Quantitative Aptitude PYQs

3x² - 25x + 28 = 0

3x² - 21x - 4x + 28 = 0

3x(x - 7) - 4(x - 7) = 0

(3x - 4) (x - 7) = 0

x = 4/3, 7

Only II and III follows

From equation I

9x2 - 31x + 12 = 0

9x² - 27x - 4x + 12 = 0

9x(x - 3) - 4(x - 3) = 0

(9x - 4)(x - 3) = 0

x = 4/9, 3

From equation II

3x² - 8x - 3 = 0

3x² - 9x + x - 3 = 0

3x(x - 3) + 1(x - 3) = 0

(x - 3)(3x + 1) = 0

x = 3, -1/3

Now,

Product of smaller roots of equation I and bigger roots of equation II = 4/9 x 3 = 4/3 = 1(1/3)

Let their initial investment be Rs. 2x, Rs. 3x, and Rs. 4x respectively

Let they withdrew Rs. y, Rs. 2y and Rs. 3y respectively after 4 months

A’s total investment = (4 × 2x) + 8 × (2x – y) = 8x + 16x – 8y = 24x – 8y

B’s total investment = (4 × 3x) + 8 × (3x – 2y) = 12x + 24x – 16y = 36x – 16y

C’s total investment = (4 × 4x) + 8 × (4x – 3y) = 16x + 32x – 24y = 48x – 24y

∵ Ratio of profits = Ratio of total investment

⇒ 4 ∶ 5 ∶ 6 = (24x – 8y) ∶ (36x – 16y) ∶ (48x – 24y)

Hence, we can write,

⇒ (24x – 8y)/(36x – 16y) = 4/5

⇒ 8(3x – y)/4(9x – 4y) = 4/5

⇒ (3x – y)/(9x – 4y) = 2/5

⇒ 15x – 5y = 18x – 8y

⇒ 3x = 3y

⇒ x = y

∴ Part of investment withdrew by C = 3y/4x = ¾

[Formula:

The simple interest is given by S.I = \(\frac{{P \times {\rm{r}} \times {\rm{n}}}}{{100}};,;\)

Where P is the principle amount, r is rate of interest and n is number of years]

It is given that principle is Rs. 25200

Rate of interest per annum is 4%

Number of year is 6

Interest earn by Mr Phanse will be \(= \frac{{P \times {\rm{r}} \times {\rm{n}}}}{{100}}\)

\(= \frac{{25200 \times 4 \times 6}}{{100}};\)[∵ substituting the values]

= 6048

At the end of 6 year Mr.Phanse will get with interest = Rs (25200 + 6048) = Rs 31248

 If he invests the whole amount again at 4% simple interest for 4 years then interest will be for these years \(= \frac{{31248 \times 4 \times 4}}{{100}}\)

= 4999.68

≈ 5000

Hence the answer is Rs 5000

Given:

A can do a work in 24 days, B can do the same work in 48 days and C can do the same work in 72 days.

Calculation:

In first case

A can do the work in 24 days

B can do the work in 48 days

C can do the work in 72 days

⇒ the ratio of their efficiency will be A : B : C = 1/24 : 1/48 : 1/72 = 6 : 3 : 2

⇒ A will get 6/(6 + 3 + 2) = 6 / 11

of the total amount which will be equal to Rs. 2400

But in second case only A and C work

⇒ the efficiency of A : C = 1 / 24 : 1 / 72 = 3 : 1

So A will get now 3/(1 + 3) = (3/4 )^th of the total amount which will be Rs. 3300

∴ extra money earned by A = Rs. (3300 - 2400) = Rs. 900

Solution:

Number of bikes purchased by company E = 120

Number of bikes purchased by company F = (120 - 40) = 80

Number of cars purchased by the company E = 160

Number of cars purchased by company F = (160 + 40) = 200

Required ratio = (120 + 80): (160 + 200) = 200: 360 = 5: 9

Solution:

Average number of bikes purchased by all the companies = (100 + 80 + 100 + 200 + 120)/5 = 120

Average number of cars purchased by all the companies = (120 + 140 + 60 + 180 + 160)/ 5 = 132

Required difference = (132 - 120) = 12

Solution:

Initial number of bikes and cars purchased in the company D = (200 + 180) = 380

After increasing x% in bike purchase number of bikes and cars purchased in company D = (210 ×  2) = 420

Number of bikes purchase increased = 40

To find the value of x

100 × (240 - 200)/200 = 100 × 40/200 = 20%

Solution:

Number of bikes purchased by the company C = 100

1/5 th of bikes purchased by company C = 100/5 = 20

Number of cars purchased by company C = 60

1/3 rd of cars purchased by company C = 60 × 1/3 = 20

Required difference = (20 - 20) = 0

Solution:

Number of yamaha bikes purchased by company A = 100 × 20/100 = 20

Number of yamaha bikes purchased by company B = 80 × (30/100) = 24

Number of Tata cars purchased by company A = 120 × (25/100) = 30

Number of Tata cars purchased by company B  = 140 × (20/100) = 28

Required total = (20 + 24 + 30 + 28) = 102

Applying BODMAS rule to solve the problem,

Step - 1: Parts of the equation enclosed in ‘Brackets’ must be solved first.

Step - 2: Any mathematical ‘Of’ or ‘Exponent’ must be solved next.

Step - 3: Next the part of the equation which contain ‘Division’ and ‘Multiplication’ must be solved.

