Olympiad Mathematics - Linear Equations In One Variable
Exam Duration: 45 Mins Total Questions : 25
Solve for x: \({(3x+1)\over 16}+{(2x-3)\over 7}={(x+3)\over 8}+{(3x-1)\over 14}\)
- (a)
5
- (b)
10
- (c)
-14
- (d)
12
We have, \({(3x+1)\over 16}+{(2x-3)\over 7}={(x+3)\over 8}+{(3x-1)\over 14}\)
\(\Rightarrow {7(3x-1)+16(2x-3)\over 112}={14(x+3)+8(3x-1)\over 112}\)
\(\Rightarrow {21x+7+32x-48\over 112}={14x-42+24x-8\over 112}\)
\(\Rightarrow {53x-41\over 112}={38x+34\over 112}\Rightarrow\) 53x - 41 = 38x + 34
\(\Rightarrow\) 15x = 75 \(\Rightarrow x={75\over 15}=5\)
A number is 56 greater than the average of its third, quarter and one-twelfth. Find the number
- (a)
85
- (b)
64
- (c)
72
- (d)
40
Let the number be x. Then, One-third of x = \({x\over 3}\), Quarter of x = \({x\over 4}\), One-twelfth of x = \({x\over 12}\)
Average of third, quarter and one-twelfth of
\(x={({x\over 3}+{x\over 4}+{x\over 12})\over 3}={1\over 3}({x\over 3}+{x\over 4}+{x\over 12})\)
According to question, we have
\(\therefore x={1\over 3}({x\over 3}+{x\over 4}+{x\over 12})+56\Rightarrow x={x\over 9}+{x\over 12}+{x\over 36}+56\)
\(\Rightarrow x-{x\over 9}-{x\over 12}-{x\over 36}=56\Rightarrow \) 36x - 4x - 3x - x = 36 x 56
\(\Rightarrow\) 28x = 36 x 56 \(\Rightarrow\) x = \({36\times 56\over28}\Rightarrow\) x = 72
Hence, the number is 72.
If \({1\over 3}\) of a number is 10 less than the original number, then the number is ________.
- (a)
30
- (b)
15
- (c)
10
- (d)
27
Let the number be x.
According to question, we have x = \({1\over 3}x+10\)
\(\Rightarrow x-{1\over 3}x=10\Rightarrow {3x-x\over 3}=10\)
\(\Rightarrow {2x\over 3}=10\Rightarrow x={10\times 3\over 2}=15\)
Hence, the number is 15.
Solve for x: 6(3x + 2) - 5(6x - 1) = 6(x - 3) - 5(7x- 6) + 12x
- (a)
-1
- (b)
1
- (c)
0
- (d)
2
We have, 6 (3x + 2) - 5 (6x - 1)
= 6 (x - 3) - 5 (7x - 6) + 12x
\(\Rightarrow\) 18x + 12 - 30x + 5 = 6x - 18 - 35x + 30 + 12x
\(\Rightarrow\) 5x = -5 \(\Rightarrow\) x = -1
The number 299 is divided into two parts in the ratio 5: 8. The product of the numbers is ______.
- (a)
21140
- (b)
21294
- (c)
21160
- (d)
31294
Ratio of two parts = 5: 8
Smaller number = \({5\over 13}\times 299=115\)
Larger number = \({8\over 13}\times 299=184\)
So, the product of the number = 115 x 184 = 21160.
If \(({2\over 3})^{rd}\) of a number is 20 less than the original number, then the number is _________.
- (a)
60
- (b)
40
- (c)
80
- (d)
120
Let the number be x.
According to question, we have \({2x\over 3}=x-20\)
\(\Rightarrow {2x\over 3}-{x\over 1}=20\Rightarrow {2x-3x\over 3}=-20\)
\(\Rightarrow {-x\over 3}=-20\Rightarrow {2x-3x\over 3}=-20\)
\(\Rightarrow {-x\over 3}=-20\Rightarrow x=20\times 3=60\)
The perimeter of a rectangle is numerically equal to the area of rectangle. If width of rectangle is \(2{3\over 4}\) cm, then its length is _________.
