Question 1.
Two numbers are in the ratio of 2 : 3 and the difference of their squares is 320. The number are : [1 Mark, Nov. 2006]
(a) 12, 18
(b) 16, 24
(c) 14, 21
(d) None
Answer:
Tricks : Go by choices (a); (b) & (c) all are in ratio 2:3 But For option
(a) 18² – 12² ≠ 320
(b) 24² – 16² = (24+ 16) (24 – 16)
= 40 × 8 = 320

Question 2.
If p : q is the sub-duplicate ratio of p – x²: q – x², then x² is : [1 Mark, Nov. 2006]
(a) pp+q
(b) qp+q
(c) qpp−q
(d) None
Answer:
Detail Method:
p−x2√q−x2√=pq
Squaring on both side; we get
p−x2q−x2=p2q2
or pq² – q²x² = p²q – p²x²
or p²x² – q²x² = p²q – pq²
or x²(p² – q²) = pq(p – q)
or x² (p + q)(p-q) = pq (p – q)
(d) is correct
Tricks : Go by choices.

Question 3.
An alloy is to contain copper and zinc in the ratio 9:4. The zinc required to melt with 24 kg of copper is : [1 Mark, Nov. 2006]
(a) 1023kg
(b) 1013kg
(c) 923kg
(d) 9 kg
Answer:
Let Zinc = x kg
(a) ∴ 94=24x ∴ x = 4×249=323
= 1023kg
∴ (a) is correct

Question 4.
Two numbers are in the ratio 7: 8. If 3 is added to each of them, their ratio becomes 8 : 9. The numbers are: [1 Mark, Feb. 2007]
(a) 14,16
(b) 24,27
(c) 21,24
(d) 16,18
Answer:
Tricks : Go by choices
(b) and (d) are not in the ratio 7 : 8 So (b) & (d) are not answer For (a) it is added then
14+310+3=1719≠89
(a) is not answer
(c) is answer Detail Method:
Let x is common in the ratio
∴ Numbers are 7x & 8x
Now 7x+38x+3=89
or 64x + 24 = 63x + 27
or 64x – 63x = 27-24
or x = 3
1st number = ix = 7×3 = 21
2nd number = 8x = 8×3 = 24
(c) is correct

Question 5.

A box contains ₹ 56 in the form of coins of one rupee, 50 paise and 25 paise. The number of 50 paise coin is double the number of 25 paise coins and four times the numbers of one rupee coins. The numbers of 50 paise coins in the box is : [1 Mark, Feb. 2007]
(a) 64
(b) 32
(c) 16
(d) 14
Answer:
Tricks : Go by choices
No. of 25 paise coins = 12 No. of 50 Paise
No. of Rupee coins = 14 No. of 50 Paise Coins
For (a) No. of coins of ₹ 1;50 Paise & 25 Paise
644; 64; 642
Total Value = 16 × 1 + 64 × 0.50 + 32 × 0.25
= ₹ 16 + 32 + 8 = ₹ 56
Which is equal to given value (a) is Correct

Detail Method
Let No. of 50 Paise coins = x
∴ No. of ₹ 1 coins = x4
and No. of 25 Paise coins = x2
∴ Total Value =
x4 × 1 + x × 0.50+ x2 × 0.25 = 56
or 0.25x + 0.50x + 0.125x
or 0.875x = 56
or x = 560.875 = 64
∴ (a) is correct

Question 6.
Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally entire cost of the car, then the share of each of the remaining persons increased by : [1 Mark, May 2007]
(a) 1/9
(b) 1/8
(c) 1/7
(d) 7/8
Answer:
Tricks : Per Person share increase = 17 of total share