Step - 4: Last but not the least, the parts of the equation that contains ‘Addition’ and ‘Subtraction’ must be solved.

Now the given expression,

\(\begin{array}{l} \Rightarrow \frac{1}{8} + \frac{{16}}{3} \times \frac{9}{2} - \frac{{13}}{4} \times \frac{{23}}{6} = ? \times \frac{5}{3}\ \Rightarrow \frac{1}{8} + 24 - \frac{{299}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{3 + 576 - 299}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{280}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{280 \times 3}}{{24 \times 5}} = ?\ \Rightarrow ? = \frac{{840}}{{120}} \end{array}\)

⇒ ? = 7

Given:

43 × 45 + 43 × 55 - 43 × 30 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

43 × 45 + 43 × 55 - 43 × 30 = ?

⇒ 43(45 + 55 - 30) = ?

⇒ 43(100 - 30) = ?

⇒ 43 × 70 = ?

⇒ 3010

∴ The value of ? is 3010.

Follow BODMAS rule to solve this question, as per the order given below,

Calculation:

(175 - 25) + √25 - 10 × 6 = ?

⇒ 150 + 5 – 60 = ? 

⇒ ? = 95

∴ The required value is 95

Given:

(16900 ÷ 26) + (53 of 10 ÷ 25) = ? × 70

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(16900 ÷ 26) + (5³ of 10 ÷ 25) = ? × 70

⇒ 650 + (125 of 10 ÷ 25) = ? × 70

⇒ 650 + (1250 ÷ 25) = ? × 70

⇒ 650 + 50 = ? × 70

⇒ 700 = ? × 70

⇒ ? = 10

∴ The value of ? is 10.

Given:

427 - 11² + (32)^2/5 + (9)^3/2 – 35 × 4 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

427 - 11² + (32)^2/5 + (9)^3/2 – 35 × 4 = ?

⇒ 427 – 121 + 4 + 27 – 140 = ?

⇒ 427 + 31 – 261 = ?

⇒ 458 – 261 = ?

⇒ ? = 197

∴ The value of ‘?’ is 197.

Additional Information

(32)^2/5 = (2 × 2 × 2 × 2 × 2)^2/5

⇒ (2⁵)^2/5

⇒ (2)^{5 × (2/5)}

⇒ (2)²

⇒ 4

(9)^3/2 = (3 × 3) ^3/2

⇒ (3²)^3/2

⇒ (3)^{2 × (3/2)}

⇒ (3)³

⇒ 27

Given:

150/25 –  √625 + 183/3 + 1.2 × 50 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

150/25 –  √625 + 183/3 + 1.2 × 50 = ?

⇒ 150/25 –  25  + 183/3 + 1.2 × 50 = ?

⇒ 6 –  25 + 61 + 1.2 × 50 = ?

⇒ 6 – 25 + 61 + 60 = ?

⇒ ? = 102

∴ The value of ‘?’ is 102.

Given:

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

⇒ (√1024/10000) + (√2401/10000) + (√1225/10000) – (√6400/10000) = (?)²

⇒ (32/100) + (49/100) + (35/100) – (80/100) = (?)²

⇒ (116 – 80)/100 = (?)²

⇒ 36/100 = (?)²

⇒ ? = (√36/100)

⇒ ? = 6/10

⇒ ? = 0.6

∴ The value of ‘?’ is 0.6.

 

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

⇒ 0.32 + 0.49 + 0.35 – 0.8 = (?)²

⇒ 1.16 - 0.8 = (?)²

⇒ 0.36 = (?)²

⇒ ? = √0.36

⇒ ? = 0.6

∴ The value of ‘?’ is 0.6.

Given:

(256)3/4 ÷ [1/(8)1/3]  ×  640.5 = 2?

Concept Used:

(a^m)ⁿ = a^{m × n}

a^m × aⁿ = a^{m + n}

If a^m = aⁿ then m = n

Calculation:

(256)3/4 ÷ [1/(8)1/3]  ×  640.5 = 2?

⇒ (4)3 × 2 × (8²)1/2 = 2 ?

⇒ (2²)3 × 2 × 2³ = 2 ?

⇒ 26 × 2 × 23 = 2 ?

⇒ 26 + 1 + 3 = 2 ?

⇒ 6 + 1 + 3 = ?

⇒ ? = 10

∴ The value of (?) is 10

Given:

125% of 120 + 55% of 460 - 21 ÷ 7 = 250% of ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

125% of 120 + 55% of 460 - 21 ÷ 7 = 250% of ?

⇒ (125/100) × 120 + (55/100) × 460 - 3 = (250/100) × ?

⇒ (5/4) × 120 + (11/20) × 460 - 3 = (5/2) × ?

⇒ 150 + 253 - 3 = (5/2) × ?

⇒ 400 = (5/2) × ?

⇒ ? = 160

∴ 160 should come in place of the question mark (?).

Given:

(? ÷ 7 ÷ 5) × 25 + 125 = 750

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(? ÷ 7 ÷ 5) × 25 + 125 = 750

⇒ (? ÷ 7 ÷ 5) × 25 = 625

⇒ (? ÷ 7 ÷ 5) = 25

⇒ ? × (1/7) × (1/5) = 25

⇒ ? = 25 × 35

⇒ ? = 875

∴ The value of ? is 875.

Don't calculate (? ÷ 7/5) = 25

⇒ (? × 5/7) = 25

⇒ ? = 35

This is the wrong way to calculate the value.