Practice Ratio and proportion - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) When a particular number is subtracted from each of 7, 9, 11 and 15, the resulting numbers are in proportion. The number to be subtracted is :
(a)
(b)
(c)
(d)
Let the number to be subtracted be x.
According to the question,
${7 - x}/{9 - x} ={11 - x}/{15 - x}$
Now, check through options
Clearly, putting x = 3,
Each ratio = $2/3$.
Note : Solve such questions orally by mental exercise.
Using Rule 32Let 'x' be a number which is subtracted from a, b, c and d to make them proportional, thenx = ${ad - bc}/{(a+d) - (b+c)}$Let 'x' be a number which is added to a, b, c and d to make them proportional, thenx = ${bc - ad}/{(a+d) - (b+c)}$Here, a, b, c and d should always be in ascending order.
The number will be x
= ${ad - bc}/{(a+d) - (b+c)}$
= ${7 × 15 - 9 × 11}/{(7 + 15) - (9 + 11)}$
= ${105 - 99}/{22 - 20} = 6/2$ = 3
Q-2) The number to be added to each of the numbers 7, 16, 43, 79 to make the numbers in proportion is
(a)
(b)
(c)
(d)
From the given options number = 5, because
${7 + 5}/{16 + 5} = {43 + 5}/{79 + 5}$
$12/21 = 48/84$
[check other options likewise]
Using Rule 32,
Here, a = 7, b=16, c = 43, d= 79
Required number
x= ${bc - ad}/{(a+d) - (b+c)}$
= ${16 × 43 - 7 × 79}/{(7 + 79) - (16 + 43)}$
= ${688 - 553}/{86 - 79}$
= $35/7$ = 5
Q-3) Which number when added to each of the numbers 6, 7, 15, 17 will make the resulting numbers proportional ?
(a)
(b)
(c)
(d)
Let required number be x.
${6 +x}/{7 + x} = {15 + x}/{17 + x}$
$102 + 17x + 6x + x^2$
= $105 + 7x + 15x + x^2$
23x - 22x = 105 - 102
x = 3
Note : It is convenient to solve it orally using options
${6 + 3}/{7 + 3} = {15 + 3}/{17 + 3}$
= $9/10 = 18/20$
Using Rule 32Let 'x' be a number which is subtracted from a, b, c and d to make them proportional, thenx = ${ad - bc}/{(a+d) - (b+c)}$Let 'x' be a number which is added to a, b, c and d to make them proportional, thenx = ${bc - ad}/{(a+d) - (b+c)}$Here, a, b, c and d should always be in ascending order.
Required Number
= ${bc - ad}/{(a+d) - (b+c)}$
Where a = 6, b = 7, c = 15, d = 17
= ${7 × 15 - 6 × 17}/{(6 + 17) - (7 + 15)}$
= ${105 - 102}/{23 - 22}$ = 3
Q-4) What number should be subtracted from both terms of the ratio 15 : 19 in order to make it 3 : 4 ?
(a)
(b)
(c)
(d)
Let x be subtracted from each term of $15/19$.
${15 - x}/{19 - x} = 3/4$
57 - 3x = 60 - 4x
x = 3
Q-5) What number should be added to each of 6, 14, 18 and 38 so that the resulting numbers make a proportion ?
(a)
(b)
(c)
(d)
${6 + x}/{14 + x} ={18 +x}/{38 + x}$
From the given alternatives
${6 + 2}/{14 + 2} = {18 + 2}/{38 + 2}$
$1/2 = 1/2$
Using Rule 32,
Here, a = 6, b = 14, c = 18, d = 38
Required number x
= ${bc - ad}/{(a+d) - (b+c)}$
= ${14 × 18 - 6 × 38}/{(6 + 38) - (14 + 18)}$
= ${252 - 228}/{44 - 32}$
= $24/12$ = 2
Q-6) There are 225 consisting of one rupee, 50 paise and 25 paise coins. The ratio of their numbers in that order is 8 : 5 : 3. The number of one-rupee coins is :
(a)
(b)
(c)
(d)
Ratio of the number of coins
= 8 : 5 : 3
Ratio of their values
= $8 : 5/2 : 3/4$ = 32 : 10 : 3
Sum of the ratios
= 32 + 10 + 3 = 45
Value of one rupee coins
= $32/45 × 225$ = Rs.160
Number of one rupee coins = 160
Q-7) If 378 coins consist of rupees, 50 paise and 25 paise coins, whose values are in the ratio of 13 : 11 : 7, the number of 50 paise coins will be :
(a)
(b)
(c)
(d)
The ratio of values of rupee, 50 paise and 25 paise coins
= 13 : 11 : 7
Ratio of their numbers
= 13 × 1 : 11 × 2 : 7 × 4
= 13 : 22 : 28
Sum of the ratios
= 13 + 22 + 28 = 63
Required number of 50 paise coins
= $22/63 × 378$ = 132
Q-8) A bag contains three types of coins-rupee-coins. 50p-coins and 25 p-coins totalling 175 coins. If the total value of the coins of each kind be the same, the total amount in the bag is
(a)
(b)
(c)
(d)
Let the number of coins of 1- rupee coin be x.
Total value of the coins of each kind is same, then the number of 50 paisa coins
= 2x and the number of 25 paisa coins = 4x.