- (a)
\({11\over 3}cm\)
- (b)
\({22\over 3}cm\)
- (c)
11 cm
- (d)
10 cm
Let the length of rectangle be x cm.
Width of rectangle = \(2{3\over 4}cm={11\over 4}cm\)
\(\therefore\) Area of rectangle = length x width = \(x\times {11\over 4}={11x\over 4}sq.cm\)
Also, perimeter of rectangle = 2(length + width)
\(=2({x-{11\over 4})=({2x+{11\over 2}})cm}\)
Now, according to question,
\({11x\over 4}=2x+{11\over 2}\Rightarrow {11x\over 4}-2x={11\over 2}\)
\(\Rightarrow {11x-8x\over 4}={11\over 2}\Rightarrow {3x\over 4}={11\over 2}\)
\(\Rightarrow 6x=4\times 11\Rightarrow x={4\times 11\over 6}={22\over 3}cm\)
A number whose seventh part exceeds its eighth part by 1, is ________.
- (a)
58
- (b)
56
- (c)
64
- (d)
68
Let the number be x.
According to question, we have
\({x\over 7}-{x\over 8}=1\Rightarrow {8x-7x\over 56}=1\Rightarrow x=56\)
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Then the original number is ________.
- (a)
53
- (b)
45
- (c)
92
- (d)
63
Let units place digit be x. So tens place digit = 9 - x.
\(\therefore\) The original number = 10(9 - x) + x
= 90 - 10x + x = 90 - 9x
Now, after interchanging the digits, the number formed
= 10(x) + 9 - x = 9x + 9
\(\therefore\) According to question, we have 90 - 9x - 27 = 9x + 9
\(\Rightarrow\) - 18x = - 54 \(\Rightarrow\) x = 3
So, the original number = 90 - 9 x 3 = 63.
The denominator of a rational number is greater than its numerator by 3. If 3 is subtracted from the numerator and 2 is added to its denominator, then the new number becomes 1/5. The original rational number is ___________.
- (a)
\(-{5\over 8}\)
- (b)
\({5\over 8}\)
- (c)
\({3\over 8}\)
- (d)
\(-{3\over 8}\)
Let numerator be x, then denominator is x + 3.
\(\therefore\) The original rational number = \({x\over x+3}\)
Now, according to question
\({x-3\over x+3+2}={1\over 5}\Rightarrow {x-3\over x+5}={1\over 5}\)
On cross multiplying, we get 5(x - 3) = x + 5
\(\Rightarrow\) 5x - 15 = x + 5 \(\Rightarrow\) 5x - x = 5 + 15
\(\Rightarrow 4x=20\Rightarrow x={20\over 4}=5\)
Hence, the original rational number is \({5\over 5+3}={5\over 8}\)
If \(x-(2x-{5x-1\over 3})={x-1\over 3}+{1\over 2}\) then, x is equal to ________.
- (a)
\({3\over 2}\)
- (b)
\({4\over 7}\)
- (c)
\({7\over 3}\)
- (d)
\({9\over 2}\)
We have, \(x-(2x-{5x-1\over 3})={x-1\over 3}+{1\over 2}\)
\(\Rightarrow x-({6x-5x+1\over 3})={2x-2+3\over 6}\Rightarrow {3x-x-1\over 3}={2x+1\over 6}\)
\(\Rightarrow\) 6(2x - 1) = 3(2x + 1) \(\Rightarrow\) 12x - 6 = 6x + 3
\(\Rightarrow 6x=9\Rightarrow x={9\over 6}={3\over 2}\)
A two digit number is less than 20. The sum of the digits is double that of their product. What is the number?
- (a)
12
- (b)
15
- (c)
13
- (d)
11
Since, 2 digit number is less then 20 so tens digit = 1.
Let ones digit of the number be x
According to question, we have
1 + x = 2(1 \(\times\) x) \(\Rightarrow\) 1 + x = 2x \(\Rightarrow\) x = 1
\(\therefore\) Required number = 11.
Find two parts of 34 such that \(({4\over 7})^{th}\) of one part is equal to \(({2\over 5})^{th}\) of the other.
