Olympiad Mathematics - Rational Numbers
Exam Duration: 45 Mins Total Questions : 25
Divide the sum of \({65\over 12}\) and \({12\over 7}\) by their difference.
- (a)
\({599\over 311}\)
- (b)
\({680\over 216}\)
- (c)
\({642\over 133}\)
- (d)
\({501\over 301}\)
Let \(x={65\over 12}\) and \(y={12\over 7}\)
According to question,
\((x+y)+(x-y)=({65\over 12}+{12\over 7})\div({65\over 12}-{12\over 7})\)
\(=({455+144\over 84})\div({455-144\over 84})\)
\(=({599\over 84})\div({311\over 84})={599\over 84}\times {84\over 311}={599\over 311}\)
The sum of the additive inverse and multiplicative inverse of \({1\over 5}\) is ________.
- (a)
\({24\over 5}\)
- (b)
\(-{24\over 5}\)
- (c)
25
- (d)
-25
Additive inverse of \({1\over 5}\) is \(-{1\over 5}\) and multiplicative inverse of \({1\over 5}\) is 5.
So, their sum = \({-1\over 5}+{5\over 1}={-1+25\over 5}={54\over 5}\)
The product of two rational numbers is \({-28\over 81}\) If one of the numbers is \({14\over 27}\) then find the other number.
- (a)
\({2\over 5}\)
- (b)
\({8\over 17}\)
- (c)
\(-{2\over 3}\)
- (d)
\(-{4\over 3}\)
Product = \({-28\over 81}\)
One of the two numbers = \({14\over 27}\)
\(\therefore\) Other number = \({Product\over First\ number}={-28\over 81}\div {14\over 27}\)
\(={-28\over 81}\times {27\over 14}={-2\over 3}\)
If a = 7, then the value of \(-({1-2a\over a-5})\) is _________.
- (a)
\(-{13\over 2}\)
- (b)
\(-{15\over 2}\)
- (c)
\({13\over 2}\)
- (d)
\({15\over 2}\)
We have, a = 7
So, \(-({1-2a\over a-5})=-({1-14\over 7-5})={13\over 2}\)
Which of the following statements is TRUE?
- (a)
Every point on the number line represents a rational number.
- (b)
The product of a rational number and its reciprocal is 0.
- (c)
(17 x 12)-1 = 17-1 x 12
- (d)
Reciprocal of \({1\over a},a\ne0\) is a.
(A) Every point on the number line is not only rational number, it can be irrational number also.
(B) The product of a rational number and its reciprocal is 1.
(C) (17 x 12)-1 = \({1\over (17\times 12)}\) = 17-1 x 12-1
(D) Reciprocal of \({1\over a},a\ne 0\) is a.
The multiplicative inverse of \(-{a\over b}\) is __________.
- (a)
\({a\over b}\)
- (b)
\({b\over a}\)
- (c)
\(-{b\over a}\)
- (d)
None of these
Multiplicative inverse of \((-{a\over b})\) is \((-{a\over b})^{-1}=-{b\over a}\)
Which of the following properties of rational numbers is given below?
\({7\over 4}\times ({-8\over 3}+{-13\over 12})={7\over 4}\times {-8\over 3}+{7\over 4}\times {-13\over 12}\)
- (a)
Commutativity of addition
- (b)
Associativity of multiplication
- (c)
Distributivity of multiplication over addition
- (d)
Distributivity of addition over multiplication
If \(x={2+3\times 2\over -5}\) then |-x| is equal to _______.
- (a)
\({8\over 5}\)
- (b)
\(-{8\over 5}\)
- (c)
0
- (d)
1
We have, \(x={2+3\times 2\over -5}={2+6\over -5}={-8\over 5}\)
So, \(|-x|=|{8\over 5}|={8\over 5}\)
Which of the following options is true?
