Mix Test - 1

Let rate \(= R \%\) and time \(= R\) years.

Then, \(\left(\frac{1200 \times R \times R}{100}\right)=432\)

\(\Rightarrow 12 R ^{2}=432\)

\(\Rightarrow R ^{2}=36\)

\(\Rightarrow R =6\)
Hence, the correct option is (B).

Let unit’s digit be x and ten’s digit by y

∴ Number formed = 10y + x …(i)

According to the question, 

2y – 1 = x …(ii) 

Also, (New number) – (Original number) 

= Original number – 20 

(10x + y) – (10y + x) = (10y + x) – 20 

10x + y – 10y – x = 10y + x – 20 

9x – 9y = 10y + x – 20 …(iii) 

On substituting the value of x = 2y – 1 from equation (i), we get 

9 (2y – 1) – 9y = 10y + 2y – 1 – 20

= 18y – 9 – 9y = 10y + 2y – 1 – 20 

= 3y = 12 

∴ y = 4 

On putting the value of y in equation (ii), we get 

x = 2 × 4 – 1 

= 8 – 1 = 7 

∴ Required number 

= 10y + x

= 10 × 4 + 7 

= 47

Hence, the correct option is (C).

The possible combinations are (5,3,2) (2,4,4),(3,3,4),(2, 2, 6), (7,2,1),(8,1,1)

Product of the combinations 

⇒ 5 × 3 × 2 = 30

⇒ 2 × 4 × 4 = 32

⇒ 4 × 3 × 3 = 36

⇒ 2 × 2 × 6 = 24

⇒ 7 × 2 × 1 = 14

⇒ 8 × 1 × 1 = 8

 36 is the maximum product among the combinations.

∴ 36 is the maximum product of (a × b × c).

Hence, the correct option is (C).

\(30 \%\) discount on \(2000=30 \%\) of \(2000=\) Rs. 600

\(25 \%\) discount on \(2000=25 \%\) of \(2000=\) Rs. 500

Remaining amount \(=2000-500=\) Rs. 1500

Second discount of \(5 \%=5 \%\) of \(1500=\) Rs. 75

Total discount \(=500+75=\) Rs. 575

So difference in discounts \(=\) Rs. \(600-\) Rs. \(575=\) Rs. 25
Hence, the correct option is (B).

According to the question:

By selling cloth at Rs. \(9\) per meter, a shopkeeper loses \(10 \%\) 

Therefore, Selling Price\(=90 \%\) of Cost Price

\(\Rightarrow\) Cost Price\(=\frac{(9 \times 100)}{90}\)

\(\Rightarrow\) Cost Price \(=\) \(10\) Rs. per meter 

To earn a profit of \(15 \%\), 

Selling Price \(=115 \%\) of \(10\) 

\(\Rightarrow\) Selling Price \(=11.50 \) Rs. per meter

Hence, the correct option is (D).

Let the sum be Rs. \(x\). Then,

C.I. \(=\left[x\left(1+\frac{4}{100}\right)^{2}-x\right]\)

\(=\left(\frac{676}{625} x-x\right)\)

\(=\frac{51}{625} x\)

S.I. \(=\left(\frac{x \times 4 \times 2}{100}\right)\)

\(=\frac{2 x}{25}\)

\(\therefore \frac{51 x}{625}-\frac{2 x}{25}=1\)

\(\Rightarrow x=625 \)
Hence, the correct option is (A).

Given:

The eldest three children got chocolates in the ratio 3 : 11 : 7.

Let the children be P, Q, R and S and Father be F.

Chocolates with P : Q : R = 3 : 7 : 11

Let the number of chocolates be 3k, 7k and 11k.

Total chocolates with three eldest children = 21k

Chocolate with F and S = 3 × 21k = 63k

Total chocolates = (21k + 63k) = 84k

Chocolate with F : (P + Q + R + S) = 3 : 4

Total 7 units of chocolate = 84 k

1 unit = 12k

Chocolate with F = 3 × 12k = 36k

Chocolate with S =  (63k – 36k) = 27k

27 k = 81

k = 3

Total number of chocolates = 84k 

= 84 × 3 

= 252

Hence, the correct option is (B).

Given.

Income = Expenditure + Saving

Adarsh: 12x = 15y + 3x (3x = 25% of 12x)

Satpal: 9x = 9y + (9x – 9y)

Rahim: 7x = 8y + (7x – 8y)

Therefore, 12x – 3x = 15y

\(\Rightarrow \frac{x}{y}=\frac{5}{3}\)

\(y=\frac{3 x}{5}\)

Therefore, savings = (income – expenditure)

Adarsh = 12x – 9x 

= 3x

Satpal = 9x – 9y 

\(=9 x-\frac{27}{5} x\)

\(=\frac{18}{5} x\)

Rahim = 7x – 8y

\(=7 x-\frac{24}{5} x\)

\(=\frac{11}{5} x\)

i.e., the ratio of savings of Adarsh : Satpal : Rahim

\(=3 x: \frac{18}{5} x: \frac{11}{5} x\)

\(=15 : 18 : 11\)

Hence, the correct option is (A).

Given:

Total numbers \(=35\)

\(\mathrm{P}(\mathrm{E})=\frac{\text{ Favourable outcomes}}{\text{Total outcomes}}\), here \(\mathrm{E}\) is an event.

Prime numbers between 1 to 35 , including 1 and 35 are \(2,3,5,7,11,13,17,19,23,29\) and 31.

\(\Rightarrow\) Total number of prime numbers \(=11\)

\(\Rightarrow\) Favourable outcomes \(=11\)

\(\Rightarrow\) Total numbers \(=35\)

\(\Rightarrow\) Total outcomes \(=35\)

\(\Rightarrow \mathrm{P}(\)Getting a prime number\()=\frac{\text{ Favourable outcomes}}{\text{Total outcomes}}\)

\(\therefore \mathrm{P}\) (Getting a prime number) \(=\frac{11}{ 35}\)

Hence, the correct option is (B).

Given: 

\(-\frac{1}{9}\) and \(\frac{1}{5}\)

Convert the rational numbers into equivalent rational numbers with the same denominator.

LCM of 9 and 5 is 45.

\(-\frac{1}{9}=\frac{-1 \times 5}{9 \times 5}=\frac{-5}{45}\)

and \(\frac{1}{5}=\frac{1 \times 9}{5 \times 9}=\frac{9}{45}\)

The integers between \(-5\) and 9 are \(-4,-3,-2,-1\), \(0,1,2,3,4,5,6,7,8\)

The corresponding rational numbers are,

\(\frac{-4}{45}, \frac{-3}{45}\) \(\frac{-2}{45}, \frac{-1}{45}, \frac{0}{45}, \frac{1}{45}, \frac{2}{45}, \frac{3}{45}, \frac{4}{45}, \frac{5}{45}, \frac{6}{45}, \frac{7}{45}, \frac{8}{45}\)

On selecting any 9 of them, we get 9 rational numbers between \(-\frac{1}{9}\) and \(\frac{1}{5}\):

\(\frac{-4}{45}, \frac{-3}{45}, \frac{-2}{45}, \frac{-1}{45}, \frac{2}{45}, \frac{3}{45}, \frac{4}{45}, \frac{5}{45}, \frac{8}{45}\)

Hence, the correct option is (D).