- (a)
16, 18
- (b)
14, 20
- (c)
15, 19
- (d)
None of these
Let one part be x
Then, other part be 34 - x
According to question, we have
\({4\over 7}(x)={2\over 5}(34-x)\Rightarrow {4x\over 7}={68\over 5}-{2x\over 5}\Rightarrow {4x\over 7}+{2x\over 5}={68\over 5}\)
\(\Rightarrow {20x+14x\over 35}={68\over 5}\Rightarrow {34\over 35}\times x={68\over 5}\times {35\over 34}\)
\(\Rightarrow \) x = 2 x 7 \(\Rightarrow \) x = 14
\(\therefore\) One part = 14 and
Other part = 34 - 14 = 20.
If the angles of a triangle are in the ratio 2: 3: 4, then the difference between the greatest and the smallest angle is _________.
- (a)
10°
- (b)
20°
- (c)
30°
- (d)
40°
Let the angles of the triangle be 2x, 3x and 4x.
As sum of angles of a triangle is 180
\(\therefore\) 2x + 3x + 4x = 180° \(\Rightarrow\) 9x = 180° \(\Rightarrow x={180^\circ\over 9}=20^\circ\)
So, the angles are 2x = 2 x 20° = 40°
3x = 3 x 20° = 60°
4x = 4 x 20° = 80°
\(\therefore\) Difference between the greatest and smallest angles
= 80° - 40° = 40°
One-sixth of a number, when subtracted from the number itself gives 25. The number is __________.
- (a)
30
- (b)
32
- (c)
35
- (d)
28
Let the number be x
According to question, we have
\(x-{1\over 6}x=25\Rightarrow {6x-x\over 6}=25\)
\(\Rightarrow 5x=25\times 6\Rightarrow x={25\times 6\over 5}\Rightarrow x=30\)
So, the number is 30.
There were only two candidates in an election. One got 62% votes and was elected by a margin of 144 votes. The total number of voters were _________.
- (a)
500
- (b)
600
- (c)
700
- (d)
800
If there were only two candidates and one of them got 62% votes. So, other got (100 - 62)% = 38% votes.
Win margin of first candidate = (62 - 38)% = 24%
Now, let the total number of voters be x.
According to question, we have
24% of x = 144
\(\Rightarrow {24x\over 100}={144}\Rightarrow 24x=144\times 100\Rightarrow x={144\times 100\over 24}=600\)
So, the total number of voters = 600.
Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times that of Ashima's age. Find the sum of their ages two years ago.
- (a)
40 years
- (b)
42 years
- (c)
36 years
- (d)
38 years
Let age of Ashima be x years.
Then, age of Sunita is 2x years.
According to question, we have
4(x - 6) = 2x + 4
\(\Rightarrow\) 4x - 24 = 2x + 4 \(\Rightarrow\) 2x = 28 \(\Rightarrow\) x = 14
\(\therefore\) Age of Ashima = 14 years
Age of Sunita = 2 x 14 = 28 years
Sum of their ages = 42 years
Hence, two years ago they were 42 - 4 = 38 years.
At a party, colas, squash and fruit juice were offered to guests. One-fourth of the guests drank colas, One-third drank squash, two-fifths drank fruit juice and just three did not drink anything. How many guests were there in all?
- (a)
240
- (b)
180
- (c)
144
- (d)
190
Let total number of guests at the party be x
Number of guests who drank colas = \({1\over 4}x\)
Number of guests who drank squash = \({1\over 3}x\)
Number of guests who drank fruit juice = \({2\over 5}x\)
Number of guests who did not drank anything = 3
According to question, we have
\({1\over 4}x+{1\over 3}x+{2\over 5}x+3=x\Rightarrow {15x+20x+24x+180\over 60}=x\)
\(\Rightarrow\) 59x + 180 = 60x \(\Rightarrow\) x = 180
Hence, there were 180 guests at the party.
Two years ago, Mohit was three times as old as his son and two years hence, twice of Mohit's age will be equal to five times that of his son. Then the present age of Mohit is ____.