Question 7.
A bag contains ₹ 187 in the form of 1 rupee, 50 paise and 10 paise coins in the ratio 3:4:5. Find the number of each type of coins: [1 Mark, May 2007]
(a) 102, 136, 170
(b) 136, 102, 170
(c) 170, 102, 136
(d) None
Answer:
Tricks I: Go by choices
For (a) Coins are in the ratio 3:4:5
∴ 1023=1364=1705 = 34
It satisfies 1st condition
Now 102 × 1 + 136 × 0.50 + 170 × 0.10
= 102 + 68 + 17 = ₹ 187
∴ (a) is correct
Tricks II
Common factor = 1873×1+4×0.50+5×0.10
= 1875.50 = 34
∴ No. of 1 Rupee coins =3 × 34 = 102
No. of 50 Paise coins =4 × 34 = 136
No. of 10 paise coins = 5 × 34 = 170 (a) is correct

Detail Method
Let x is common in the ratio
No. of 1 Rupee ; 50 Paise and 10 Paise Coins are 3x ; 4x and 5x
∴ 3x × 1 + 4x × 0.50 + 5x × 0.10 = 187
or 5.50x = 187
∴ x = 1875.50 = 34
∴ No. of 1 Rupee coins = 3x = 3 × 34 = 102
No. of 50 Paise coins = 4x = 4 × 34 = 136
No. of 10 Paise coins = 5x = 5 × 34 = 170
∴ (a) is correct

Question 8.
Ratio of earnings of A and B is 4 : 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earning becomes 8 : 7. What is A’s earning ?
(a) ₹ 21,000
(b) ₹ 26,000
(c) ₹ 28,000
(d) Data inadequate [1 Mark, Aug. 2007]
Answer:
Detailed Method Let x is common in the ratio
∴ A’s and B’s present earnings are 4x and 7x respectively
From question 4x+4x×0.507x−7x×0.25=87
or 6x5.25x=87
x cannot be found.
Data is inadequate
∴ (d) is Correct

Question 9.
P, Q and R are three cities. The ratio of average temperature between P and Q is 11 : 12 and that of between P and R is 9: 8. The ratio between the average temperature of Q and R is : [1 Mark, Aug. 2007]
(a) 22: 27
(b) 27: 22
(c) 32: 33
(d) None
Answer:
∴ P: Q= 11 : 12 ∴ Q :P= 12 : 11
QP×PR=1211×98
QR=2722 Q :P= 12 : 11
(b) is Correct

Question 10.
₹ 407 are to be divided among A, B and C so that their shares are in the ratio 14:15:16
The respective shares of A, B, C are: [1 Mark, Nov. 2007]
(a) ₹ 165, ₹ 132, ₹ 110
(b) ₹ 165, ₹ 110, ₹ 132
(c) ₹ 132, ₹ 110, ₹ 165
(d) ₹ 110, ₹ 132, ₹ 165.
Answer:
A:B:C = 14:15:16 × LCM of denominators = 60
= 15 : 12 : 10
∴ A’s share = 40715+12+10 × 15 = Rs. 165
B’s share = 40737 × 12 = Rs. 132
C’s share = 40737× 10 = Rs. 110
(a) is Correct
Tricks : Go by Choices.

Question 11.
The incomes of A and B are in the ratio 3 :2 and their expenditures in the ratio 5 : 3. If each saves ₹ 1,500, then B’s income is : [1 Mark, Nov. 2007]
(a) ₹ 6,000
(b) ₹ 4,500
(c) ₹ 3,000
(d) ₹ 7,500
Answer:
Detail Method Let x is common in the ratio.
A’s income = 3x
B’s income = 2x
∴ 3x−15002x−1500=53
or 10x – 7500 = 9x – 4500
or 10x – 9x = 7500 – 4500
or x = 3000
B’s income = 2x =2 × 3000
= ₹ 6000.
(a) is Correct Tricks : Go by choices For (a)
AB Expenditure = 60002×3−15006000−1500=75004500=53
(a) is Correct

Question 12.