- (a)
\({5\over 7}<{7\over 9}<{9\over 11}<{11\over 13}\)
- (b)
\({11\over 13}<{9\over 11}<{7\over 9}<{5\over 7}\)
- (c)
\({5\over 7}<{11\over 13}<{7\over 9}<{9\over 11}\)
- (d)
\({5\over 7}<{9\over 11}<{11\over 13}<{7\over 9}\)
We have, \({5\over 7},{7\over 9},{9\over 11},{11\over 13}\)
L.C.M. of 7, 9, 11, 13 = 9009
\(\therefore {5\over 7}={5\times 9\times 11\times 13\over 7\times 9\times 11\times 13}={6435\over 9009}\)
\({7\over 9}={7\times7\times11\times13\over 9\times7\times11\times13}={7007\over 9009};{9\over 11}={9\times7\times9\times13\over 11\times7\times9\times13}={7371\over 9009}\)
\({11\over 13}={11\times7\times9\times11\over 13\times7\times9\times11}={7623\over 9009}\)
\(\therefore{6435\over 9009}<{7007\over9009},{7371\over9009}<{7623\over9009}\)
\(\therefore {5\over7}<{7\over9}<{9\over11}<{11\over13}\)
The rational number which is not lying between \({5\over 16}\) and \({1\over 2}\) is _______.
- (a)
\({3\over 8}\)
- (b)
\({7\over 16}\)
- (c)
\({1\over 4}\)
- (d)
\({13\over 32}\)
Simplify: \(({3\over 11}\times {5\over 6})-({9\over 12}\times {4\over 3})+({5\over 13}\times {6\over 15})\)
- (a)
\(-{177\over 286}\)
- (b)
\(-{303\over 40}\)
- (c)
\({286\over 492}\)
- (d)
\({17\over 24}\)
We have, \(({3\over 11}\times {5\over 6})-({9\over 12}\times {4\over 3})+({5\over 13}\times {6\over 15})\)
\(={5\over 22}-1+{2\over 13}={65-286+44\over 286}=-{177\over 286}\)
What should be subtracted from \(({3\over 4}-{2\over 3})\) to get \({-1\over 6}?\)
- (a)
\(-{6\over 13}\)
- (b)
\({1\over 4}\)
- (c)
\({2\over 7}\)
- (d)
\(-{1\over 8}\)
Let the required number be x.
Then, \(({3\over 4}-{2\over 3})-x=-{1\over 6}\)
\(\Rightarrow x=({3\over 4}-{2\over 3})+{1\over 6}={9-8+2\over 12}={3\over 12}={1\over 4}\)
\(\therefore\) Required number = \({1\over 4}\)
Simplify: \({3\over 8}+{7\over 2}+({-3\over 5})+{9\over 8}+({-3\over 2})+{6\over 5}\)
- (a)
\({-2\over 3}\)
- (b)
\({-41\over 10}\)
- (c)
\({39\over 5}\)
- (d)
\({41\over 10}\)
\({3\over 8}+{7\over 2}+({-3\over 5})+{9\over 8}+({-3\over 2})+{6\over 5}\)
\(={3\over 8}+{7\over 2}-{3\over 5}+{9\over 8}-{3\over 2}+{6\over 5}\)
\(={15+140-24+45-60+48\over 40}={164\over 40}={41\over 10}\)
If \(x={2\over 3}\) and \(y={3\over 2}\) then find the value of (x + y) + (x - y).
- (a)
\({15\over 2}\)
- (b)
\(-{13\over 5}\)
- (c)
\(17\over 6\)
- (d)
\(-{11\over 6}\)
We have, \(x={2\over 3}, y={3\over 2}\)
(x + y) + (x - y) = \(({2\over 3}+{3\over 2})\div({2\over3}-{3\over 2})\)
\(={13\over 6}\div(-{5\over 6})={13\over 6}\times ({6\over -5})={-13\over 5}\)
If \(x=-{4\over 11}\) then which of the following rational number lies between x and |x|?
