Comparing Quantities Class 7 MCQ with Answers

Answer: (c) 44 %

Explanation: To convert a Fraction into Percentage, we need to multiply the Fraction by 100, and assign the Percentage (%) sign to it
that is, ((44/100)x 100 ) % = 44 %

Answer: (b) 22/25

Explanation: In order to convert a Percentage into a Fraction, we divide the Percentage by 100, and remove the sign of Percentage (%).
We have,
88 % = (88/100)

On Simplifying, 88/100

HCF of 88 and 100 is 4

Divide both 88 and 100 by their HCF
(88÷4)/(100÷4) = 22/25

Answer: (c) 32 %

Explanation: To convert a given decimal, into Percentage, we have to multiply the given decimal by 100 and put the Percentage sign to the answer (%).

i.e.,
0.32 = ( 0.32 x 100 ) % = 32 %

Hence, 0.32 = 32 %

Answer: (a) 1 : 5

Explanation: In order to express a Percentage as a ratio, we need to write

First term /Second term = Given Percentage /100 = 20/100

On Simplifying, 20/100

HCF of 20 and 100 is 20

Divide both 20 and 100 by their HCF

(20÷20)/(100÷20) = 1/5 = 1 : 5

Answer: (a) 10 %

Explanation: To convert the given ratio into Percentage we have to follow following steps:

Step 1 – Convert the ratio into Fraction

12 : 120 = 12/120

Step 2 – Multiply the Fraction by 100 and put the Percentage sign to the result (%)

( (12/120) x 100 ) %

On solving the given Fraction, we get,

= 10 %

Answer: (a) 28

Explanation: Step 1 – Convert, the given Percentage into Fraction, by dividing the given number by 100

i.e., 7 % = 7100

Step 2 – Multiply the Fraction with given whole number.

i.e., (7/100) x 400 = 28

Hence, 7 % of 400 = 28

Answer: (c) 0.35

Explanation: In order to convert given Percentage into into decimal we have to perform the following steps:

Step 1 – Convert the Percentage into Fraction, by dividing the given Percentage by 100

i.e., 35 % = 35/100

Step 2 – Convert the Fraction obtained at Step 1, into a Decimal form , by dividing numerator by Denominator

i.e., 35/100 = 0.35

Hence, 35 % = 0.35

Answer: (c) ₹ 40000

Explanation: The problem can be solved using linear equation.

Let the unknown sum of money be ₹ a

We know, that 65 % of ₹ a = ₹ 26000

65 % x a = ₹ ( 26000 )

(65/100) x a = ₹ ( 26000 )

a = ₹ ( (26000×100)/65 )

= ₹ 40000

Therefore, the unknown sum of money available with the man is ₹ 40000

Answer: (b) 4950 metres

Explanation: Step 1 – Convert the Percentage into Fraction

i.e., 99 % = 99/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ((99/100) x 5000 ) metres = 4950 metres

Hence, 99 % of 5000 metres = 4950 metres

Answer: (c) 15 litres

Explanation: Step 1 – Convert the Percentage into Fraction

i.e., 5 % = 5/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ( (5/100) x 300 ) litres = 15 litres

Hence, 5 % of 300 litres = 15 litres

Answer: (c) 1128 kg

Explanation: Step 1 – Convert the Percentage into Fraction

i.e., 47 % = 47/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ((47/100) x 2400 ) kg = 1128 kg

Hence, 47 % of 2400 kg = 1128 kg

Answer: (c) ₹ 276

Explanation: Step 1 – Convert the Percentage into Fraction

i.e., 23 % = 23/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ₹ (( 23/100) x 1200 ) = ₹ 276

Hence, 23 % of ₹ 1200 = ₹ 276

Answer: (c) 35 %

Explanation: Let a % of 400 m = 140 m

Step 1 – Convert the Percentage into Fraction

i.e., a % = a/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ((a×400)/100 ) m = ( 140 ) m

((a×400)/100 ) = 140

a = (140×100)/400

a = 35

Hence, 140 m is 35 % of 400 m.

Answer: (c) 20 %

Explanation: Let a % of 280 kg = 56 kg

Step 1 – Convert the Percentage into Fraction

i.e., a % = a/100

Step 2 – Multiply the Fraction with given quantity.
i.e., ( (a×280)/100 ) kg = ( 56 ) kg

( (a×280)/100 ) = 56

a = (56×100)/280

a = 20
Hence, 56 kg is 20 % of 280 kg.