In 40 litres mixture of glycerine and water, the ratio of glycerine and water is 3:1. The quantity of water added in the mixture in order to make this ratio 2:1 is : [1 Mark, Feb. 2008 ]
(a) 15 litres
(b) 10 litres
(c) 8 litres
(d) 5 litres
Answer:
Glycerine = 403+1 x 3 = 30 litres.
Water = 404 x 1 = 10 litres 4
Let x litres of water is added to the mixture
Then 3010+x=21
or, 2x + 20 = 30
or x = 5
∴ (d) is Correct
Tricks : Go by Choices

Question 13.
The third proportional between (a² – b²) and (a + b)² is : [1 Mark, Feb. 2008]
(a) a+ba−b
(b) a−ba+b
(c) (a−b)2a+b
(d) (a+b)3a−b
Answer:
3rd Proportion = ( Mean prop. )2 lst Proportional 
= {(a+b)2}2a2−b2=(a+b)43(a+b)(a+b)=(a+b)3a−b
(d) is Correct

Question 14.
In what ratio should tea worth ₹ 10 per kg. be mixed with tea worth ₹ 14 per kg., so that the average price of the mixture may be ₹ 11 per kg.? [1 Mark, June 2008]
(a) 2 : 1
(b) 3 : 1
(c) 3 : 2
(d) 4 : 3
Answer:
∴ (b) is Correct

Question 15.
The ages of two persons are in the ratio 5:7. Eighteen years ago their ages were in the ratio of 8:13, their present ages (in years) are : [1 Mark, June 2008]
(a) 50; 70
(b) 70,50
(c) 40,56
(d) None
Answer:
Tricks : Go by choices (a) & (c) are in the ratio 5 : 7 not (b)
For (a) 18 year ago
So, (a) is Correct

Question 16.
If A, B and C started a business by investing ₹ 1,26,000, ₹ 84,000 and ₹ 2,10,000. If at the end of the year profit is ₹ 2,42,000 then the share of each is : [1 Mark, Dec. 2008]
(a) ₹ 72,600; ₹ 48,400 ; ₹ 1,21,000
(b) ₹ 48,400 ; ₹ 1,21,000 ; ₹ 72,600
(c) ₹ 72,000 ; ₹ 49,000 ; ₹ 1,21,000
(d) ₹ 48,000 ; ₹ 1,21,400 ; ₹ 72,600
Answer:
Investment ratio is
A : B : C = 126,000 : 84,000 : 2,10,000 ÷ 14,000
= 9: 6 : 15 ÷ 3
= 3: 2: 5
A’s share = ₹242,0003+2+5 × 3 =₹ 72, 600
B’s share = 242,00020 × 2 = ₹48, 400
C’s share = 24200010 × 5 = ₹ 1,21,000
So, (a) is Correct

Question 17.
If pq=−23 then the value of 2p+q2p−q is: [1 Mark, June 2009]
(a) 1
(b) −17
(c) 17
(d) 7
Answer:
∵ pq=−23
Tricks
2p+q20−q=2(−2)+32(−2)−3=−4+3−4−3=−1−7=17
(c) is correct

Question 18.
Fourth proportional to x, 2x, (x +1) is :
(a) x + 2
(b) (x + 2)
(c) (2x + 2)
(d) (2x – 2)
Answer:
Let Fourth Proportional is K.
∴ x2x=x+1K
or k.x = 2x (x + 1)
or k = 2 (x + 1) = 2x +2
(c) is correct

Question 19.
What must be added to each term of the ratio 49 : 68 so that it becomes 3 : 4? [1 Mark, June 2010]
(a) 3
(b) 5
(c) 8
(d) 9
Answer:
Detail Method:
Let x is added to each term
Then 49+x68+x=34
or 196 +4x = 204 + 3x
or 4x – 3x = 204 – 196
or x = 8
(c) is Correct
Tricks : Go by Choices
1st Find 34 = 0.75 (By Calculator)
Question 20.

The students of two classes are in the ratios 5 : 7, if 10 students left from each class, the remaining students are in the ratio of 4 : 6, then the number of students in each class was : [1 Mark, June 2010]

(a) 30, 4
(b) 25, 24
(c) 40, 60
(d) 50, 70
Answer:
Tricks : Go by choices:
(a); (b) and (c) are not in the ratio 5 : 7
∴ (d) is Correct.