- (a)
14 years
- (b)
38 years
- (c)
32 years
- (d)
34 years
Let present age of Mohit be x years
Two years ago, his age = (x - 2) years
So, his son's age = \(({x-2\over 3})\) years
So, present age of son = \(({x-2\over 3}+2)\) years
After 2 years, son's age = \(({x-2\over 3}+2+2)\) years
and that of Mohit = (x + 2) years
So, according to question, we have
\(2(x+2)=5({x-2\over 3}+2+2)\Rightarrow 2x+4={5x-10\over 3}+20\)
\(\Rightarrow 2x-({5x-10\over 3})=20-4\Rightarrow {6x-5x+10\over 3}=16\)
\(\Rightarrow\) x + 10 = 48 \(\Rightarrow\) x = 48 - 10 = 38 years
\(\therefore\) Present age of Mohit = 38 years.
A steamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km/hr, find the speed of the steamer in still water.
- (a)
12 km/hr
- (b)
11 km/hr
- (c)
13 km/hr
- (d)
14 km/hr
Let the speed of the steamer in still water be x km/hr.
Speed of the stream = 1 km/hr
Downstream speed = (x + 1) km/hr
Upstream speed = (x - 1) km/hr
According to question, 5(x + 1) = 6(x - 1)
\(\Rightarrow\) 5x + 5 = 6x - 6
\(\Rightarrow\) 5 + 6 = 6x - 5x \(\Rightarrow\) x = 11
Hence, the speed of the steamer in still water is 11 km/hr.
Fill in the blanks.
(i) The solution of the equation ax + b = a is, ________
(ii) The shifting of a number from one side of an equation to other is ________
(iii) If a and b are positive integers then the solution of the equation ax = b has to be always ________
(iv) Linear equation in one variable has only one variable with power _________.
- (a)
(i) (ii) (iii) (iv) x = b/a commutativity positive 1 - (b)
(i) (ii) (iii) (iv) x = -b/a commutativity negative 2 - (c)
(i) (ii) (iii) (iv) x = b/a transposition negative 2 - (d)
(i) (ii) (iii) (iv) x = -b/a transposition positive 1
(i) We have, ax + b = 0
\(\Rightarrow ax=-b\Rightarrow x={-b\over a}\)
(ii) The shifting of a number from one side of an equation to other is called transposition.
(iii) We have, ax = b
\(\Rightarrow x={b\over a}\) Which is the required solution.
Given that a and b are positive integers.
\(\therefore\) Solution of the equation ax = b has to be always positive.
(iv) Linear equation is of the form ax + b = 0 i.e., only one variable with power 1.
Which of the following statements is CORRECT?
Statement -1: \(x={1\over 2}\) is the solution of \(({2x-3\over 4})-({2x-1\over 2})={x-2\over 3}\)
Statement -2: \(x={63\over 2}\) is the solution of \({2x-17\over 2}-({x-{x-1\over 3}})=12.\)
- (a)
Only Statement - 1
- (b)
Only Statement - 2
- (c)
Both Statement - 1 and Statement - 2
- (d)
Neither Statement - 1 nor Statement - 2
Statement-1: \(({2x-3\over 4})-({2x-1\over 2})={x-2\over 3}\)
\(\Rightarrow {2x-3-4x+2\over 4}={x-2\over 3}\Rightarrow {-2x-1\over 4}={x-2\over 3}\)
\(\Rightarrow\) - 6x - 3 = 4x - 8 \(\Rightarrow\) - 6x - 4x = - 8 + 3
\(\Rightarrow\) -10x = - 5 \(\Rightarrow\) x = \({1\over 2}\)
So, Statement-1 is true.
Statement-2: \({2x-17\over 2}-({x-{x-1\over 3}})=12.\)
\(\Rightarrow {2x-17\over 2}-({3x-x+1\over 3})=12\Rightarrow {3(2x-17)-2(2x+1)\over 6}=12\)
\(\Rightarrow\) 6x - 51 - 4x - 2 = 12 x 6 \(\Rightarrow\) 2x - 53 = 72
\(\Rightarrow\) 2x = 72 + 53 \(\Rightarrow\) x = \({125\over 2}\)
So, Statement-2 is false.
State 'T' for true and 'F' for false.