Answer: (b) 30 %

Explanation: Let a % of 50 litres = 15 litres

Step 1 – Convert the Percentage into Fraction

i.e., a % = a/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ( (a×50)/100 ) litres = ( 15 ) litres

(a×50)/100 = 15

a = (15×100)/50

a = 30

Hence, 15 litres is 30 % of 50 litres.

Answer: (a) 110

Explanation: Here we take the help of linear equation. First we will form the equation.

Let the number be ‘a’

70 % of ‘a’ = (70/100)a

Since, the equation is:

a – (70/100)a = 33

Taking 100 as LCM on LHS

(100a−70a)/100 = 33

(30/100)a = 33

a = (33×100)/30

a = 110

Hence, the number is 110.

Answer: (a) 80

Explanation: The problem can be solved using linear equation.

Let the number be a’

60 % of a’ = (60/100)a

We are given that a + (60/100)a = 128

Taking 100 as LCM on LHS

(100a+60a)/100 = 128

(160/100)a = 128

a = (128×100)/160

a = 80

Hence, the number is 80.

Answer: ₹ 1200

Explanation: Given:

P = ₹ 15000

R = 8% p.a.

T = 1 years

Where,

P = Principal

R = Rate

T = Time

As we know that,

Simple Interest = (P×R×T)/100

Simple Interest = ₹ (15000×8×1)/100

Simple Interest = ₹ 1200

Hence, Simple Interest = ₹ 1200.

Answer: (b) ₹ 75

Explanation: We know that

SI = (P×R×T)/100

Where,

P = Principal in Rs.

R = Rate ( In percentage per annum or percentage per month )

T = Time ( In Years, if Rate is given in Years or months, if Rate is given per month )

We are given that

P = ₹ 5000

R = 6% p.a.

T = 3 months

Since the rate of interest has been given on a yearly basis, we need to convert the time into Years

T = 3/12 years

T = 1/4 years

We know that,

SI = (P×R×T)/100

SI = ₹ [ 5000 x 6 x (1/4) x (1/100) ]

SI = ₹ 75

Hence, SI = ₹ 75.

Answer: (c) ₹ 32

Explanation: Given:

SI = (P×R×T)/100

Where,

P = Principal (in Rs.)

R = Rate

T = Time

P = ₹ 1000

R = 8% p.a.

T = 146 days

Since the rate of interest has been given on a yearly basis, we need to convert the time into Years

T = (146/365) years

T = (2/5) years

We know that,

Simple Interest = (P×R×T)/100

Simple Interest = ₹ [ 1000 x 8 x (2/5) x (1/100) ]

Simple Interest = ₹ 32

Hence, Simple Interest = ₹ 32.

Answer: (a) ₹ 12840

Explanation: We know that

Amount Payable at the end of Period = Principal + Simple Interest

Further

Simple Interest = (P×R×T)/100

Where,

P = Principal

R = Rate

T = Time

Given that

P = ₹ 12000

R = 7% p.a.

T = 1 years

Simple Interest = (P×R×T)/100

Simple Interest = ₹ (12000×7×1)/100

Simple Interest = ₹ 840

We know that

Amount Payable at the end of Period = Principal + Simple Interest

= ₹ ( 12000 + 840 )

= ₹ 12840

Hence, the Amount that would be payable would be ₹ 12840.

Answer: (a) ₹ 10200

Explanation: We know that

Amount Payable at the end of Period = Principal + Simple Interest

Further

Simple Interest = (P×R×T)/100

Where,

P = Principal

R = Rate

T = Time

P = ₹ 10000

R = 8% p.a.

T = 3 months

Since the rate of interest has been given on a yearly basis, we need to convert the time into Years

T = (3/12) year

T = (1/4) year

We know that,

Simple Interest = (P×R×T)/100

Simple Interest = ₹ [ 10000 x 8 x (1/4) x (1/100) ]

Simple Interest = ₹ 200

As we know that,

Amount Payable at the end of Period = Principal + Simple Interest

= ₹ ( 10000 + 200 )

= ₹ 10200

Hence, the Amount is ₹ 10200

Answer: (a) ₹ 4128

Explanation: We know that

Amount Payable at the end of Period = Principal + Simple Interest

Further

Simple Interest = (P×R×T)/100

Where,

P = Principal

R = Rate

T = Time

P = ₹ 4000

R = 8% p.a.