According to the question.
x + 2x + 4x = 175
7x = 175
$x = 175/7 = 25$
Total amount in bag
= 25 + 25 + 25 = Rs.75
Q-9) A man has in all 640 in the denominations of one-rupee, five rupee and ten-rupee notes. The number of each type of notes are equal. What is the total number of notes he has ?
(a)
(b)
(c)
(d)
Let the number of each type of notes be x
x + 5x + 10 x = 640
16 x = 640
x = 40
Total number of notes
= 3 × 40 = 120
Q-10) 180 contained in a box consists of one rupee, 50 paise and 25 paise coins in the ratio 2 : 3 : 4. What is the number of 50 paise coins?
(a)
(b)
(c)
(d)
Using Rule 1It does not change the ratio, when we multiply or divide antecedent and consequent of the ratio by a same non–zero number as–e.g. a : b = $a/b = {a×c}/{b×c}$ = ac : bc = a : b
Ratio of the values
= $2 : 3/2 : 4/4$ = 4 : 3 : 2
Value of 50 paise coins
= $3/9 ×$ 180 = Rs.60
Numbers of 50 paise coins = 120.
Q-11) The ratio between Sumit’s and Prakash’s age at present is 2 : 3. Sumit is 6 years younger than Prakash. The ratio of Sumit’s age to Prakash’s age after 6 years will be
(a)
(b)
(c)
(d)
Sumit’s present age
= 2x years
Prakash’s present age
= 3x years
3x - 2x = 6 ⇒ x = 6
Required ratio
= (2 × 6 + 6) : (3 × 6 + 6)
= 18 : 24 = 3 : 4
Q-12) The ratio of the age of Ram and Rahim 10 years ago was 1 : 3. The ratio of their age five years hence will be 2 : 3. Then the ratio of their present age is
(a)
(b)
(c)
(d)
Let the age of Ram and Rahim 10 years ago be x and 3x years respectively.
After 5 years from now,
${x + 15}/{3x +15} = 2/3$
6x + 30 = 3x + 45
3x = 45 - 30 = 15
x = 5
Ratio of their present age
= (x + 10) : (3x + 10)
= 15 : 25 = 3 : 5
Q-13) If the sum of two quantities is equal to three times their difference, then the ratio of the two quantities is
(a)
(b)
(c)
(d)
x + y = 3 (x - y)
x + y = 3x - 3y
2x = 4y
$x/y = 2/1$
x : y = 2 : 1
Q-14) Two numbers whose sum is 84 can not be in the ratio
(a)
(b)
(c)
(d)
According to the question,
The number 84 must be a multiple of sum of the terms of ratio.
For ratio 3 : 2,
Sum of the terms of ratio
= 3 + 2 = 5
which is not a factor of 84.
Q-15) If the square of the sum of two numbers is equal to 4 times of their product, then the ratio of these numbers is :
(a)
(b)
(c)
(d)
$(x + y)^2 = 4xy$
$x^2 + y^2 + 2xy - 4xy = 0$
$(x - y)^2$ = 0 ⇒ x = y
x : y = 1 : 1
Q-16) The present age of A and B are in the ratio 4 : 5 and after 5 years they will be in the ratio 5 : 6. The present age of A is
(a)
(b)
(c)
(d)
Let the present age of A and B be 4x and 5x years respectively,
According to the question,
${4x + 5}/{5x + 5} = 5/6$
25x + 25 = 24x + 30
x = 30 - 25 = 5
A’s present age
= 4x = 4 × 5 = 20 years
Q-17) Four years ago, the ratio of A’s age to B’s age was 11 : 14 and four years later their age will be in the ratio 13 : 16. The present age of A is
(a)
(b)
(c)
(d)
Let the age of A and B four years ago be 11x and 14x years respectively.
According to the question,
After 4 years from now,
${11x + 8}/{14x + 8} = 13/16$
176x + 128 = 182x + 104
182x - 176x = 128 - 104
6x = 24 ⇒ $x = 24/6$ = 4
A’s present age = (11x + 4) years
= 11 × 4 + 4 = 48 years
Q-18) The present ages of A and B are in the ratio 5 : 6 respectively. After seven years this ratio becomes 6 : 7. Then the present age of A in years is :
(a)
(b)
(c)
(d)
A’s present age = 5x years
B’s present age = 6x years
According to the question,
After 7 years,
${5x + 7}/{6x + 7} = 6/7$
36x + 42 = 35x + 49
36x - 35x = 49 - 42
x = 7
A’s present age
= 5x = 35 years
Q-19) The ratio of the present age of Rahul and Rashmi is 2 : 1. The ratio of their age after 30 years will be 7 : 6. What is the present age of Rahul ?
(a)
(b)
(c)
(d)
Let the present age of Rahul and Rashmi be 2x and x years respectively.
After 30 years,
${2x + 30}/{x +30} = 7/6$
12 x +180 = 7x + 210
12 x - 7x = 210 - 180
5 x = 30 ⇒ $x = 30/5 = 6$
Rahul’s present age
= 2x = 2×6 = 12 years
Q-20) The ratio of the ages of A and B at present is 3:1. Four years earlier the ratio was 4:1. The present age of A is
(a)
(b)
(c)
(d)
A’s present age = 3x years
B’s present age = x years
4 years ago,
${3x - 4}/{x -4} = 4/1$
4x - 16 = 3x - 4
4x - 3x = 16 - 4
x = 12
A’s present age
= 3x = 3 × 12 = 36 years