Question 21.
If A: B= 2: 5, then (10A + 3B) : (5A + 2B) is equal to: [1 Mark, Dec. 2010]
(a) 7: 4
(b) 7: 3
(c) 6: 5
(d) 7: 9
Answer:
It A : B = 2 : 5 Then
10A+3B5A+2B=10×2+3×55×2+2×5=3520=74
= 7: 4
(a) is Correct

Question 22.
In a film shooting, A and B received money in a certain ratio and B and C also received the money in the same ratio. If A gets ₹ 1,60,000 and C gets ₹ 2,50,000. Find the amount received by B ? [1 Mark, June 2011]
(a) ₹ 2,00,000
(b) ₹ 2,50,000
(c) ₹ 1,00,000
(d) ₹ 1,50,000
Answer:
Detail Method
A : B = B : C
So, B² = AC ;
so, B = AC−−−√=1,60,000×2,50,000−−−−−−−−−−−−−−−−√
= 400 × 500 = 2,00,000

Question 23.
The ratio compounded of 4:5 and sub-duplicate of “ a” : 9 is 8:15. Then value of “a” is: [1 Mark, Dec. 2011]
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 45×a9−−√=815
or 45×a√3=815
∴ √a = 2 ⇒ a = 4
(c) is Correct

Question 24.
If X varies inversely as square of Y and given that Y=2 for X = 1, then the value of X for Y =6 will be: [1 Mark, Dec. 2011]
(a) 3
(b) 9
(c) 1/3
(d) 6
Answer:
(d) is Correct
x × 1y2 ⇒ x = K.1y2 x = ky2, where k = proportional constant
where k = proportional constant
When x = 1 Then y = 2
1 = k22 ⇒ k = 4 ∴ x = 4y2
When y = 6, Then x = 462=19
x = 19

Question 25.
Which of the numbers are not in proportion ?
(a) 6, 8, 5, 7
(b) 7, 14, 6, 12
(c) 18, 27,12, 18
(d) 8, 6, 12, 9
Answer:
(a) Go by choices
For (a) 68=34≠57
(a) is not in proportion

Question 26.
Find two numbers such that mean proportional between them is 18 and third proportional between them is 144: [1 Mark, Dec. 2012]
(a) 9 ; 36
(b) 8 ; 32
(c) 7 ; 28
(d) 6 ; 14
Answer:
(a) is correct Tricks : Go by choices
For (a) Mean Proportional of 9 and 36
= 9×36−−−−−√ = 18
It satisfies 1st condition.
If 144 is its 3rd condition.
36² = 9 × 144
It also satisfies the 2nd Condition.

Question 27.
Triplicate ratio of 4 : 5 is: [1 Mark, June 2013]
(a) 125 : 64
(b) 16 : 25
(c) 64 : 125
(d) 120 : 46
Answer:
(c) Triplicate ratio of 4:5 = 4³ :5³ = 64: 125

Question 28.
The mean proportion between 24 and 54 is _______. [1 Mark, June 2013]
(a) 33
(b) 34
(c) 35
(d) 36
Answer:
(d) Mean – Proportion = 24×54−−−−−−√ = 36

Question 29.
The ratio of numbers is 1:2:3 and sum of their squares is 504 then the numbers are: [1 Mark, Dec. 2013]
(a) 6, 12, 18
(b) 3, 6, 9
(c) 4, 8, 12
(d) 5, 10, 15
Answer:
(a) is correct
Tricks : Go by choices
Tricks : See Quicker BMLRS

Question 30.
If P is 25% less than Q and R is 20% higher than Q the Ratio of R and P: [1 Mark, Dec. 2013]
(a) 5:8
(b) 8:5
(c) 5:3
(d) 3:5
Answer:
(b) is correct
Let Q = 100, So, P = 100 – 025 = 75
&R = 100 + 20= 120
RP=12075=85