T = 146 days

Since the rate of interest has been given on a yearly basis, we need to convert the time into Years

T = (146/365) years

T = (2/5) years

We know that,

Simple Interest = (P×R×T)/100

Simple Interest = ₹ [ 4000 x 8 x (2/5) x (1/100) ]

Simple Interest = ₹ 128

Amount Payable at the end of Period = Principal + Simple Interest

= ₹ ( 4000 + 128 )

= ₹ 4128

Hence, the Amount is ₹ 4128.

Answer: (c) 10% p.a.

Explanation:
Given:

Principal = ₹ 12000

Amount = ₹ 13200

Time = 1 years

Simple Interest = ?

Rate = ?

We know that,

SI = Amount – Principal

SI = ₹ ( 13200 – 12000 )

SI = ₹ 1200

Thus,

Principal = ₹ 12000

SI = ₹ 1200

Time = 1 years

As we know that,

SI = (P×R×T)/100

or in other words

R = ( (100×SI)/(P×T))% p.a.

R = ( (100×1200)/(12000×1))% p.a.

R = 10% p.a.

Hence, the required rate of interest is 10% p.a.

Answer: (c) 10% p.a.

Explanation: Given:

P = ₹ 2000

Simple Interest = ₹ 200

T = 1 years

R = ?

Where,

P = Principal

R = Rate

T = Time

SI = Simple Interest

We know that,

SI = (P×R×T)/100

then,

R = (( 100×SI)/(P×T))% p.a.

R = ( (100×200)/(2000×1))% p.a.

R = 10% p.a.

Hence, required rate of interest is 10% p.a.

Answer: (b) 1 years

Explanation: Given:

Principal = ₹ 12000

Amount = ₹ 12960

Rate = 8% p.a.

Simple Interest = ?

Time = ?

We know that,

SI = Amount – Principal

SI = ₹ ( 12960 – 12000 )

SI = ₹ 960

Thus,

Principal = ₹ 12000

SI = ₹ 960

Rate = 8% p.a.

We know that,

SI = (P×R×T)/100

or in other words

T = ( (100×SI)/(P×R)) years

T = ( (100×960)/(12000×8)) years

T = 1 years

Hence, the time period is 1 years.

Answer: (b) 1 years

Explanation: Given that:

Principal = ₹ 5000

Simple Interest = ₹ 600

Rate of Interest = 12% p.a.

Time = ?

We know that,

Simple Interest =( P×R×T)/100

then,

Time = (100×SI)/(P×R)

Time = ( (100×600)/(5000×12))years

Time = 1 years

Hence, required time is 1 years

Answer: (a) 15% p.a.

Explanation: Let the sum be ₹ a

then, Amount = ₹ 4a

thus,

Principal = ₹ a

Amount = ₹ 4a

SI = Amount – Principal

= ₹ ( 4a – a )

= ₹ 3a

Time = 20 years

Thus,

P = ₹ a

SI = ₹ 3a

T = 20 years

We know that,

SI = (P×R×T)/100

or in other words

R = ( (100×SI)/(P×T))% p.a.

R = ( (100×3a)/(a×20))% p.a.

R = ( 300a/20a)% p.a.

R = 15% p.a.

Hence, the required rate of interest at which the given sum would double itself in 20 years 15% p.a.

Answer: (a) ₹ 215

Explanation: Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which the Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

In the present case,

CP = ₹ 675

SP = ₹ 890

We know that,

Gain = SP – CP

Gain = ₹ ( 890 – 675 )

Gain = ₹ 215

Hence, Gain = ₹ 215

Answer: (b) ₹ 140

Explanation: Cost price(CP) – Price at which an article is purchased.

Selling price(SP) – Price at which an article is sold.

