500 should not be counted
so number of divisors = 48 - 2 = 46
The sum of the factors of 19600 is –
Properties of the number 19600
Factorization
-
2 * 2 * 2 * 2 * 5 * 5 * 7 * 7
Divisors- 1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 25, 28, 35, 40, 49, 50, 56, 70, 80, 98, 100, 112, 140, 175, 196, 200, 245, 280, 350, 392, 400, 490, 560, 700, 784, 980, 1225, 1400, 1960, 2450, 2800, 3920, 4900, 9800, 19600
Count of divisors- 45
Sum of divisors - 54777
The product of divisors of 7056 is –
The number of odd factors (or divisors) of 24 is –
24 can be divided by two odd numbers which are 1 and 3 only.
The number of even factors (or divisors) of 24 is –
Factors of 24 are 2, 3 ,4 6 8 12 and 24 itself
therefore , no of even factors are 6
In how many ways can 576 be expressed as a product of two distinct factors?
576 = 26 x 32
∴ Total number of factors = (6 + 1) (2 + 1) = 21
So the number of ways of expressing 576 as a product of two distinct factors = 
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