HTET TGT Quantitative Aptitude Questions Solved Problems with Detailed Solutions [Free PDF]

The correct answer is option 1.

Given:

The price of tea being increased by 20%, a man reduces his consumption by 20%. 

Calculation:

⇒ Let assume initial price of tea = Rs. 100 per kg

And man consumption was 1 kg.

So, Initial expenses for tea = 100 x 1 = Rs.100

Now price is increased by 20% = 120% of 100 = Rs. 120 per kg

And consumption is reduced by 20% = 80% of 1 kg = 0.8 kg

Now, man expenses for tea = 120 x 0.8 = Rs. 96

Percentage reduction in expenses = 

 = - 4%

So, his expenses for tea  will be decreased by 4%.

Given:

An engineering student has to secure 15% marks to pass.

A student gets 55 marks and fails by 20 marks.

Calculation:

Let the total marks be x

Passing marks = x × 15%

Passing marks = 15x/100      ----(1)

According to the question,

Passing marks = 55 + 20 = 75      ----(2)

From equation (1) and (2) we get,

⇒ 15x/100 = 75

⇒ x = 7500/15

⇒ x = 500

∴ The maximum marks of the student is 500.

Shortcut Trick

Passing percentage = 15%

Passing marks = 55 + 20 = 75

Maximum marks = (75/15) × 100 = 500

∴ The maximum marks of the student is 500.

GIVEN:

The cost price of an article = Rs. 6000

MP of an article = Rs. 7800

Profit percentage = 4%

CALCULATION:

The cost price of the article = Rs. 6000

Profit percentage = 4%

SP of the article

⇒ CP × (100 + Profit%)/100

⇒ 6000 × 104/100 = Rs. 6240

∵ Discount = MP – SP

= 7800 – 6240 = Rs. 1560

Let the discount percentage = x%

∴ (7800 × x)/100 = 1560

= x = 1560/78 = 20%

∴ Discount percent = 20%

Alternate Method

(100 – Discount%)/ (100 + Profit%) = CP/MP

Ratio of CP and MP = 6000/7800 = 10/13

∵ (100 – Discount%)/ (100 + Profit%) = CP/MP

⇒ (100 – x)/(100 + 4) = 10/13

⇒ 1300 – 13x = 1000 + 40

⇒ 1300 – 1040 = 13x

⇒ 260 = 13x

∴ x = 20%

∴ Discount% = 20%