Loss – Where the Cost Price is greater than the Selling Price, the difference between the SP and CP is the Loss

In present case,

CP = ₹ 663

SP = ₹ 523

We know that,

Loss = CP – SP

Loss = ₹ ( 663 – 523 )

Loss = ₹ 140

Answer: (c) 25 %

Explanation: Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which the Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

Gain % = (( Gain/CP) x 100 ) %

In the present case,

CP = ₹ 1200

SP = ₹ 1500

We know that,

Gain = SP – CP

Gain = ₹ ( 1500 – 1200 ) = ₹ 300

Further,

Gain % = ( ( Gain/CP) x 100 ) %

Gain % = ( (300/1200) x 100 ) % = 25 %

Hence, Gain % = 25 %

Answer: (b) 10 %

Explanation: Cost price(CP) – Price at which an article is purchased.

Selling price(SP) – Price at which an article is sold.

Loss – Where the Cost Price is greater than the Selling Price, the difference between the SP and CP is the Loss

Loss% = ( (Loss/CP) x 100 ) %

In present case,

CP = ₹ 200

SP = ₹ 180

We know that,

Loss = CP – SP

Loss = ₹ ( 200 – 180 )

Loss = ₹ 20

Further, we also know that,

Loss% = ((Loss/CP) x 100 ) %

Loss% = (( 20/200) x 100 ) %

Loss% = 10 %

Hence, Loss% = 10 %

Answer: (a) Gain = ₹ 50 and Gain % = 10

Explanation: CP of an Article = ₹ 500

SP of an Article = ₹ 550

Since, SP > CP

So, there is Gain

We know that,

Gain = SP – CP

Gain = ₹ 550 – ₹ 500

Gain = ₹ 50

Further

Gain % = ( (Gain/CP) x 100 ) %

Gain % = (( 50/500) x 100 ) %

Gain % = 10 %

Hence, Gain = ₹ 50 and Gain % = 10 %

Answer: (c) ​Loss = ₹ 150 and Loss% = 20%​

Explanation: CP of an Article = ₹ 750

SP of an Article = ₹ 600

Since, SP < CP

So, there is Loss

We know that,

Loss = CP – SP

Loss = ₹ 750 – ₹ 600

Loss = ₹ 150

Further

Loss% = (( Loss/CP) x 100 ) %

Loss% = ( (150/750) x 100 ) %

Loss% = 20 %

Hence, Loss = ₹ 150 and Loss% = 20 %

Answer: (a) ₹ 4250

Explanation: Cost price(CP) – Price at which an article is purchased.

Selling price(SP) – Price at which an article is sold.

Loss – Where the Cost Price is greater than the Selling Price, the difference between the SP and CP is the Loss

In present case,

CP = ?

SP = ₹ 3400

Loss% = 20 %

We know that,

CP = { ( 100/(100−Loss) ) x SP }

CP = ₹ { ([ 100/(100−20)] 100 ) x 3400 }

CP = ₹ {(100/80) x 3400 }

CP = ₹ 4250

Hence, CP = ₹ 4250

Answer: (b) ₹ 4000

Explanation: Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which the Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

CP = ?

SP = ₹ 4600

Gain% = 15 %

We know that,

CP = { [ 100/(100+Gain )] x SP }

CP = ₹ {[100/(100+15 ) ]x 4600 }

CP = ₹ {(100/115) x 4600 }

CP = ₹ 4000

Hence, CP = ₹ 4000

Answer: (b) ₹ 3168

Explanation: Cost price(CP) – Price at which an article is purchased.

Selling price(SP) – Price at which an article is sold.

Loss – Where the Cost Price is greater than the Selling Price, the difference between the SP and CP is the Loss

Loss % = (( Loss/CP) x 100 ) %

12 % = [(Loss/3600) x 100] %

Loss = ₹ (12/100) x 3600

= ₹ 432

CP = ₹ 3600

SP = ?

Loss = 432

We know that,

Selling Price (SP) = Cost Price – Loss

= 3600 – 432

= 3168

Hence, SP = ₹ 3168

Alternative Method:

In present case,

CP = ₹ 3600

SP = ?

loss% = 12 %

As we know that,

SP = { [( 100/100) – loss% ] x CP }

SP = ₹ { [( 100/100) – (12/100) ] x 3600 }

SP = ₹ { (88/100) x 3600 }

SP = ₹ 3168

Hence, SP = ₹ 3168

Answer: (b) ₹ 2013

Explanation: Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which the Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

Gain % = ( (Gain/CP) x 100 ) %

22 % = (Gain/1650) x 100 %

Gain = ₹(22/100) x 1650

= ₹ 363

CP = ₹ 1650

SP = ?