Question 31.
A person has assets worth ₹ 1,48,200. He wish to divide it amongst his wife, son and daughter in the ratio 3:2:1 respectively. From this assets the share of his son will be: [1 Mark, June 2014]
(a) ₹ 74,100
(b) ₹ 37,050
(c) ₹ 49,400
(d) ₹ 24, 700
Answer:
(c) is correct
Share of son = 23+2+1 × 1,48,200
= ₹ 49,400

Question 32.
If x : y = 2 : 3 then (5x+2y): (3x -y) = [1 Mark, June 2014]
(a) 19 : 3
(b) 16: 3
(c) 7 : 2
(d) 7: 3
Answer:
(b) is correct
5x+2y3x−y=5×2+2×33×2−3=163

Question 33.
The first, second and third month salaries of aperson are in the ratio 2:4:5. The difference between the product of the salaries of first 2 months & last 2 months is ₹ 4,80,00,000. Find the salary of the second month [1 Mark, Dec. 2014]
(a) ₹ 4,000
(b) ₹ 6,000
(c) ₹ 12,000
(d) ₹ 8,000
Answer:
(d) is correct
Let x is common in the ratio.
1st, 2nd and 3rd month salaries of a person = 2x ; 4x ; 5x
From Qts.
4x × 5x – 2x × 4x = 4,80,00,000.
or, 12x² = 4,80,00,000.
or, x² = 4000000
x = 2000.
2nd month salary = 4x = 4×2000
= ₹ 8000

Question 34.
(2p² – q²) = 7pq, where p, q are positive then p : q. [1 Mark, June 2015]
(a) 5:6
(b) 5:7
(c) 3:5
(d) 3:7
Answer:
(a) is correct 15(2p² – q²) = 7pq
Tricks : Go by choices
For (a) put p = 5; q = 6 we get
15[2 × 5² – 6²] = 3 × 5 × 6
or 15 × 14 = 210
or 210 = 210

Question 35.

If one type of rice of cost ₹ 13.84 is mixed with another type of rice of cost ₹ 15.54, the mixture is sold at ₹ 17.60 with a profit of 14.6% on selling price then in which proportion the two types of rice mixed ? [1 Mark, June 2015]
(a) 3:7
(b) 5:7
(c) 7:9
(d) 9:1
Answer:
Cost of mixture per kg = 17.60 – 14.6% = 15.0304 = 15.03 (approx.)
By rules of Alligation

51: 119 = 3: 7
Go by choices
(a) is correct (approx)

Question 36.
Find the ratio of third proportional of 12 ; 30 and mean proportional of 9; 25 : [1 Mark, Dec. 2015]
(a) 7: 2
(b) 5 : 1
(c) 9 : 4
(d) None of these
Answer:
3rd proportional = 30212 = 75
Mean Proportional = 9×25−−−−−√ =15
Ratio = 7515 = 5:1
(b) is correct

Question 37.
What must be added to each of the numbers 10, 18, 22, 38 to make them proportional: [1 Mark, Dec. 2015]
(a) 5
(b) 2
(c) 3
(d) 9
Answer:
(b) is correct
let x be added.
∴ 10+x18+x=22+x38+x
Tricks: Go by choices.
∴ x = 2 satisfies it.

Question 38.
x, y, z together starts a business, if x invests 3 times as much as y invests and y invests two third of what z invests, then the ratio of capitals of x, y, z is: [1 Mark, June 2016]
(a) 3 : 9 : 2
(b) 6: 3: 2
(c) 3 : 6 : 2
(d) 6: 2: 3
Answer:(d)
Tricks: Go by choices
6 = 3 × 2 and 2 = 3 × 23

Question 39.
A bag contains 23 number of coins in the form of 1 rupee, 2 rupee and 5 rupee coins. The total sum of the coins is ?₹ 43. The ratio between 1 rupee and 2 rupees coins is 3 : 2. Then the number of 1 rupee coins is: [1 Mark, Dec. 2016]
(a) 12
(b) 8
(c) 10
(d) 16
Answer:
(a)
Tricks : Go by choices
Let option (a) is correct.
Let x is common in the ratio.
So, ₹ 1 coins = 3x = 12 ; So, x = 4
No. of ₹ 2 coins = 2 × 4 = 8
Hence no. of coins of ₹ 5 coins = 23 – 12 – 8= 3
Total money = 12 × 1 + 8 × 2 + 3 × 5 = ₹ 43
Satisfied. So (a) is correct.