I. An altitude of a triangle is five-third the length of its corresponding base. If the altitude be increased by 4 cm and the base be decreased by 2 cm, the area of the triangle would remain the same. The base and the altitude of the triangle respectively is 12 cm and 20 cm.
II. The perimeter of a rectangle is 140 cm. If the length of the rectangle is increased by 2 cm and its breadth decreased by 2 cm, the area of the rectangle is increased by 66 sq. cm. The length and breadth of the rectangle respectively is 35 cm and 30 cm.
III. The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other number, then one of the numbers will be 1411.
- (a)
I II III F F F - (b)
I II III F T T - (c)
I II III T F F - (d)
I II III T F T
I. Let length of the base of a triangle corresponding to the altitude be x cm.
Altitude of the triangle = \({5\over 3}x\) cm.
Area of the triangle = \({1\over 2}\times x\times {5\over 3}x={5\over 6}x^2\)
According to question, we have \({1\over 2}(x-2)({5\over 3}x+4)={5\over 6}x^2\)
\(\Rightarrow ({1\over 2}x-1)({5\over 3}x+4)={5\over 6}x^2\Rightarrow {5\over 6}x^2+2x-{5\over 3}x-4={5\over 6}x^2\)
\(\Rightarrow {5\over 6}x^2+2x-{5\over 3}x={5\over 6}x^2+4\Rightarrow {6x-5x\over 3}=4\)
\(\Rightarrow {x\over 3}=4\Rightarrow x=12\)
\(\therefore\) Length of base of the triangle = 12 cm
Altitude of the triangle = \({5\over 3}\times 12=20cm\)
II. Let length of the rectangle be x cm
Perimeter of the rectangle = 140 cm
\(\Rightarrow\) 2(Length + Breadth) = 140 cm
\(\Rightarrow\) (x + Breadth) = \({140\over 2}=70\Rightarrow\) Breadth = (70 - x) cm
\(\therefore\) Area of the rectangle = Length x Breadth
= x \(\times\) (70 - x) = 70x - x2
According to question, we have
(x + 2) (70 - x - 2) = 70x - x2 + 66
\(\Rightarrow\) (x + 2) (68 - x) = 70x - x2 + 66
\(\Rightarrow\) 68x - x2 + 136 - 2x = 70x - x2 + 66
\(\Rightarrow\) 66x - x2 - 70x + x2 = 66 - 136
\(\Rightarrow\) -4x = -70 \(\Rightarrow\) x = 17.5
\(\therefore\) Length of the rectangle = 17.5 cm
Breadth of the rectangle = (70 - 17.5) cm = 52.5 cm.
III. Let the one number be x
Then, the other number be 2490 - x
6.5% of the first number = \(x\times {6.5\over 100}\)
8.5% of the second number = (2490 - x) x \({8.5\over 100}\)
According to question, we have
\({6.5\over 100}\times x=(2490-x)\times {8.5\over 100}\Rightarrow {6.5\over 100}\times {8.5\over 100}x={21165\over 100}\)
\(\Rightarrow {15x\over 100}={21165\over 100}\Rightarrow x={21165\over 15}\Rightarrow x=1411\)
Which of the following statements is INCORRECT?
- (a)
Kusum buys some chocolates at the rate of Rs 10 per chocolate. She also buys an equal number of candies at the rate of Rs 5 per candy. She makes a 20% profit on chocolates and 8% profit on candies. At the end of the day, all chocolates and candies are sold out and her profit is Rs 240. Therefore, Kusum buys 100 chocolates.
- (b)
A carpenter charged Rs 2500 for making a bed. The cost of materials used is Rs 1100 and the labour charges are Rs 200/hr. So, the carpenter will work for 7 hours.
- (c)
On dividing Rs 200 between A and B such that twice of A's share is less than 3 times B's share by 200. So, B's share is Rs 120.
- (d)
Madhulika thought of a number, double it and added 20 to it. On dividing the resulting number by 25, she gets 4. Hence, the required number is 45.
Let the number of chocolates purchased be x.