Gain = ₹ 363

As we know that,

Selling Price (SP) – = Cost Price + Gain

= ₹ ( 1650 + 363 )

= ₹ ( 2013 )

Hence, SP = ₹ 2013

Answer: (b) 3 : 4

Explanation: HCF of 75 and 100 is 25

Since, 75 : 100

= 75/100

Dividing Both 75 and 100 by their HCF = (75÷25)/(100÷25)

= 3/4

= 3 : 4

Hence, the simplest form of 75 : 100 is 3 : 4

Answer: (d) 3 : 20

Explanation: Taking both the quantities in same unit, we have

₹ 3 = ( 3 x 100 ) = 300 paise

The equation now becomes 45 paise : 300 paise

or

= 45/300

Dividing both the numbers by their HCF, i.e 15

= (45÷15)/(300÷15)

= 3/20 = 3 : 20

Hence, the required ratio is 3 : 20

Answer: (a) 1 : 5

Explanation: Taking both the quantities in same unit, we have

2.1 m = ( 2.1 x 100 ) = 210 cm

The equation now becomes, 42 cm : 210 cm

or

= 42/210

Dividing both the numbers by their HCF, i.e.,

= (42÷42)/(210÷42)

= 1/5= 1 : 5

Hence, the required ratio is 1 : 5

Answer: (a) 1 : 5

Explanation: Taking both the quantities in same unit, we have

4 hours = ( 4 x 60 ) = 240 min

The equation now becomes 48 min : 240 min

or

= 48/240

Dividing both the numbers by their HCF, i.e., 48

= (48÷48)/(240÷48)

= 1/5= 1 : 5

Hence, the required ratio is 1 : 5

Answer: (b) 3 : 10

Explanation: Taking both the quantities in same unit, we have

0.7 kg = ( 0.7 x 1000 ) = 700 g

The equation now becomes 210 g : 700 g

or

= 210/700

Dividing both the numbers by their HCF, i.e., 70

= (210÷70)/(700÷70)

= 3/10= 3 : 10

Hence, the required ratio is 3 : 10

Answer: (b) 20/45

Explanation: On multiplying or dividing each term of a ratio by the same non zero number, we get a ratio equivalent to the given ratio

For, 20/45

Both numerator and denominator of given fraction is multiplied by same non zero number i.e., 5

(4×5)/(9×5) = 20/45

20/45 is an equivalent ratio of 4/9

9/4 is not an equivalent ratio of 4/9 As both 4 and 9 are not multiply by same non zero number.

21/27 is not an equivalent ratio of 4/9 As both 4 and 9 are not multiply by same non zero number.

20/27 is not an equivalent ratio of 4/9 As both 4 and 9 are not multiply by same non zero number.

Answer: (b) 42 and 54

Explanation: Let the required number be 7a and 9a

Since the sum of these two numbers is given, we can say that

7a + 9a = 96

16a = 96

a = 96/16

a = 6

So, the first number is 7a = 7 x 6

= 42

Second number is 9a = 9 x 6

= 54

Hence, two numbers are 42 and 54

Answer: (b) X = 5600 , Y = 800

Explanation: Total money = ₹ 6400

Given ratio = 7 : 1

Sum of ratio terms = ( 7 + 1 )

= 8

Give: (7/8) part of ₹ 6400 to X

Give: (1/8) part of ₹ 6400 to Y

that is,

X ‘s share = ₹ ( 6400 x (7/8)) = ₹ 5600

Y ‘s share = ₹ ( 6400 x (1/8)) = ₹ 800

Answer: (c) X = 3500 , Y = 2000 , Z = 2500

Explanation:

Answer: (c) ( 10 : 9 ) < ( 13 : 6 )

Explanation: We can write

( 10 : 9 ) = (10/9) and ( 13 : 6 ) = (13/6)

Now, let us compare 10/9 and 13/6

LCM of 9 and 6 is 18

Making the denominator of each fraction equal to 18

We have, (10/9) = (10×2)/(9×2) = 20/18

and (13/6) = (13×3)/(6×3) = 39/18

In case of Like fractions, the number whose numerator is greater is larger. Hence we can say (20/18) < (39/18)

That is (10/9) < (13/6)

Hence, ( 10 : 9 ) < ( 13 : 6 )

Answer: (c) 21

Explanation: Let the missing number be y

(42/24) = (y/12)

On Cross Multiplication,

42 x 12 = 24y

24y = 42 x 12

y = (42×12)/24

y = 21

Hence, (42/24) = (21/12).

