A bag contains 50 p, 25 P and 10 p coins in the ratio 3: 5: 7, amounting to Rs 345. Find the number of coins of each type.

(a.) 300, 200,600

(b.) 300,500,700

(c.) 200,600,180

(d.) 600,200,700

(e.) None of these

Let number of coins of 50P, 25P, and 10 p are respectively 3c, 5c and 7c.

3c x 50 + 5c x 25 + 7c x 10 =34500

150c + 125c +70c = 34500

345c= 34500

c = 100

Number of coins = 300, 500, 700

Two number P and Q are such that sum of 60% of P and 30% of Q is equal to the 50% of sum of P and Q. Find the ratio of Q : P.

(a.) 1 : 2

(b.) 2 :1

(c.) 3 : 2

(d.) 2 : 3

(e.) 2 : 5

Hi Tin,

Please find the answer

60% of P + 30% of Q =50%(P+Q)

6P+3Q=5P+5Q

P=2Q

P:Q=2:1

Consider n a 2 digit number such that the ratio of the number to the sum of its digit is maximum. **How many such numbers exist?**

Let the two digit number be N = 10X+Y ,

(10X + Y)/(X+Y) should be maximum ,

i.e [9X + (X + Y)]/(X+Y) should be maximum

i.e. 9X/(X+Y) should be maximum

For the expression to be maximum y must be 0. And X can take any values from 1 to 9.

Hence , 9 two digit numbers.

10,20,30,..... and 90.

Sheetal has only three types of earrings- Gold, Diamond and Silver. All her earrings except X are Diamond, all except Y are Gold and all except Z are Silver. If the ratio of X:Y:Z = 25:12:17, then what is the ratio of the number of Gold earrings to the number of Silver earrings?

Hello Samyak!

X = Gold + Silver = 25x

Y = Silver + Diamond = 12x

Z = Diamond + Gold = 17x

( X + Y + Z ) = 2 ( Diamond + Gold + Silver ) = 54x

Diamond + Gold + Silver = 27x

Gold = 27x - Y = 15x

Silver = 27x - Z = 10x

Gold : Silver = 15x : 10x = 3 : 2 .

Three brothers A, B and C, work together in a company and their total monthly salary is Rs. 25000 per month. They spend 75%, 80%, and 90% of their income respectively. And their saving are in the ratio of 25:18:6.

find individual salary of all three?

This question is an excellent application of ratios. We can solve it by assuming the salaries of the three brothers to be a, b and c.

Savings:

A= 0.25 a

B= 0.2 b

C= 0.1 c

According to the question

0.25 a = 25 k

0.2 b = 18 k

0.1 c = 6 k

which transforms to

a = 100 k

b = 90 k

c = 60 k

As we know,

a + b + c = 250 k

or

25000 = 250 k

or

k = 100

Individual Salaries

A = 100 k = INR 10000

B = 90 k = INR 9000

C = 60 k = INR 6000

Sir please tell me the solution of this problem.

Hello Samyak !

Let the number of balls with :

Abhinav = 5a

Bipin : 12b

Chetan = 13c

The sum of the number of balls should be equal to 60 , as no ball is remaining in the boxes .

5a + 12b + 13c = 60

the only solutions satisfying the above equation are ( 7, 1 , 1 ) and ( 2 , 2, 2 )

a < 7 because 5a is less than or equal to 30.

Hence the answer is 10 , 24 and 26 . 🙂

In two alloys, respective ratios of copper and zinc are 4 : 1 and 1 : 3. Ten kg of the 1st alloy, 16 kg of the 2nd alloy and some quantity of pure copper are melted together. In the resultant alloy the ratio of copper to zinc is 3 : 2. Find the weight (in kg) of the new alloy.

How to handle such problems?

Calculate the quantity of zinc in mixture of both the alloys -

10 × 1/5 + 16 × 3/4 = 14

This 14kg of zinc is 2/5 of the new alloy, hence total weight of alloy is 14× 5/2

=35kg

Sir , kindly share the solution to this problem-

At the NSB demonstration , supporters of Ms. P outnumbered the police by 9:1. The police arrested 135 NSB supporters averaging 5 for every 3 policemen. How many supporters of NSB were there in the demonstration?

Number of police involved : (135/5 ) x 3 = 81

Ratio of police involved : supporters of Ms. P = 1 : 9

Hence => 81 x 9 = 729.

Let the number of apples be A and number of Mangoes be B

Now CP of each apple => x , SP of each apple => 2x .

CP of Each mango => 2x , SP of each mango => 6x

Profit = Total SP - Total CP = (2Ax + 6Bx) - ( Ax + 2Bx) =150% of ( Ax + 2Bx)

Ax + 4Bx = 3/2 ( Ax + 2Bx )

Ax = 2Bx

A : B = 1 : 2

Sahil employ 200 men to build a building. They finish 5/6of work in 10 weeks. Because of some natural calamity not only does the work remain suspended for 4 weeks but also half of work already done is washed away. After the calamity , when work is resumed, only 140 men turn up. The total time in which contractor is able to complete the work assuming that there are no further disruptions in the schdule is

a 25 weeks

b 26weeks

c 24weeks

d 20 weeks

apply (men*days)/work

after calamity half of the work done is destroyed which is 5/12 and now remaining work to be done=1-5/12=7/12

(200*10)/(5/6)=140*weeks/(7/12)

weeks =10

so total time taken=10+4+10 =24 weeks

concentrations of three types of milks X,Y and Z are 10%, 20%and 30% respectively. They are mixed in ratio 2:3:P resulting in a 23% concentration solution. Find P

Take 200ml of first solution , 300 ml of 2nd solution and 100p ml of third solution :

Now , solve the following equation

20 + 60 + 30p = (23/100) ( 200 + 300 + 100p)

80 + 30p = 23 ( 5 + p)

p = 7

Alternate approach

you can solve by average

(10*2 + 20*3 +30*p)/(2+3+p) =23

80 +30p=115+23p

7p=35

p=5

The cost of an article (which is composed of raw materials and wages) was 3 times the value of the raw materials used. The cost of raw materials oncreased in the ratio 3:7 and wages increased in the ratio 4:9 . Find the present cost of article if its original cost was 18

a 41

b 30

c 40

d 46

Original Cost : 18

Cost of raw material used : 1/3 of 18 = 6

so , raw material : wages = 6 : 12

Cost of raw material increasesd by 3 : 7 so

New cost of raw material : 6/3 x 7 = 14

Wages increased by 4 : 9 so

increased wages : 12/4 x 9 = 27

Present cost : 14 + 27 = 41

Option (A)