- (a)
\({7\over 13}\)
- (b)
\(-{11\over 15}\)
- (c)
\(-{2\over 11}\)
- (d)
\({5\over 8}\)
We have, x = -4/11
Rational numbers between \({-4\over 11}\) and \(|{-4\over 11}|\)
i.e., between \({-4\over 11}\) and \({4\over 11}\) are \({-3\over 11},{-2\over11},{-1\over 11},0,{1\over11},{2\over11},{3\over11}\)
There are 42 students in a class. Out of these \({3\over 4}\) of the boys and \({2\over 3}\) of the girls come to school by bus. The total number of boys and girls of the same class who come to school by bus is 30. How many boys are there in the class?
- (a)
20
- (b)
24
- (c)
26
- (d)
16
Let number of boys in the class be x.
\(\therefore\) Number of girls in the class = 42 - x
According to question, \({3\over 4}x+{2\over 3}(42-x)=30\)
\(\Rightarrow {3\over 4}x-{2\over 3}x+28=30\Rightarrow {x\over 12}=2\Rightarrow x=24\)
\(\therefore\) Number of boys in the class = 24.
Mrs Priya earns Rs 18000 per month. She spends \({7\over 12}\) on household Items and \({1\over 8}\) on rest of the things. The amount she saves is ________.
- (a)
Rs 7120
- (b)
Rs 5250
- (c)
Rs 5520
- (d)
Rs 6562.50
Savings = Earnings - Surn of expenditures
= Rs 18000 - Rs \([({7\over 12}+{1\over 8})\times 18000]\)
= Rs 18000 - Rs \([{17\over 24}\times 18000]\) = Rs 5250.
One fruit salad recipe requires \({1\over 2}\) cup of sugar. Another recipe for the same fruit salad requires 2 tablespoons of sugar. If 1 tablespoon is equivalent to \({1\over 16}\) cup, then how much more sugar does the first recipe require?
- (a)
\({4\over 5}\) cup
- (b)
\({6\over 5}\) cup
- (c)
\({3\over 8}\) cup
- (d)
\({5\over 8}\) cup
We have, 1 tablespoon = \({1\over 16}\) cup
\(\Rightarrow\) 2 tablespoons = \({2\over 16}\) cup = \({1\over 8}\) cup
So, the amount of sugar requires in second recipe = \({1\over 8}\) cup
Also, amount of sugar requires in first recipe = \({1\over 2}\) cup
\(\therefore\) Difference = \({1\over 2}-{1\over 8}={4-1\over 8}={3\over 8}\) cup
So, \({3\over 8}\) cup more sugar requires for first recipe.
The wingspans of different species of birds is given below.
Species of birds | Blue jay | Golden eagle | Seagull | Albatross |
---|---|---|---|---|
Length of wingspans |
\({41\over 100}\)m | \(2{1\over 2}\)m | \(1{7\over 10}\)m | \(3{3\over 5}\)m |
How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay?
- (a)
\({209\over 100}cm\)
- (b)
\({209\over 100}m\)
- (c)
\({9\over 100}m\)
- (d)
\({215\over 100}cm\)
Wingspan of Blue jay = \({41\over 100}m\)
Wingspan of Golden eagle = \(2{1\over 2}m={5\over 2}m\)
Wingspan of a Golden eagle is longer than wingspan of Blue jay by \(({5\over 2}-{41\over 100})m=({250-41\over 100})m={209\over 100}m\)
There are few adults and children in a restaurant. If \({3\over 8}\) of the people in the restaurant are adults and there are 90 more children than adults, then how many children are there in the restaurant?
- (a)
180
- (b)
200
- (c)
225
- (d)
230
Let total number of people in the restaurant be x.
Then, number of adults = \({3\over 8}x\)
\(\therefore\) Number of children = \(=({3\over 8}x+90)\)
According to question, \({3\over 8}x+({3\over 8}+90)=x\)
\(\Rightarrow x-{6x\over 8}=90\Rightarrow {2x\over 8}=90\Rightarrow x=360\)
\(\therefore\) Number of children = \({3x\over 8}\) + 90 = 135 + 90 = 225.
Which of the following options is INCORRECT?