\(\therefore\) The number of candies purchased = x
Cost of x chocolates = Rs 10x
and cost of x candies = Rs 5x
According to question,
20% of 10x + 8% of 5x = Rs 240
\(\Rightarrow {20\over 100}\times 10x+{8\over 100}\times 5x=240\Rightarrow {200x+40x\over 100}=240\)
\(\Rightarrow 240x=24000 \Rightarrow x={24000\over 240}\Rightarrow x=100\)
Hence, the number of chocolates purchased is 100.
(B) Let the carpenter worked for x hours.
\(\therefore\) Labour charges for x hours = Rs 200x
According to question, 200x + 1100 = 2500
\(\Rightarrow\) 200x = 2500 - 1100 \(\Rightarrow\) 200x = 1400
\(\Rightarrow x={1400\over 200}\Rightarrow x=7\)
Hence, the carpenter worked for 7 hours.
(C) Let B's share be Rs x.
Since, 2 x A's share = 3 x B's share - 200
\(\Rightarrow\) A's share = \({3x-200\over 2}\)
According to question, \({3x-200\over 2}-x=200\)
\(\Rightarrow {3x\over 2}-{200\over 2}+x+200\Rightarrow {3x\over 2}-100-x=200\)
\(\Rightarrow {3x\over 2}+x=200+100\Rightarrow {3x+2x\over 2}=300\)
\(\Rightarrow {5x\over 2}=300\Rightarrow x={2\times 300\over 5}\Rightarrow x=120\)
Therefore, B's share = Rs 120
(D) Let the number be x.
Double the number and added 20 to it, we get 2x + 20
According to the question, \({1\over 25}(2x+20)=4\)
\(\Rightarrow\) 2x + 20 = 4 x 25 \(\Rightarrow\) 2x + 20 = 100
\(\Rightarrow\) 2x = 100 - 20 \(\Rightarrow\) 2x = 80
\(\Rightarrow x={80\over 2}\Rightarrow x=40\)
Match the following:
Column-I | Column-II |
---|---|
P. If \({5m\over 6}+{3m\over 4}={19\over 12}\) then m = | (i) \({1\over 6}\) |
Q. If \(2x+{3\over 4}={x\over 2}+1\) then x = | (ii) 36 |
R. If \({z\over 2}-{3z\over 4}+{5z\over 6}=21\) then z = | (iii) \({27\over 10}\) |
S. If \({y\over 2}-{1\over 5}={y\over 3}+{1\over 4}\) then y = | (iv) 1 |
- (a)
P \(\rightarrow\) (iii); Q \(\rightarrow\) (iv); R \(\rightarrow\) (i); S \(\rightarrow\) (ii)
- (b)
P \(\rightarrow\) (iv); Q \(\rightarrow\) (ii); R \(\rightarrow\) (iii); S \(\rightarrow\) (i)
- (c)
P \(\rightarrow\) (ii); Q \(\rightarrow\) (i); R \(\rightarrow\) (iii); S \(\rightarrow\) (iv)
- (d)
P \(\rightarrow\) (iv); Q \(\rightarrow\) (i); R \(\rightarrow\) (ii); S \(\rightarrow\) (iii)
P. \({5m\over 6}+{3m\over 4}={19\over 12}\)
\(\Rightarrow {10m+9m\over 12}={19\over 12}\Rightarrow 19m=19\Rightarrow m=1\)
Q. \(2x+{3\over 4}={x\over 2}+1\Rightarrow {4x-x\over 2}={4-3\over 4}\)
\(\Rightarrow 4(3x)=2\Rightarrow 12x=2\Rightarrow x={2\over 12}={1\over 6}\)
R. \({z\over 2}-{3z\over 4}+{5z\over 6}=21\Rightarrow {6z-9z+10z\over 12}=21\)
\(\Rightarrow 7z=21\times 12\Rightarrow z={21\times 12\over 7}\Rightarrow z=36\)
S. \({y\over 2}-{1\over 5}={y\over 3}+{1\over 4}\Rightarrow {y\over 2}-{y\over 3}={1\over 4}+{1\over 5}\)
\(\Rightarrow {3y-2y\over 6}={5+4\over 2}\Rightarrow {y\over 6}={9\over 20}\Rightarrow y={9\times 6\over 20}={9\times 3\over 10}={27\over 10}\)