Question 40.
If a: b = 2: 3, b : c = 4: 5, c: d = 6: 7 then a : d is: [1 Mark, June 2017]
(a) 24 : 35
(b) 8 : 15
(c) 16 : 35
(d) 7 : 15
Answer:
Option (c) is correct.
Multiply all ratios.
= 23×45×67=1635

Question 41.
The ratio of the number of five rupee coins to number of ten rupee coins is 8: 15. If the total value of five rupee coins is 360, then the no. of ten rupee coins is _______. [1 Mark, Dec. 2017]
Answer:
Option (d) is correct.
Total No. of ₹ 5 coins = 360/5 = 72
Let x is common in the ratio.
So, ₹ 5 coins = 8x = 72 ; So, x = 9
No. of ₹ 10 coins = 15 × 9 = 135

Question 42.
If 12,13,15,1x are in proportion then x = . [1 Mark, Dec. 2017]
(a) 152
(b) 315
(c) 215
(d) 115
Answer:
Option (a) is correct.
Product of middle two terms = Product of extremes
So, 12x=115; x = 15/2

Question 43.
If (a + b): (b + c): (c + a) = 7 : 8 : 9 and a + b + c = 18 then a : b : c = . [1 Mark, June 2018]
(a) 5 : 4 : 3
(b) 3 : 4 : 5
(c) 4 : 3 : 5
(d) 4: 5 : 3
Answer:
(c) 4 : 3 : 5 is correct
Tricks: Go by choices.
(c) Let a : b : c = 4 : 3 : 5
It is in ratio. So, it should must satisfy given ratio (a + b): (b + c): (c + a) = 7 : 8 : 9
i.e. (4 + 3): (3 + 5): (5 + 4) = 7 : 8 : 9 (True) Avoid 2nd condition.
In detail it will take too much time.

Question 44.
If p : q is the sub-duplicate ratio of p – x²:q – x², then x² is : [1 Mark, May 2018]
(a) pp+q
(b) qp+q
(c) qpp−q
(d) None
Answer:

Question 45.
The mean proportional between 24 and 54 is : [1 Mark, May 2018]
(a) 33
(b) 34
(c) 35
(d) 36
Solution:
Formula
Mean Proportion of a & b = ab−−√
(d) = 24×54−−−−−−√ = 36

Question 46.
3x−25x+6 is the duplicate ratio of 23 then find the value of x : [1 Mark, Nov. 2018]
(a) 6
(b) 2
(c) 5
(d) 9
Answer:
(a)
Given 3x−25x+6=(23)2=49
Tricks : Go by choices
for option (a) putting x = 6 in LHS; we get
3×2−25×2+6=49
∴ (a) is correct.

Question 47.
If x : y : z = 7 : 4 : 11 then is: [1 Mark, Nov. 2018]
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a)
x+y+zz=7+4+1111 = 2

Question 48.
If the ratio of two numbers is 7 : 11. If 7 is added to each number then the new ratio will be 2 : 3 then the numbers are. [1 Mark, June 2019 ]
(a) 49,77
(b) 42,45
(c) 43,42
(d) 39,40
Answer:
Tricks:- GBC (Go by Choices)
(a)
49 ÷ 7 = 7
77 ÷ 11 = 7 both must be equal.
Here it is correct.
Now:
49+777+7=5684=23
Divide 56 by numerator (2) and 84 by Denominator (3) we get same value “28”
Note:- No need to solve ; only check.
By Calculator.