- (a)
The rational number 0 is the additive identity for rational numbers
- (b)
The additive inverse of the rational number a/b is -a/b and vice-versa
- (c)
Rational numbers are closed under the operations of subtraction, multiplication and division
- (d)
There are infinite rational numbers between any two rational numbers
Match the following.
Column-I | Column-II |
---|---|
(P) Product of a rational number and its reciprocal is | -1 |
(Q) If \({12\over 30}\) and \({x\over 5}\) are equivalent, then x = | 0 |
(R) \([{8\over 21}-({-32\over 39}\div {16\over 13})]\times {7\over 4}=\) | 2 |
(S) Sum of a rational number and its additive inverse is | 1 |
- (a)
(P) \(\rightarrow\) (iv); (Q) \(\rightarrow\) (iii); (R) \(\rightarrow\) (i); (S) \(\rightarrow\) (ii)
- (b)
(P) \(\rightarrow\) (i); (Q) \(\rightarrow\) (iii); (R) \(\rightarrow\) (iv); (S) \(\rightarrow\) (ii)
- (c)
(P) \(\rightarrow\) (iv); (Q) \(\rightarrow\) (iii); (R) \(\rightarrow\) (ii); (S) \(\rightarrow\) (i)
- (d)
(P) \(\rightarrow\) (i); (Q) \(\rightarrow\) (iv); (R) \(\rightarrow\) (iii); (S) \(\rightarrow\) (ii)
(P) Product of a rational number and its reciprocal is 1.
(Q) \({12\over 30}={x\over 5}\times {6\over 6}\Rightarrow {12\over 30}={6x\over 30}\Rightarrow x=2\)
(R) We have, \([{8\over 21}\div ({-32\over 39}\div {16\over 13})]\times {7\over 4}=[{8\over 21}\div ({-32\over 39}\times {13\over 16})]\times {7\over 4}\)
\(=[{8\over 21}\div({-2\over 3})]\times {7\over 4}=[{8\over 21}\times {3\over -2}]\times {7\over 4}={-4\over 7}\times {7\over 4}=-1\)
(S) Sum of a rational number and its additive inverse is 0.
Fill in the blanks.
(i) 0 is neither ___P___ nor __Q____.
(ii) ___R____ has/have no reciprocal.
(iii) The rational numbers ____S____ and ____T____ are equal to their reciprocal.
- (a)
P Q R S T Positive negative 1 1/2 -1/2 - (b)
P Q R S T Integer rational 0 -1 0 - (c)
P Q R S T Positive negative 0 1 -1 - (d)
P Q R S T Natural integer -1 1 -1
Which of the following options holds?
Statement - 1: Rational numbers are closed under division.
Statement - 2: The value of \(({-7\over 18}\times {15\over -7})-(1\times {1\over 4})+({1\over 2}\times {1\over 4})\) is \({17\over 24}\)
- (a)
Both Statement - 1 and Statement - 2 are true
- (b)
Statement - 1 is true and Statement - 2 is false
- (c)
Statement - 1 is false but Statement - 2 is true
- (d)
Both Statement - 1 and Statement - 2 are false
Non-zero rational numbers are closed under division.
So, Statement-1 is false.
Now, \(({-7\over 18}\times {15\over -7})-(1\times {1\over 4})+({1\over 2}\times {1\over 4})={5\over 5}-{1\over 4}+{1\over 8}\)
\(={20-6+3\over 24}={17\over 24}\)
So, Statement -2 is true.
State 'T' for true and 'F' for false.
(i) The rational number \({-8\over -3}\) lies neither to the right nor to the left of zero on the number line
(ii) The rational numbers \({1\over 2}\) and \(-{5\over 2}\) are on the opposite sides of 0 on the number line.
(iii) 0 is the smallest rational number.
(iv) For every rational number x, x + 1 = x.
- (a)
(i) (ii) (iii) (iv) F T T F - (b)
(i) (ii) (iii) (iv) T T F F - (c)
(i) (ii) (iii) (iv) F T F F - (d)
(i) (ii) (iii) (iv) T T F F