Answer: (b) 21 , 14

Explanation: Let (27/9) = (y/7)

Then, 27 x 7 = 9y

9y = 27 x 7

y = (27×7)/9

y = 21

Since, (27/9) = (21/7)

Again, let (21/7) = (42/z)

Then, (21/z) = 42 x 7

z = (42×7)/21

z = 294/21

z = 14

Since, (21/7) = (42/14)

Hence, (27/9) = (21/7) = (42/14)

Answer: (b) No

Explanation: We have 80 cm : 160 cm

= 80 : 160

= 80/160

=(80÷80)/(160÷80) (HCF of 80 and 160 is 80 )

= 1/2

90 m : 150 m

= 90 : 150

= 90/150

=(90÷30)/(150÷30) ( HCF of 90 and 150 is 30 )

= 3/5

Since, the ratios 80 cm : 160 cm and 90 m : 150 m are equal to (1/2) and (3/5)​. So, they are not in proportion.

Answer: (b) No

Explanation: We have, 9 : 27 = 9/27 = (9÷9)/(27÷9)= (1/3)

and 10 : 20 = 10/20 = (10÷10)/(20÷10) = (1/2)

Since, 9 : 27 ≠ 10 : 20

Hence, 9 , 27 , 10 , 20 are not in Proportion

Alternative method: Product of extremes = Product of means

Here, Means are 27 and 10

Extremes are 9 and 20

Product of extremes = 9 x 20 = 180

Product of means = 27 x 10 = 270

Since, Product of extremes ≠ Product of means

Hence, 9 , 27 , 10 , 20 are not in Proportion

Answer: (b) 23

Explanation: We know that, Product of means = Product of extremes

In the given numbers, we can say that 12 , y are means and 46 , 6 are extremes

12 x y = 46 x 6

y = (46×6)/12

y = 23

Hence, y = 23

Answer: (a) 8

Explanation: Clearly, Product of means = Product of extremes

y x y = 16 x 4

y² = 16 x 4

y² = 64

Hence, y = 8

Answer: (b) 99

Explanation: 11 , 33 , y are in proportion

Which means 11 , 33 , 33 , y are in proportion

i.e, 11 : 33 : : 33 : y

Product of Means = Product of Extremes

here,

Means = 33 and 33

Extremes = 11 and y

33 x 33 = 11 x y

1089 = 11 x y

y = 1089/11

Hence, y = 99

Answer: (a) 7 : 6​

Explanation : Number of Girls = 35

Number of Boys = 30

Ratio of Number of Girls : Ratio of Number of Boys

(Number of Girls / Number of Boys) = 35/30

On Simplifying,

Taking HCF of 35 and 30 is 5

(Number of Girls / Number of Boys) = (35÷5)/(30÷5)

= 7/6 = 7 : 6

Hence, the ratio of number of Girls , to the number of Boys is 7 : 6

Answer: (a) 6 : 2 : 3

Explanation: Let,

1A = 3B = 2C = k

This implies that 1A = k

A = k/1

Also if 3B = k

B = k/3

Further, if 2C = k

C = k/2

A : B : C = (k/1) : (k/3) : (k/2)

LCM of 1 , 3 , 2 is 6

Multiplying each of the ratio by 6k we get the ratios as

= ( (k/1) x (6/k) ) : ( (k/3) x (6/k) ) : ( (k/2) x (6/k) )

= 6 : 2 : 3

Hence, A : B : C = 6 : 2 : 3

Answer: (d) 24 : 18 : 8

Explanation: Given A : B = 8 : 6

and B : C = 9 : 4

To find A : B : C we have to make the value of common term in both the ratios equal

that is, B = 6

For this B : C = 1 : (4/9) ( On dividing each term by 9 )

B : C = ( 6 : (4/9) x 6 ) (On multiplying each term by 6 )

B : C = 6 : (24/9)

B : C = 6 : (24÷3)/(9÷3) (HCF of 24 and 9 is 3 )

We get, B : C = 6 : (8/3)

Since, A = B = 8 : 6 and B : C = 6 : (8/3)

Therefore, A : B : C = 8 : 6 : (8/3)

Hence, A : B : C = 24 : 18 : 8 (Multiplying each term by 3 )

Answer: (b) 16

Explanation: Let the required number to be added be ‘ a ‘

then, ( 4 + a ) : ( 9 + a ) = 4 : 5

therefore, (4+a)/(9+a) = 4/5

5 ( 4 + a ) = 4 ( 9 + a )

20 + 5a = 36 + 4a

5a – 4a = 36 – 20

1a = 16

Hence, the required number is 16

Answer: (c) 11 : 15

Explanation: We have:

a :b = 1 : 3

(a/b) = (1/3)

To Find:

5a + 2b : 3a + 4b = (5a+2b)/(3a+4b)

(On dividing numerator and denominator by b )

= ((5a/b)+2)/((3a/b)+4)

Put the value of (a/b) = (1/3)

= ((5×1/3)+2)/((3×1/3)+4)

= ((5/3)+2)/((3/3)+4)

= ((5+6)/3)/((3+12)/3)

= (11×3)/(15×3)

= 33/45

Taking, HCF of 33 and 45 is 3

= (33÷3)/(45÷3) = 11/15

Hence, 5a + 2 b : 3a + 4b = 11 : 15

Answer: (b) 7 : 13

Explanation: Present age of Mohit = 42 years

Present age of Grandfather = 78 years

To find the ratio of present age of Mohit to the present age of Grandfather

(Present age of Mohit / Present age of Grandfather) = 42/78

On simplifying,

Taking, HCF of 42 and 78 is 6

(Present age of Mohit / Present age of Grandfather) = (42÷6)/(78÷6)

= 7/13 = 7 : 13

Hence, the ratio of the present age of Mohit to Grandfather is 7 : 13

Answer: (b) 6 : 7

Explanation: Present age of Mohit = 50 years

After 10 years Mohit’s age = 50 + 10 = 60 years

Present age of Ramesh = 60 years

After 10 years Ramesh’ age = 60 + 10 = 70 years

Ratio of age of Mohit and Ramesh after 10 years

Taking, HCF of 60 and 70 is 10

(After 10 years Mohit′s age / After 10 years Ramesh′s age) = (60÷10)/(70÷10)

= 6/7 = 6 : 7

Hence, the ratio of Mohit’s age to Ramesh’ age after 10 years is 6 : 7

Answer: (b) 6 : 5

Explanation: Present age of Mukesh = 50 years

Mukesh’s age 2 years ago = 50 – 2 = 48 years

Present age of Rahul = 42 years

Rahul’s age 2 years ago = 42 – 2 = 40 years

Ratio of age of Mukesh and Rahul 2 years ago

(Mukesh′s age 2 year ago / Rahul′s age 2 year ago) = 48/40

Since, HCF of 48 and 40 is 8

(Mukesh′s age 2 year ago / Rahul′s age 2 year ago) = (48÷8)/(40÷8)

= 6/5 = 6 : 5

Hence, the ratio of Mukesh’s age to Rahul’s age is 6 : 5

Answer: (d) 25

Explanation: To find third proportional of any two numbers let say a and b be two numbers and c is in third proportion with a and b

Then,

a : b = b : c

Let, the third proportion to 9 and 15 be a

9 : 15 :: 15 : a

(Product of Extremes= Product of Means)

Here, Extremes are= 9 and a

Means are = 15 and 15

9 x a = 15 x 15

a = (15×15)/9 = 25

Hence, the value of ‘a’ is 25

Answer: (b) 30

Explanation: Let, the mean proportional between 15 and 60 be ‘ a ‘

15 : a :: a : 60

(Product of extremes = Product of means)

Here, Extremes are 15 and 60

Means are a and a

15 x 60 = a x a

a² = 15 x 60 = 900

a = 30

Hence, the value of ‘a’ is 30

Answer: (b) 90

Explanation: If a, b, c are in continued proportion

Then,

a : b : : b : c

Since, it is given that 10 , 30 , a are in continued proportion.

So,

10 : 30 :: 30 : a

(Product of extremes = Product of means)

Here, Extremes are 10 and a

Means are 30 and 30

10 x a = 30 x 30

Now, on simplifying we get a = (30×30)/10 = 90

Hence, the value of ‘a’ is 90